18.745: LECTURE 25

PROFESSOR: VICTOR KAČ SCRIBE: MAKSIM MAYDANSKIY

Let 𝔤 be as in the last lecture - finite dimesional semisimple lie algebra. Let 𝔥 be a Cartan subalgebra, and Π = {α_{1}, …, α_{r}} ⊂ Δ_{+} ⊂ Δ, as before, a system of simple roots. We have the triangular decomposition 𝔤 = 𝔫_{−} + 𝔥 + 𝔫_{+}, 𝔟 = 𝔥_{+} + 𝔫_{+}, with 𝔟 - a Borel subalgebra, and [𝔟, 𝔟] = 𝔫_{+}. Let (·, ·) be a nondegenerate invariant symmetric bilinear form on 𝔤, let ρ = \frac{1}{2}Σ_{α∈Δ_{+}}α. Let \{E_{i}, H_{i}, F_{i}\}

be the Chevalley generators satisfying H_{i} = \frac{2ν^{−1}(α_{i})}{(α_{i},α_{i})}, E_{i} ∈ 𝔤_{α}, F_{i} ∈ 𝔤_{−α} and such that < E_{i}, H_{i}, F_{i} > form the standard basis of sl_{2}(𝔽). Recall that E_{i}’s (respectively F_{i}’s) generate 𝔫_{+} (respectively 𝔫_{−}) and H_{i}’s form a basis of 𝔥. We have the weight lattice P = {λ ∈ 𝔥^{∗}|λ(H_{i}) ∈ Z for all i = 1, … r}. Note that Q ⊂ P, since α_{i}(H_{j}) ∈ Z. Define the subset P_{+} = {λ ∈ 𝔥^{∗}|λ(H_{i}) ∈ Z_{+} for all i = 1, … r}, called the set of dominant integral weights.

Theorem 25.1. (Cartan) The 𝔤-modules {L(Λ)}_{Λ∈P_{+}} are, up to isomorphism, all irreducible finitedimensional 𝔤-modules. (Recall from previous lectures that L(Λ) is the irreducible heigest weight module with heighest weight λ.)

Theorem 25.2. (Weyl’s dimension formula) dimL(Λ) = Π_{α∈Δ} \frac{(Λ+ρ,α)}{(ρ,α)} provided that Λ ∈ P_{+}

Note. ρ(H_{i}) = 1 for all i. Indeed, s_{i}(ρ) = s_{i} (\frac{1}{2}α_{i} + Σ_{α∈Δ_{+}\\{α_{i}\}}α) = −\frac{1}{2}α_{i} + Σ_{α∈Δ_{+}\\{α_{i}\}}α = ρ − α_{i}. But for any λ we have s_{i}(λ) = λ − λ(H_{i})α_{i}. Hence ρ(H_{i}) = 1

Example 25.3. 𝔤 = sl_{2}. All finite dimensional irreducible sl-modules are L(Λ(mρ)) for m ∈ Z_{+} (observe that mρ(H) = m). dimL(mρ) = m + 1.

Example 25.4. 𝔤 = sl_{3}. We have Π = \{α_{1}, α_{2}\}, (α_{i}, α_{j}) = ( \begin{matrix}2 & −1 \\ −1 & 2 \end{matrix} ), ρ = α_{1} + α_{2}, ρ(α_{i}) = 1,

Λ = k_{1}Λ_{1} + k_{2}Λ_{2}, where (Λ_{i}, α_{j}) = δ_{ij}. By Cartan’s theorem, dimL(Λ) < ∞ iff k_{1}, k_{2} ∈ Z_{+}. We compute (Λ + ρ, α_{1}) = k_{1} + 1, (Λ + ρ, α_{2}) = k_{2} + 1, and (Λ + ρ, α_{1} + α_{2}) = k_{1} + k_{2} + 1, so the Weyl’s dimension formula gives dimL(Λ) = \frac{(k_{1}+1)(k_{2}+1)(k_{1}+k_{2}+1)}{2}.

