If you find a mistake, omission, etc., please let me know by e-mail.

The orange ball marks our current location in the course, and the current problem set.

Another point of view on real representations (§3.5): If $W$ is an irreducible real representation of a finite (or more generally compact) group $G$ then ${\rm End}_G(W)$ (a.k.a. ${\rm Hom}_G(W,W)$) is a finite-dimensional real division algebra, call it $D$. There are only three such: $\bf R$ itself, ${\bf C},$ and the Hamilton quaternions ${\bf H}.$ These three cases correspond to the three cases of Theorem 3.37, with $W = V$ in the real case, $W \cong V \oplus V'$ in the complex case, and $W \cong V \oplus V$ in the quaternionic case. Subfields $K$ of $\bf C$ other than $\bf R$ (and $\bf C$ itself) may have a larger repertoire of division rings, making the classification more difficult.

In the proof of Lemma 3.35 (page 40), it might not be obvious that $\lambda$ is a positive real. To show this, recall that $H$ may be assumed positive definite, and let $x$ be any nonzero vector. Then $H(\varphi^2 x, x)$ is the complex conjugate of $H(\varphi(x),\varphi(x))$ [plug in $(\varphi(x),x)$ for $(x,y)$ in the displayed identity]. But that is a positive real number, as is $H(x,x)$. Hence $\lambda$, being the ratio $H(\varphi^2 x,x)/H(x,x)$ of these two positive reals, is also a positive real number.

We have shown that an element $g$ of a finite group $G$ is conjugate with its inverse $g^{-1}$ if and only if $\chi(g)$ is real for every character $\chi$ of $G$. Here is a similar characterization, with a similar proof, of group elements each of whose characters is an integer:

Theorem: Let $g$ be an element of order $n$ in a finite group $G$. Then $\chi(g) \in {\bf Z}$ for every character $\chi$ of $G$ if and only if $g^d$ is conjugate with $g$ for each integer $d$ coprime to $n$.

(Equivalently: if and only if $g$ is conjugate with every generator of the cyclic group $\langle g \rangle$.)

Proof (sketch): It is enough to show that the criterion is equivalent to the rationality of each $\chi(g)$, because character values are automatically algebraic integers (because $\chi(g)$ is the sum of the eigenvalues of $\rho(g)$, each of which is a root of unity). Diagonalize the action of $g$ on any representation. The eigenvalues are $n$-th roots of unity. Conclude that $\chi(g^d)$ is the image of $\chi(g)$ under the Galois automorphism sending each $n$-th root of unity to its $d$-th power. This equals $\chi(g)$ for each $\chi$ if and only if $g^d$ is conjugate to $g$.

Example: Every character of the symmetric group $S_d$ takes integer values at all permutations of $d$ letters.

We shall see that $S_d$ actually satisfies a stronger condition: all its representations are defined over $\bf Q$ (and hence over $\bf Z$).

To expand on “(and hence over $\bf Z$)”: If $G$ acts on $\bf Q$-vector space $V,$ choose any basis $v_1,\ldots,v_d$, and let $L$ be the $\bf Z$-span of all vectors $g v_i$ with $g \in G$ and $1 \leq i \leq d.$ Then $L$ is a finitely generated additive subgroup of $V$, so it is isomorphic with ${\bf Z}^{\dim V}\!;$ and $GL = L.$

Recall that for any finite group $G$ we have the formula $|G| = \sum_V (\dim V)^2 = \sum_\chi (\chi(1))^2$ where $V$ ranges over the (isomorphism classes of) irreducible representations of $G$ (with associated character $\chi$). In our case of $G = S_d$, each $\dim(V)$ is the number of standard tableaux in the Young diagram corresponding to $V$, so the sum of $\dim(V)^2$ is the number of pairs of standard tableaux of order $d$ with the same shape. Can the fact that this number equals $d!$ be proved combinatorially, by exhibiting a bijection between such pairs and permutations of $d$ letters? Yes, but nontrivially: that’s the RSK (Robinson–Schensted–Knuth) correspondence.

The Hardy-Ramanujan asymptotic formula $p(d) \sim (1/\alpha d) e^{\beta\sqrt{d}}$ cited at the top of p.45 implies that $p(d)$ grows faster than any polynomial in $d,$ but slower than any exponential $C^d$ ($C>1$). Both parts of this could be seen a priori. Indeed the number of partitions with $k$ parts already grows as a multiple of $d^{k-1}$ (which multiple?), so $p(d) \gg d^n$ for all $n$; and if $p(d) > C^d$ for infinitely many $d$ then the power series $\sum_d p(d) t^d$ diverges at $t = 1/C,$ whence $C \leq 1.$ For much more information and numerous references on $p(d)$ you can start with the extensive OEIS entry for the sequence $\{p(d)\}$.

A nice example of the hook-length formula (4.12, p.50) for the dimension of an irreducible representation of $S_d$ is the case of a partition of the form $(m,m)$ where $d=2m$. The Young diagram for this partition is a rectangle of height 2 and width $m$. The dimension obtained from the hook-length formula is the $m$-th Catalan number $(2m)! / (m! (m+1))!$. For $m=2$ and 3, we obtain representations of dimension 2 and 5 of $S_4$ and $S_6$ respectively, coming from the exceptional homomorphisms from those groups onto $S_3$ and $S_6$. These two (and the representation of $S_6$) associated with the partition (2,2,2) conjugate to (3,3)) are the only representations of $S_d$ of dimension less than $d$ other than the trivial, sign, standard, and standard-tensor-sign representations (Exercise 4.14*). [The Catalan numbers also appear with some regularity as the solutions of various combinatorial enumeration problems, such as the number of triangulations of an $(m+2)$-gon.]

These exceptional small representations are also connected with the exceptional homomorphisms $S_d \to S_e$ with $3 \leq e \leq d$. (We exclude $e=2$ because the sign representation is likewise the nontrivial representation of $S_2$ composed with the sign homomorphism $S_d \to S_2$.) Namely, the 2-dimensional irreducible representation of $S_4$ associated to the partition (2,2) comes from the 2-dimensional irr.rep. of $S_3$ composed with a surjection $S_4 \to S_3$ that made possible the solution in radicals of the general quartic equation once the cubic case was solved. (Incidentally, note that for $d=3$ the representations $V$ and $V \otimes \epsilon$ are isomorphic.) The extra 5-dimensional irr.reps. of $S_6$ come from the usual ones via an outer automorphism of $S_6$. This automorphism, though not as familiar as $S_4 \to S_3$, appears in many places in the story of sporadic simple groups; for example, ${\rm Aut}(S_6)$ is a maximal subgroup of the Mathieu group $M_{12}$. There are various was to obtain an outer automorphism of $S_6$ from an exceptional isomorphism among small linear or permutation groups: $A_5 \cong {\rm SL}_2({\bf F}_4) \cong {\rm PSL}_2({\bf F}_5),$ $S_5 \cong \Sigma{\rm L}_2({\bf F}_4) \cong {\rm PGL}_2({\bf F}_5),$ $A_6 \cong {\rm PSL}_2({\bf F}_9),$ $S_6 \cong {\rm Sp}_4({\bf F}_2).$ [For the last of these, consider the coordinate permutations of the quotient of the $5$-dimensional space $\{(x_1,\ldots,x_6)\} \in {\bf F}_2^5 : x_i+\cdots+x_6=0\}$ by $\langle(1,1,1,1,1,1)\rangle$.] The ${\rm Sp}_4({\bf F}_2)$ route also connects with the exceptional isomorphism between Lie algebras ${\mathfrak{sp}}_4$ and ${\mathfrak{so}}_5,$ to which we shall return in a few weeks.

Two sanity checks on the Frobenius formula (4.10): the polynomial $\Delta(\vec x) \prod_j P_j(\vec{x})^{{\bf i}_j}$ is homogeneous of dimension $l_1 + \cdots + l_k;$ and artificially extending $\lambda$ by one or more zeros does not change the formula. For the former: $\Delta$ has degree $k \choose 2$, and each factor $P_j(\vec{x})^{{\bf i}_j}$ has degree $j {\bf i}_j,$ so the product has degree ${k \choose 2} + \sum_j j {\bf i}_j = {k \choose 2} + d$ — which equals $l_1 + \cdots + l_k$ because

In Lemma 4.23(1) we cannot expect $a_\lambda \cdot x \cdot b_\mu = 0$ for all $\lambda \neq \mu.$ Indeed if $\mu$ is the partition $d$ then $Q$ is just $\{1\}$ and $b_\mu$ is the identity element of $A = {\bf C}[S_d],$ so $a_\lambda \cdot b_\mu \neq 0$ for all $\lambda$. Of course this $\mu$ is the top partition in our lexicographic order, so this calculation does not contradict Lemma 4.23(1) because $d$ has no partition $\lambda > \mu.$

The final step in the proof of Lemma 4.25 (p.54), and thus of Theorem 4.3 (the description of irr.reps. of $S_d$ in terms of Young symmetrizers), hinges on the fact that if $W$ is a (one-sided) ideal of the group algebra $A$ then $W=0$ if and only if $W^2=0$. Here’s another way to see this, which works in any semisimple algebra $A$. Since $A$ is a direct sum of matrix algebras ${\rm End}(V)$, it is enough to prove the result for $A = {\rm End}(V).$ But we know the ideals of ${\rm End}(V)$ are just ${\rm Hom}(V,V')$ as $V'$ ranges over subspaces of $V$. This ideal is zero if and only if $V'$ is. But ${\rm Hom}(V,V')$ contains projections from $V$ to $V'$. A projection is idempotent (equal to its own square); thus if $W^2=0$ then the projection itself is zero, whence $V'$ is the zero subspace and $W=0$ as claimed.

The F-H text uses a construction of $V_\lambda$ that generalizes our construction of $V = V_{(d-1,1)}$ as a subrepresentation of the permutation group $U$ (which now becomes $U_{(d-1,1)}$). We start from a permutation representation associated to λ, which contains a unique copy of $V_\lambda$ and also one or more copies of $V_{\lambda_1}$ for each $\lambda_1 > \lambda$; we then remove those other components to isolate $V_\lambda$. We shall use a similar process to get at representations of semisimple Lie algebras. See Problem 4.47* for an overview of the alternative construction via Specht modules. (This time the star, indicating further information (here on page 519), leads only to two references.) This also connects with another important appearance of the hook-number formula: $V_\lambda$ has a basis indexed by the standard Young tableaux of shape λ; that is, a labeling of the boxes by the integers in $[1,d]$ for which both columns and rows are increasing (reading right-to-left and top-to-bottom, so for starters the top left entry must be 1). It follows that the number of such tableaux must equal $\dim V_\lambda$ and is thus given by the hook-length formula. Note that $d$ must appear in one of the boxes that are the last in both their row and their column, and thus can be removed to leave a partition of $d-1$; this is consistent with the “branching theorem” (see the paragraph of the textbook that spans pages 58–59): the restriction of $V_\lambda$ to $S_{d-1}$ decomposes as the sum of $V_{\lambda_1}$ over partitions $\lambda_1$ of $d-1$ whose diagram is contained the diagram of $\lambda$.

The “Young subgroup” of (4.27) (page 54, the start of section 1.4.3) is the same as what we called $P_\lambda$ last week.

We did not cover induced representations (§3.3 of Chapter 1, starting on page 32). This topic is often taught in courses that introduce representation theory of finite groups. Anyway we do not need this theory here: if $H$ is a subgroup of a finite group $G$ then the induced representation of the trivial representation of H is just the permutation representation of $G$ acting on the cosets of $H$ (Example 3.13 on p.33, which is the first example of §3.3). For example, if $\lambda=(d)$ then $P_\lambda = S_d$ and we get the trivial representation; the next case, $\lambda = (d-1, 1)$, leads to $P_\lambda = S_{d-1}$ and the permutation representation. Going a few steps further down the lexicographic ordering (assuming $d \geq 4$):

• The next $\lambda$ is $(d-2,2)$, so $P_\lambda = S_{d-2} \times S_2$ and we get the permutation representation of $S_d$ acting on unordered pairs, which is ${\rm Sym}^2 U \ominus U$; since ${\rm Sym}^2 U = {\rm Sym}^2 ({\bf 1} + V) = {\bf 1} \oplus V \oplus {\rm Sym^2} V,$ the representation $U_{(d-2,2)}$ of $S_d$ is isomorphic with ${\rm Sym}^2 V.$ But the hook-length formula tells us $\dim V_{(d-2,2)} = d(d-3)/2$ which is $\dim({\rm Sym}^2 V) - 1;$ indeed ${\rm Sym}^2 V$ has a one-dimensional fixed subspace (because $V$ is defined over $\bf R$!) and $V_{(d-2,2)} = {\rm Sym}^2 V \ominus {\bf 1}.$
• Next comes $\lambda = (d-2,1,1)$, for which $P_\lambda = S_{d-2}$ and $U_\lambda$ is the permutation representation of $S_d$ acting on ordered but distinct pairs. This is isomorphic with $(U \otimes U) \ominus U,$ and contains both $V_{(d-2,2)}$ and $\bigwedge^{\!2} \! V = V_{(d-1,1)}\!,$ each with multiplicity 1. What are the multiplicities of $\bf 1$ and $V$ in $U_\lambda$?

