18.745: LECTURE 15

PROFESSOR: VICTOR KAČ SCRIBE: MAKSIM MAYDANSKIY

Recall that for n = 2r we defined sp_{2r}(𝔽) = {A ⊂ sl_{2r}(𝔽)|A^{t}J + JA^{t} = 0}, where J was any skew symmetric, but via a change of basis may be taken to be

J = ( \begin{matrix}0 & 0 & 0 & ⋯ & & 1 \\ 0 & 0 & & & ... \\ 0 & & ... & 1 & & ⋮ \\ ⋮ & & −1 & ... & & 0 \\ & ... & & & 0 & 0 \\ −1 & & ⋯ & 0 & 0 & 0 \end{matrix} )

Exercise 1. Prove that sp_{2r}(𝔽) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) |d = −a′, b = b′, c = c′\}, where ′ is the transposition

with respect to the antidiagonal of an r × r matrix.

Note. For so_{2r}(𝔽) we have b′ = −b, c′ = −c.

Solution. Denote the antidiagonal r × r matrix by j. Then M = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ sp_{2r}(𝔽) ⇔

( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}0 & j \\ −j & 0 \end{matrix} ) + ( \begin{matrix}0 & j \\ −j & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & d \end{matrix} ) = 0 ⇔

( \begin{matrix}a^{t} & c^{t} \\ b^{t} & d^{t} \end{matrix} ) ( \begin{matrix}0 & j \\ −j & 0 \end{matrix} ) + ( \begin{matrix}jc & jd \\ −ja & −jb \end{matrix} ) = 0 ⇔

( \begin{matrix}−c^{t}j + jc & a^{t}j + jd \\ −d^{t}j − ja & b^{t}j − jb \end{matrix} ) = 0 ⇔ \begin{matrix}−c^{t}j + jc = 0 \\ a^{t}j + jd = 0 \\ −d^{t}j − ja = 0 \\ b^{t}j − jb = 0 \end{matrix} ⇔ \begin{aligned}c & = & −j^{−1}c^{t}j \\ d & = & j^{−1}a^{t}j \\ a & = & −j^{−1}d^{t}j \\ b & = & j^{−1}b^{t}j \end{aligned}

Now for A = {a_{i,j}}_{i,j} we have jA = {a_{n+1−i,j}}_{i,j} and Aj = {a_{n+1−i,n+1−j}}_{i,j}. Since j^{−1} = j, we get j^{−1}A^{t}j = {a_{n+1−i,n+1−j}}_{i,j} = A′, and so indeed M ∈ sp_{2r}(𝔽) ⇔ c = c′, b = b′, a = −d′.

□

Inside sp_{2r}(𝔽) we have

h = ( \begin{matrix}a_{1} \\ & . & . & . \\ & & & & a_{r} \\ & & & & & −a_{r} \\ & & & & & & . & . & . \\ & & & & & & & & & −a_{1} \end{matrix} ) .

as a Cartan subalgebra. Indeed, in this basis sp_{2r}(𝔽) contains a matrix with distinct eigenvalues, and so by a remark in one of the previous lectures the intersectio of sp_{2r}(𝔽) and the diagonal subalgebra is a Cartan subalgebra. In light of Exercise 1, this intersection is as above.

The vectors F_{i,j} = E_{i,j} − E_{n+1−i,n+1−j} for 1 ≤ i, j ≤ r, i ≠ j

and F_{i,j} = E_{i,j}−E_{n+1−j,n+1−i} for r+1 ≤ j ≤ n, 1 ≤ i ≤ r are root vectors. The corresponding roots are Δ_{sp} = {ε_{i} − ε_{j} for i, j = 1, …, r, with i ≠ j; ε_{i} + ε_{j} for i, j = 1, …, r; −ε_{i} − ε_{j} for i, j = 1, …, r}. Note that ±2ε_{i} are roots.

Note. This set is indecomposable for all r ≥ 1, hence sp_{2r}(𝔽) is simple.