In general, we may write Λ = Σ_{i}k_{i}Λ_{i}, where Λ_{i}(H_{j}) = δ_{ij}. Then dimL(Λ) < ∞ iff k_{i} ∈ Z_{+}. These

k_{i} are called labels of the heighest weight. They are depicted on the Dynkin diagram: [figure: k_{1}; k_{2}]

Operations on modules can then often be described by manipulations of such labeled Dynkin diagrams.

Proof of Theorem 25.1. First suppose that V is a finite dimensional irreducible 𝔤-module. By Lie’s theorem, since 𝔟 is solvable, there exists a non-zero vector v ∈ V and λ ∈ 𝔟^{∗} such that b(v) = λ(b)v for any b ∈ 𝔟. Now if n ∈ 𝔫_{+} = [𝔟, 𝔟], then λ(n) = Σ_{i}λ([b_{i}, b′_{i}]) = Σ_{i}λ(b_{i})λ(b′_{i})−λ(b′_{i})λ(b_{i}) = 0. So we have hv = λ(h)v and 𝔫_{+}v = 0. Also U(𝔤)v = V, since V is irreducible, and U(𝔤)v is a non-zero submodule (it contains v). Hence V = L(λ). Why is λ ∈ P_{+}? This is because V is a finite-dimeensional module with respect to < E_{i}, H_{i}, F_{i} >≅ sl_{2}, and since E_{i}v = 0, H_{i}v = λ(H_{i})v, by the key lemma on sl_{2}- modules we conclude that λ(H_{i}) ∈ Z_{+}. It remains to show that dimL(Λ) < ∞ if Λ ∈ P_{+}. This follows from the Weyl’s dimension formula, which we will in turn deduce from the Weyl’s character formula. □

Definition. Let V be a 𝔤-module, such that V = ⨁_{μ∈𝔥^{∗}} V_{μ}, where V_{μ} = {v|hv = μ(h)v for all h ∈ 𝔥} (we assume dimV < ∞). Then the (formal) character of V is chV = Σ_{μ∈𝔥^{∗}}(dimV_{μ})e^{μ}, where e^{μ} are formal symbols, which obey the property of exponentials, e^{λ}e^{μ} = e^{λ+μ}, e^{0} = 1.

Theorem 25.5. (Weyl’s character formula.) Let R = Π_{α∈Δ_{+}}(1 − e^{−α}). Provided that Λ ∈ P_{+}, one has: e^{ρ}R chL(Λ) = Σ_{w∈W} (detw)e^{w(Λ+ρ)}

Example 25.6. sl_{2}. We get e^{\frac{α}{2}} (1 − e^{−α}) chL(mρ) = e^{(m+1)ρ} − e^{−(m+1)ρ}, so that chL(mρ) = \frac{e^{(m+1)ρ}−e^{−(m+1)ρ}}{e^{ρ}−e^{−ρ}} = e^{mρ} + e^{(m−2)ρ} + … + e^{−mρ}, which exactly corresponds to what we would expect from last lecture, since applying F means subtracting α = 2ρ.

Derivation of Theorem 25.2 from Theorem 25.5. Given v ∈ 𝔥^{∗}, consider the linear map F_{ν} characterised by e^{λ} ↦ e^{t(ν,λ)}. It maps linear combinations to linear combinations and products to products. Applying F_{ρ} to both sides of Weyl’s character formula:

e^{t(ρ,ρ)}Π_{α∈Δ_{+}} (1 − e^{−t(ρ,α)}) Σ_{λ} ( dimL(Λ)_{λ}) e^{t(ρ,λ)} = Σ_{w∈W} (detw)e^{t(ρ,w(Λ+ρ))} =
= Σ_{w∈W} (detw)e^{t(w^{−1}(ρ),Λ+ρ)} = Σ_{w∈W} (detw)e^{t(w(ρ),Λ+ρ)} = F_{Λ+ρ} (Σ_{w∈W} (detw)e^{w(ρ)}) .