The character of our permutation representation is easy enough to describe: the character of $g \in S_d$ is the number of set partitions of $\{1,\ldots,d\}$ into subsets of sizes $\lambda_1,\ldots,\lambda_k$ with each subset consisting of a union of cycles of $g$. Still, one might reasonably question whether it is actually “easy to compute” (as the textbook claims on line 6 of page 55) when the answer is a sum over doubly indexed tuples $r_{p,q}$ satisfying a list of linear conditions… Such sums are the kind of thing that generating functions can help us analyze, and we shall go as far as proving (4.32,4.33). We shall say very little about the rest of this section (and its needed background in Appendix A) of the textbook.

We’ll finish our selection from Chapter 1 (at least for now — we may return for “Lecture 6” on the Schur functors etc.) by connecting the Catalan case of the second construction of Problem 4.47* with a description of moduli spaces of 4 points in ${\bf P}^1$ (the modular curve ${\rm X}(2)$), 6 points in ${\bf P}^1$ (the Segre cubic in ${\bf P}^4$), and 6 points in ${\bf P}^2$ (double cover of ${\bf P}^4$ branched on the Igusa quartic). We shall say more about the moduli of $n$ unordered points in ${\bf P}^1$ when we describe the invariants of the action of ${\mathfrak{sl}}_2$ (and thus of ${\rm{SL}}_2$) on binary forms of degree $n$.

About ${\rm (S)O}_{k,l}({\bf R})$ (page 96): any nondegenerate quadratic form $Q$ on a finite-dimensional real vector space $V$ can be written as $\sum_{i=1}^k x_i^2 - \sum_{j=1}^l y_j^2$ for some $k,l \geq 0$ with $k+l = n$ (use Gram-Schmidt to find an orthogonal basis $e_1,\ldots,e_n$ and scale each basis vector $e_m$ to get $Q(e_m) = \pm 1$). It may not be obvious that $(k,l)$ is an invariant of $Q$: certainly if $(k,l) = (n,0)$ or $(0,n)$ then $Q$ is positive- or negative-definite, but otherwise it might seem we can only determine $l\bmod 2$ (and thus also $k \bmod 2$) from $\det Q$. One way to prove this result (“Sylverster’s law of inertia”) is to define $k^*$ and $l^*$ intrinsically as the maximal dimensions of positive- and negative-dimension subspaces of $V$. Then $k^* + l^* \leq n,$ else the two subspaces would have a nontrivial intersection on which $Q$ is both positive and negative, which is impossible. Diagonalization gives subspaces of dimension $k \leq k^*$ and $l \leq l^*$ with $k+l = n,$ so $k = k^*$ and $l = l^*$ and we’re done. The same proof applies to the signature $(p,q)$ of a nondegenerate sesquilinear form, whose automorphism group is ${\rm U}_{p,q}$ (mentioned together with ${\rm SU}_{p,q}$ on page 98).
[“sesqui-“ is the Latin prefix for 1½, as in sesquicentennial = pertaining to 150 years (a century and a half); a sesquilinear form is a function of two complex vectors that is linear in one variable and conjugate-linear (“semilinear”) in the other, such as $(w,z) = \sum_{i=1}^k \bar w_i z_i - \sum_{i=k+1}^n \bar w_i z_i.$ The sesqui- prefix is thought to have originally been “semis que”.]

Over a finite field $k$ of odd characteristic, for each $n \geq 1$ there are two equivalence class of nondegenerate quadratic forms, distinguished by whether the determinant of a Gram matrix $G = (e_i,e_j)_{i,j=1}^n$ is a square or not. (There is always a $k^\times / {k^\times}^2$ invariant, because a change of basis changes $G$ to ${}^t \! A G A$ which has determinant $(\det A)^2 \det G$.) For other ground fields the question may be more subtle. For examle, over $\bf Q$ the form $x^2+y^2$ is equivalent to $5(x^2 + y^2) = (2x-y)^2 + (x+2y)^2$ but not to $3(x^2+y^2)$; we shall not delve further into such matters in Math 222.

Alternating forms, though less familiar than quadratic forms, actually have a simpler theory: a vector space $V$ of dimension $m$ over any field $F$ admits a nondegenerate alternating pairing $(\cdot,\cdot)$ if and only if $n$ is even, in which case the pairing is unique up to isomorphism. Note that “alternating” is defined by the condition that $\forall x \in V: (x,x) = 0,$ which implies $\forall x,y \in V: (x,y) + (y,x) = 0$ by the usual technique of expanding $0 = (x+y,x+y).$ [The two conditions are equivalent if and only if $F$ does not have characteristic 2.] We prove this by induction on $n$. The case $n=0$ is vacuous, and for $n=1$ the only alternating form is zero which is degenerate. For $n \geq 2$ take any nonzero $x \in V$ and (by nondegeneracy) choose $y \in V$ such that $(x,y)=1.$ Then $x,y$ span a subspace $U_1$ on which $(\cdot,\cdot)$ is nondegenerate; hence $V$ is the direct sum of $U_1$ and its complement $$ V_1 := \{ v \in V: \forall u \in U_1, (u,v) = 0 \} = \{ v \in V: (x,v) = (y,v) = 0 \}. $$ Moreover, the pairing is nondegenerate on $V_1$ iff it is nondegenerate on $V$. Since $\dim V_1 = \dim V - 2$ we are done. This also shows that if $n = 2m$ then $V$ has a basis $\{x_i, y_i : 1 \leq i \leq m\}$ with $(x_i,y_i) = -(y_i,x_i) = 1$ and all other pairings zero. We usually take such a basis in the order $x_1,\ldots,x_m, y_1,\ldots,y_m$ so that the Gram matrix has the block form $\Bigl(\begin{array}{cl}0\!&\!I_m\cr-I_m\!&\!0\end{array}\Bigr).$

The statement of Proposition 7.9 (p.101) includes the claim that $\ker\varphi$ is in the center of $H$; the proof does not seem to address this. If $z \in \ker\varphi$ then $hzh^{-1} \in \ker\varphi$ for all $h \in H;$ since $H$ is connected we can get from $1_H$ to $h$ by a path $\alpha : [0,1] \to H$ and observe that $c: t \mapsto \alpha(t) z \alpha^{-1}$ is a continuous function to the discrete set $\ker\varphi$ and is thus constant. Hence $hzh^{-1} = c(1) = c(0) = z$ and we are done. (In particular it follows that $\pi_1(G)$ is abelian; this is usually proved by using the group structure to construct a homotopy between $\alpha \circ \beta$ and $\beta \circ \alpha$ for any loops $\alpha,\beta$.)

The orthogonal groups ${\rm SO}_n({\bf R}), {\rm SO}_n({\bf C})$ have fundamental group ${\bf Z} / 2 {\bf Z}$ (as claimed on page 102) only for $n \geq 3;$ for $n=2$ the real group is the circle, and the complex group is ${\bf C}^\times,$ both of which have $\pi_1 = {\bf Z}$ (though one can still call the unique ${\bf Z} / 2 {\bf Z}$ cover a real or complex spin group).

In the course of proving that $[X,Y] = XY-YX$ in ${\rm GL}(V)$ [p.107], Fulton and Harris implicitly use the following formula: let $g$ be a differentiable map from an interval in $\bf R$ to ${\rm GL}(V)$, with derivative $g'$; then the derivative of $g^{-1}$ (the composition of $g$ with the inverse map on ${\rm GL}(V)$) equals $-g^{-1} g' g^{-1}$. This generalizes the familiar formula $d(1/u) = -du/u^2$ of elementary calculus, and can be proved in much the same way: differentiate the identity $gg^{-1} = e$ and use the product rule.

A possibly simpler way to identify ${\mathfrak sl}(V)$ with the trace-zero subspace of ${\rm End}(V)$ is to use the formula $\det(\exp X) = \exp({\rm tr}\,X)$ which holds for any complex $n \times n$ matrix $X$ (proof: choose a basis in which $X$ is upper-triangular). [One sometimes sees “traceless” used for “trace zero”; see for example Exercise 7.17 on page 103 of the text.]

A couple of warnings about the exponential map exp from $\mathfrak g$ to $G$:

• exp is in general not a group homomorphism! We shall see from the Campbell-Hausdorff formula that exp is a homomorphism iff the connected component of the identity in $G$ is abelian iff $[X,Y] = 0$ for all $X,Y$ in $G$. The restriction of exp to any one-dimensional subspace of $\mathfrak g$ is a homomorphism, though
• The operation $X*Y := \log(\exp(X) \, \exp(Y))$ is not bilinear — it's claimed to be bilinear on p.117 of the textbook, but that “bilinear” must be a typo for “binary”. Also, in general $X*Y$ is not defined for all $X,Y$ in $\mathfrak g$, only for $X,Y$ sufficiently small.

The key result of the final section (§8.3) of Lecture 8 is the if-part of the “Second Principle” (page 119): if $G,H$ are Lie groups with $G$ simply connected, and ${\mathfrak g}, {\mathfrak h}$ are their Lie algebras, then a linear transformation $\alpha: {\mathfrak g} \to {\mathfrak h}$ lifts to a group homomorphism $G \to H$ if and only if $\alpha$ is a map of Lie algebras (that is, $\alpha[X, Y] = [\alpha X, \alpha Y]$ for all $X,Y \in {\mathfrak g}$). This is done by embedding ${\mathfrak g}$ and ${\mathfrak h}$ into matrix algebras and applying the BCH (Baker–Campbell–Hausdorff) formula. The former is Ado’s theorem, proved in Appendix E.2 (pages 500–503). We shall not dwell on this result because our main interest will be in Lie algebras that are already given as matrix algebras or have trivial center and thus have a faithful adjoint representation ${\mathfrak g} \to {\mathfrak{gl}}({\mathfrak g}),$ $X \mapsto {\rm ad}_X.$ For the latter (BCH), we need the existence of a formula for $X * Y$ as an infinite sum of “Lie polynomials” that converges to $X * Y$ for $X,Y$ sufficiently small. Various proofs of this are known, but I could not think of one that I would expect a reader to find when approaching it as an “Exercise” (8.36 on page 117, even not requiring convergence). One approach is via the formula $$ \frac{d}{dt} \exp X(t) = \exp X(t) \frac{1-e^{-{\rm ad}_X}}{{\rm ad}_X} \frac{dX(t)}{dt} $$ for the derivative of the exponential map $\exp: {\mathfrak g} \to G$. (Here the fraction “$(1-e^{-{\rm ad}_X}) / {\rm ad}_X$” is interpreted as the power series for $(1-e^{-z})/z$ evaluated at $z = {\rm ad}_X$.) This gives rise to the Poincaré integral formula $$ X * Y = X + \left(\int_0^1 \psi\bigl(e^{{\rm ad}_X} e^{t \, {\rm ad}_Y}\bigr) \, dt \right) Y $$ where $\psi(z)$ is the power series for $(z \log z) / (z-1)$, which gives $X * Y$ as an infinite sum of terms $c a_1 a_2 \ldots a_k Y$ ($k \geq 0$), with each $a_i = {\rm ad}_X$ or ${\rm ad}_Y,$ that is readily shown to converge for $X,Y$ in a small enough neighborhood of $0$. Another proof of convergence, using the “method of majorants”, is given in these notes from a 2007 course given at UCLA by Varadarajan, which also prove the formula for differeniating $\exp A(t).$

We mentioned already in class that for large $n$ there are lots of Lie algebras of dimension $n$, even two-step nilpotent ones. This is basically the same construction that produces $p^{2n^3/27-O(n^2)}$ finite groups of order $p^n$. Fix vector spaces $V, W$ of dimensions $2m, m$ respectively. Then a Lie group $G$ with center $Z(G) = W$ and quotient $G/W = V$ encodes an alternating pairing $(\cdot,\cdot): V \times V \to W:$ to compute $(v,v'),$ lift $v,v'$ from $V$ to $\bar v, \bar v' \in G,$ and check that the commutator $[\bar v, \bar v']$ is an element of $W$ independent of the lift. But we did not prove that every bilinear pairing arises this way. For Lie algebras this is easy: since the pairing encodes commutators, all we need do is make $V \oplus W$ a Lie algebra $\mathfrak g$ by defining $[(X,Y), (X',Y')] = (0, (X,X'))$ where $X,X' \in V$ and $Y,Y' \in W,$ and $(X,X')$ is the pairing. (The Jacobi identity holds automatically because $[X,[Y,Z]]=0$ for all $X,Y,Z \in \mathfrak{g}$.) Again there is a $(2m^3-m^2)$-dimensional space of pairings, most of which have trivial kernels (so $W$ is indeed the center of $\mathfrak g$), and only an $O(m^2)$-dimensional space of possible equivalences among them. The correspondence between Lie algebras and simply connected Lie groups then gives us the same profusion of $n$-dimensional Lie groups that are two-step nilpotent and simply connected. (Of course if $n = 3m \pm 1$ we get much the same result by changing $\dim W$ to $m \pm 1$.)

Lecture 9: Much of the terminology and constructions (commutator, the lower central and derived series, etc.) commutes with the map from Lie groups to Lie algebras; for instance, the “center” of a Lie algebra $\mathfrak g$ is defined to be $\{X \in \mathfrak{g} : \forall Y \in \mathfrak{g}, [X,Y] = 0\}$ so that if $\mathfrak{g}$ is the Lie algebra of a connected Lie group $G$ then the center $Z(\mathfrak{g})$ of $\mathfrak{g}$ is the Lie algebra of $Z(G)$. [As it happens $Z(\mathfrak{g})$ could be equivalently defined as $\{X \in \mathfrak{g} : \forall Y \in \mathfrak{g}, [X,Y] = [Y,X]\};$ but that misses the point.] A notable exception is “ideal”, which as the text notes corresponds to “normal subgroup”; I don’t know why this isn’t called a “normal subalgebra”.

