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Let $\mathfrak{g}$ be a Lie algebra, and $V,W$ be two $\mathfrak{g}$ modules. Then one can define a $\mathfrak{g}$ module structure on $V\bigotimes W$ by:

$x\cdot (v\otimes w)=(x\cdot v)\otimes w+v\otimes (x\cdot w)$, whenever $x\in\mathfrak{g},\;v\in V\;w\in W$. My question is - what bilinear maps should be used to write down a universal property for this construction?

If $f:V\times W\to U$ is a bilinear map, let's check what are the conditions for the induced linear map $V\bigotimes W\to U$ to be a $\mathfrak{g}$-module homomorphism. Now $x\cdot (v\otimes w)=xv\otimes w+v\otimes xw$ maps to $f(xv,w)+f(v,xw)$, so that our condition is $f(xv,w)+f(v,xw)=xf(v,w).$

My basic questions is - what is this condition? I know that if $U$ happens to be the base field $\mathbb{F}$, viewed as a trivial $\mathfrak{g}$ module, then this condition is called "invariant form" and is precisely the condition that a bilinear map $V\times W\to \mathbb{F}$ induces a $\mathfrak{g}$ module homomorphism $V\to W^{*}$. What happens when $\mathbb{F}$ is replaced with an arbitrary module $U$? Is there a charachterization\definition of such maps?.

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    $\begingroup$ The map you mention is also not linear, so it could never be a homomorphism of modules, regardless of how the action was defined. So I don't see why that is any hindrance for the usual universal property. $\endgroup$ – Tobias Kildetoft May 31 '20 at 13:54
  • $\begingroup$ Thank you, I edited accordingly. $\endgroup$ – Espace' etale May 31 '20 at 17:48
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    $\begingroup$ Still seems odd to me. The universal property starts with a bilinear map, not a linear one. $\endgroup$ – Tobias Kildetoft May 31 '20 at 18:09
  • $\begingroup$ Thank you, I've edited again and now it should be fine. $\endgroup$ – Espace' etale May 31 '20 at 19:19
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Note that every bilinear map $f: V\times W \rightarrow U$ induces (and conversely, can be recovered from) a linear map

$$\tilde f:V \rightarrow \mathrm{Hom}(W,U)$$ $$v \mapsto [w \mapsto f(v,w)].$$

Now when all of $V,W,U$ are $\mathfrak g$-modules, I claim that your condition

$f(xv,w)+f(v,xw)= xf(v,w)$

is exactly the same as demanding that

$\tilde f$ is a homomorphism of $\mathfrak{g}$-modules;

for which I have to define a $\mathfrak g$-module structure on the full set of linear homomorphisms $\mathrm{Hom}(W,U)$.

Namely, for any two $\mathfrak g$-modules $W,U$, one makes $\mathrm{Hom}(W,U)$ into a $\mathfrak g$-module by defining $x \cdot l$ (for $x\in \mathfrak g$, $l \in \mathrm{Hom}(W,U)$) as the map $$w\mapsto x(l(w))-l(xw).$$ Note that the underlying vector space is really the full set of linear homs; this is made so that those homomorphisms which are $\mathfrak{g}$-equivariant are exactly the ones fixed by this action. Note that in the special case you describe in the question, where $U$ is the ground field (with trivial $\mathfrak g$-action), this action gives the standard definition of the dual representation. So all this neatly generalises this special case, where, as you write, the condition translates to $\tilde f$ being a homomorphism of $\mathfrak g$-modules $V \rightarrow W^\ast$. (And if one wants more motivation for the general case, I guess this action is kind of the derived version of what on a group level would be something like $g(f(g^{-1}\cdot))$, but I'm not sure about that.)

So one could define the tensor product of two $\mathfrak g$-modules $V, W$ together with the bilinear map $\phi_{V,W}: V\times W \rightarrow V\otimes W$ categorically as:

For any bilinear map $f: V \times W \to U$ to any $\mathfrak g$-module $U$ such that $\tilde f:V \rightarrow \mathrm{Hom}(W,U), v \mapsto [w \mapsto f(v,w)]$ is a homomorphism of $\mathfrak g$-modules, there is a unique homomorphism of $\mathfrak g$-modules $\hat{f}: V \otimes W \to U$ such that $f = \hat{f} \circ \phi_{V,W}$.

By the way, going to the tensor product now turns this into the usual adjoint relation between tensor product and homs,

$$\mathrm{Hom}_{\mathfrak g}(V\otimes W, U) \simeq \mathrm{Hom}_\mathfrak{g}(V, \mathrm{Hom}(W,U)),$$

again with the aforementioned action on the full linear $\mathrm{Hom}(W,U)$ inside the RHS; and to be honest, first looking at this and insisting "this must be true, so what's the $\mathfrak g$-action on that Hom set?" gave me the idea for this answer.

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  • $\begingroup$ Thank you, the adjunction was what I was meeting. By the way, could you please reference\broaden your point about the action on $Hom(W,U)$ - does it come from some action on lie groups? Why this is precisely the action, and the best answer would be categorical. In any way, thank you for the answer!. $\endgroup$ – Espace' etale Jun 2 '20 at 10:59
  • $\begingroup$ Look e.g. at Bourbaki's treatise on Lie Groups and Algebras, book I, §3, in particular no. 2--4. Another way to look at it, which they use there, is to translate representations into plain old modules over the universal enveloping algebra $U:=U(\mathfrak g)$. Now since this is a noncommutative ring, a set of linear homs $Hom(V,W)$ is not a priori a module over $U$, but over $U^{op}\otimes U$; now magically, $x \mapsto -x$ is an iso $U \simeq U^{op}$ and, unravelled, gives the above definition back Details in loc. cit. $\endgroup$ – Torsten Schoeneberg Jun 3 '20 at 1:45
  • $\begingroup$ As for the group case, I'm much surer now that the corresponding action there is as described. A categorical approach to that which honestly I don't understand is given in math.stackexchange.com/q/2334514/96384. I like that again, that $G$-action is made so that the fixed set under the action are exactly the $G$-equivariant homs. $\endgroup$ – Torsten Schoeneberg Jun 3 '20 at 1:49
  • $\begingroup$ In short, I admit that some of the definitions of the actions on tensors or homs might look strange at first, but after understanding how they are all interrelated, if one accepts any one of them (on tensor product of group representations, on homs of group representations, on tensor product of algebra rep, on homs of algebra rep.), one has to accept all three others. Now choose which one is most intuitive. $\endgroup$ – Torsten Schoeneberg Jun 3 '20 at 1:50

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