# Cartan subalgebras of classical Lie algebras are diagonal matrices

I want to prove that Cartan subalgebras of $$A_l, B_l, C_l, D_l$$ consist of the respective diagonal matrices.

As for $$A_l$$: let $$\mathfrak{h}$$ the set of diagonal matrices of $$\mathfrak{sl}(l+1)$$. We have that $$\mathfrak{h}$$ is of course toral; to prove that it is maximal, I showed that if $$A \in \mathfrak{sl}(l+1)$$ and $$A$$ commutes with every element of $$\mathfrak{h}$$, then $$A$$ must be diagonal. (In particular I chose $$A = E_{ij}$$ from the basis of $$\mathfrak{sl}(l+1)$$, computed $$[A, H]$$ with $$H \in \mathfrak{h}$$ and $$H$$ having $$1$$ at $$i$$-th line and $$-1$$ at the $$j$$-th line and found that $$[A, H] \neq 0$$). However, I noticed that this concludes the proof for every other classical Lie algebra, since every diagonal matrix in $$\mathfrak{sp}(2l), \mathfrak{so}(2l), \mathfrak{so}(2l+1)$$ must have trace zero. Is this sufficient to complete the proof or if $$\mathfrak{h}$$ is the set of diagonal matrices in each classical algebra, do I have to check that $$[A, \mathfrak{h}] \neq 0$$ for each $$A$$ in the basis of those algebras? I think this is not necessary since every element of those bases is a linear combination of elementary matrices. Am I missing something?

Also, what about uniqueness of the Cartan subalgebra in these cases? Since it consists of diagonal matrices, could we argue that it is unique?

• See this post. Commented Jan 1, 2021 at 14:22
• Imprecise question. You can prove that the diagonal matrices in the respective standard matrix representations of those classical Lie algebras give Cartan subalgebras, which one might call "standard" CSAs. Cf. math.stackexchange.com/a/3771729/96384. You can not prove that all CSAs consist of diagonal matrices, because that is wrong. Note that every automorphism sends CSAs to CSAs, and there are many. E.g. $\{ \pmatrix{0 & c \\ c & 0} \vert \; c \in \mathbb C \}$ is a CSA in $\mathfrak{sl}_2(\mathbb C)$. And $\{ \pmatrix{0 & 17c \\ c & 0} \vert \; c \in \mathbb C \}$ is another. Commented Jan 1, 2021 at 17:55

As Dietrich Burde points out, of the various characterisations of Cartan subalgebras in semisimple Lie algebras (cf. 1, 2), the one that likely works best here is that of a maximal toral subalgebra (toral = abelian and consisting of $$ad$$-semisimple elements).

Then, we should agree on what Lie algebras we even mean here: Namely, the split simple forms of types $$A_n, B_n, C_n, D_n$$ (over a field $$K$$, let's assume characteristic zero to be safe). For $$A_n$$, everybody knows that it is $$\mathfrak{sl}_{n+1}(K)$$, the $$n+1 \times n+1$$-matrices over $$K$$ with trace $$0$$. As per 3, the others are:

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 $$C_n$$,

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

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

Now what you want to show is that in each of these algebras, those diagonal matrices which sit inside it form a CSA. (Note that these are never all diagonal matrices. E.g. in all of them, the trace of every matrix has to be $$0$$. Further, e.g. $$\pmatrix{1&0&0&0&0\\0&-1&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0} \notin \mathfrak{so}(5)$$ etc.)

It is nearly trivial that diagonal matrices describe $$ad$$-semisimple elements. It is well known that they commute with each other. Leaves to show the maximality. Indeed, one can show (seemingly) more, namely, that in each case, the diagonal matrices inside the given algebra are their own centraliser.