Exercise 9.1 is not quite correct as stated: if $G$ is not connected then $Z(G)$ may be strictly smaller than $\exp Z(\mathfrak{g})$. An example is ${\rm O}_2({\bf R}).$ That is why we required $G$ to be connected in the previous paragraph.

On page 122, note that we do not count the 1-dimensional Lie algebra as “simple”; the corresponding Lie groups are somewhat reminiscent of finite cyclic groups $Z_p$ of prime order, which though simple behave quite differently from the non-abelian finite simple groups. In our Lie setting it’s even worse: if we counted a 1-dimensional Lie algebra as simple then some direct sums of simple algebras would not be semisimple (using the “no nonzero solvable ideals” definition on p.123); then some short exact sequences of Lie algebras involving such direct sums would fail to split (see the above construction of large families of nilpotent Lie algebras), as would some short exact sequences of representations of a single Lie algebra (the text already gives an example for the Lie algebra of dimension 1). Anyhow a 1-dimensional Lie group is never simple since it is commutative and has nontrivial discrete subgroups.

Page 123: To see that semisimplicity is not inherited by subalgebras, consider that $\mathfrak{sl}_n$ contains $\mathfrak{n}_n$ (and take $n \geq 3$ if you think a 1-dimensional Lie algebra ought to count as simple).

Note that Engel’s Theorem 9.9 (page 125), requires a Lie subalgebra of $\mathfrak{gl}(V)$, not just a subspace. There are nilpotent subspaces that are not contained in any nilpotent subalgebra; this does not happen for $\dim V \leq 2$ (*), but if $V$ has basis $e_1,e_2,e_3$ then the maps $e_1 \mapsto e_2 \mapsto e_3 \mapsto 0,$ $e_3 \mapsto e_2 \mapsto -e_1 \mapsto 0$ generate a counterexample. (This is the case $m=1$ of a construction from a paper “Linear spaces of nilpotent matrices” (B. Mathes, M. Omladič, and H. Radjavi; Linear Algebra and its Applications 149 (1991), 215–225) that for each $m > 0$ produces a subspace $V \subset \mathfrak{gl}_{3m}$ of dimension $m^2+1$ that consists of nilpotent matrices but is not contained in any ${\rm GL}_{3m}$ conjugate of $\mathfrak{n}_{3m}$. It is known that $\mathfrak{n}_n$ maximizes the dimension of a nilpotent subspace of $\mathfrak{gl}_n$, and that the only nilpotent subspaces attaining this maximal dimension $(n^2-n)/2$ are the conjugates of $\mathfrak{gl}_n$; see for instance this Math Overflow item.)

(*) Suppose $X,Y$ are nonzero $2 \times 2$ matrices all of whose linear combinations are nilpotent. Without loss of generality $Y = \bigl({0 \; 1 \atop 0 \; 0}\bigr).$ Write $X = \bigl({a \; \phantom-b \atop c \; -a}\!\bigr)$ with $a^2 + bc = 0.$ Since $\det(X+tY)=0$ for all $t$ we find $c=0$ and then $a=0$. Thus $X$ is a multiple of $Y$.

In the opening paragraph of the proof of Engel’s Theorem, it is shown that if $X$ is a nilpotent element of ${\rm End}(V)$ then ${\rm ad}(X)$ is a nilpotent linear transformation of $\mathfrak{gl}(V)$, and hence on any Lie subalgebra of $\mathfrak{gl}(V)$. Another, possibly simpler, way to see this is to note that for any $Y \in \mathfrak{gl}(V)$ we may write ${\rm ad}(X)^m(Y)$ as a linear combination of the endomorphisms $X^a Y X^b$ of $V$ with $a+b=m.$ [Digression: what are the coefficients of this linear combination?] If $X$ is nilpotent then $X^n=0$ for some $n$; hence ${\rm ad}(X)^m(Y) = 0$ once $m>2n-2,$ so ${\rm ad}(X)$ is nilpotent as claimed. (Yet another approach: $X$ is nilpotent iff all its eigenvalues are zero, and we know that if $X$ has eigenvalues $\lambda_1,\ldots,\lambda_n$ then ${\rm ad}(X)$ has eigenvalues $\lambda_i - \lambda_j$ with $1 \leq i,j \leq n$.)

Page 126, last paragraph of the proof of Engel: of course $W$ must be defined as “the subspace of all such vectors $v \in V$ together with $0$”, because $v$ is required to be nonzero…

In Lie’s Theorem 9.11, the hypothesis that the solvable Lie subalgebra of $\mathfrak{gl}(V)$ be complex is necessary: over $\bf R$, the span of the $2 \times 2$ matrix $\bigl({\phantom-0\;1 \atop -1\; 0}\bigr)$ (a.k.a. $\mathfrak{so}_2$) in $\mathfrak{gl}({\bf R}^2)$ would be a counterexample. For the same reason, this hypothesis is needed (though not stated) for Exercise 9.16. Also, while the paragraph preceding the statement of Lie’s theorem suggests that it depends on Engel’s Theorem 9.9, the proof seems not to require Engel (though it proceeds along similar lines).

Lie’s Theorem, unlike Engel’s, also requires that our ground field be of characteristic zero. This is needed at the end of the proof, where the trace is used to prove that if a commutator $[X,Y]$ of square matrices is a multiple of the identity then $[X,Y] = 0.$ Indeed Lie’s Theorem becomes false in characteristic $p \gt 0$. It is well known that there are $p \times p$ matrices $X,Y$ with entries in ${\bf Z} / p{\bf Z}$ such that $XY-YX = I;$ then $X,Y,I$ span a Lie subalgebra that is solvable and even nilpotent (isomorphic to $\mathfrak{n}_3 \bmod p)$ but not contained in a conjugate of $\mathfrak{b}_p$ (because in $\mathfrak{b}_p$ a commutator that is a multiple of the identity does have to vanish).

A digression on “unitary trick”: As Serre notes, Weyl chose (for reasons unknown) to call it the “unitarian trick” even though that word is usually associated with the domain of religion rather than mathematics.

While we’re at it, the proof of Lie’s Theorem 9.11 is one of the few places where the traditional use of $\lambda$ for an eigenvalue seems particularly apt: the map $\lambda: \mathfrak{g} \to {\bf C}$ is λinear :-)

We shall not cover any more of Lecture 10 than we did at the end of the lecture of Wednesday (February 28) when we showed that a perfect 3-dimensional Lie algebra over $\bf C$ must be isomorphic with $\mathfrak{sl}_2({\bf C})$. I nevertheless urge you to at least read through it once to see how some remarkable phenomena appear already in dimensions 2 and 3, some of which you may encounter again in or near mathematics even if we shall not need them again in Math 222. Note that in this Lecture the “rank” of a Lie algebra $\mathfrak g$ is the dimension of $[\mathfrak{g}, \mathfrak{g}];$ this is different from the notion of “rank” that we shall use when we come to the general theory of semisimple Lie algebras. For example, $\mathfrak{sl}_2$ has rank 3 in this sense, but rank 1 in the usual sense. (What made it easy to prove this early that $\mathfrak{sl}_2({\bf C})$ is the unique perfect 3-dim. complex Lie algebra is that such an algebra must have $dim([\mathfrak{g}, \mathfrak{g}]) = 3 = \dim(\bigwedge^{\!2}\!\mathfrak{g}),$ which for starters requires $Z(\mathfrak{g}) = 0$.)

The calculation in the first display of page 142 is our first case of the observation that if $X$ and $Y$ are eigenvectors of ${\rm ad}_H,$ with eigenvalues $\lambda$ and $\mu$ respectively, then $[X,Y]$ is also a (possibly zero) eigenvector of ${\rm ad}_H,$ with eigenvalue $\lambda + \mu.$ We shall use this fact several times in our study of Lie algebras and their representations. (The first is the “Fundamental Calculation” at the top of page 148.)

Lecture 11 obtains the irreducible finite-dimensional representations of ${\mathfrak{sl}}_2$ while introducing some of the tools that we shall use to get at the structure and finite-dimensional representations of semisimple Lie algebras in general. For ${\mathfrak{sl}}_2$ the irreducible finite-dimensional representations are the symmetric powers ${\rm Sym}^n V$ of the 2-dimensional defining representation $V$, or equivalently the spaces of homogeneous polynomials $P(x,y)$ of degree $n$ ($n=0,1,2,3,\ldots$). The Lie group ${\rm SL}_2$ then acts by linear substitutions, taking $P(x,y)$ to $P(ax+cy,bx+dy).$ The character of this action on ${\rm Sym}^n V$ is determined by the character of ${\rm diag}(\lambda,\lambda^{-1}),$ because the conjugates of such matrices are dense in ${\rm SL}_2({\bf C})$); and the monomial basis for ${\rm Sym}^n V$ diagonalizes ${\rm diag}(\lambda,\lambda^{-1}),$ letting us compute the character $$ \sum_{k=0}^n \lambda^{n-2k} = \lambda^n + \lambda^{n-2} + \lambda^{n-4} + \cdots + \lambda^{-n} = \frac{\lambda^{n+1} - \lambda^{-(n+1)}}{\lambda - \lambda^{-1}}, $$ which we shall see is the first case of the Weyl character formula.

The rest of the lecture then uses the complete description of finite-dimensional representations of ${\mathfrak{sl}}_2$ to answer various “plethysm” questions such as the decomposition into irreducibles of the tensor product $V \otimes W$ of irreducible representations. We shall cover a bit of this in class and then use the character formula to do some ${\rm SL}_2$ invariant theory.

Warning: §11.1 uses $V$ both for a general irreducible representation of ${\mathfrak{sl}}_2$ and for the 2-dimensional defining representation. In each instance it should be clear from context which $V$ is intended.

For that matter, $W$ is used for ${\rm Sym}^2 V,$ which is also the adjoint representation (${\mathfrak{sl}}_2$ acts on itself by $X \mapsto {\rm ad_X}$), and the defining representation of ${\mathfrak{so}}_3$. This also means that once we show that the representation ${\rm Sym}^n V$ is irreducible for all $n$, the case $n=2$ proves that ${\mathfrak{sl}}_2$ is semisimple and simple.

Wikipedia ascribes to Littlewood the introduction into mathematics of the somewhat ungainly word “plethysm”, at the suggestion of M. L. Clark; this is supported by Richard Stanley’s Enumerative Combinatorics (Vol.2, Appendix 2, cited by this page, which is also the source cited by the “plethysm” entry in this site of “Earliest Known Uses of Some of the Words of Mathematics”). The term is “after the Greek word plethysmós meaning ‘multiplication’”, and should thus be related with the fancy word “plethora” for “a great multiplicity, profusion” (and more distantly with “complete” and all the poly- words, see the Etymonline entry for plethora).

In quantum physics one often applies the “Fundamental Calculation” to a Hamiltonian operator $H$ whose eigenvalues and eigenvectors are the energy levels and energy states of some system. In this setting an operator $A_\pm$ satisfying $[H,A_\pm] = \pm c A_\pm$ may be called a “ladder operator” (with $A_+$ and $A_-$ being respectively “raising” and “lowering operators”) because they increment or decrement the energy level by $c$. Turn the diagram on page 148 by 90° counterclockwise to see the “ladder” that one ascends or descends by applying $X$ or $Y$ respectively.

A nice companion to the result of Exercise 11.22 is the following geometrical description of the map from the exterior square of ${\rm Sym}^2 V$ to ${\rm Sym}^2 V$. The generic element of ${\rm Sym}^2 V$ can be described both as a pair of points in the Riemann sphere ${\bf P}^1({\bf C})$ and as the involution in ${\rm SL}_2({\bf C})$ whose fixed points in ${\bf P}^1({\bf C})$ are that pair. [Any pair is equivalent to $\{0,\infty\},$ which is the fixed pair of the involution $z \leftrightarrow -z$ and of no other involution in ${\rm SL}_2({\bf C})$.] Now for any two disjoint pairs of points there exists a unique involution switching both pairs. But this involution is itself identified with a pair of points. This extends to a map from ${\rm Sym}^2 V \times {\rm Sym}^2 V$ to ${\rm Sym}^2 V$, which is none other than our identification of ${\rm Sym}^2 V$ with its exterior square! That is, if $P,Q$ are quadrics without repeated or common roots then so is their Jacobian $R = J(P,Q),$ and the roots of $R$ are the fixed points of the involution that switches the roots of $P$ and switches the roots of $Q$. To prove this, it is enough to check the case that this involution is $z \leftrightarrow -z.$ But then $P,Q$ are polynomials of the form $x^2 + c y^2$ for distinct choices of $c$, and we readily check that their Jacobian is a multiple of $xy$ as desired. (See also Proposition A in the Appendix of my ANTS-III paper where this result is used.)