Proof. For r ≥ 2 we have the chains ε_{a} + ε_{b}, −ε_{b} + ε_{c}, ε_{c} + ε_{b} if c ≠ b and ε_{a} + ε_{b}, −ε_{b} + ε_{a}, ε_{c} + ε_{b} if a ≠ b. This shows that all ε_{i} + ε_{j}’s are connected (if r = 1 this statement is vacuously true).

The chain ε_{a} + ε_{b}, ε_{a} − ε_{b} for a ≠ b shows that all ε_{i} − ε_{j}’s are connected to the ε_{i} + ε_{j}’s (and hence among themselves).

The −ε_{i} − ε_{j}’s are connected to ε_{i} + ε_{j}’s by obvious chains −ε_{i} − ε_{j}, ε_{i} + ε_{j}.

This shows that any two roots are connected, and Δ is indecomposable.

□

Proposition 15.1. If Δ is decomposable, i.e. Δ = Δ′ ∪ Δ′′, with α′ + α′′ ∉ Δ ∪ {0} for all α′ ∈ Δ, α′′ ∈ Δ then we have a corresponding decomposition of 𝔤 into direct sum of ideals 𝔤 = 𝔤′ ⊕𝔤′′ where

𝔤′ = 𝔥′ ⊕ ( ⨁_{α′∈Δ′} 𝔤_{α′})

for 𝔥′ = span{ν^{−1}(Δ′)} and

𝔤′′ = 𝔥′′ ⊕ ( ⨁_{α′′∈Δ′′} 𝔤_{α′′})

for 𝔥′′ = span{ν^{−1}(Δ′′)}.

Proof. Because α′ + α′′ ∉ Δ ∪ {0}, we get, [𝔤_{α′}, 𝔤_{α′′}] ⊂ 𝔤_{α′+α′′} = 0.

Since 𝔽[𝔤_{α}, 𝔤_{−α}] = 𝔽ν^{−1}(Δ′) we conclude, say by Jacobi identity, [𝔥′, 𝔤_{α′′}] = 0 and [𝔥′′, 𝔤_{α′}] = 0, so [𝔤′, 𝔤′′] = 0. It remains to show that 𝔤′ and 𝔤′′ are subalgebras. First note that β′ ∈ Δ′ implies −β′ ∈ Δ′. Indeed, we have −β′ ∈ Δ, and supposing −β′ ∈ Δ′′ leads to β′ + (−β′) ∈ Δ ∪ {0}, a contradiction. Now if α′, β′ ∈ Δ′ then γ = α′ +β′ ∉ Δ′′, because that would mean γ +(−β′) = α′ ∈ Δ for γ ∈ Δ′′ and −β′ ∈ Δ′, another contradiction. So either α′, β′ ∈ Δ′ ∪ 0 and [𝔤_{α′}, 𝔤_{β′}] = 𝔤_{α′+β′} ∈ 𝔤′ or α′, β′ ∉ Δ ∪ 0 and [𝔤_{α′}, 𝔤_{β′}] = 0. It is now clear that 𝔤′ is a subalgebra. Similarly, 𝔤′′ is also a subalgebra. □

Definition. An abstract root system is a pair (V, Δ), where V is a finite dimensional vector space over ℝ of dimension r, and Δ is a finite set of vectors in V, such that the following axioms hold

(1) 0 ∉ Δ, V = span{Δ}

(2) If α ∈ Δ then kα ∈ Δ iff k = 1 or −1 for all k ∈ Z

(3) (string property) If α, β ∈ Δthen {β + jα|j ∈ Z} ∩ (Δ ∪ 0) = {β + jα| − p ≤ j ≤ q} where p, q ∈ Z_{+} and p − q = \frac{2(α,β)}{(α,α)}.

Elements of Δ are called roots, and rank of the root system is rank(V, Δ) = r.

In what follows we will often call abstract root systems simply root systems.

