# Why these Lie algebras have all Cartan subalgebras with the same dimension?

This question is about the following passage in the book "Quantum Field Theory for Mathematicians" by R. Ticciati:

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

For any one of the Lie algebras of immediate interest to us, $${\frak u}(n)$$, $${\frak{su}}(n)$$, $${\frak so}(n)$$, and $${\frak so}(1,3)$$, 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.

If I correctly understand the definition, given the Lie algebra $${\cal G}$$ a Cartan subalgebra is an abelian Lie subalgebra $$H\subset {\cal G}$$ such that if $$H'\subset {\cal G}$$ is another abelian Lie subalgebra with $$H\subset H'$$ then $$H'=H$$.

Now honestly I admit that I don't see what it is easy to see that all these quoted Lie algebras have the property that all Cartan subalgebras have the same dimension.

I've tried to think in terms of generators. Let $$H$$ be a Cartan subalgebra of $${\cal G}$$. We can take any basis of $$H$$, say $$\{h_i\}$$, and complete it to a basis of $${\cal G}$$ by adding some vectors $$\{v_i\}$$. For each $$v_i$$ there must be at least one $$h_j$$ with $$[v_i,h_j]\neq 0$$ otherwise $$H$$ would not be maximal. From this perspective I think that all Cartan subalgebras having the same dimension is tantamount of saying that given any basis $$\{e_i\}$$ of $${\cal G}$$ the maximal number of vectors which commute among themselves is the same. But again I don't find it easy to see why this is the case for these Lie algebras.

So why is the dimension of all Cartan subalgebras of the quoted Lie algebras the same? Why should that be considered as something easy to see? Moreover, what is the common feature of all these Lie algebras underlying this result?

• This is awkward. True, one finds different definitions of CSA'a which are equivalent in some specified settings (math.stackexchange.com/q/3537312/96384, math.stackexchange.com/q/1071417/96384, math.stackexchange.com/q/3480970/96384, math.stackexchange.com/q/3125858/96384). But this here is the worst definition I have ever seen. It might match the standard one for compact real LA's, but I'm sure that in $\mathfrak{so}(1,3)$, this definition would exhibit something as Cartan subalgebra which is not a CSA in any of the above, far more standard, definitions. Sep 8, 2020 at 0:17
• Concretely, it's well-known (although confusing) that $\mathfrak{so}(1,3)$ is isomorphic to $\mathfrak{sl}_2(\mathbb C)$ viewed as a six-dimensional real Lie algebra; now in that, look at the two-(real-)dimensional subspace $\{ \pmatrix{0&a+bi\\0&0}: a,b \in \mathbb R\}$. That's an abelian subalgebra of dimension $2$, I bet it's maximal (even if not, it were contained in a maximal abelian subalgebra), but it's full of non-zero nilpotent elements which should never ever be in a CSA, and according to all definitions quoted above, are not. Sep 8, 2020 at 0:27

First, the way the definition is stated is literally incorrect, without some (implicit?) assumptions. For one, the Cartan subalgebra $$\mathfrak h$$ is required to act semi-simply (by adjoint) on the Lie algebra $$\mathfrak g$$. Dropping this inadvertently allows certain (abelian) subalgebras of unipotent radicals of simple algebras. E.g., in $$\mathfrak g =\mathfrak s\mathfrak l_{2n}$$, the matrices $$\mathfrak n$$ with zeros except at the $$i,j$$ place with $$1\le i \le n$$ and $$n+1\le j\le 2n$$ (that is, all zeros except the upper-right $$n$$-by-$$n$$ block) form an abelian subalgebra of dimension $$n^2$$, which for $$n\ge 2$$ is much larger than $$2n-1$$, the dimension of the standard, diagonal Cartan subalgebra. To be clear, this $$\mathfrak n$$ is not a Cartan subalgebra.
The operational reason that acting semi-simply is required is that we want to decompose the Lie algebra into simultaneous eigenspaces (called "root spaces") of $$\mathfrak h$$. If the action were not semi-simple, even a single individual might not have a basis of eigenvectors. And the unipotent examples show that semi-simple action is not automatic.
The way that I myself know these ranks and/or "standard" Cartan subalgebras is simply that someone told me, or I read it (long ago), and gave some sort of explanation of checking once you have a good guess. E.g., having the side-channel information that "geometric algebra" is relevant is almost enough to be able to answer these questions. Even over $$\mathbb R$$, knowing Sylvester's inertia theorem, and analogues, and Witt's theorem about extension of "form"-preserving automorphisms, is enough to recover the determination of rank (=dimension of any Cartan subalgebra).
The fact that the ranks of certain classical groups over $$\mathbb R$$ or $$\mathbb C$$ are the same is fairly meaningless, in general, for basic purposes, so far as I know. For subtler purposes, this can mean something...
• Re last paragraph: I think the claim in the book is not about non-isomorphic LAs having the same rank, but e.g. that all CSAs in $\mathfrak{su}(n)$ have the same dimension, i.e. the rank being well-defined to begin with. That is generally proved via non-trivial conjugacy theorems which give an insight into the structure as well. I do not know what "easy" argument the book has in mind for the examples it mentions (especially given that its definition deviates from the standard one). Sep 8, 2020 at 2:07