We can already see how a good part of the $\mathfrak{sl}_3$ picture (at least through Proposition 12.11) will work for $\mathfrak{sl}_n$ for all $n \gt 2.$ In general:

1. $\mathfrak{h}$ is the $n-1$ dimensional subspace of $\mathfrak{sl}_n$ consisting of diagonal matrices of trace zero. Such matrices all commute, so this is an abelian Lie subalgebra of $\mathfrak{sl}_n$. The corresponding abelian Lie subgroup of ${\rm SL}_n$ is the diagonal matrices of determinant 1.
   While we’re at it: The footnote on page 161 is not quite right: a diagonal matrix $H \in \mathfrak{sl}_n$ cannot have “entries independent over $\bf Q$” (i.e. its diagonal entries linearly independent over $\bf Q$) because they sum to zero. The correct condition is that $H = {\rm diag}(a_1,\ldots,a_n)$ where $\sum_i c_i a_i = 0$ ($c_i \in \bf Q$) if and only if the $c_i$ are all equal. Still, Fulton and Harris are surely correct in asserting that it would be perverse to develop the theory starting from such an $H$ rather than the full subalgebra $\mathfrak{h}$.
2. As on page 163, the dual space $\mathfrak{h}^*$ is the quotient of the span of $L_1,\ldots,L_n$ by its one-dimensional subspace ${\bf C}(L_1 + \cdots + L_n),$ where each $L_i$ is the functional on $\mathfrak{h}$ taking ${\rm diag}(a_1,\ldots,a_n)$ to $a_i$.
3. $\mathfrak{sl}_n$ decomposes as a direct sum of eigenspaces $\mathfrak{g}_\alpha$ of the action of $\mathfrak{h}$. Namely, $\mathfrak{g}_0$ is $\mathfrak{h}$ itself, and all other nonzero eigenspaces are the one-dimensional subspaces $\mathfrak{g}_{L_i - L_j}$ $(i \neq j)$ consisting of $n \times n$ matrices supported on the $(i,j)$ entry. (You should have done this calculation already; see the comment below this numbered list.)
4. The “Fundamental Calculation” applies to this decomposition of $\mathfrak{g}$ (page 164), and more generally to an $\mathfrak{h}$-eigenspace $V_\alpha$ in any representation $V$ of $\mathfrak{g}$ (page 165), showing that ${\rm ad}_\alpha(V_\beta) \subseteq V_{\alpha+\beta}.$
5. We say “more generally” because the decomposition $\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{i\neq j} \mathfrak{g}_{L_i - L_j}$ is the special case where $V$ is the adjoint representation of $\mathfrak{g}$. These $\alpha = L_i - L_j$ ($i \neq j$) are called the roots of $\mathfrak{sl}_n$, and each $\mathfrak{g}_\alpha$ is the root space corresponding to the root $\alpha$ (see the paragraph spanning pages 165–166). By the “Fundamental Calculation”, $[\mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha}]$ is contained in of $\mathfrak{g}_0 = \mathfrak{h}$; we calculate that $[\mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha}] \neq 0,$ so $[\mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha}]$ is a one-dimensional subspace of $\mathfrak{g}_0 = \mathfrak{h}$. This subspace together with $\mathfrak{g}_\alpha$ and $\mathfrak{g}_{-\alpha}$ spans a subalgebra of $\mathfrak{g}$ isomorphic with $\mathfrak{sl}_2$, namely the trace-zero matrices supported on rows $i,j$ and columns $i,j$. This algebra is called $\mathfrak{s}_{L_i-L_j}$ on page 170.
6. For an arbitrary representation $V$ it follows that if some nonzero $v$ is in $V_\alpha$ then the subrepresentation of $\mathfrak{g}$ generated by $v$ is contained in $\oplus_{\beta \in \alpha + \Lambda_R} V_\beta,$ where $\Lambda_R$ is the lattice in $\mathfrak{h}^*$ spanned by the roots $L_i - L_j.$ Therefore Observation 12.6 (page 165) applies: any irreducible finite-dimensional representation $V$ is a direct sum of eigenspaces $V_\alpha$ for $\alpha$ ranging over some coset of $\Lambda_R$ in $\mathfrak{h}^*$.
7. [See pages 170–172 for the case $n=3$.] Suppose $V$ is a finite-dimensional representation of $\mathfrak{sl}_n$ (not necessarily irreducible), and suppose $V_\alpha \neq \{0\}$ for some $\alpha = \sum_i a_i L_i \in \mathfrak{h}^*.$ For $i \neq j$ we can restrict $V$ to a representation of the $\mathfrak{sl}_2$ supported on rows $i,j$ and columns $i,j$ and use our analysis of representations of $\mathfrak{sl}_2$ to find that $a_i - a_j \in {\bf Z}.$ Now $\alpha$ is defined only up to translation by multiples of $(1,\ldots,1);$ the condition $a_i - a_j \in {\bf Z}$ means that $\alpha$ is in the lattice $\Lambda_W \subset \mathfrak{h}^*$ generated by $L_1,\ldots,L_n,$ a lattice that contains $\Lambda_R$ with index $n$ and cyclic quotient. This necessary condition is also sufficient, in the sense that if $\alpha = \sum_i a_i L_i$ with each $a_i \in {\bf Z}$ then $\mathfrak{sl}_n$ has a finite-dimensional representation $V$ such that $V_\alpha \neq \{0\}.$ Indeed the $n$-dimensional representation has $V = \oplus_{i=1}^n V_{L_i},$ and then tensor powers (or even symmetric powers) reach the full lattice $\Lambda_W$. The subscript $W$ stands for “weight”, and $\Lambda_W$ is called the “weight lattice”; its elements — which are the functionals on $\mathfrak{h}$ that can appear as eigenvalues in a finite-dimensional representation of $\mathfrak{sl}_n$ — are called “weights”.
8. As in the case $n=2$ we recover a well-defined formula $\prod_{i=1}^n \lambda_i^{a_i}$ for the eigenvalue of a diagonal matrix ${\rm diag}(\lambda_1,\ldots,\lambda_n) \in \exp\mathfrak{h} \subset {\rm SL}_n({\bf C})$ acting on $V_\alpha.$ [NB this is well-defined, even though $(a_1,\ldots,a_n)$ is defined only up to multiples of $(1,\ldots,1)$, because $\prod_{i=1}^n \lambda_i = \det({\rm diag}(\lambda_1,\ldots,\lambda_n)) = 1$.] The sublattice $\Lambda_R$ yields eigenvalues of the diagonal subgroup of ${\rm SL}_n$ that are invariant under multiplying each $\lambda_i$ by the same $n$-th root of unity, and thus descend to ${\rm PSL}_n({\bf C})$; we shall see that this is true also of the representations of ${\rm SL}_n({\bf C})$ whose $\mathfrak{h}$-eigenvalues lie in $\Lambda_R$.
9. For $n=2$ a finite-dimensional representation was characterized by the largest eigenvalue of $H$. For $n \gt 2$ the eigenvalues of $\mathfrak{h}$ are in a space of dimension $\gt 1,$ so we must make a further choice to define the highest weight. In fact we already made such a choice for $n=2$ because $H$ and $-H$ are equivalent under an automorphism of $\mathfrak{sl}_2$. For arbitrary $n \gt 1$ we use the choice of linear functional $l$ on $\mathfrak{h}^*$ (see page 166), which is to say $l \in \mathfrak{h}.$ We could choose $l$ to be irrational as Fulton-Harris do, but all that we need is that the hyperplane $\ker(l)$ contains none of the roots and thus separates them with positive roots (those with $l(\alpha) \gt 0$) on one side and negative roots (with $l(\alpha) \lt 0$) on the other. That is, $l = {\rm diag}(a_1,\ldots,a_n)$ for some pairwise distinct $a_i$ (with $\sum_i a_i = 0$ as usual). Then the root $L_i - L_j$ is positive iff $a_i \gt a_j.$ This amounts to breaking the $S_n$ symmetry of coordinate permutations by choosing a total order. Having already numbered the coordinates $1,\ldots,n$ we already have such an order: we choose $a_i$ with $a_1 \gt a_2 \gt \cdots \gt a_n$, and then the positive roots are $L_i - L_j$ with $i \lt j.$ The root spaces $\mathfrak{g}_\alpha$ with $\alpha > 0,$ together with $\mathfrak{h},$ generate the solvable subalgebra $\mathfrak{b}_n \subset \mathfrak{sl}_n$ of upper-triangular matrices; removing $\mathfrak{h}$ leaves its nilpotent subalgebra $\oplus_{\alpha>0} \mathfrak{g}_\alpha = \mathfrak{n}_n$ of strictly upper-triangular matrices.
10. Once $n \geq 3$ there are roots $\alpha,\alpha'$ for which the root spaces $\mathfrak{g}_\alpha, \mathfrak{g}_{\alpha'}$ do not commute even though $\alpha + \alpha' \neq 0.$ Since $[\mathfrak{g}_\alpha, \mathfrak{g}_{\alpha'}] \subseteq \mathfrak{g}_{\alpha + \alpha'},$ this requires that $\alpha + \alpha'$ be a root as well; equivalently, $\{\alpha, \alpha'\} = \{L_i - L_j, L_j - L_k\}$ for some distinct $i,j,k$ (that’s why we need $n \geq 3$), and then $\alpha + \alpha' = L_i - L_k.$ In this case we see (by explicit computation of the commutator) that indeed $\mathfrak{g}_\alpha$ and $\mathfrak{g}_{\alpha'}$ do not commute. In other words (and formulas): $[\mathfrak{g}_\alpha, \mathfrak{g}_{\alpha'}] = \mathfrak{g}_{\alpha + \alpha'}$ for any roots $\alpha,\alpha'$ such that $\alpha + \alpha' \neq 0.$
11. It follows that $\mathfrak{n}_n$ is generated as a Lie algebra by the spaces $\mathfrak{g}_\alpha$ where $\alpha = L_i - L_{i+1}$ $(1 \leq i \leq n-1)$; these $\alpha$ are the $n-1$ simple roots, i.e. the positive roots that cannot be written as $\beta+\beta'$ for positive roots $\beta,\beta'.$ Then $\mathfrak{b}_n$ is generated as a Lie algebra by the same $\mathfrak{g}_\alpha$ together with $\mathfrak{h}$, and the full algebra $\mathfrak{sl}_n$ is generated by $\mathfrak{g}_\alpha,\mathfrak{h},\mathfrak{g}_{-\alpha}$ with $\alpha$ ranging over the simple roots.

We can now give the proof of Claim 12.10 (page 167, proved on pages 168–169) and Proposition 12.11 (page 169) for representations $V$ of $\mathfrak{sl}_n$. Let $v \in V$ be a nonzero eigenvector of $\mathfrak{h}$ such that $\mathfrak{g}_\alpha v = 0$ for all simple roots $\alpha$, and thus for all positive roots $\alpha$. Define subspaces $V_k \subset V$ ($k \in {\bf Z}$) as follows: $V_0 = {\bf C} v,$ and $V_k = \{0\}$ if $k \lt 0;$ For $k \gt 0$ we define $V_k$ inductively as the space generated by $\mathfrak{g}_{-\alpha} V_{k-1}$ with $a$ ranging over the simple roots. By induction (and the “Fundamental Calculation”) each $V_k$ is generated by its $\mathfrak{h}$-eigenspaces, so $\mathfrak{h} V_k \subseteq V_k.$ Moreover, denoting by $w \in \mathfrak{h}^*$ the eigenvalue of $v,$ we see that $V_k$ is contained in the union of $V_{w'}$ over $w'$ such that $w - w'$ is a linear combination of simple roots, say $w - w' = \sum_i a_i (L_i - L_{i+1}),$ whose coefficients $a_i$ sum to $k$. Since the simple roots are linearly independent (indeed a basis for $\mathfrak{h}^*$), it follows that the vector space sum $\bigoplus_{k \in {\bf Z}} V_k$ is direct.

We claim: $\mathfrak{g}_\alpha V_k \subseteq V_{k-1}$ for all $k \in {\bf Z}$ and each simple root $\alpha$. It will follow that $\bigoplus_{k \in {\bf Z}} V_k$ is the subrepresentation of $V$ generated by $v$.

The claim is clear for $k \leq 0$ (trivially for $k \lt 0,$ and by our hypothesis on $v$ for $k=0$). for $k \gt 0$ we use induction. (This is much the same argument that Fulton and Harris call only “a sort of induction” towards the bottom of page 168; as far as I can see it is induction.) Suppose we have proved the claim for some $k \geq 0.$ To prove the $k+1$ case, we must show that $\mathfrak{g}_{\alpha'} \mathfrak{g}_{-\alpha} v \subseteq V_k$ for all $v \in V_k$ and all simple roots $\alpha,\alpha'.$ Indeed for $X \in \mathfrak{g}_{\alpha'}$ and $Y \in \mathfrak{g}_{-\alpha}$ we have $XYv = YX v + [X,Y] v,$ and $Xv \in V_{k-1}$ by the inductive assumption so $YX v \in Y V_{k-1} \subseteq V_k.$ As for $[X,Y] v,$ if $\alpha = \alpha'$ then $[X,Y] \in \mathfrak{h}$ so $[X,Y] v \in V_k.$ Otherwise we claim that $\alpha - \alpha'$ is not a root, and thus $[X,Y]$, being in $V_{\alpha - \alpha'},$ must vanish, so $[X,Y] v = 0 \in V_k$ and we are done. This claim can be seen by writing $\alpha - \alpha' = (L_i - L_{i+1}) - (L_{i'} - L_{i'+1})$ for some distinct $i,i';$ alternatively, if $\alpha - \alpha'$ were a root then it would be either positive or negative, and either way we would obtain the simple root $\alpha$ or $\alpha'$ as the sum of two positive roots, which is a contradiction. This proves our claim that $\bigoplus_{k \in {\bf Z}} V_k$ is the subrepresentation of $V$ generated by $v$.