Example 15.1. Let 𝔤 be any semisimple Lie algebra over a chosen field 𝔽 of characteristic 0, Δ ∈ 𝔥^{∗} the set of roots, V = (span over ℝ of Δ) = ℝ ⊗_{ℚ} 𝔥_{ℚ}^{∗}, where 𝔥_{ℚ}^{∗} is the span of Δ over ℚ, and (·, ·) - the bilinear extension of K |_{𝔥_{ℚ}} (which we know has rational values on 𝔥_{ℚ}) to 𝔥_{ℝ}- the span of Δ over ℝ. Then by Theorems 3 and 4 from preceding lectures (V, Δ) is an abstract root system called 𝔤-root system.

Example 15.2. dim V = 1, Δ = {α, −α} [figure: α; −α]

This is the only root system of rank 1, by the first two axioms. The string property obviously holds here.

Definition. An abstract root system is called indecomposable if there is no decomposition into a union of two nonempty sets Δ = Δ′ ∪ Δ′′ such that α′ + α′′ ∉ Δ ∪ {0} for all α′ ∈ Δ′, α′′ ∈ Δ′′.

Proposition 15.2. A root system (V, Δ) is decomposable iff V = V′ ⊕ V′′ where V′ ⊥ V′′ and Δ = Δ′ ∪ Δ′′ where Δ′ = V′ ∩ Δ and Δ′′ = V′′ ∩ Δ.

Proof. First, suppose that Δ = Δ′ ∪ Δ′′ such that α′ + α′′ ∉ Δ ∪ {0} for all α′ ∈ Δ′, α′′ ∈ Δ′′. Then by the string property q = 0. But −α′′ ∈ Δ′′ (since if −α′′ ∈ Δ′ then α′′ + (−α′′) = 0 ∈ Δ ∪ {0}, a contradiction), hence α′ − α′′ ∉ Δ ∪ {0}, so, by the string property p = 0 as well, and hence (α′, α′′) = 0. Taking V′ = span Δ′, V′′ = span Δ′′ we get the desired decomposition of V.

Conversely, suppose V = V′ ⊕ V′′ and Δ = Δ′ ∪ Δ′′ for Δ′ = V′ ∩ Δ, Δ′′ = V′′ ∩ Δ and (α′, α′′) = 0 for all α′ ∈ Δ′, α′′ ∈ Δ′′. We shall show that α′ + α′′ ∉ Δ ∪ {0} for all α′ ∈ Δ′, α′′ ∈ Δ′′, so that Δ = Δ′ ∪ Δ′′ is a decomposition of Δ. Note that α′ + α′′ ≠ 0 because α′ ∈ V′, α′′ ∈ V′′ and V′ ∩ V′′ = 0. If α′ + α′′ ∈ Δ then either α′ + α′′ ∈ Δ′ or Δ′′. Without loss of generality, suppose α′ + α′′ ∈ Δ′′. But then 0 = (α′, α′ + α′′) = (α′, α′), which is impossible. □

Conclusion. Thus (V, Δ) uniquely decomposes into a sum of indecomposable root systems V = ⊕_{j}V_{j}, where V_{i} ⊥ V_{j} for i ≠ j, and Δ = ∪_{j}Δ_{j} with Δ_{j} ⊂ V_{j}.

This reduces the study of general root systems to indecomposable ones.

Definition. Two indecomposable abstract root systems (V_{1}, Δ_{1}) an (V_{2}, Δ_{2}) are isomorphic if there exists a vector space isomorphism φ : V_{1} ↦ V_{2} such that φ(Δ_{1}) = Δ_{2} and (φ(a), φ(b)) = γ(a, b) for all a, b ∈ V_{1} and a constant γ ∈ ℝ^{+}.

Note. If (·, ·) is replaced with γ(·, ·) for some γ ∈ ℝ^{+} we get an isomorphic root system.

Proposition 15.3. If (V, Δ) is an indecomposable abstract root system, and (·, ·) its bilinear form, then for any other positive definite symmetric bilinear form (·, ·)_{1} for which the string property (the third axiom) holds, there exists a δ ∈ ℝ^{+}, such that (α, β)_{1} = δ(α, β).

