In the book "QFT for Mathematicians" by Robin Ticciati the author says in the Chapter 6 about representations of the Lorentz group:

Definition 6.2.28: A Cartan subalgebra of a Lie algebra $\mathcal{G}$ is a maximal commuting Lie subalgebra of $\mathcal{G}$.

For any one of the Lie algebras of immediate interest to us, $\mathfrak{u}(n),\mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$, it is easy to show that its Cartan subalgebras all have the same dimension. This result shows that the dimension of a Cartan subalgebra often depends only on the Lie algebra and is therefore characteristic of the Lie algebra.

Definition 6.2.29: If all the Cartan subalgebras of a Lie algebra have the same dimension, then this dimension is called the rank of the Lie algebra.

Homework 6.2.30: Find Cartan subalgebras of $\mathfrak{u}(n),\mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$. Show that these subalgebras have dimensions $n,n-1,[n/2]$ (the greatest integer less than $n/2$) and $2$ respectively.

Now the author doesn't say how one should proceed in finding a Cartan subalgebra. He just states these definitions and leave as an exercise to find some Cartan subalgebras.

Now I'm wondering how we should go in finding such Cartan subalgebras. I've came up with the following idea.

A Cartan subalgebra must always be abelian. In that case the problem of constructing such an algebra is the same of finding a set of linearly independent vectors which commute among themselves.

So my idea is: take a basis $\{E_a\}$ for the Lie algebra. Set arbitrarily $X_1 = E_1$ as the first vector in the basis of the Cartan subalgebra. Consider the operator $T_1 : \mathcal{G}\to \mathcal{G}$ given by $T_1(Z)=[X_1,Z]$. If we find $\ker T$ we find a vector subspace on which everything commutes with $X_1$.

Now, I imagine that although we are granted that everything there commutes with $X_1$, there might be non-commuting vectors. We thus arbitrarilly take $X_2\in \ker T$ some vector distinct than $X_1$. We know consider the operator $T_2 : \ker T_1 \to \ker T_1$ defined by $T_2(Z) = [X_2,Z]$. If we find $\ker T_2\subset \ker T_1$ we find a set of vectors which all commute with $X_2$, but since this is a subspace of $\ker T_1$ they also commute with $X_1$.

We would them iterate this until we get a maximal set. My idea is that we reach the maximal set when $\ker T_n =\{0\}$.

If we always expand this in basis of $\mathcal{G}$ and later of the appropriate $\ker T_k$ these become just homogeneous systems of linear equations.

The only issue is that this seems not very practical. My question is: is this the correct approach to find a Cartan subalgebra? If not, what is the correct way to do it?


1 Answer 1


What is very useful here is the following linear algebra fact:

If two operators $A$ and $B$ acting on a finite dimensional vector space V commute, then they have the same eigenspace.

Without loss of generality, you can suppose $A$ is in Jordan normal form, and then understanding which operators commutes with it is a simpler computation.

Now we have to understand the possible Jordan normal forms of $A$. My hint is try to find a CSA in the very basic cases of $\mathfrak{su}(2)$ and $\mathfrak{su}(3)$ and then try to generalize your conclusions looking at the structure of common eigenspaces fort CSA in su(n) and the other Lie algebras.


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