To finish the proof of Proposition 12.11 (that this subrepresentation is irreducible) we need the result that finite-dimensional representations of a semisimple Lie algebra are direct sums of irreducibles, which we have not had to use yet. Using that result, the proof in Fulton–Harris (p.169) for $\mathfrak{sl}_3$ works equally for $\mathfrak{sl}_n$ once we know that $\mathfrak{sl}_n$ is actually semisimple. In fact $\mathfrak{sl}_n,$ like $\mathfrak{sl}_2$, is simple. That is, the adjoint representation $\mathfrak{g}=\mathfrak{sl}_n$ has no subrepresentations other than $\{0\}$ and $\mathfrak{g}$ itself. Indeed if a subrepresentation contains any nonzero $X \in \mathfrak{g}$ then it contains a nonzero eigenvector of $\mathfrak{h}$, so may assume that either $X \in \mathfrak{h}$ or $X$ is in some root space. In either case we readily recover all of $\mathfrak{g}$ by repeatedly applying ${\rm ad}_Y$ for $Y$ in various root spaces and then taking linear combinations.

A nonzero $v \in V_0$ is called a “highest weight vector” in $V,$ and its eigenvalue $w \in \mathfrak{h}^*$ is called the “highest weight” of $V.$ Some basic examples of irreducible finite-dimensional representations $V$ of $\mathfrak{sl}_n,$ and their highest weights $w \in \Lambda_W$ and subspaces $V_k$:

• If $V$ is the trivial representation $\bf C$, then of course $w=0$ and $V=V_0.$
• If $V$ is the defining representation, then $V_0$ is the first coordinate line, with $w=L_1,$ and then $V_k$ is the $(k+1)$-st for $0 \lt k \lt n,$ with $V_k = \{0\}$ for $k \geq n.$
• If $V$ is the dual of the defining representation, then $V_0$ is generated by the $n$-th dual basis vector, with $w=-L_n,$ and then $V_k$ is the line generated by the $(n-k)$-th dual basis vector for $0 \lt k \lt n,$ again with $V_k = \{0\}$ for $k \geq n.$
• Finally (for now), if $V$ is the adjoint representation $\mathfrak g$, then $V_0 = \mathfrak g_{L_1 - L_n}$ with $w = L_1 - L_n,$ and then each $V_k$ is the union of the root spaces $\mathfrak g_{L_i - L_j}$ with $j-i = n-1-k,$ which has dimension $n-|n-1-k|$ for each integer $k \in [0, 2n-2],$ except that $V_{n-1} = \mathfrak{h}$ which has dimension $n-1$.

I forgot to note this in class: If $w$ is the highest weight of a finite-dimensional representation, we know from the representation theory of $\mathfrak{sl}_2$ that $w(\alpha)$ must be a nonnegative integer for each positive root $\alpha$. Such $w$ is called a dominant weight. It is again sufficient to require this for each simple root $\alpha$. If $w = \sum_i a_i L_i,$ this condition means that $a_i \geq a_j$ for $i \leq j,$ and it is enough to require $a_i \geq a_{i+1}$ for each $i=1,\ldots,n-1.$ That is, $w$ is a nonnegative integer combination of $L_1, L_1+L_2, L_1+L_2+L_3, \ldots, \sum_{i=1}^{n-1} L_i$ (the last of which is also $-L_n$). We shall soon identify these fundamental weights $w_j := \sum_{i=1}^j L_i$ ($1 \leq j \leq n-1$) with the basis of $\mathfrak{h}^*$ dual to the basis $L_1-L_2, L_2-L_3, \ldots, L_{n-1}-L_n$ under a natural inner product (which for $\mathfrak{sl}_n$ comes from the inner product $(X,Y) = {\rm tr}(XY)$ on $\mathfrak{h}$).

To prove that each $w$ is the highest weight of some irreducible representation of $\mathfrak{sl}_n$ we need only find some finite-dimensional representation $W$ with highest weight $w$, because we know by now that any highest-weight vector in such a representation generates the desired irreducible. For $n=3,$ we have $w = d_1 w_1 + d_2 w_2 = d_1 L_1 - d_2 L_3,$ so $W = {\rm Sym}^{d_1} V \otimes {\rm Sym}^{d_2} V^*$ works. For $n \gt 3$ we need a new idea: if $w = \sum_{j=1}^{n-1} d_j w_j$ for some integers $d_j \geq 0$, take $$ W = \bigotimes_{j=1}^{n-1} {\rm Sym}^{d_j} (\wedge^j V). $$ The point is that $\wedge^j V$ is a representation with highest weight $\sum_{i=1}^j L_i = w_j.$ Note that $\wedge^{n-1} V \cong V^*$ as representations of ${\rm SL}_n$ (and thus of $\mathfrak{sl}_n$), so for $n=3$ we recover ${\rm Sym}^{d_1} V \otimes {\rm Sym}^{d_2} V^*.$

Once $n \gt 2$ there is at least one irreducible representation for which some eigenspace of $\mathfrak{h}$ has dimension $\gt 1$ (namely the zero eigenspace of the adjoint representation, which is its $(n-1)$-dimensional subspace $\mathfrak{h}$). For each $n$ there are still infinitely many irreducible representations for which all eigenspaces do have dimension $0$ or $1.$ The symmetric powers of the defining representation $V$ satisfy this condition: the degree-$d$ monomials in the $n$ coordinates constitute an eigenbasis for ${\rm Sym}^d V,$ with all eigenvalues distinct. Likewise symmetric powers of $V^*$. This includes the case $n=2,$ for which we know that ${\rm Sym}^d V \cong {\rm Sym}^d V^*$ are the only irreducible representations. For $n=3,$ these representations ${\rm Sym}^d V$ and ${\rm Sym}^d V^*$ are the only ones for which there are no eigenspaces of dimension $\gt 1$; the representations $\wedge^j V$ ($1 < j < n-1$) provide a few further examples for each $n \gt 3$.

The classification of root systems and their diagrams appears also in mathematical contexts far removed from the theory of Lie groups and algebras; it has also been extended in various directions by relaxing or modifying some of the axioms (1)–(4) (page 320) of a root system. We indicate some of these connections and generalizations during the commentary on this Lecture.

Axiom 2: In some sources, the condition $k^2 \neq 1 \Rightarrow k\alpha \notin R$ is dropped; root systems that do satisfy this condition are then said to be “reduced”. This does not much change the picture (though non-reduced root systems are not directly relevant to Lie theory). There is only one additional family of irreducible root systems that are not reduced: the systems “$BC_n$” that are obtained by combining the roots of $B_n$ and $C_n$ (see below), i.e. $\pm L_i, \pm 2L_i, \pm L_i \pm L_j$ $(i\neq j)$ where $L_1,\ldots,L_n$ is an orthonormal basis for $\mathbb{E}^n$.

In any case we must require $0 \notin R$; there is no reflection $W_0!$

Axioms 3,4: The reflection $W_\alpha$ takes any $x \in \mathbb{E}^n$ to $x$ minus twice the projection of $x$ to ${\bf R}\alpha$: $$ W_\alpha(x) = x - 2 \frac{(x,\alpha)}{(\alpha,\alpha)} \alpha. $$ Thus the definition $n_{\beta\alpha} = 2(\beta,\alpha) / (\alpha,\alpha)$ in Axiom 4 is not as unmotivated as it may seem. (This coefficient $n_{\beta\alpha}$ in $W_\alpha(\beta) = \beta - n_{\beta\alpha} \alpha$ already appeared towards the end of Lecture 14, see the displayed formula on page 208 following Exercise 14.28.)

Examples of root systems:
$\circ$ $A_{n-1}$, consisting of the roots $L_i-L_j$ of $\mathfrak{sl}_n$.
$\circ$ $B_n,$ which consists of $\pm L_i$ and $\pm L_i \pm L_j \ (i\neq j)$ where $L_1,\ldots,L_n$ is an orthonormal basis for $\mathbb{E}$.
$\circ$ $C_n,$ which consists of $\pm 2L_i$ and $\pm L_i \pm L_j \ (i\neq j)$ where $L_1,\ldots,L_n$ is an orthonormal basis for $\mathbb{E}$. Note that in $B_n$ and $C_n$ not all roots have the same norm.
$\circ$ $D_n$ ($n \geq 2$), which consists of just $\pm L_i \pm L_j \ (i\neq j)$ where $L_1,\ldots,L_n$ is an orthonormal basis.
$\circ$ if $R$ is a root system, so is $cR$ for any $c \in {\bf R}^\times.$ We usually identify $cR$ with $R,$ and choose the smallest $|c|$ that makes all the inner products $(\alpha,\beta)$ integral. (Exception: we scale $A_1$ so that $(\alpha,\alpha) = 2,$ not 1.) This choice is generally not the Killing form but some fraction of it (see Exercise 14.36*).
$\circ$ The “orthogonal direct sum of two root systems” (bottom of page 321) is defined as follows. If $R_1,R_2$ are root systems in Euclidean spaces $\mathbb{E}_1, \mathbb{E}_2$ then we get a root system $R_1 \oplus R_2$ in $\mathbb{E}_1 \oplus \mathbb{E}_2,$ defined by $R_1 \oplus R_2 := (R_1,0) \cup (0,R_2).$ (For example, if $\alpha \in R_1$ then $W_{(\alpha,0)}$ acts as $W_\alpha$ on $\mathbb{E}_1$ and as multiplication by $-1$ on $\mathbb{E}_2$.) This also means that if $R_1,R_2$ are associated to semisimple Lie algebras $\mathfrak{g}_1, \mathfrak{g}_2$ then the direct sum of $R_1,R_2$ corresponds to $\mathfrak{g}_1 \oplus \mathfrak{g}_2,$ which as we know is also the Lie algebra of $G_1 \times G_2$ if $G_1,G_2$ are Lie groups with algebras $\mathfrak{g}_1, \mathfrak{g}_2$.

A root system is irreducible if it cannot be written as $R_1 \oplus R_2$ for positive-dimensional $R_1,R_2.$

The Weyl group $\mathfrak{W}$ of $R$ is the subgroup of ${\rm O}(\mathbb{E})$ generated by the reflections $W_\alpha$. [The ostentatiously ornate symbol $\mathfrak{W}$ is a Fraktur capital letter W, and will thus be written W on the blackboard.] This group is finite because it injects into the group of permutations of $R$. The Weyl group of $A_{n-1}$ is the symmetric group $S_n$; for both $R=B_n$ and $R=C_n$ the same Weyl group is the hyperoctahedral group (a.k.a. signed permutation group) of order $2^n n!$ which is the semidirect product of $S_n$ with $\{\pm1\}^n$. For $D_n$ the reflection $W_{L_i-L_j}$ switches coefficients $a_i,a_j$ of a vector $\sum_{i=1}^n a_i L_i$, while $W_{L_i+L_j}$ takes $a_i,a_j$ to $-a_j,-a_i$; hence $\mathfrak{W}$ is the group of $2^{n-1} n!$ signed permutations with an even number of $-1$ signs (see page 271). If $R_1,R_2$ are root systems with Weyl groups $\mathfrak{W}_1, \mathfrak{W}_2$ then $R_1 \oplus R_2$ has Weyl group $\mathfrak{W}_1 \times \mathfrak{W}_2.$ Note that in general $\mathfrak{W}$ may be strictly smaller than ${\rm Aut}(R)$, though $\mathfrak{W}$ is always a normal subgroup of ${\rm Aut}(R)$. For example, ${\rm Aut}(R)$ always contains $-I$, but the Weyl group of $A_{n-1}$ does not once $n \gt 2.$

Finite subgroups of ${\rm O}_n$ that are generated by reflections are not common; the only irreducible reflection groups for $n \geq 3$ are the Weyl groups of root systems and the symmetry groups of the regular icosahedron (equivalently: the regular dodecahedron) for $n=3$ and of the 120-cell (equivalently: its dual polytope, the 600-cell) for $n=4.$ The irreducible finite subgroups of ${\rm U}_n$ generated by reflections were classified by G.C. Shephard and J.A. Todd (Finite unitary reflection groups, Canad. J. Math. 6 (1954), 274–304; see Table VII on page 301, and also the Wikipedia page for such groups). Here a “reflection” is a linear transformation $T,$ not necessasrily of order 2, such that $I-T$ has rank 1.

Suppose $\alpha,\beta \in R$ with $\beta \neq \pm \alpha,$ and let $\vartheta$ be the angle between these roots (as in the discussion in Fulton–Harris starting at the bottom of page 320). Then $W_\alpha W_\beta$ fixes the orthogonal complement of the plane ${\rm Span}(\alpha,\beta),$ and rotates that plane by an angle $2\vartheta$. […]

On page 337, the display preceding Exercise 21.24 can be written more compactly as $({\rm ad}(X_i))^{1 - n_{ij}} X_j = ({\rm ad}(Y_i))^{1 - n_{ij}} Y_j = 0.$

We noted already that the diagram involution of $A_{n-1}$ (for $n \geq 3$) corresponds to an outer automorphism of $\mathfrak{sl}_n$ such as the involution $X \mapsto -X^t$ (where $t$ denotes transpose), corresponding to the inverse-transpose automorphism $A \mapsto (A^t)^{-1}$ of the Lie group ${\rm SL}_n$. The diagram involution of $D_n$ cannot be described this way, because ${\rm SO}_{2n}$ already consists of matrices $A$ equal to their inverse transpose. Instead the outer automorphism is conjugation by an orthogonal matrix $M$ of determinant $-1$, for example $M = {\rm diag}(-1,1,1,\ldots,1)$. This does not work for ${\rm SO}_{2n+1}$ because $-I_{2n+1}$ has determinant $-1$, so conjugation by $M$ is the same as conjugation by $-M \in SO_{2n+1}$.

