# Center of of a subalgebra of a lie algebra: Under what conditions is $Z(\mathfrak{h})=Z(\mathfrak{g})\cap \mathfrak{h}$

I am aware this is about the same question as this question. But I do not believe this is a duplicate post. I am looking to understand some properties about the center and am using this question is a good example. I want to know:

(1) Under what conditions is do we have for $$\mathfrak{h}\subset \mathfrak{g}$$ to imply $$Z(\mathfrak{h})=Z(\mathfrak{g})\cap \mathfrak{h}$$

(2) How does the argument I link to conclude that $$Z(\mathfrak{sl}(n,F))=Z(\mathfrak{gl}(n,F))\cap \mathfrak{sl}(n,F)$$

Since (1) is true in the case of $$\mathfrak{sl}(n,F)\subset\mathfrak{gl}(n,F)$$ I feel trying to understand an argument which shoes this will help tease out what conditions are needed for the general case. There is a question in Humphreys that asks us:

To show that $$\mathfrak{sl}(n,F)$$ (matrices with trace zero) has center $$0$$, unless $$\operatorname{char}F$$ divides $$n$$, in which case the center is $$\mathfrak{s}(n,F)$$ (scalar multiples of the identity).

Some facts that will be useful are that the $$Z(\mathfrak{gl}(n,F))$$ is $$\mathfrak{s}(n,F)$$. Also $$\mathfrak{gl}(n,F)=\mathfrak{sl}(n,F)+\mathfrak{s}(n,F)$$ as vector spaces.

Here is an argument taken from this document of solutions which claims that $$Z(\mathfrak{sl}(n,F))=Z(\mathfrak{gl}(n,F))\cap \mathfrak{sl}(n,F)$$.: I will rewrite the argument to show my confusion. If $$c\in Z(\mathfrak{sl}(n,F)$$ then $$[x,c]=0$$ for all $$x\in \mathfrak{sl}(n,F)$$. Obviously $$c\in \mathfrak{gl}(n,F)=\mathfrak{sl}(n,F)+\mathfrak{s}(n,F)$$, but I do not see why this means that $$c\in Z(\mathfrak{gl}(n,F)$$. I do not see then why it follows that $$Z(\mathfrak{sl}(n,F))=Z(\mathfrak{gl}(n,F))\cap \mathfrak{sl}(n,F)$$.

Let's first look at the question in your last paragraph. I claim you can easily generalise the argument to:

(*) If $$\mathfrak g = \mathfrak h + \mathfrak a$$ such that $$\mathfrak a$$ commutes with $$\mathfrak h$$, then $$Z(\mathfrak h) \subseteq Z(\mathfrak g)$$ (and the opposite inclusion is always true anyway).

Namely, let $$c \in Z(\mathfrak h)$$ and $$x \in \mathfrak g$$; by assumption, we can write $$x=h+a$$ for $$h \in \mathfrak h, a \in \mathfrak a$$ and we have $$[c, x]=\underbrace{[c,h]}_{0 \text{ bc. } c\in Z(\mathfrak h)}+\underbrace{[c,a]}_{0 \text{ bc. } \mathfrak a \text{ comm. w. } \mathfrak h}=0.$$

This (*) is a sufficient but not necessary criterion which settles question (2).

As for the general question (1), first note that $$Z(\mathfrak h) \supseteq Z(\mathfrak g) \cap \mathfrak h$$ is always true for $$\mathfrak h \subseteq \mathfrak g$$, and of course $$Z(\mathfrak h) \subseteq \mathfrak h$$, so the question boils down to when

$$Z(\mathfrak h) \stackrel{?}\subseteq Z(\mathfrak g).$$

(Examples where this is not the case abound. E.g. take any non-zero $$\mathfrak g$$ which has centre $$0$$, and $$\mathfrak h =$$ the one-dimensional (hence abelian!) subalgebra spanned by a non-zero element.)

Inspecting our argument for (*) from the beginning shows that actually we do not need that $$\mathfrak a$$ commutes with all of $$\mathfrak h$$, but only with $$Z(\mathfrak h)$$; further, we don't need $$\mathfrak a$$ to be a subalgebra, we just need to write every element $$x \in \mathfrak g$$ as

(something in $$\mathfrak h$$ + something that commutes with $$Z(\mathfrak h)$$).

So a less restrictive sufficient criterion for what we want is:

There is a vector space complement $$A$$ of $$\mathfrak h$$ in $$\mathfrak g$$ such that every element of $$A$$ commutes with every element of $$Z(\mathfrak h)$$.

Note this is true for one vector space complement iff it is true for every vector space complement. Namely, as soon as there is $$x \in \mathfrak g \setminus \mathfrak h$$ and $$z \in Z(\mathfrak h)$$ such that $$[x, z] \neq 0$$, we have $$z \notin Z(\mathfrak g)$$. Another way to express it is to see that $$\mathfrak h$$ naturally acts on the (vector space) quotient $$V:=\mathfrak g/\mathfrak h$$ and to say

$$Z(\mathfrak h)$$ acts trivially on $$\mathfrak g/\mathfrak h$$.

So there's a criterion. If it's nicer/easier/more useful than just writing $$Z(\mathfrak h) = Z(\mathfrak g)$$ depends on taste/context.

Finally, as an example for my claim that what we used at the beginning for (2), criterion (*) which we can now phrase as

(*) $$\mathfrak h$$ acts trivially on $$\mathfrak g/\mathfrak h$$,

is a sufficient but not necessary criterion: Take $$\mathfrak g = \mathfrak{sl}_{n\ge 2}(\mathbb C)$$ and $$\mathfrak h =$$ upper triangular matrices in $$\mathfrak g$$. You'll find that actually $$Z(\mathfrak h) =Z(\mathfrak g) = 0$$, but every non-zero element of $$\mathfrak h$$ acts non-trivially on $$\mathfrak{g}/\mathfrak h$$ (which can be identified with the strict lower triangular matrices).