# Why is a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ equal to $\mathfrak{sl}(n,\mathbb{C})$?

That question came to me, by reading a proof, where the author shows that a semisimple Lie-algebra $$\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$$ contains the full subalgebra of diagonal matrices of trace $$0$$. He does this by showing, that $$[E_{1j}, E_{j1}]= E_{11}-E_{jj} \in \mathfrak{g}$$ for $$j=2,...,n$$. He than states, that a semisimple Lie-subalgebra of $$\mathfrak{sl}(n,\mathbb{C})$$ of rank $$n-1$$ is equal to $$\mathfrak{sl}(n,\mathbb{C})$$ which is, what i want to verify.

What i understand as the rank of a finite-dimensional semisimple Lie algebra (over an algebraically closed field of characteristic zero) is the (unique) dimension of a Cartan subalgebra. Equivalently, it is defined to be the dimension of the maximal abelian subalgebra, or in the context of subalgebras of $$\mathfrak{sl}(n,\mathbb{C})$$, the largest number of (linear combinations of) generators which commute with each other.

I know that for $$\mathfrak{sl} (n,\mathbb{C} )$$, the Cartan subalgebra may be taken to be the diagonal subalgebra of $$\mathfrak{sl} (n,\mathbb{C} )$$, consisting of diagonal matrices whose diagonal entries sum to zero. So if this subalgebra has dimension $$n-1$$ then $$\mathfrak{sl} (n,\mathbb{C} )$$ has rank $$n-1$$.

So basically the two questions i have are:

Q1: Why has the Subalgebra of $$\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$$ of diagonal trace $$0$$ matrices, dimension $$n-1$$? (answered in the comments)

Q2: Why is a semisimple Lie-subalgebra of $$\mathfrak{sl}(n,\mathbb{C})$$ of rank $$n-1$$ equal to $$\mathfrak{sl}(n,\mathbb{C})$$?

• Your question is somewhat confusing; perhaps too many pronouns without proper antecedents, such as the word "this" in the last paragraph where on first reading I thought you were asserting that every matrix in $\mathfrak{sl} (n,\mathbb{C} )$ is diagonal. Commented Oct 12, 2022 at 14:10
• Question 1: There are $n$ linear independent diagonal matrices of size $n$. However, by requiring trace zero, there remain only $n-1$. So $\dim H=n-1$. Commented Oct 12, 2022 at 19:09
• No, for example $\mathfrak{sl}_3(\Bbb C)$ and $\mathfrak{sl}_2(\Bbb C )\oplus \mathfrak{sl}_2(\Bbb C )$ both have rank $2$. Of course, the root system is different. Commented Oct 12, 2022 at 20:15
• For the second question, could you show that if $\mathfrak h \subset \mathfrak g$ are semisimple LAs, and $\mathfrak c$ is a Cartan subalgebra of $\mathfrak h$, then there exists a CSA of $\mathfrak g$ which contains $\mathfrak c$? And once that is granted, would you be able to conclude from there? Commented Oct 12, 2022 at 21:55
• This is true and can be derived from Dynkin's classification of maximal subalgebras. Commented Oct 13, 2022 at 2:44

Let $$\mathfrak{g}$$ be a simple complex Lie algebra and let $$\mathfrak{h}\leq \mathfrak{g}$$ be a semisimple subalgebra. Firstly, we should note that the maximum possible rank of a semisimple subalgebra of $$\mathfrak{g}$$ is the rank of $$\mathfrak{g}$$. This follows from the observation that Torsten made which is that any Cartan subalgebra of $$\mathfrak{h}$$ is contained in a Cartan subalgebra of $$\mathfrak{g}$$. One way to see this would be to convince your self that semisimple elements of $$\mathfrak{h}$$ must be semisimple elements of $$\mathfrak{g}$$.
So a maximal rank subalgebra must have a Cartan subalgebra $$\mathfrak{c}$$ which is also a Cartan subalgebra of $$\mathfrak{g}$$. Note that this forces $$\mathfrak{h}$$ to be the span of $$\mathfrak{c}$$ together with some of the root spaces. In fact these roots must form a subsystem of our original root system. So the question becomes a combinatorial one: what are the maximal rank root subsystems of the root system of $$\mathfrak{g}$$?
Well there's a nice way to find these. Draw out the Dynkin diagram of $$\mathfrak{g}$$ then add a node to turn it into the extended diagram (This node corresponds to the lowest root and if there are two root lengths we can also extend by the lowest short root instead). Then remove any node and the lines attached to it. The remaining diagram is the Dynkin diagram of a maximal rank subalgebra. You can repeat this process recursively to find all the maximal rank subalgebras. If you try this with an $$A_n$$ root system you should see that you end up with another $$A_n$$ root system so that the only maximal rank subalgebra of $$\mathfrak{sl}_{n+1}$$ is itself. With other types you can get other possibilities though such as $$C_i \times C_{n-i} \subset C_n$$, $$B_4 \subset F_4$$ or $$A_2\times A_2 \times A_2 \subset E_6$$.