# Tensoring with induced representation

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?

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I'm not sure the isomorphism takes $(y\otimes a)\otimes b$ to $y\otimes(a\otimes b)$. Are you? –  Jyrki Lahtonen Mar 11 '14 at 16:22

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 $x \in \mathfrak{g}$).

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I was just adding that to my answer as you commented :) –  mt_ Mar 11 '14 at 18:14
Thanks! I have always wondered what is the correct formalism to think about the $U(\mathfrak{g})$-action on the tensor of two $U(\mathfrak{g})$-modules. (Obviously, an analog does not exist for arbitrary associative ring other than $U(\mathfrak{g})$). Your answer tells me exactly that - $U(\mathfrak{g})$ is a Hopf algebra! –  waikit Apr 7 '14 at 2:55

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.

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 $f$?

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Normally you expect to move from the group side to Lie algebra side by taking the derivative. The presence of the universal enveloping algebra makes this a bit cumbersome to see the correct translation of the above $f$ right away. –  Jyrki Lahtonen Mar 11 '14 at 17:07
Thanks. I didn't know there is this analog in the case of finite groups. Indeed, (formally) taking derivatives does indeed yield the isomorphism written by Matthew. –  waikit Apr 7 '14 at 2:50