Letting Λ = 0 in the Weyl character formula, we get the Weyl denominator identity:

e^{ρ}R · 1 = Σ_{w∈W} (detw) e^{w(ρ)}.

Substituting in, we obtain

F_{Λ+ρ} (Σ_{w∈W} (detw)e^{w(ρ)}) = e^{t(Λ+ρ,ρ)}Π_{α∈Δ_{+}} (1 − e^{−t(Λ+ρ,α)}) .

Together with the first eguality, this gives

e^{t(ρ,ρ)}Σ dimL(Λ)_{λ}e^{−t(ρ,α)} = e^{t(Λ+ρ,ρ)}Π_{α∈Δ_{+}} \frac{1 − e^{−t(Λ+ρ,α)}}{1 − e^{−t(ρ,α)}}

Taking the limit as t goes to 0, we get, by L’Hopital’s rule, dimL(Λ) = lim_{t→0} Π_{α∈Δ_{+}} \frac{(Λ+ρ,α)e^{−t(Λ+ρ,α)}}{(ρ,α)e^{−t(ρ,α)}} =

Π_{α∈Δ_{+}} \frac{(Λ+ρ,α)}{(ρ,α)} □

Proof of the Weyl’s character formula. The Weyl group acts on the linear combinations of formal exponentials in an obvious way: w (Σ_{λ}c_{λ}e^{λ}) = Σ_{λ}c_{λ}e^{w(λ)}.

Lemma 1. If Λ(H_{i}) ∈ Z_{+}, then chL(Λ) is r_{i}-invariant.

Proof. By the key sl_{2} lemma, F_{i}^{Λ(H_{i}+1)}v_{Λ} is a singular vector of L(Λ) (it is killed by E_{i} by the key lemma, and by E_{j} for j ≠ i since F_{i} and E_{j} commute). As L(Λ) is irreducible, we conclude that F_{i}^{Λ(H_{i})+1}v_{Λ} = 0. But L(Λ) = U(𝔤)v_{Λ}. Since (adF_{i})^{N} u = 0 for all N ≫ 0, given u ∈ U(𝔤), we conclude that F_{i}^{N}v = 0 for N ≫ 0, given v ∈ V. It is easy to deduce, using Weyl’s complete reducibility theorem, that L(Λ) is isomorphic to a direct sum of irreducible sl_{2} =< E_{i}, H_{i}, F_{i} >- modules, say V_{j} : L(Λ) = ⊕_{j}V_{j}. But for each V_{j} the lemma holds, since V_{j} ≅ sl_{2}-module L(mρ). Hence the lemma holds for L(Λ) as well. □

Lemma 2. For the Verma module M(Λ) we have R chM(Λ) = e^{λ}

Proof. By Proposition 2 from the last lecture, vectors E_{−β_{1}}^{k_{1}} … E_{−β_{N}}^{k_{N}} form a basis of M(Λ). Hence chM(Λ) = Σ_{(k_{1},...k_{N})∈Z_{+}^{N}} e^{Λ−k_{1}β_{1}...−k_{N} β_{N}} = e^{Λ}Π_{β∈Δ_{+}}(1 + e^{−β} + e^{−2β} + …). Multiplying both sides by R we get the desired result. □

Lemma 3. w(e^{ρ}R) = (detw)e^{ρ}R

Proof. Since s_{i}’s generate W, it suffices to prove s_{i}(e^{ρ}R) = −e^{ρ}R. Indeed, s_{i}(e^{ρ}R) = s_{i}(e^{ρ}(1 − e^{−α_{i}}) (Π_{α∈Δ_{+}\α_{i}}(1 − e^{−α})) = e^{ρ−α_{i}}(1 − e^{α_{i}})Π_{α∈Δ_{+}\α_{i}}(1 − e^{−α}) = −e^{ρ}R, as wanted. □

Lemma 4. Let Λ ∈ 𝔥_{ℝ}^{∗} and let V be a heighest weight module with heighest weight Λ. Let D(Λ) = {Λ − Σk_{i}α_{i}, k_{i} ∈ Z_{+}}. Then chV = Σ_{λ∈B(Λ)}a_{λ} chL(λ), where a_{Λ} = 1, a_{λ} ∈ Z_{+}, and B(Λ) = {λ ∈ D(Λ)|(Λ + ρ, Λ + ρ) = (λ + ρ, λ + ρ)}.

