4
$\begingroup$

Let $k$ be a field of characteristic zero, $\mathfrak{g}$ a finite-dimensional Lie algebra over $k$, and let $A,B$ associative $k$-algebras.

Suppose that $\mathfrak{g}$ acts on $A$ and $B$, and that each of these actions admit a quantum moment map. Then the associative algebra $A\otimes_kB$ also has a $\mathfrak{g}$ action and admits a quantum moment map.

In this case, is it true that

$$(A\otimes_kB)//\mathfrak{g}\cong (A//\mathfrak{g})\otimes_k(B//\mathfrak{g})$$

where "$//\mathfrak{g}$" is the operation of quantum Hamiltonian reduction?

$\endgroup$
2
  • $\begingroup$ What is a quantum moment map? $\endgroup$ Commented Feb 3, 2020 at 17:53
  • $\begingroup$ For e.g. $A$, a quantum moment map is an algebra morphism $\mu:\mathcal{U}\mathfrak{g}\to A$ such that for $a\in\mathfrak{g}$, $b\in A$, we have $[\mu(a),b]=a\cdot b$ (where $a\cdot b$ is the action of $a$ on $b$). $\endgroup$ Commented Feb 3, 2020 at 17:58

1 Answer 1

3
$\begingroup$

This is not true.

It is analogous to asking if $(X \times Y)/G \cong (X/G) \times (Y/G)$ for $G$ a group acting on spaces $X$ and $Y$, which is almost never the case. For example, take $X=Y=G$ with the action by left translation. Then the left hand side is isomorphic to $G$, but the right hand side is a point.

The corresponding counterexample in your setting is the following. Suppose $G$ is a connected complex algebraic group acting on $X=G$ by left translations, then set $A=B=\mathcal D_X$. Then the left hand side is isomorphic to $\mathcal D_X$ but the right hand side is isomorphic to $\mathbb C$. To be even more concrete you could set $G$ to be the additive group $\mathbb C$ acting on itself by left translation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .