# Is Cartan subalgebra of Complex semisimple Lie algebra the maximal Abelian subalgebra? I found two places give the different answers.

In wiki https://en.wikipedia.org/wiki/Cartan_subalgebra Example 4, it says that Cartan subalgebra of complex semisimple Lie algebra is not maximal Abelian subalgebra.

However in Brian C. Hall's GTM222 Page162, 10th line from the bottom, it says Cartan subalgebra of complex semisimple is maximal Abelian subalgebra.

How to explain these? Who is right?

• That is not quite what it says in Hall's book. Take another look. – Tobias Kildetoft Jan 19 '16 at 9:55
• @TobiasKildetoft In page 162 The 10th line from bottom, there is a saying "h is a maximal commutative subalgebra" – 346699 Jan 19 '16 at 9:59
• Ahh, sorry, indeed it does. Rather, you misquoted the wikipedia article. It says that the Cartan might not have largest possible dimension among abelian subalgebras, which is not the same as not being maximal. – Tobias Kildetoft Jan 19 '16 at 10:03
• @TobiasKildetoft Why is the abelian subalgebra with the largest possible dimension different from the maximal abelian subalgebra? – 346699 Jan 19 '16 at 10:07
• There is no "the" maximal abelian subalgebra. Maximal here is with respect to inclusion. – Tobias Kildetoft Jan 19 '16 at 10:09

If $\mathfrak{g}$ is a complex, semisimple, finite-dimensional Lie algebra and $\mathfrak{h} \subseteq \mathbb{g}$ a Cartan subalgebra, then $\mathfrak{h}$ is abelian and self-centralizing, i.e. $Z_\mathfrak{g}(\mathfrak{h}) = \mathfrak{h}$ (this is, for example, show in Humphrey’s book).

If $\mathfrak{h}$ was not a maximal abelian subalgebra with respect to the inclusion, then $\mathfrak{h}$ would be properly contained is some maximal abelian subalgebra $\mathfrak{a}$ of $\mathfrak{g}$. But then $\mathfrak{a} \subseteq Z_\mathfrak{g}(\mathfrak{h})$, contradicting $\mathfrak{h}$ being self-centralizing. So if $\mathfrak{g}$ is finite-dimensional it is true that any Cartan subalgebra is a maximal abelian subalgebra. (I don’t know what happens in the infinite-dimensional case.)

The example in Wikipedia, namely $$\mathfrak{a} = \left\{ \begin{pmatrix} 0 & A \\ 0 & 0 \end{pmatrix} \in \mathfrak{sl}_{2n}(\mathbb{C}) \,\middle|\, A \in \mathfrak{gl}_n(\mathbb{C}) \right\},$$ shows that while Cartan-subalgebras are maximal abelian subalgebras with respect to the inclusion, they are not necessarily of maximal dimension among all abelian subalgebras: Every Cartan subalgebra of $\mathfrak{sl}_{2n}(\mathbb{C})$ has dimension $2n-1$ (for example the traceless diagonal matrices), but $\mathfrak{a}$ has dimension $n^2$.

What this tells us that $\mathfrak{a}$ does not contain a Cartan subalgebra of $\mathfrak{sl}_{2n}(\mathbb{C})$ (strictly speaking we only get this for $n > 1$, but for $n = 1$ this is also easy to see).

Both are right. First, the book is right: the notion of a maximal abelian subalgebra refers to inclusion, and a Cartan subalgebra of a complex semisimple Lie algebra certainly is maximal abelian.

The second notion is the one of the maximal dimension of an abelian subalgebra is an interesting invariant, say $\alpha(L)$, for a Lie algebra $L$. For simple complex Lie algebras it is given as follows (see here); $$\begin{array}{c|c|c|} \mathfrak{s} & \dim (\mathfrak{s}) & \alpha(\mathfrak{s}) \\ \hline A_n,\,n\ge 1 & n(n+2) & \lfloor (\frac{n+1}{2})^2 \rfloor \\ \hline B_3 & 21 & 5 \\ \hline B_n,\, n\ge 4 & n(2n+1) & \frac{n(n-1)}{2}+1 \\ \hline C_n,\,n\ge 2 & n(2n+1) & \frac{n(n+1)}{2} \\ \hline D_n,\,n\ge 4 & n(2n-1) & \frac{n(n-1)}{2} \\ \hline G_2 & 14 & 3 \\ \hline F_4 & 52 & 9 \\ \hline E_6 & 78 & 16 \\ \hline E_7 & 133 & 27 \\ \hline E_8 & 248 & 36 \\ \end{array}$$

The wikipedia references says the following: " The dimension of a Cartan subalgebra is not in general the maximal dimension of an abelian subalgebra, even for complex simple Lie algebras." This is also true, and an example is given there.

Hence both statements are correct, but mean something different.