group of automorphisms of vector spaces

Question

let α is an automorphism of G as a vector space and β is an automorphism of H as a vector space as well, then the tensor product of G and H is an automorphism of G tensor H? Let suppose that G and H are lie algebras, I want to compute the the automorphism group of tensor product of two lie algebras. Can we say that automorphism group of tensor product of lie algebras has a subgroup isomorphic to the direct product Aut(G) × Aut(H)?

Answer 1

Abelian groups are in particular $\Bbb Z$-modules, and the tensor product operation is defined for arbitrary $R$-modules. From the categorical viewpoint, the tensor product of modules is a (provably unique up to isomorphism) object satisfying the appropriate universal properties. The usual way to show existence is to explicitly construct it as a quotient of the free module on the underlying set of the direct product. All of this is discussed over at the Wikipedia article on the tensor product.

Free modules are huge relative to the sets we build them off of. Even though we are quotienting it, the resulting object (the tensor product) will still generally be big compared to the direct product of modules. For example, $\dim(V\otimes W)=\dim(V)\cdot\dim(W)$ for vector spaces $V,W$, whereas the direct product has $\dim(V\times W)=\dim(V)+\dim(W)$.

The exception to this rule is when the two modules are incompatible in a specific way, namely if their torsion/divisibility are in conflict. For example $\Bbb Z/n\Bbb Z\otimes_{\Bbb Z}\Bbb Z/m\Bbb Z\cong \Bbb Z/(n,m)\Bbb Z$ are isomorphic as $\Bbb Z$-modules, where $(n,m)$ is shorthand for $\gcd(n,m)$ used in elementary number theory (and beyond). I discuss more on this in more detail in this answer.

So far this only covers abelian groups. If you try to naively define $G\otimes H$ via the universal property for arbitrary (not necessarily abelian) groups $G,H$, we obtain the free group on the set $G\times H$ quotiented by the relations $g_1g_2\otimes h=(g_1\otimes h)(g_2\otimes h)$ and $g\otimes h_1h_2=(g\otimes h_1)(g\otimes h_2)$. The resulting group will be simply $G^{\rm ab}\otimes_{\Bbb Z}H^{\rm ab}$ as a group, where $-^{\rm ab}$ denotes the abelianization operation, defined by $G^{\rm ab}:=G/[G,G]$, and $\otimes_{\Bbb Z}$ denotes the tensor product of two abelian groups.

Since the above is a rather boring situation, a generalized type of "nonabelian tensor product" operation has been defined for groups which is not so trivial. The definition seems to takes some inspiration from semidirect products, as it invokes actions of $G$ and $H$ on each other and encodes this in the relations we are quotienting alongside conjugation. (For comparison, the semidirect product $G\times_\varphi H$ is essentially the free product $G*H$ quotiented by the relations $\varphi_h(g)=hgh^{-1}$ for all $g\in G,h\in H$ (and $\varphi_{\square}:H\to{\rm Aut}(G)$)). GroupProps (as expected) has an article that introduces the nonabelian tensor product. It seems the motivations for it (discussed elsewhere) involve relatively advanced math.

In summary, tensor products are quotients of free modules/groups built from the direct product, they are generally much bigger except when certain parts conspire to annihilate each other.