Root spaces are 1-dimensional

Recap

If 𝔤 is the complexified Lie algebra of a compact semisimple group K then 𝔤 = 𝔥 ⊕ ⊕ α ∈ R 𝔤 α where:

The figure below shows the root diagram of 𝔰 ⁢ 𝔩 ⁢ ( 3 , 𝐂 ) . We've been gradually discovering that various nice features of this Lie algebra and its root diagram carry over to root diagrams in this more general context. These nice features include:

  • If α is a root then so is - α .

  • If X ∈ 𝔤 α and Y ∈ 𝔤 - α then X , Y and H α = [ X , Y ] span a subalgebra S α ≅ 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) .

  • Partway through the proof of the previous point, we saw that [ X , Y ] = K ⁢ ( X , Y ) ⁢ α ♯ ∈ 𝔥 , where α ♯ is Killing-dual to α . This depends on X and Y only through their Killing pairing K ⁢ ( X , Y ) . We also saw that if we pick X and Y suitably (so that K ⁢ ( X , Y ) = 2 / K ⁢ ( α ♯ , α ♯ ) ) then H α = [ X , Y ] = 2 ⁢ α ♯ K ⁢ ( α ♯ , α ♯ ) satisfies the commutation relations [ H α , X ] = 2 ⁢ X , [ H α , Y ] = - 2 ⁢ Y , [ X , Y ] = H α .

    • In this video, we will show that:

    Proposition:

    All the root spaces 𝔤 α are 1-dimensional.

    Proof

    Fix a root α and consider the line through α . We will sum some of the root spaces along this line:

    Sum of root spaces along line through alpha

    The result is a subspace V = 𝐂 ⋅ H α ⊕ ⊕ k ∈ 𝐙 ∖ { 0 } 𝔤 k ⁢ α

    Lemma:

    V is preserved by the action of the subalgebra S α . Moreover, it is irreducible as a representation of S α .

    Remark:

    Irreducibility is the key thing, because this implies that the root spaces 𝔤 k ⁢ α are all 1-dimensional.

    Proof:

    We need to show that ad X , ad Y and ad H α preserve V . Let's just do ad X (the others are similar/easier).

    • We have ad X ⁢ H α = [ X , H α ] = - [ H α , X ] = - 2 ⁢ X ∈ 𝔤 α ⊂ V , so ad X sends 𝐂 ⋅ H α to something in V .

    • We have ad X : 𝔤 k ⁢ α → 𝔤 ( k + 1 ) ⁢ α because X ∈ 𝔤 α .

      • If k + 1 ≠ 0 we have 𝔤 ( k + 1 ) ⁢ α ⊂ V and we're done.

      • If k + 1 = 0 then 𝔤 ( k + 1 ) ⁢ α = 𝔤 0 , which is not contained in V . In fact, we have V ∩ 𝔤 0 = 𝐂 ⋅ H α , so we need to show that ad X ⁢ Y is a multiple of H α for any Y ∈ 𝔤 - α . But we saw that if X ∈ 𝔤 α and Y ∈ 𝔤 - α then [ X , Y ] = K ⁢ ( X , Y ) ⁢ α ♯ , and α ♯ is just a multiple of H α , so ad X ⁢ ( 𝔤 - α ) = 𝐂 ⋅ H α ⊂ V and we're done.

    How do we see that this representation is irreducible? Let's understand the weight space decomposition of V under the action of S α . I claim that 𝔤 k ⁢ α is a weight space with weight 2 ⁢ k for the action of H α .

    This is because if Z ∈ 𝔤 k ⁢ α , we have ad H α ⁢ Z = k ⁢ α ⁢ ( H α ) ⁢ Z and, since H α = 2 ⁢ α ♯ / K ⁢ ( α ♯ , α ♯ ) , we have α ⁢ ( H α ) = 2 ⁢ α ⁢ ( α ♯ ) / K ⁢ ( α ♯ , α ♯ ) , but α ⁢ ( α ♯ ) = K ⁢ ( α ♯ , α ♯ ) , so ad H α ⁢ Z = 2 ⁢ k ⁢ Z and Z has weight 2 ⁢ k under the action of H α as required.

    So our weight diagram looks like this: ⋯ ⊕ 𝔤 - 2 ⁢ α ⊕ 𝔤 - α ⊕ 𝐂 ⋅ H α ⊕ 𝔤 α ⊕ 𝔤 2 ⁢ α ⊕ ⋯ in weights ⋯   - 4 , - 2 , 0 , 2 , 4 , ⋯ respectively. In particular, the weight space with weight zero is 𝐂 ⋅ H α , which is 1-dimensional. If we decompose V into irreducibles then these irreducibles all have even highest weight (there are only even weights) and in particular they will all have 1-dimensional weight space in weight zero. But there is only room for one such irreducible piece, so V must itself be irreducible.