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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 $Z(\mathfrak{g})\subseteq Z(\mathfrak{g'})$?

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    $\begingroup$ 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. $\endgroup$
    – Joshua Mundinger
    Sep 24 at 15:59
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Sep 25 at 5:59

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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 $\mathbb{C}$.

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