The proof of Claim 21.25 (page 338) parallels the proof of the uniqueness half of Theorem 14.18 (Lecture 14.1, page 205). The cryptic-looking symbol $\mathfrak{k}$ (as in “The kernel of the second projection is $\mathfrak{k} \oplus 0$”) is a Fraktur lower-case k, presumably chosen to suggest “kernel”.

In Lecture 23 (top of p.375) $\Lambda$ is introduced as shorthand for the weight lattice $\Lambda_W$; we naturally regard the character of a representation of a semisimple Lie group as a nonnegative element of the group ring ${\bf Z}[\Lambda]$. This is the context for “Writing $x_i = e(L_i)$ in ${\bf Z}[\Lambda]$ as in the preceding lecture …” (top of p.400); for $L \in \Lambda$ the notation “$e(L)$” just means $L$ considered as an element of that group ring, so $e: \Lambda \to {\bf Z}[\Lambda]^\times$ is a homomorphism.

We already saw that the “complete symmetric polynomial” $H_d$ (see page 404) is the character of ${\rm Sym}^d({\bf C}^n)$; this is also the case $r=1$ of formula 24.10 on that page. Earlier we found that this is a Schur polynomial for $n=2$ (in class — it’s just a geometric series) and $n=3$ (in the fourth problem set). One way to do this in general is to start from the generating function $\sum_{d=0}^\infty H_d t^d = \prod_{i=1}^n (1-x_i t)^{-1}$ (given on page 453, at the start of Appendix A on symmetric polynomials, with $d=j$; to prove it, expand each factor $(1-x_i t)^{-1}$ in a geometric series, distribute, and collect $t^d$ terms). This yields a linear recursion for the $H_d$: let $s_k$ ($0 \leq k \leq n$) be the $k$-th elementary symmetric function of $x_1,\ldots,x_n$, so $\prod_{i=1}^n (1-x_i t) = \sum_{k=0}^n (-1)^k s_k t^k$; then $\sum_{k=0}^n (-1)^k s_k H_{d-k} = 0$ for each $d \gt 0$, with $H_0 = 1$ and $H_d = 0$ for $d\lt 0$. Now check that the Schur polynomials $S_{d,0,0,\ldots,0}$ satisfy the same recursion (we only need $H_d=0$ for the values $1-n \leq d \leq -1$ for which the Schur determinant has two equal rows). That follows from the observation that those Schur determinants differ only in their first columns $(x_1^{d+n-1},\ldots,x_n^{d+n-1})^t$ — and those columns satisfy the same linear recursion for all $d \in {\bf Z}.$

As for the character of $\wedge^d({\bf C}^n)$, which is the elementary function $s_d$ itself: Consider the Vandermonde determinant $V_{n+1}(z,x_1,\ldots,x_n)$, and factor it as $V_n(x_1,\ldots,x_n) \prod_{i=1}^n (z-x_i)$. Now expand the determinant $V_{n+1}(z,x_1,\ldots,x_n)$ by minors of the first column (powers of $z$), divide by $V_n(x_1,\ldots,x_n)$, and compare $z^{n-d}$ coefficients to identify $s_d$ with a Schur polynomial.

You should now be able to show that for the adjoint representation of $\mathfrak{sl}_n$ the Schur polynomial is $s_1 s_{n-1} - s_n$. In our setting $s_n = \prod_{i=1}^n x_i = 1$, so $$ s_1 s_{n-1} - s_n = s_1 \frac{s_{n-1}}{s_n} - 1 = \Bigl( \sum_{i=1}^n x_i \Bigr) \Bigl( \sum_{j=1}^n x_j^{-1} \Bigr) - 1 = n-1 + \mathop{\sum\!\!\sum}_{i\neq j} x_i^{\phantom.} x_j^{-1}, $$ which agrees with the decomposition $\mathfrak{sl}_n = \mathfrak{h} \oplus \bigoplus_{i\neq j} \mathfrak{g}_{L_i-L_j}$.

Once we know that the quotient $A_{\lambda+\rho} / A_\rho$ in the Weyl character formula (WCF, bottom of page 400) is in ${\bf Z}[\Lambda]$, we soon deduce that $A_{m'\rho} / A_{m\rho} \in {\bf Z}[\Lambda]$ for any integers $m,m'$ with $m|m'$. Thus the sequence $\{A_{m\rho}\}_{m\in{\bf Z}}$ is a “divisibility sequence” satisfying a linear recurrence whose order is the number of roots. For the $A_1$ root system, this sequence is familiar, as are specializations like the Fibonacci numbers (technically we’d have to use the Fibonacci numbers of even order, because $F_m = (\varphi^m - \bar\varphi^m) / (\varphi-\bar\varphi)$ where $\varphi$ is the golden ratio, and $\varphi \bar\varphi = -1$ rather than $+1$). For some larger root systems similar examples are known but less familiar; for example, specializing the $A_2$ root system’s $x_1,x_2,x_3$ to the solution of $x^3 + x^2 - 1$ yields the “sixth-order linear divisibility sequence” $$ 1, -1, 1, -1, -1, 5, -8, 7, 1, -19, 43, -55, 27, 64, -211, 343, -307, -85, 911, -1919, \ldots $$ (OEIS Sequence A005120); using the same $x_1,x_2,x_3$ and the $G_2$ root system yields a 12th-order sequence $$ 1, -1, 1, -5, -1, 35, -64, 35, 19, -209, 989, -1925, 729, 4096, -12871, 29155, -42059, -11305, 208619, -527725, \ldots, $$ with generating function $$ \frac {t \, (1 + 4t^2 - 4t^3 + 7t^4 - 7t^6 + 4t^7 - 4t^8 - t^{10})} {(1+2t+2t^2-t^3+4t^4-t^5+t^6) \, (1-t+4t^2-t^3+2t^4+2t^5+t^6)}, $$ which is the product of the previous example with the sixth-order Sequence A001945 $$ 1, 1, 1, 5, 1, 7, 8, 5, 19, 11, 23, 35, 27, 64, 61, 85, 137, 133, 229, 275, \ldots. $$ Lecture 26

We introduce Weyl’s proof of WCF with the most familiar case of ${\rm SL}_n$ and $\mathfrak{sl}_n$, for which we already have many of the ingredients and can most easily fill in the rest. While Weyl’s proof requires more analysis than we have used so far, the quotients $A_{\lambda+\rho} / A_\rho$ appear much more naturally than in the combinatorial proof in Lecture 25. First the product formula for the common denominator $A_\rho$ appears in the projection from uniform measure on ${\rm SU}_n$ to a measure on a Cartan torus; then the numerators are seen to be orthogonal with respect to the uniform measure on that torus because they are Fourier polynomials with disjoint support. Not only is this consistent with orthogonality of irreducible characters of a compact group, but we shall see that this necessary condition is also sufficient to determine the characters uniquely, given what we know already about highest-weight representations.

The plan is as follows:

1. Thanks to the “unitary/unitarian trick” we can work with the compact form of the real Lie algebra $\mathfrak{sl}_n({\bf R})$, i.e. the unitary group ${\rm SU}_n$ (which gives its name to the “trick”). The associated Lie algebra is $\mathfrak{u}_n$ and consists of the $n \times n$ anti-Hermitian matrices $X = -\bar{X}^{\rm t}$.
2. For a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{u}_n$ we can take the pure imaginary diagonal matrices. The corresponding Cartan subgroup $T = \exp\mathfrak{h}$ of ${\rm SU}_n$ is then the group of diagonal matrices with diagonal entries of norm 1, which is an $(n-1)$-dimensional torus (hence the name $T$, replacing $H$ for the Cartan subgroup of ${\rm SL}_n$). This exponential map, unlike the one for ${\rm SL}_n$, has a nontrivial kernel: $\exp X = 1$ iff each entry of the diagonal matrix $X$ is in $2\pi i{\bf Z}$ iff $X \in 2\pi i \Lambda_R$.
3. Every $g \in {\rm SU}_n$ is ${\rm SU}_n$-conjugate to some $h \in T$. This is even better than the situation for ${\rm SL}_n$ and other adjoint-form semisimple Lie groups, in which the semisimple elements form a dense but proper subset (though it still contains a dense open set). [Recall that in ${\rm SL}_n$ “semisimple” means the same as “diagonalizable”; in ${\rm SU}_n$ it is known that every $g$ is diagonalizable and has an orthonormal basis, which means it can be diagonalized by conjugation by a unitary matrix.] Moreover, $h$ is uniquely determined up to permutation (that is, up to the action of the Weyl group $\mathfrak{W}$), and the choice of permutation is unique in a dense open subset of $\rm{U}_n$, namely all $g \in {\rm SU}_n$ with distinct eigenvalues.
4. Therefore a character of a representation $V$ of ${\rm SU}_n$ can be regarded as a function on $T$. Moreover, this function is $e({\rm Char}(V))$ where $e$ is the ring homomorphism taking any $\lambda \in \Lambda := \Lambda_W$ to the function $T \to {\bf C}$, $\exp X \mapsto \exp((\lambda,X))$. We have seen a real-exponential version of this map for ${\rm SL}_n$, but the unitary case is more rigid: since $\Lambda$ is the dual lattice of $\Lambda_R$, the map is well-defined (recall that $X \cong \mathfrak{h} / 2\pi i \Lambda_R$ where $\mathfrak{h}$ is the space of traceless pure imaginary diagonal matrices); moreover $e(\Lambda)$ is the group of continuous characters $T \to {\bf C}^*$, which also means that $e({\bf Z}[\Lambda])$ is dense in the space $C(T,{\bf C})$ of continuous maps $T \to {\bf C}$ by Fourier analysis on the torus.
5. The image of the homomorphism ${\rm Char}$ from the representation ring $R(\mathfrak{g})$ to ${\bf Z}[\Lambda]$ is contained in the invariant ring $({\bf Z}[\Lambda])^{\mathfrak{W}}\!$. Theorem 23.24 (page 376) identifies $({\bf Z}[\Lambda])^{\mathfrak{W}}$ with a polynomial ring ${\bf Z}[{\rm Char}(\Gamma_1),\ldots,{\rm Char}(\Gamma_n)]$ where each $\Gamma_j$ is the irreducible representation whose highest weight is the $j$-th fundamental weight $\omega_j$. We have not yet proved the existence of $\Gamma_j$ in general, but we do know this for ${\rm SL}_n$ and thus for ${\rm SU}_n$. Namely, $\Gamma_j$ is the $j$-th exterior power of the defining representation ${\bf C}^n$. For a general semisimple Lie algebra we can at least see that $({\bf Z}[\Lambda])^{\mathfrak{W}}$ is a polynomial ring with $\dim(\mathfrak{h})$ generators, which can be taken to be the sums of the $\mathfrak{W}$-orbits of the $e(\omega_j$). We can also see that ${\rm Char}: R(\mathfrak{g}) \to {\bf Z}[\Lambda]^{\mathfrak{W}}$ is injective, and is surjective iff each $\omega_j$ is indeed the highest weight of some representation.
6. Multiplication by $A_\rho$ is an isomorphism from ${\bf Z}[\Lambda]^{\mathfrak{W}}$ to the ${\bf Z}[\Lambda]^{\mathfrak{W}}\!$-submodule of ${\bf Z}[\Lambda]$ that is anti-invariant under each root reflection $W_\alpha$ and is thus sign-invariant under $\mathfrak{W}$. The functions $e(A_{\lambda+\rho})$ thus consistute an orthogonal topological basis for the sign-invariant functions on $T$, each with norm $\left|\mathfrak{W}\right|$ (which is $n!$ in the case of ${\rm U}_n$).
7. The characters of irreducible representations are orthogonal with respect to the measure that averages over ${\rm SL}_n$. For conjugation-invariant functions this corresponds to the measure on $T$ that multiplies by $\left|\mathfrak{W}\right|^{-1}e(A_\rho)$ and then averages. The is the ${\rm U}_n$ case of Weyl’s integration formula, (26.19) on page 443.

The basis $(B,C,H)$ for $\mathfrak{su}_2$ obtained at the top of page 432 identifies this Lie algebra with ${\bf R}^3$ together with the usual cross product on an oriented three-dimensional Euclidean space: $[B,C] = H,\; [C,H] = -[H,C] = B,$ and $[B,C] = H.$ You should check that this also identifies $\mathfrak{su}_2$ with the real Lie algebra $\mathfrak{so}_3$ of skew-symmetric $3 \times 3$ matrices.

At the start of the proof of Proposition 26.4 we must of course take nonzero $H \in \mathfrak{h}_0$. Likewise for $Z \in \mathfrak{l}_\alpha$ later in the paragraph. We already know that ${\rm ad}(H)^2$ vanishes on $H$ itself and acts on each root space $\mathfrak{g}_\alpha$ by multiplication by $\alpha(X)^2$.

In the first displayed equation of the subsection “Split Forms and Compact Forms” (page 434), $\mathfrak{i}_\alpha$ seems to be a typo for the $\mathfrak{j}_\alpha$ that appears several times in the ensuing text (i.e. Fraktur i should be Fraktur j).

