Centers of universal enveloping algebra of complex Lie algebras

Let $$\mathfrak{g}$$ and $$\mathfrak{g'}$$ be complex Lie algebras such that $$\mathfrak{g}$$ is a subalgebra of $$\mathfrak{g'}$$. Let $$Z(\mathfrak{g})$$ and $$Z(\mathfrak{g'})$$ be the centers of the universal enveloping algebras.

1. Is there any relation between $$Z(\mathfrak{g})$$ and $$Z(\mathfrak{g'})$$?

2. Is there any example where one has $$Z(\mathfrak{g})\subseteq Z(\mathfrak{g'})$$?

• What kind of relationship do you expect or desire? One shouldn't expect much of a relation in general--if $A \subseteq B$ are two associative algebras, then $Z(A)$ and $Z(B)$ can be quite different. Sep 24 at 15:59
• Technically, $\mathfrak{g}=\mathfrak{g}'$ answers Question 2. Maybe it is more meaningful to ask when the inclusion $Z(\mathfrak{g})\subseteq Z(\mathfrak{g}')$ holds.
– YCor
Sep 25 at 5:59

For an example to the second question, it is enough to consider a subalgebra $$\mathfrak{g}$$ contained in the center of a Lie algebra $$\mathfrak{g}'$$. For an example showing the possibility of the reverse inclusion, consider as $$\mathfrak{g}'$$ the free Lie algebra of rank $$>1$$ over $$\mathbb{C}$$. Then the universal enveloping algebra of $$\mathfrak{g}'$$ is a free associative algebra and so its center is just $$\mathbb{C}$$. Now, consider any 1-dimensional subalgebra $$\mathfrak{g}$$ of $$\mathfrak{g}'$$. Then the universal enveloping algebra of $$\mathfrak{g}$$ is isomorphic to the polynomial algebra in one indeterminate over $$\mathbb{C}$$.