Proof. Fix α ∈ Δ. Then for any β ∈ Δ there exist a sequence α = γ_{0}, γ_{1}, …, γ_{k} = β such that γ_{i} + γ_{i+1} ∈ Δ ∪ {0} and (γ_{i}, γ_{i+1}) ≠ 0 for all i = 0, …, k − 1.

To see that, for a root α define C = {γ ∈ Δ| there exists a sequence of γ’s as above } and B = Δ \ C. Then γ ∈ C and β ∈ B imply (γ, β) = 0. Indeed, (γ, β)¬0 by string property means either γ +β ∈ Δ or γ −β ∈ Δ, and so the strings γ, β or γ, −β, β, respectively, imply β ∈ C, a contradiction. Now C and B are orthogonal subsets of Δ and C is nonempty, since it contains α. By proposition 15.2, and since Δ is indecomposable, we conclude that B is empty, as wanted.

Now, define δ by (α, α)_{1} = δ(α, α). We will show that (β, β)_{1} = δ(β, β) for the same δ. We know

\frac{2(α, γ_{1})}{(α, α)} = \frac{2(α, γ_{1})_{1}}{(α, α)_{1}} = p − q

so (α, γ_{1}) = δ(α, γ_{1}), and as

\frac{2(α, γ_{1})}{(γ_{1}, γ_{1})} = \frac{2(α, γ_{1})_{1}}{(γ_{1}, γ)_{1}} = p − q

and (α, γ_{1}) ≠ 0 we get (γ_{1}, γ_{1})_{1} = δ(γ_{1}, γ_{1}).

Similarly (γ_{2}, γ_{2})_{1} = δ(γ_{2}, γ_{2}) and so on, until γ_{k} = β.

Finally the string property implies (α, β)_{1} = δ(α, β) for all roots, and as the roots span V the same is true for all a, b. □

We have 4 series of abstract root systems:

Type A_{r}: We take V = {a ∈ ℝ^{r+1} | Σa_{i} = 0} and for the standard basis ε_{i} of ℝ^{r+1} we let Δ_{A_{r}} = {ε_{i} − ε_{j} | i, j = 1, …, r + 1}. The inner product is the standart one on ℝ^{n+1} i.e. (ε_{i}, ε_{j}) = δ_{i,j}. To check that this is a root system we only need to verify the string property (the other two axioms obviously hold). The explicit computation for α = ε_{i} − ε_{j}, β = ε_{k} − ε_{l} is given by the following cases (note that (α, α) = 2 for all roots α):

(1) i ≠ k, l; j ≠ k, l. Then (α, β) = 0 and α + β, α − β ∉ Δ ∪ {0}, so p = q = 0 and the string property holds.

(2) α = ε_{i} − ε_{j}, β = ε_{j} − ε_{k}. Then (α, β) = −1 and q = 1, p = 0, so the string property holds again.

(3) α = ε_{i} − ε_{j}, β = −ε_{i} + ε_{k}. Then (α, β) = −1 and q = 1, p = 0, so the string property holds.

(4) α = ε_{i} − ε_{j}, β = −ε_{i} − ε_{j}. Then (α, β) = −2 and q = 2, p = 0, so the string property holds.

(5) α = ε_{i} − ε_{j} = β. Then (α, β) = 2 and q = 0, p = 2, so the string property holds.

Exercise 2. Taking V = ℝ^{r} with the standard basis ε_{i}, i = 1, … r we define

Δ_{B_{r}} = Δ_{so_{2r+1}} = {ε_{i} − ε_{j}, −ε_{i} − ε_{j}, ε_{i} + ε_{j}; ε_{i}, −ε_{i} | i, j = 1, …, r
Δ_{C_{r}} = Δ_{sp_{2r}} = {ε_{i} − ε_{j}, −ε_{i} − ε_{j}, ε_{i} + ε_{j}; 2ε_{i}, −2ε_{i} | i, j = 1, …, r}
Δ_{D_{r}} = Δ_{so_{2r}} = {ε_{i} − ε_{j}, −ε_{i} − ε_{j}, ε_{i} + ε_{j} | i, j = 1, …, r}

Prove that these are root systems, i.e. that the string property holds by explicitly verifying it.