Proof. By induction on dimV = Σ_{λ∈B(Λ)}dimV_{λ} < ∞, since B(Λ) is finite (by Proposition 1 from last lecture). If v_{λ} is the only singular vector of V, then V = L(Λ) and we are done. If we have another singular vector v_{λ} then by Proposition 1 from last time λ ∈ B(Λ). Let U = U(𝔤)v_{λ}, a heighest weight submodule of V. Then we have an exact sequence of 𝔤-modules 0 → U → V → V/U → 0. Then chV = chU + chU/V, and we apply the induction assumption to both summands. □

Lemma 5. In the assumptions of Lemma 4 and presuming V is irreducible, we have chV = Σ_{λ∈B(λ)}b_{λ}chM_{λ}, where b_{Λ} = 1 and b_{λ} ∈ Z.

Proof. By Lemma 4 we have for any μ ∈ M(Λ): chM_{μ} = Σ_{λ∈B(μ)}a_{λ,μ}chL(λ). Now B(Λ) = {λ_{1}, …, λ_{s}}, where λ_{i} − λ_{j} ∉ {Σk_{i}α_{i}|k_{i} ∈ Z_{+}} if i > j. We therefore have a system of linear equations chM_{μ} = Σ_{λ∈B(μ)}a_{λ,μ}chL(λ), for which the matrix (a_{ij})_{ij} is upper triangular matrix of integers with ones on the diagonal, and so its inverse, which expresses chL(Λ)’s in terms of chM(μ)’s for μ ∈ B(Λ) is a matrix of integers with ones on the diagonal as well, and we are done. □

Now, to deduce the theorem from the lemmas, observe that by Lemma 5 chV = Σ_{λ∈B(λ)}a_{λ}chM_{λ}, where a_{Λ} = 1 and a_{λ} ∈ Z We multiply both sides by e^{ρ}R and use Lemma 2 to obtain e^{ρ}R chL(Λ) = Σ_{λ∈B(λ)}a_{λ}e^{ρ+λ}, a_{Λ} = 1, a_{λ} ∈ Z. By Lemma 1 chL(Λ) is W -invariant, and by Lemma 3 e^{ρ}R is W -anti-invariant (i.e. multiplied by the determinant). Hence the left hand side of the equation is anti-invariant, and therefore so is the right hand side. We have

e^{ρ}R chL(Λ) = Σ_{w∈W} (detw)e^{w(Λ+ρ)} + Σ_{λ∈B(Λ)\{Λ},λ+ρ∈P_{+}}a_{λ}Σ_{w∈W} (detw)e^{w(λ+ρ)}

It remains to show that the second term in this sum is empty, i.e. that there are no λ with λ+ρ ∈ P_{+}, λ = Λ − α for α = Σk_{i}α_{i}, k_{i} ∈ Z_{+}, α ≠ 0 and (λ + ρ, λ + ρ) = (Λ + ρ, Λ + ρ). Indeed, for such a λ we would have 0 = (Λ + ρ, Λ + ρ) − (λ + ρ, λ + ρ) = (λ + Λ + 2ρ, α) > 0, since (Λ, α_{i}) = \frac{2Λ(H_{i})}{(α_{i},α_{i})} ≥ 0 and similarly (λ + ρ, α_{i}) ≥ 0, and (ρ, α) > 0 since \frac{2}{(α_{i},α_{i})} > 0. This gives a contradictionand complets the proof.

□