1
$\begingroup$

Let $\frak G$ be a Lie algebra and let $M$ be a $\frak G$-module generated by a vector $v$ satisfying some set of defining relations denoted by $R$. I mean, $M = U(\frak G)/\langle R \rangle$, where $\langle R \rangle$ is the $U(\frak G)$-submodule generated by $R$ and $U(\frak G)$ denotes the universal enveloping algebra of $\frak G$. In this case, $v = \overline 1$.

Suppose that $\frak B \subset \frak G$ is an ideal such that ${\frak B} M = 0$. So, we can naturally regard $M$ as a $\frak G/\frak B$-module.

Can we say that $M$ is a $\frak G/\frak B$-module given by one generator and relations coming from the defining relations of $M$? How to formalize this fact?

I feel that it is some sort of consequence of the Theorem of Isomorphisms, but I don't know how to write it in a formal way.

$\endgroup$
6
  • $\begingroup$ I'm not quite sure what sort of formalization you want, but in terms of the enveloping algebra, the fact that $M$ is generated by a single vector $v$ is equivalent to the fact that that map $ U(\mathfrak G) \to M $, $X \mapsto X.v$ is surjective. The fact that $\mathfrak B.M=0$ is equivalent to the fact that the left ideal $ U(\mathfrak G) \cdot U(\mathfrak B)^+ $ is contained in the annihilator of $M$. (Here, $U(\mathfrak B)^+$ denotes the augmentation ideal of $U(\mathfrak B)$). $\endgroup$ Jul 23, 2012 at 15:58
  • $\begingroup$ (Here I am assuming you mean that $\mathfrak B$ is a Lie subalgebra of $\mathfrak G$) $\endgroup$ Jul 23, 2012 at 16:00
  • $\begingroup$ I agree with your comment. Based on this, it follows that $M$ is generated by a single vector satisfying relations which are not exactly the same defining relations for $M$, because we took a quotient of the base Lie algebra, right? So we have to rewrite this relations excluding from this relations the elements which are in $U(\frak B)$, is not it? Do you understand where I am in "trouble"? $\endgroup$
    – Matt
    Jul 23, 2012 at 16:06
  • $\begingroup$ Generally the closest one gets to "defining relations" for a cyclic $\mathfrak G$-module $M$ is finding a generating set of the annihilator of $M$, which will be some left ideal in $U(\mathfrak G)$ (these ideals are called primitive ideals). Usually this is nontrivial, so I don't know if there's more that can be said; but I'm not an expert in primitive ideals. You might want to look at Joseph's paper "Primitive Ideals in Enveloping Algebras" which you can find here: mathunion.org/ICM/ICM1983.1/Main/icm1983.1.0403.0414.ocr.pdf $\endgroup$ Jul 23, 2012 at 17:02
  • $\begingroup$ Also, I'm not sure what you mean by a $\mathfrak G / \mathfrak B$-module, because $\mathfrak G / \mathfrak B$ will not in general be a Lie algebra since $\mathfrak B$ need not be a Lie ideal of $\mathfrak G$. $\endgroup$ Jul 23, 2012 at 17:06

1 Answer 1

1
$\begingroup$

If I understand you correctly, you want to write $M$ as $U(\frak G/\frak B)/\langle R'\rangle$ where $R'$ is somehow determined by $S$.

This can indeed be done. There is a natural map $U(\frak G) \to U (\frak G/\frak B)$, the unique ring homomorphism that sends elements of $G$ to the corresponding elements of $G/B$. Or alternately, you can see this map as coming from the universal property. It is clear that this map is surjective, as every element of $U(\frak G/\frak B)$ can be written in terms of elements of $\frak G/\frak B$ which can themselves be written in terms of elements of $\frak G$.

Thus, it is the quotient by some ideal $I$. If we can show that $I$ is contained in $\langle R\rangle$ then we are done because $M=U(\frak G )/\langle R\rangle=U(\frak G )/\langle I\rangle / \langle R'\rangle=U(\frak G/\frak B)/\langle R'\rangle$, where $R'$ is the image of $R$ in $U(\frak G)/\langle I\rangle=U(\frak G/\frak B)$.

But it is easy to check that $I$ is just the ideal of $U(\frak G)$ generated by $\frak B$. Take a basis for $\frak G$ which includes a basis for $\frak B$ and use this to define a basis for $U(\frak G)$, then the map to $U(\frak G/\frak B)$ is just removing terms which contain an element of the basis of $\frak B$, which is the same as taking the quotient by the ideal generated by $\frak B$.

So we are done.

$\endgroup$
1
  • $\begingroup$ @Will Sawin You understood correctly my question and answered correctly as well. It was exactly what I was looking for. Thank you so much! $\endgroup$
    – Matt
    Jul 23, 2012 at 18:22

Your Answer

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

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