The classification of real forms of a complex group $G$ is also one manifestation of non-abelian group cohomology: if complex conjugation acts by $g \mapsto \bar g$ then the real forms are classified by the first cohomology set $H^1(\{\pm1\}, G)$ where $-1$ acts on $G$ by $g \mapsto \bar g$. Note that when $G$ is non-abelian $H^1(\{\pm1\}, G)$ is not a group but just a “pointed set”, with a distinguished element but no group law. The general picture is as follows. Suppose $G$ and $\Gamma$ are groups and $\Gamma$ acts on $G$ by automorphisms (so we have a homomorphism $\Gamma \to {\rm Aut}(G)$). Form the semidirect product $\overline G$, which can be defined as the set $G \times \Gamma$ with the group law $(g,\gamma) (g',\gamma') = (g \gamma(g'), \gamma \gamma')$. Thus $\overline G$ has subgroups $\{(g,1) : g \in G\}$ and $\{(1,\gamma) : \gamma \in \Gamma \}$ isomorphic with $G,\Gamma$ respectively, and every $(g,\gamma) \in \overline G$ is $(g,1)(1,\gamma)$ while conjugation by $(1,\gamma)$ takes any $(g,1)$ to $(1,\gamma) (g,1) (1,\gamma^{-1}) = (1,\gamma)(g,\gamma^{-1}) = (\gamma(g),1)$. Now the same $\overline{G}$ can be such a semidirect product in several inequivalent ways, corresponding to splittings $\sigma: \Gamma \to \overline{G}$ of the short exact sequence $1 \to G \to \overline{G} \to \Gamma \to 1$. But $\sigma$ is a splitting if and only if there is some function $s: \Gamma \to G$ such that $\sigma(\gamma) = (s(\gamma),\gamma)$ for all $\gamma \in \Gamma$ and $(s(\gamma\gamma'),\gamma\gamma') = (s(\gamma),\gamma) (s(\gamma'),\gamma')$ for all $\gamma,\gamma' \in \Gamma$; by the formula for the group law in a semidirect product, $\gamma \mapsto (s(\gamma),\gamma)$ is a group homomorphism iff $s(\gamma\gamma') = s(\gamma) \, \gamma(s(\gamma'))$ for all $\gamma,\gamma' \in \Gamma$. That is exactly the condition for $s$ to be a (non-abelian) 1-cocycle. Two splittings $\sigma_1,\sigma_2$ are equivalent iff they are related by $G$-conjugation, that is, iff $a \sigma_2(\cdot) = \sigma_1(\cdot) a$ for some $a \in G$. This unwinds to $a s_2(\gamma) = s_1(\gamma) \gamma(a)$ for all $\gamma \in \Gamma$, which is precisely the equivalence relation for non-commutative $1$-cocycles. So, $H^1(\Gamma, G)$ classifies inequivalent realizations of $\overline G$ as a semidirect product of $G$ with $\Gamma$, with the distinguished element of $H^1(\Gamma, G)$ (which is the equivalence class of the 1-cocycle $s(\cdot)=1$) corresponding to the semidirect product for the action we began with.

If $G$ is a complex Lie group with Lie algebra $\mathfrak{g}$, a conjugate-linear involution $\sigma$ of $\mathfrak{g}$ (see the bottom of page 436) yields an anti-holomorphic involution of $G$ and thus a splitting of our sequence $1 \to G \to \overline{G} \to \Gamma \to 1$ where $\Gamma = {\rm Gal}({\bf C}/{\bf R})$.

If $G$ is actually commutative then the 1-cocycles are the same as for ordinary group cohomology, and the equivalence condition is congruence modulo coboundaries, so we recover the usual $H^1(\Gamma,G)$ which fits into a series $H^0, H^1, H^2, \ldots$ with the usual accoutrements of products, connecting homomorphisms, long exact sequences, etc., as well as some additional structure. For noncommutative $G$ we still have $H^0(\Gamma,G) = G^\Gamma$ (the subgroup fixed by $\Gamma$); moreover, if $1 \to N \to G \to Q \to 1$ is a short exact sequence of groups with consistence $\Gamma$-actions then there's a connecting homomorphism $Q^\Gamma = H^0(\Gamma,Q) \to H^1(\Gamma,N)$ that fits into a “long exact sequence” of cohomology sets (though the condition “image at each step = preimage of distinguished element at the next step” is much less powerful without a group structure on the $H^1$'s); but this sequence stops at $H^1(\Gamma,Q)$ for lack of an $H^2(\Gamma,N)$. If $N$ (but not $G$) is commutative then there is a connecting map $H^1(\Gamma,Q) \to H^2(\Gamma,N)$ that extends the long exact sequence one more step, which sometimes is exactly what we need to complete some argument.

Re Proposition 26.23 (page 445): In fact for every semisimple algebra $\mathfrak{g}$ all representations $V$ of the split form $\mathfrak{g}_0$ are real. It should be possible to construct a real subspace by starting from any highest weight vector $v\in V$ and looking at the real span of the vectors $Y_{\alpha_1} Y_{\alpha_2} \cdots Y_{\alpha_k} v$ where each $\alpha_i$ is a fundamental root (so $Y_{\alpha_i} \in \mathfrak{g}_{-\alpha_i}$); recall that these are our generators of the $k$-th summand $V_k$ in the inductive construction of the highest weight (sub)representation generated by $v$. This requires checking that for each $k$ the space of linear relations on these $Y_{\alpha_1} Y_{\alpha_2} \cdots Y_{\alpha_k} v$ is real — in fact it is rational — which should work the same way as in the construction of $\mathfrak{g}$ itself from its root system. But I gather from the (absence of an) answer to this Math Overflow question that proving it that way might not be easy.

On further thought, it does seem to be obviously true … We have seen (albeit with a somewhat mysterious proof) that if $V,V'$ are two irreducible representations with the same highest weight, and $v,v'$ are highest weight vectors, then there is a unique isomorphism $V \to V'$ taking $v$ to $v'$. This means that the spaces of linear relations among our generators of $V_k$ and $V'_k$ are the same. Now the actions of $\mathfrak{g}_0$ on $V$ and $V'$ give $V,V'$ the structure of representations of this Lie algebra if and only if $[X,Y]=XY-YX$ holds in ${\rm End}(V)$ for all $X,Y$ in a basis of $\mathfrak{g}_0$. But we also claimed (though barely outlining a proof) that $\mathfrak{g}_0$ has a multiplication table with rational coefficients; that is, each $[X,Y]$ has rational coordinates with respect to our basis. This forces the space of linear relations to be rational as well, because if it weren’t then some Galois automorphism would change $V$ to a nonisomorphic representation $V'$ with the same highest weight.

In the compact case one might expect that we would classify representations into real, complex, or quaternionic using the Weyl character and integral formulas and the Schur index $\int_G \chi_V^{\phantom.}(g^2)$ (see Exercise 26.29 on page 449). This works for $\mathfrak{su}_2$, fitting nicely with the formula we used some weeks back (in the context of invariant theory for $\rm{SL}_2$ and $\mathfrak{sl}_2$) to recover the dimension of the fixed subspace of a representation of $\mathfrak{sl}_2$ from its character. But it is not clear how this might be done in general. Fortunately other tools are available.

Just in case you have not seen this already: ${\rm Sp}_{2n}({\bf C}) \subseteq {\rm SL}_{2n}({\bf C})$ (with equality iff $n=1$). Indeed $A \in {\rm Sp}_{2n}({\bf C})$ iff $A$ preserves the form $\omega \in \wedge^2 V$ corresponding to $Q$; but then $A$ preserves $\omega^n = \omega \wedge \omega \wedge \cdots \wedge \omega = n! \, e_1 \wedge e_2 \wedge \cdots \wedge e_{2n}$, so $\det A = 1$ as claimed. It follows that $\mathfrak{sp}_{2n}({\bf C}) \subseteq \mathfrak{sl}_{2n}({\bf C})$, again with equality only for $n=1$.

“Clearly, the isomorphism classes of the abstract group [${\rm Sp}_{2n}({\bf C})$] and the Lie algebra [$\mathfrak{sp}_{2n}({\bf C})$] do not depend on the particular choice of [nondegenerate, skew-symmetric bilinear form] $Q$; …” (bottom of page 238) — at least once we know that all such forms are equivalent. This is actually true over any field, even fields of characteristic 2 if we’re careful to use the definition $\forall v \in V : Q(v,v) = 0$ of “skew-symmetric” (a.k.a. “alternating”). This is one way in which symplectic groups and Lie algebras are structurally simpler than orthogonal ones, even though they are not as familiar to most of us.

The Gram matrix (page 239) $M = \bigl(Q(e_i,e_j)\bigr)_{i,j=1}^{2n} = \Bigl({\begin{array}{cc}0&\!I_n\!\cr \!-I_n\!&0\end{array}}\Bigr)$ of a standard symplectic basis $e_1,\ldots,e_{2n}$ is often called $J$. I’ll try to stick with the $M$ chosen by Fulton–Harris. (Also with their notation ${}^t X$ for the transpose of $X$, which I usually write as $X^t$ or $X^{\sf T}$ or $X^*$.) And I’ll try to avoid potential confusion due to the use of $A$ for both a generic element of ${\rm Sp}_{2n}({\bf C})$ and the top left block of a generic element of $\mathfrak{sp}_{2n}({\bf C})$ . . .

“In sum, then, the roots of the Lie algebra $\mathfrak{sp}_{2n}({\bf C})$ are the vectors $\pm L_i \pm L_j \in \mathfrak{h}^*$” (page 240) … except, of course for the case $i=j$ of $\pm L_i \mp L_j$, which gives rise to the zero element of $\mathfrak{h}^*$ — and indeed those $X_{i,j}$ (see the first display of page 240) would be in $\mathfrak{h}$.

[ on page 241, around formula (16.3), I’d have written $H' = \sum_i a'_i H_i^{\phantom.}$, not $H' = \sum_i b_i H_i$, to correspond to $H = \sum_i a_i H_i$.]

Check that the dimension formula $$\dim \Gamma_\lambda = \sum_{\alpha\in R^+} \frac{(\lambda+\rho,\alpha)}{(\rho,\alpha)}$$ (Corollary 24.6, page 402) gives ${2n \choose k} - {2n \choose k-2}$ when $\lambda$ is the $k$-th fundamental weight $L_1 + \cdots + L_k$ of $\mathfrak{sp}_{2n}({\bf C})$. [If you’re more ambitious, you might check the full Weyl character formula.] This means that $\Gamma_\lambda$ must be all of $\wedge^k V \ominus \wedge^{k-2} V$. (That’s Exercise 24.21 on page 406; the “contraction from $\wedge^k V$ to $\wedge^{k-2} V$” is dual to the map $\psi \mapsto \omega\wedge\psi$. The map is a bijection for $k=n+1$, and a surjection for $k \gt n+1$, so the series naturally stops at $k=n$.)

Lecture 18.1 and 19.2/19.4

We noted already that the long roots of the $C_n$ root system form a $D_n$ root system. This corresponds to $\mathfrak{so}_{2n+1} \supset \mathfrak{so}_{2n+1}$. That’s similar to the long $G_2$ roots forming a copy of $A_2$ which lets us construct the Lie algebra $\mathfrak{g}_2$ from $\mathfrak{sl}_3$, though for $\mathfrak{so}_{2n+1} \supset \mathfrak{so}_{2n+1}$ this does not tell us much that we didn’t know already. You might still see just how the direct sum of $\mathfrak{so}_{2n}$ and its $2n$-dimensional defining (a.k.a. standard) representation form a copy of $\mathfrak{so}_{2n+1}$.

Theorem 19.2: the decomposition of $\wedge^n V$ can be explained by constructing an $\mathfrak{so}_{2n}$-involution $\sigma$ of $\wedge^n V$ as follows. We know that the invariant quadratic form identifies $V$ with $V^*$, and thus identifies $\wedge^k V$ with $(\wedge^k V)^*$ for each $k$. On the other hand, having chosen a generator of the top exterior power $\wedge^{2n} V$ we have a perfect pairing $\wedge^k V \times (\wedge^{2n-k} V)^* \to {\bf C}$, and thus an identification of $\wedge^k V$ with $(\wedge^{2n-k} V)^*$. For $k=n$ this gives two canonical identifications of $\wedge^n V$ with its dual, and thus a canonical map $\Sigma$ from $\wedge^n V$ to itself. We can check that $\Sigma^2$ is a multiple of the identity; our involution $\sigma$ is a suitable multiple of $\Sigma$ (there is a choice of sign, but changing $\sigma$ to $-\sigma$ simply switches the two subspaces which are the $+1$ and $-1$ eigenspaces of $\sigma$).

For the orthogonal groups, the exterior powers of the defining representation are irreducible (except for $\wedge^n V$ for $\dim V = 2n$), but the symmetric powers decompose; the irreducible representation with highest weight $k L_1$ is the complement of $Q(x,x) {\rm Sym}^{k-2} V$ in ${\rm Sym}^k V$, which is the subspace of spherical harmonics. For one take on these polynomials, which also connects with infinite-dimensional representations of $\mathfrak{sl}_2$ (with a lowest weight but no highest weight), see these notes from a course I taught here in 2019, starting on page 2.

Our general picture of $\Lambda_R$, $\Lambda_W$, etc. suffices to see that each of the spin representations of $\mathfrak{so}_{2n}$ is the sum of $2^{n-1}$ one-dimensional eigenspaces for $\mathfrak{h}^*$ with eigenvalues $\frac12 \sum_{j=1}^n \pm L_j$ with the $\pm$ signs ranging over half of the sign choices, while the spin representation of $\mathfrak{so}_{2n+1}$ is the sum of $2^n$ one-dimensional eigenspaces, again with eigenvalue $\frac12 \sum_{j=1}^n \pm L_j$ but this time with all $2^n$ sign choices. It may still be amusing to check consistency with the Weyl dimension formula. For example, for $\mathfrak{so}_{2n}$ we have $\rho = \sum_{j=0}^{n-1} j L_{n-j}$, so the dimension formula gives $\prod_{0 \leq j < k \leq n-1} (j+k+1)/(j+k)$; for each $k=1,\ldots,n-1$ we get a telescoping product that simplifies to $2k/k = 2$, and these $n-1$ factors multiply to $2^{n-1}$.

More about root lattices and their Weyl groups

Root lattices and their sphere packings

A sphere packing in a Euclidean space $E$ is a configuration $C$ of congruent spheres in $E$ with disjoint interiors. The density, call it $d(C)$, of a sphere packing $C$ is the fraction of $E$ that is covered by $C$. [More precisely, $\delta(C)$ is the limit, if it exists, of ${\rm Vol}(C \cap B) / {\rm Vol}(B)$ over boxes $B$ of sides approaching $\infty$. Using boxes rather than balls lets us use box tilings to prove that for any sphere packing $C$ there is a sphere packing $C'$ whose density is the lim sup of ${\rm Vol}(C \cap B) / {\rm Vol}(B)$.] To any lattice $L \subset E$ is associated a sphere packing $\{B_r(x) : x \in L\}$ where $2r$ is the minimal length of a nonzero vector in $L$; its density is ${\rm Vol}(B_r(x))$ [any $x$, since all $r$-balls are congruent] divided by the covolume ${\rm Vol}(E/L)$.

The densest lattice packing was determined in dimension up to 8 by Blichfeldt in 1935 (building on earlier work in dimensions up to 5). In each dimension the best $L$ are scalings of a root lattice (for which $r = 2^{-1/2}$), namely $A_1, A_2, A_3 (\cong D_3), D_4, D_5 (\cong E_5), E_6, E_7, E_8.$ Each of those is conjectured to be the densest packing in its dimension even without the lattice requirement; this is known only dimension 1 (trivial), 2 (Fejes Tóth 1943), 3 (Hales 1998), and 8 (Viazovska 2014, published 2017 in Ann. Math.).

Somewhat shockingly, almost 90 years after Blichfeldt the densest 9-dimensional sphere packing is not yet known (though it is conjectured to be the laminated lattice in that dimension). It is known that no root lattice of rank $\geq 9$ yields an optimal lattice in its dimension. The only other dimension in which the best sphere packing is known is 24, for which Cohn and Kumar (2004, published 2009 in Ann. Math.) proved the long-standing conjecture that the Leech lattice yields the optimal lattice packing. The lattice requirement was removed by Viazovska et al. soon after Viazovska’s breakthrough in dimension 8; this, too, was published in the Annals in 2017.

The Weyl groups of the exceptional root systems

The root systems $A_n,B_n,C_n,D_n$ of the classical Lie algebras all have familiar or nearly familiar Weyl groups: symmetric groups $S_{n+1}$ for $A_n$, signed permutation (a.k.a. hyperoctahedral groups) groups for $B_n$ and $C_n$, and evenly signed permutation groups for $D_n$. The exceptional root systems, other than $G_2$ (the 12-element dihedral group), are less familiar but quite interesting.

• $F_4$: Recall that the lattice $\Lambda_R = \Lambda_W$, which is also the $D_4$ lattice, can be identified with the Hurwitz quaternions $H$ (with the pairing $(q,q') = {\rm Tr}(q \bar q')$ where ${\rm Tr}$ is the “reduced trace” taking $a+bi+cj+dk$ to $2a$). The $24$ short roots are then identified with the units in $H$, which form a group $Q_{24}$ under quaternionic multiplication; the long roots are the elements of norm $2$ in $H$. We have a homomorphism $Q_{24} \times Q_{24} \to \mathfrak{W}$ taking any $u,u' \in Q_{24}$ to the symmetry $q \mapsto u^{-1} q u'$; the kernel is $\{(1,1), (-1,-1)\}$. Hence the image is a subgroup of $\mathfrak{W}$ of order $24^2 / 2 = 288$. Thus subgroup turns out to have index 4; to get the full Weyl group, add the quaternionic conjugation $q \mapsto \bar q$ (which is the root reflection for the root $1 \in H$) and conjugation $q \mapsto v q v^{-1}$ by an element $v \in H$ of norm 2 (all are equivalent under $Q_{24}$).
• $E_6$: This is the only exceptional root system whose Weyl group does not contain $-1$ (we already noted that $\mathfrak{W}$ acts trivially on $\Lambda_W / \Lambda_R$ — and for $E_6$ this quotient group is ${\bf Z}/3{\bf Z}$, on which $-1$ does not act trivially). The orientation-preserving subgroup of $\mathfrak{W}$ is the famous finite group of order 25920, which is isomorphic to linear groups in characteristics $2$ and $3$. Reduction mod 2 gives a map from $\mathfrak{W}$ to an orthogonal subgroup of ${\rm GL}_6({\bf Z}/2{\bf Z})$; reduction mod 3 can be refined to an action on $\Lambda_R / 3 \Lambda_W$ and thus to an orthogonal subgroup of ${\rm GL}_5({\bf Z}/3{\bf Z})$. In each case the target orthogonal group is just large enough to accommodate $\mathfrak{W}$, so the map is an isomorphism. The Weyl group also acts on the shortest nonzero weights, which fall in two orbits mod $\Lambda_R$ each of size 27. Either of these is the configuration of weights of a minimal nontrivial representation of (the simply connected form of) $E_6$. This configuration is also closely connected with the classical configuration of 27 lines on a smooth cubic surface. Such a surface, call it $X,$ has a Néron–Severi group ${\rm NS}(X)$ of rank 7 with an intersection pairing of discriminant 1 and signature $(1,6)$, together with a distinguished hyperplane class $h$ such that $h \cdot h = 3.$ The orthogonal complement of $h$ is $\Lambda_R$ with the pairing multiplied by $-1$; orthogonal projection to $h^\perp$ takes ${\rm NS}(X)$ to $\Lambda_W\langle-1\rangle,$ taking the 27 lines to the minimal nonzero weights in one of the cosets. [Note: usually in the lattice literature “$E_6$” is used for the root lattice of the $E_6$ root system, and the weight lattice is called $E_6^*$ because it is the lattice dual of $\Lambda_R$. Likewise for $E_7$ and $E_8$ (which is its own lattice dual).]
• $E_7$: The Weyl group contains $-1$ and, since the rank $7$ is odd, is the direct product of $\{1,-1\}$ with the orientation-preserving subgroup, call it $\mathfrak{W}_0$. Reduction mod 2 embeds $\mathfrak{W}_0$ into an orthogonal subgroup of ${\rm GL}_7({\bf Z}/2{\bf Z})$, or equivalently (via the action on $\Lambda_R / 2 \Lambda_W$) into a symplectic subgroup of ${\rm GL}_6({\bf Z}/2{\bf Z})$. [Over a finite field of characteristic 2, the groups ${\rm SO}_{2n+1}$ and ${\rm Sp}_{2n}$ are the same; in the Lie picture the orientation of a double or triple bond in the Dynkin diagram nearly disappears in characteristic 2 or 3 respectively.] Again the linear group is just large enough to accommodate $\mathfrak{W}_0$, so the map is an isomorphism. This time there is a single orbit of shortest nonzero weights, consisting of $28$ pairs $\{w,-w\}$ that again give the weights of a minimal nontrivial representation. Again this is closely connected to the Néron–Severi group of a del Pezzo surface, this time the double cover $y^2 = Q(x_0,x_1,x_2)$ of ${\bf P}^2$ branched on a smooth quartic curve $Q(x_0,x_1,x_2) = 0.$ The 28 pairs project to the 28 bitangent lines of the quartic.
• $E_8$: Again $\mathfrak{W} / \{1,-1\}$ is so large that it barely fits into the appropriate finite orthogonal group, here an orthogonal subgroup of ${\rm GL}_8({\bf Z}/2{\bf Z})$. The smallest nontrivial representation is the adjoint representation, of dimension $\dim \mathfrak{g} = \dim \mathfrak{h} + |R| = 8 + 240 = 248.$ The 240 minimal roots correspond to the minimal sections of a generic rational elliptic surface, which again is a del Pezzo surface, here $y^2 = x^3 + a(t) x + b(t)$ with $a,b$ of degrees 4,6.

The action of $\mathfrak{W}$ on $\mathfrak{h}$ (equivalently: on $\mathfrak{h}^*$) extends naturally to an action on the polynomial ring ${\bf C}[\mathfrak{h}] \equiv {\bf C}[\mathfrak{h}^*]$. Remarkably the invariant ring $({\bf C}[\mathfrak{h}])^{\mathfrak{W}} \cong ({\bf C}[\mathfrak{h}^*])^{\mathfrak{W}}$ is itself a polynomial ring in each case; that is, $({\bf C}[\mathfrak{h}^*])^{\mathfrak{W}} = {\bf C}[P_1,\ldots,P_n]$ for some homogeneous polynomials $P_1,\ldots,P_n \in {\bf C}[\mathfrak{h}^*]$ where $n = \dim {\mathfrak{h}}$ is the rank of the Lie algebra. Let $d_i = \deg P_i$. There are in general many choices for each $P_i$ (though if $\min_i d_i$ occurs for a unique $i$ then that $P_i$ is determined up to a scalar multiple), but the $d_i$ are readily seen to be independent of the choice of homogeneous generators, with $\#\mathfrak{W} = \prod_{i=1}^n d_i$. If the semisimple Lie algebra decomposes as a direct sum, with $\mathfrak{h} = \mathfrak{h}_1 \oplus \mathfrak{h}_2$ and $\mathfrak{W} = \mathfrak{W}_1 \oplus \mathfrak{W}_2$, then ${\bf C}[\mathfrak{h}^*] = {\bf C}[\mathfrak{h}_1^*] \otimes {\bf C}[\mathfrak{h}_2^*]$ and $({\bf C}[\mathfrak{h}^*])^{\mathfrak{W}} = ({\bf C}[\mathfrak{h}_1^*])^{\mathfrak{W}_1} \otimes ({\bf C}[\mathfrak{h}_2^*])^{\mathfrak{W}_2}$, so it is enough to describe the invariant rings of irreducible root systems. For the classical root systems this is a consequence of well-known facts about symmetric functions; we find that $d_i = i+1$ for $A_n$ (e.g. $P_i$ can be the $i$-th elementary symmetric function in $n+1$ variables $L_i$ that sum to zero), while $d_i = 2i$ for $B_n$ and $C_n$ (elementary symmetric functions in the $L_i^2$), and the invariant degrees of $D_n$ are $2,4,6,\ldots,2(n-1)$ and $n$ (so $n$ is repeated if $n$ is even; the first $n-1$ invariants are as for $B_n$/$C_n$, but the last invariant $\prod_i L_i^2$ is replaced by $\prod_i L_i$). Note that this is consistent with the isomorphisms $A_1 \cong B_1$, $A_1 \oplus A_1 \cong D_2$, and $A_3 \cong D_3$. For the exceptional root systems the invariant degrees are:
  $2,6$ for $G_2$,
  $2,6,8,12$ for $F_4$,
  $2,5,6,8,9,12$ for $E_6$ (note the odd degrees $5,9$, which arise because $-1 \notin \mathfrak{W}$),
  $2,6,8,10,12,14,18$ for $E_7$, and
  $2,8,12,14,18,20,24,30$ for $E_8$.
Check that in each case (classical as well as exceptional) these degrees are consistent with $\#\mathfrak{W} = \prod_{i=1}^n d_i$. We always have $d_1 = 2 < d_i$ ($i>1$) because $\mathfrak{h}$ is an irreducible real representation of $\mathfrak{W}$ so it has no invariant linear forms and a one-dimensional space of invariant quadratic forms. Much less easy, but still known (without resorting to case analysis), is that for each irreducible roots system the degrees $d_i$ are symmetric: if placed in increasing order then $d_i + d_{n+1-i}$ is the same for each $i$. Moreover, the common value is $2+h$ where $h = d_n$ is the order of the conjugacy class of Coxeter elements of  $\mathfrak{W}$ and is also equal to $\#R / n$ (check this too in each case! For example, $h = (n^2+n)/n = n+1$ for $A_n$, and $h = 240 / 8 = 30$ for $E_8$).

In fact it is known that a finite subgroup $G \subset {\rm GL}_n({\bf C})$ has a polynomial invariant ring if and only if $G$ is generated by reflections (where as before we say $T \in {\rm GL}_n({\bf C})$ is a “reflection” if $I-T$ has rank 1). The symmetry (constant $d_i + d_{n+1-i}\!$) holds also for the irreducible Euclidean reflection groups that are not crystallographic [the invariant degrees are $2,6,10$ for the dodecahedral/icosahedral subgroup of $O_3({\bf R})$, and $2,12,20,30$ for the symmetries of the 120- and 600-cell in $O_4({\bf R})$], but not for many unitary reflection groups. (A Euclidean reflection group $G$ is “crystallographic” when there is a $G$-stable lattice, in which case $G$ is a Weyl group.)