To show that, I would recommend first looking at the case of $$\mathfrak{sl}_n$$. The crucial step is learning to "see" centralisers of individual diagonal matrices. E.g. you should be able to see with a glance that

the centraliser of $$\pmatrix{3&0&0&0&0\\0&0&0&0&0\\0&0&-3&0&0\\0&0&0&0&0\\0&0&0&0&0} \in \mathfrak{sl}_5$$ is $$\pmatrix{*&0&0&0&0\\0&*&0&*&0\\0&0&*&0&0\\0&*&0&*&*\\0&0&0&*&*}$$, or that

the centraliser of $$\pmatrix{1&0&0&0&0\\0&1&0&0&0\\0&0&-1&0&0\\0&0&0&-1&0\\0&0&0&0&0} \in \mathfrak{sl}_5$$ is $$\pmatrix{*&*&0&0&0\\*&*&0&0&0\\0&0&*&*&0\\0&0&*&*&0\\0&0&0&0&*}$$, or that

the centraliser of $$\pmatrix{17&0&0&0&0\\0&-17&0&0&0\\0&0&-17&0&0\\0&0&0&17&0\\0&0&0&0&0} \in \mathfrak{sl}_5$$ is $$\pmatrix{*&0&0&*&0\\0&*&*&0&0\\0&*&*&0&0\\*&0&0&*&0\\0&0&0&0&*}$$.

All these things can be shown with direct matrix computations, but it is strongly recommended to instead learn to think of them in terms of root space decompositions, cf. 4. This allows you to quickly grasp answers to questions like 5.

Once you have that, you should be able to quickly see that indeed in all cases, the subalgebra made up of diagonal matrices (that is, those diagonal matrices that actually lie in the respective Lie algebra) are self-centralising. (Use that the centraliser of a subalgebra is the intersection of the centralisers of all its elements.)

Then in hindsight, you can actually even use something like

the centraliser of $$\pmatrix{1&0&0&0&0\\0&2&0&0&0\\0&0&-1&0&0\\0&0&0&-2&0\\0&0&0&0&0} \in \mathfrak{sl}_5$$ is $$\pmatrix{*&0&0&0&0\\0&*&0&0&0\\0&0&*&0&0\\0&0&0&*&0\\0&0&0&0&*}$$ (and hence so is its centraliser in $$\mathfrak{so}(5)$$, where there are just a few restrictions on the asterisks anyway), and that kind of thinking should lead you right into the usefulness of regular elements (which this one is, and the other ones above were not), and give you a feeling connected to your previous question.

As final words of warning, as noted in a comment, a) these algebras, even when written exactly as matrices as above, contain many other Cartan subalgebras which are not given by diagonal matrices; and b) note that at least over certain base fields, one can alternatively represent the Lie algebras by matrices which do not even contain non-trivial diagonal ones. E.g. $$\mathfrak{so}_3(\mathbb C)$$ is isomorphic to $$\{\pmatrix{0&a&b\\-a&0&c\\-b&-c&0}: a,b,c \in \mathbb C\}$$. There's infinitely many Cartan subalgebras in there, but they're hard to see ...

• I cannot thank you enough for taking your time to explain things so clearly, this was very illuminating and insightful. I have a bunch of questions now: it looks like the key to prove this in full generality is making sure that such regular elements exist (so that its centraliser is the “smallest” possible and forces the intersection of all centralisers to be the set of diagonal matrices). How can we prove this? In other words, how can we make sure that a diagonal matrix with distinct entries will always exist in $\mathfrak{sl}(n +1)$?
– cip
Commented Jan 2, 2021 at 11:11
• Also, by looking at the matrix structure of all the other Lie algebras, I realised that it may be enough to look at $\mathfrak{sl}(2n)$ to conclude the same for $\mathfrak{sp}(2n)$ and $\mathfrak{so}(2n)$ and at $\mathfrak{sl}(2n+1)$ to conclude the same for $\mathfrak{so}(2n+1)$ (for the same reason you mentioned, we only need to add some additional conditions on the $*$, which are irrelevant for our purpose)
– cip
Commented Jan 2, 2021 at 11:23
• I think you don't need to show the existence of regular elements, although they should be easy to come up with in all cases (e.g. fill your diagonal with as many mutually distinct entries as possible). Alternatively, you can write down certain "basis" elements ($E_{ii}-E_{i+1,i+1}$ for $\mathfrak{sl}$, $E_{ii}-E_{n+i+1,n+i+1}$ and $E_{ii}-E_{i+1,i+1}-E_{n+i+1,n+i+1}+E_{n+i+2,n+i+2}$ for type $B_n$, ...) and show the intersection of their centralisers is diagonal. The second approach seems more cumbersome, but finding the "right" basis and understanding it in terms of (co)roots is worthwhile. Commented Jan 2, 2021 at 17:03