Let $\mathfrak{g}$ be a semisimple Lie algebra and $M, N$ be two modules of $\mathfrak{g}$. Is it true that $M \otimes N \cong N \otimes M$? If $\mathfrak{g}$ is replaced by other algebras, $M \otimes N \cong N \otimes M$ is also true? What conditions do we need in order to have $M \otimes N \cong N \otimes M$? Thank you very much.
Lie Algebras – Tensor Product of Modules
lie-algebras
Related Solutions
What kind of ring is $U(L)$?
Since representations of Lie algebras behave like representations of groups (the category has tensor products and duals, for example), you should expect that the universal enveloping algebra $U(\mathfrak{g})$ has some extra structure which causes this, and it does: namely, it is a Hopf algebra (a structure shared by group algebras). The comultiplication is defined on basis elements $x \in \mathfrak{g}$ by $$x \mapsto 1 \otimes x + x \otimes 1$$
(this is necessary for it to exponentiate to the usual comultiplication $g \mapsto g \otimes g$ on group algebras) and the antipode is defined by $$x \mapsto -x$$
(again necessary to exponentiate to the usual antipode $g \mapsto g^{-1}$ on group algebras).
This is an important observation in the theory of quantum groups, among other things.
Thus, via the envelopping algebra Lie algebras and their represnetations cn be studied from a ring theoretic point of view. Is the cconverse true in some sense?
Not in the naive sense, the basic problem being that if $A$ is an algebra and $L(A)$ that same algebra regarded as a Lie algebra under the bracket $[a, b] = ab - ba$, then a representation of $L(A)$ does not in general extend to a representation of $A$, but to a representation of $U(L(A))$, which may be a very different algebra (take for example $A = \text{End}(\mathbb{C}^2)$).
Of course there are other relationships between ring theory and Lie theory. For example, if $A$ is a $k$-algebra then $\text{Der}_k(A)$, the space of $k$-linear derivations of $A$, naturally forms a Lie algebra under the commutator bracket. Roughly speaking this is the "Lie algebra of $\text{Aut}(A)$" in a way that is made precise for example in this blog post.
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_1$ and $B_2:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by $$ [g_1\otimes g_2, g_1'\otimes g_2']:= B_1(g_1,g_2)\otimes B_2(g_1',g_2')\quad \text{for } g_1,g_1' \in \mathfrak{g}_1 \text{ and } g_2,g_2' \in\mathfrak{g}_2.$$ For example if $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.
Best Answer
There are two things going on here. The first is that you are taking the tensor product of two vector spaces. The second is that you have Lie algebra structures.
For any two vector spaces, we have an isomorphism $V\otimes W \to W\otimes V$ given by $v\otimes w \mapsto w\otimes v$. We can easily get that this exists and is an isomorphism by using the universal property: a bilinear map from $V\times W$ is easily turned into a binlinear map from $W\times V$ by defining $\widetilde{f}(w,v)=f(v,w)$. If you have not seen this before, you should try to work out the details.
The issue that remains is whether this isomorphism of vector spaces is compatible with the Lie algebra structure. Let $\tau(v\otimes w)=w\otimes v$. Then we must show that for each $g\in \mathfrak g$, $\tau(g.v\otimes w)=g.\tau(v\otimes w)$. Note that by linearity, it suffices to check the statement on simple tensors. This is a straightforward calculation, using only the definition of the $\mathfrak g$-module structure on the tensor product.
More generally, what is going on is that the universal enveloping algebra $\mathcal U\mathfrak g$ is a Hopf algebra with comultiplication $\Delta:\mathcal U\mathfrak g\to \mathcal U\mathfrak g\otimes \mathcal U\mathfrak g$ defined on generators by the formula $\Delta(g)=g\otimes 1 + 1 \otimes g$ and extended by the condition that $\Delta$ is a map of algebras (exercise: verify that this map is well defined).
The Hopf algebra structure is what allows us to define an action on $V\otimes W$. Because a $\mathfrak g$-module is the same thing as a $\mathcal U\mathfrak g$-module, $V\otimes W$ is a priori only a module over $\mathcal U\mathfrak g\otimes \mathcal U\mathfrak g$. To get the structure of a $\mathcal U\mathfrak g$-module, we must pull back along $\Delta$. The module structure comes from the composite $$\mathcal U\mathfrak g\to \mathcal U\mathfrak g\otimes \mathcal U\mathfrak g \to \operatorname{End}(V)\otimes \operatorname{End}(W) \to \operatorname{End}(V\otimes W).$$
where all the maps are maps of algebras. The fact that $\tau$ is a map of $\mathfrak g$-modules comes from the easy to verify fact that $\tau \Delta = \Delta \tau$ (where $\tau$ in this context is the swapping map on $\mathcal U\mathfrak g \otimes \mathcal U\mathfrak g$).
In general, the category of modules over a cocommutative Hopf algebra, or comodules over a commutative Hopf algebra will have a commutative tensor product induced from the tensor product of vector spaces. More generally, one can study quantum groups, where the commutativity or cocommutativity is weakened so that instead of the category of representations being symmetric monoidal, it is braided monoidal. In this way, understanding quantum groups allows one to better understand representations of the braid group.