Root space decomposition of $C_n=\mathfrak{sp}(2n,F)$

Question

I want to find the root space decomposition of the symplectic lie algebra $\mathfrak{sp}(2n,F)=C_n$.

I use the notation from Humphreys. The root space decomposition of a semisimple lie algebra $L$ is $L=H\oplus \bigoplus_{\alpha \in \Phi} L_\alpha$. Where $H$ is a maximal total subalgebra (this is more typically called cartan subalgebra). The $L_\alpha$ are the root spaces, and $\Phi$ is the root system.

First we must determine a suitable $H$. For this it seems we can pick the diagonal matrices in $C_n$. I think that this is toral since all its elements are diagonal and hence semisimple? To see that it is maximal, suppose not i.e. $H\subset H'$ where $H'$ is the maximal toral subalgebra. There there must be some $a\in H'$ that commutes with every $ha=ah$ for every $h\in H$. But I think by picking some $h$'s cleverly this implies that $a$ must also be diagonal.

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My main confusion is about trying to find the roots, and then the root spaces. The roots are the $\alpha$ such that $L_\alpha$ is non zero. How are we supposed to find which $L_\alpha$ are non zero before finding the $\alpha$'s?

If we try to work directly from the definition we have $L_\alpha=\{x\in L \,|\, [h,x]=\alpha(h)x \quad \forall h \in H \}$, we are left with quite a complicated eigen value equation to solve. I think if we had an intuition for for the spaces should look like, we could use the fact that root spaces are one dimensional.

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I have done this calculation for $\mathfrak{sl}(n,F)$ but that feel too prototypical to help get a feel for doing these. I would like to complete this calculation for $\mathfrak{sp}(2n,F)$ and then try again myself to do the other classical lie algebras.

Answer 1

You're right, $H$, the subalgebra consisting of the diagonal matrices in ${sp}(2n, \mathbb C)$ do form a Cartan subalgebra. Clearly, $H$ is an abelian subalgebra consisting of diagonalisable elements.

But how can we see there is no bigger abelian subalgebra consisting of diagonalisable elements? I suggest you don't worry about showing that this straight away. Instead, I suggest you go ahead and show that the $L = sp(2n, \mathbb C)$ has a decomposition $$ L = H \oplus \bigoplus_{\alpha} L_\alpha,$$ where each $\alpha \in H^\star$ is non-zero, and where $[h , x] = \alpha(h) x$ for $h \in H $ and $x \in L_\alpha$. Once you know that the $\alpha$'s are non-zero, then you know that no element outside of $H$ commutes with all the elements in $H$, which means that you can't make $H$ bigger and have it remain abelian.

So how do we find generators for these $L_\alpha$'s? I agree that it's a hard eigenvalue equation to solve. But actually, it's not that hard to guess a set of generators for the $L_\alpha$'s. In fact, the most natural vector-space basis for $L$ that you can think of gives you a set of generators for these $L_\alpha$'s!

I'll do the case of $sp(4, \mathbb C)$. A convenient basis for $H$ is $$ H_1 := \begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &-1 & 0 \\ 0 &0 &0 &0\end{bmatrix}, \ \ H_2 := \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 &0 & 0 & 0 \\ 0 &0 &0 &-1\end{bmatrix}$$

And then, the $L_\alpha$'s are generated by $$ X_{1, 2} = \begin{bmatrix} 0 & 1 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &-1 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 1, \ \alpha(H_2) = -1)$$ $$ X_{2, 1} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 1 & 0 & 0 & 0 \\ 0 &0 &0 & -1 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -1, \ \alpha(H_2) = 1)$$ $$ Y_{1, 2} = \begin{bmatrix} 0 & 0 & 0 &1 \\ 0 & 0 & 1 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 1, \ \alpha(H_2) = 1)$$ $$ Z_{1, 2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &1 &0 & 0 \\ 1 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -1, \ \alpha(H_2) = -1)$$ $$ U_{1} = \begin{bmatrix} 0 & 0 &1 & 0\\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 2, \ \alpha(H_2) = 0)$$ $$ U_{2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 1 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 0, \ \alpha(H_2) = 2)$$ $$ V_{1} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -2, \ \alpha(H_2) = 0)$$ $$ V_{2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &1 &0 & 0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 0, \ \alpha(H_2) = -2)$$

It shouldn't be too hard to generalise this to higher $n$!

Answer 2

The other answer answers the question very well.

