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

  • 2
    $\begingroup$ 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. $\endgroup$ 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

1 Answer 1


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 $\mathbb{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.