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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.

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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 $U(\mathfrak B)$). – Chuck Hague Jul 23 '12 at 15:58
(Here I am assuming you mean that $\mathfrak B$ is a Lie subalgebra of $\mathfrak G$) – Chuck Hague Jul 23 '12 at 16:00
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"? – Matt Jul 23 '12 at 16:06
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: – Chuck Hague Jul 23 '12 at 17:02
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 $\mathfrak G$. – Chuck Hague Jul 23 '12 at 17:06
up vote 1 down vote accepted

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.

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@Will Sawin You understood correctly my question and answered correctly as well. It was exactly what I was looking for. Thank you so much! – Matt Jul 23 '12 at 18:22

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