I'd just like to point out that the standard matrix representations of all the classical Lie algebras over an algebraically closed field $K$ (i.e. the split forms) are chosen such that the diagonal matrices contained therein form a CSA. Namely:

For type $B_n$,

$$\mathfrak{so}(2n+1) := \{M \in M_{2n+1}(K): M^TS+SM=0\} \text{, where } S=\begin{pmatrix} O&I_n&O\\ I_n&O&O\\ O&O&1 \end{pmatrix} $$

$$=\{\left(\begin{matrix}A&B&-f^T\\C&-A^T&-e^T\\e&f&0\end{matrix}\right): A,B,C\in M_n(K), B=-B^T, C=-C^T , e, f \in M_{1\times n}(K)\}.$$

For type $D_n$,

$$\mathfrak{so}(2n) := \{M \in M_{2n}(K): M^TS+SM=0\} \text{, where } S=\begin{pmatrix} O&I_n\\ I_n&O \end{pmatrix} $$

$$=\{\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right): A,B,C\in M_n(K), B=-B^T, C=-C^T \}.$$

For type $C_n$,

$$\mathfrak{sp}(2n) := \{M \in M_{2n}(K): M^TS+SM=0\} \text{, where } S=\begin{pmatrix} O&I_n\\ -I_n&O \end{pmatrix} $$

$$=\{\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right): A,B,C\in M_n(K), B=B^T, C=C^T \}.$$

Note that in all cases, the intersection of that Lie algebra with the diagonal matrices in $M_{2n(+1)}(K)$ is a subalgebra of dimension $n$ and turns out to be a CSA. Actually, once one has understood the structure of the root spaces in $\mathfrak{sl}_n$, the above presentations allow one to see the roots in those cases, because these presentations are basically embeddings into $\mathfrak{sl}_{2n(+1)}$ which also embed the standard CSAs. This should make the calculations in the other answer more transparent. You say that you did the calculations in the prototypical case $\mathfrak{sl}$; now look how through the above conditions on the matrices, certain entries which would be individual root spaces in $\mathfrak{sl}$ get "tied together" and how that changes the relations between the roots. I did some related explicit computations in https://math.stackexchange.com/a/3629615/96384.

But this really depends on the above matrix representations. For example, another presentation of $\mathfrak{so}_n$ one often finds would use the identity matrix for $S$ instead of the above choices. That makes $\mathfrak{so}_n$ consist of the skew-symmetric matrices. Now remark that the intersection of the diagonal matrices with the skew-symmetric ones is $\{0\}$! Still, as long as our base field $K$ contains a square root of $-1$, there is a base change isomorphism from that to the above presentation, cf. An Explicit isomorphism between the orthogonal Lie algebras $\mathfrak{so}_n$ and the Lie algebras of type $B_n$ or $D_n$., Explicit isomorphism between the four dimensional orthogonal Lie algebra and the direct sum of special linear Lie algebras of dimension 3.. So there are CSAs in there, but now they are much harder to see, as they "live off the diagonal". Cf. also Two Definitions of the Special Orthogonal Lie Algebra (where the accepted answer makes that exact point, but should really be read with the caution that it assumes we're over an algebraically closed field, cf. my comment there.)

Answer 3

To build familiarity with the the classical root space decompositions there's a really nice way to think of it in terms of the "defining" representation.

So for example we think of a Cartan subalgebra of $\mathfrak{sl}_n$ as the diagonal matrices (w.r.t. some basis of $\mathbb{C}^n$). Flipping this on its head a little: a Cartan subalgebra is the same data as a decomposition of $\mathbb{C}^n$ into lines $L_1,\dots,L_n$ (i.e. a basis "up to scale"). The Cartan subalgebra is the trace-free elements which act as the scalars on these lines. In other words, trace-free linear combinations of $\mathrm{id}_{L_i}$. The root spaces are then all the possible $\mathrm{Hom}(L_i,L_j)$. This is the root space for the root $\epsilon_i - \epsilon_j$ in the usual way of writing these.

We can do the same for the other classical Lie algebras but now we require that the lines are null lines and are orthogonal to each other for the corresponding symmetric/symplectic form. So lets do that for the one in the question:

Consider $\mathbb{C}^{2n}$ equipped with a symplectic form $\omega$. Let $L_{-n} \oplus \cdots \oplus L_{-1} \oplus L_1 \oplus \cdots \oplus L_{n} = \mathbb{C}^{2n}$ such that each $ L_i $ is orthogonal to each each $L_j$ unless $j=-i$ (w.r.t. $\omega$). Choosing such a decomposition is equivalent to choosing a Cartan subalgebra of $\mathfrak{sp}_{2n}$.

We may identify $\mathfrak{sp}_{2n}$ with $S^2\mathbb{C}^{2n}$ by: $$ a \odot b \mapsto \omega(a,\cdot)b + \omega(b,\cdot)a.$$ Under this identification the Cartan subalgebra we picked is: $$ \mathfrak{h} = \bigoplus_{i=1}^n L_i \odot L_{-i}. $$ The root spaces are all the different $L_i \odot L_{j}$. Indeed we can write the root corresponding to $L_{\pm i} \odot L_{\pm j} $ as $\pm\epsilon_i \pm\epsilon_j$ to make contact with the usual way of writing roots. In particular the long roots are of the form $\pm 2\epsilon_i$ and so have root spaces $L_i \odot L_i = S^2L_i$.