Tensoring with induced representation

Question

On J. Humphreys' book "Representations of Semisimple Lie Algebras in the BGG Category O", Theorem 3.6, a Tensor Identity is quoted:

$$ (U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} L) \otimes M\simeq U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} (L \otimes M)$$

where $L$ is a $U(\mathfrak{b})$-module and $M$ is a $U(\mathfrak{g})$-module.

While this isomorphism seems clear as a $k$-module isomorphism, it doesn't seem to be a $U(\mathfrak{g})$-module homomorphism.

Apparently, an element $x\in \mathfrak{g}$ should act on the left module by $$ x\cdot (y\otimes a \otimes b) = xy\otimes a \otimes b + y\otimes a \otimes xb $$ while it acts on the right by $$ x\cdot (y\otimes a \otimes b) = xy\otimes a \otimes b $$

Am I misunderstanding sometihng?

Answer 1

Let $\Gamma$ be a sub-Hopf algebra of the Hopf algebra $\Lambda$. Let $M$ be a $\Lambda$-module and $N$ a $\Gamma$-module. Then $$ \Lambda \otimes _\Gamma (N \otimes M|_\Gamma) \cong (\Lambda \otimes_\Gamma N) \otimes M.$$ The isomorphism is given by $$ \lambda \otimes_\Gamma (n\otimes m) \mapsto \sum (\lambda_{(1)} \otimes _\Gamma n) \otimes \lambda_{(2)} m.$$

I'm using Sweedler notation here: $\Delta(\lambda) = \sum \lambda_{(1)} \otimes \lambda_{(2)}$ where $\Delta$ is the comultiplication.

Note that this does agree with Jyrki's map in the group case, for then $\Delta(g)=g\otimes g$ for $g$ a group element.

In your case, $\Delta(x) = x\otimes 1 + 1 \otimes x$ for $x \in \mathfrak{g}$. You can use that to get an explicit expression for the isomorphism, for example it sends $$ x \otimes (n \otimes m) \mapsto (x\otimes m) \otimes n + (1\otimes n) \otimes xm $$ for $x \in \mathfrak{g}$.

The inverse map is given by $$ (\lambda \otimes_\Gamma n)\otimes m \mapsto \sum \lambda_{(1)} \otimes_\Gamma (n \otimes S(\lambda_{(2)}) m) $$ where $S$ is the antipode of $\Lambda$ (in your case, $S$ is the antihomomorphism defined by $S(x)=-x$ for $x \in \mathfrak{g}$).

Answer 2

Not an answer. Just explaining why we should not expect the obvious $k$-linear mapping to work. Hopefully this leads the OP or somebody else to dig out the correct isomorphism.

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The analogous result at the level of finite groups is the following. If $H\le G$, $L$ is a $kH$-module, and $M$ a $kG$-module, then there is an isomorphism of $kG$-modules $$ f:kG\otimes_{kH}(L\otimes M)\simeq (kG\otimes_{kH} L)\otimes M $$ given at the level of elementary tensors by the recipe $$ x\otimes (\ell\otimes m)\mapsto (x\otimes \ell)\otimes (xm) $$ where $x\in G$, $\ell\in L$ and $m\in M$. This is a well-defined map, because for all $h\in H$ we have $$ (xh)\otimes(\ell\otimes m)\mapsto (xh\otimes\ell)\otimes (xhm), $$ and also $$ x\otimes (h\ell\otimes hm)\mapsto (x\otimes h\ell)\otimes (xhm)=(xh\otimes\ell)\otimes (xhm). $$ This is absolutely needed, because in the range we have (the first tensor product is over $kH$) $$ (xh)\otimes(\ell\otimes m)=x\otimes h(\ell\otimes m)=x\otimes(h\ell\otimes hm). $$

Observe that the resulting map is an isomorphism of $kG$-modules as for all $g\in G$ $$ \begin{aligned} f(g\cdot(x\otimes(\ell\otimes m)))&=f(gx\otimes(\ell\otimes m))\\ &=(gx\otimes \ell)\otimes gxm=g\cdot((x\otimes\ell)\otimes m))\\ &=g\cdot f(x\otimes(\ell\otimes m)) \end{aligned} $$ by the definition of $f$ and the action of $G$ on the induced module as well as the tensor product.

The tensor identity that you want needs to reflect this somehow, and use something other than the obvious $k$-linear mapping. The action of $U({\mathfrak g})$ on the tensor product of two modules uses the coproduct. This we can sufficiently test (as you also indicated) at the level of ${\mathfrak g}$. But what about $f$?