Solution. We have the following obvious

Observation. Suppose a pair (V, Δ) of a finite dimensional vector space over ℝ and a finite set of vectors in it satisfy the first two axioms of abstract root system, and α, β ∈ Δ. Then the string property holds for α, β iff it holds for α, −β iff it holds for −α, β.

We will carry out the proof for all three of above root systems in parallel. We have the following cases:

0, 0: α = ε_{i} − ε_{j}, β = ε_{k} − ε_{l}. In all three root systems,

(1) i ≠ k, l; j ≠ k, l. Then (α, β) = 0 and α + β, α − β ∉ Δ ∪ {0} so p = q = 0 and the string property holds.

(2) α = ε_{i} − ε_{j}, β = ε_{j} − ε_{k}. Then (α, β) = −1 and q = 1, p = 0, so the string property holds again.

(3) α = ε_{i} − ε_{j}, β = −ε_{i} + ε_{k}. Then (α, β) = −1 and q = 1, p = 0, so the string property holds.

(4) α = ε_{i} − ε_{j}, β = −ε_{i} − ε_{j}. Then α = −β and automatically q = 2, p = 0, so the string property holds.

(5) α = ε_{i} − ε_{j} = β. Then automatically q = 0, p = 2, so the string property holds. 0, 1: α = ε_{i} − ε_{j}, β = ε_{k} + ε_{l}.

(1) i ≠ k, l; j ≠ k, l. Then (α, β) = 0 and p = q = 0 for all three root systems, the string property holds.

(2) α = ε_{i} − ε_{j}, β = ε_{j} + ε_{k}. Then (α, β) = −1 and q = 1, p = 0 for all three root systems, the string property holds.

(3) α = ε_{i} − ε_{j}, β = ε_{i} + ε_{k}. Then (α, β) = 1 and q = 0, p = 1,for all three root systems, the string property holds.

(4) α = ε_{i} − ε_{j}, β = ε_{i} + ε_{j}. Then (α, β) = 0 and q = p = 1 for Δ_{C_{r}}, p = q = 0 for

Δ_{B_{r}}, Δ_{D_{r}},and the string property holds in all three. 1, 0: α = ε_{i} + ε_{j}, β = ε_{k} − ε_{l}.

(1) i ≠ k, l; j ≠ k, l. Then (α, β) = 0 and p = q = 0 for all three root systems, the string property holds.

(2) α = ε_{i} + ε_{j}, β = −ε_{j} + ε_{k}. Then (α, β) = −1 and q = 1, p = 0 for all three root systems, the string property holds.

(3) α = ε_{i} + ε_{j}, β = ε_{i} − ε_{k}. Then (α, β) = 1 and q = 0, p = 1,for all three root systems, the string property holds.

(4) α = ε_{i} + ε_{j}, β = ε_{i} − ε_{j}. Then (α, β) = 0 and q = p = 1 for Δ_{C_{r}}, p = q = 0 for

Δ_{B_{r}}, Δ_{D_{r}},and the string property holds in all three. 1, 1: α = ε_{i} + ε_{j}, β = ε_{k} + ε_{l}. Then in all three root systems

(1) i ≠ k, l; j ≠ k, l. Then (α, β) = 0 and p = q = 0, the string property holds.

(2) α = ε_{i} + ε_{j}, β = ε_{j} + ε_{k}. Then (α, β) = 1 and q = 0, p = 1, the string property holds.

(3) α = ε_{i} + ε_{j} = β Then automatically q = 0, p = 2, so the string property holds. By the Observation we get that the string property holds in all three root systems in the

cases 0,-1: α = ε_{i} − ε_{j}, β = −ε_{k} − ε_{l} -1, 0: α = −ε_{i} − ε_{j}, β = ε_{k} − ε_{l} -1, 1: α = −ε_{i} − ε_{j}, β = ε_{k} + ε_{l} 1,-1: α = ε_{i} + ε_{j}, β = −ε_{k} − ε_{l} -1,-1: α = e_{i} − ε_{j}, β = −ε_{k} − ε_{l}

This complets the proof for Δ_{D_{r}} and leaves only the cases involving "singletons" for the other two root systems.

⋄, 0: α = (2)ε_{i}, β = ε_{j} − e_{k}. (1) i ≠ j, k. Then (α, β) = 0 and p = q = 0, the string property holds.

(2) i = j. Then (α, β) = 1 for Δ_{B_{r}} and q = 0, p = 2, the string property holds and (α, β) = 2, (α, α) = 4 for Δ_{B_{r}} and q = 0, p = 1, the string property holds.

(3) i = k. Then (α, β) = −1 for Δ_{B_{r}} and q = 2, p = 0, the string property holds and

(α, β) = −2, (α, α) = 4 for Δ_{C_{r}} and q = 0, p = 2, the string property holds. ⋄, 1: α = (2)ε_{i}, β = ε_{j} + e_{k}.

(1) i ≠ j, k. Then (α, β) = 0 and p = q = 0, the string property holds.

(2) i = j. Then (α, β) = 1 for Δ_{B_{r}} and q = 0, p = 2, the string property holds and (α, β) = 2,

(α, α) = 4 for Δ_{C_{r}} and q = 0, p = 1, the string property holds. 0, ⋄: α = ε_{j} − e_{k}, β = (2)ε_{i}.

(1) i ≠ j, k. Then (α, β) = 0 and p = q = 0, the string property holds.

(2) i = j. Then (α, β) = 1 for Δ_{B_{r}} and q = 0, p = 1, the string property holds and (α, β) = 2 for Δ_{B_{r}} and q = 0, p = 2, the string property holds.

(3) i = k. Then (α, β) = −1 for Δ_{B_{r}} and q = 1, p = 0, the string property holds and

(α, β) = −2 for Δ_{B_{r}} and q = 2, p = 1, the string property holds. 1, ⋄: α = ε_{i} + e_{j}, β = (2)ε_{k}.

(1) i ≠ j, k. Then (α, β) = 0 and p = q = 0, the string property holds.

(2) i = j. Then (α, β) = 1 for Δ_{B_{r}} and q = 0, p = 2, the string property holds and (α, β) = 2

for Δ_{C_{r}} and q = 0, p = 2, the string property holds. ⋄,⋄: α = (2)ε_{i}, β = (2)ε_{j}.

(1) i ≠ j. Then (α, β) = 0 and p = q = 1 in both, the string property holds.

(2) i = j This is automatic.

By the Observation we get that the string property holds in both root systems in the cases -⋄, 0: α = −(2)ε_{i}, β = ε_{j} − e_{k} -⋄, 1: α = −(2)ε_{i}, β = ε_{j} + e_{k} ⋄,-1: α = (2)ε_{i}, β = −ε_{j} − e_{k} -⋄,-1: α = −(2)ε_{i}, β = −e_{j} − e_{k} 0,-⋄: α = ε_{j} − e_{k}, β = −(2)ε_{i} -1, ⋄: α = −ε_{i} − e_{j}, β = (2)ε_{k} 1,-⋄: α = ε_{i} + e_{j}, β = −(2)ε_{k} -1,-⋄: α = −ε_{i} − e_{j}, β = −(2)ε_{k} -⋄, ⋄: α = −(2)ε_{i}, β = (2)ε_{j} ⋄,-⋄: α = (2)ε_{i}, β = −(2)ε_{j} -⋄,-⋄: α = −(2)ε_{i}, β = −(2)ε_{j}

This exhausts the possibilities, and shows that Δ_{B_{r}} and Δ_{C_{r}} are indeed root systems. □