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The Kronecker Product is defined here: https://en.m.wikipedia.org/wiki/Kronecker_product

It is sometimes also called 'tensor product'.

Hence, I would like to know whether it satisfies the universal property of the tensor product, which is defined here (https://en.m.wikipedia.org/wiki/Tensor_product_of_modules) I.e. every R-bilinear map from $M_{n\times m}(R) \times M_{p\times q}(R)$ to an Abelian group $G$ factors uniquely via a linear map through the tensor product $M_{n\times m}(R) \otimes M_{p\times q}(R)$.

A seemingly related question is Tensor product and Kronecker Product, but here I'm specifically interested in the universal property.

Jens Wagemaker
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1 Answers1

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I am not sure about the answer for general modules over a ring. I will instead answer only for the special case of vector spaces.

In the category $\mathrm{Vec}$ of vector spaces (over the same field) and linear maps we can define the tensor product functor $\otimes : \mathrm{Vec} \times \mathrm{Vec} \to \mathrm{Vec} $ by sending two vector spaces $(V,W)$ to the tensor product $V \otimes W$ and two morphisms $T: V \to X$, $S: W \to Y$ to the morphism $T \otimes S : V \otimes W \to X \otimes Y$ defined as the linearization (as in the universal property) of the bilinear map $V \times W \to X \otimes Y$, $(v,w) \mapsto (Tv) \otimes (Sw)$.

Now if the included vectorspaces are finite dimensional then by choosing bases it is not hard to see that $T\otimes S$ is represented by the Kronecker product in an appropriate sense. See the second part of my answer here. Hence it is called the tensor product of $T$ and $S$.

Oddly in finite dimensions the above defined tensor product of linear maps also satisfies the universal property in the following sense:

Let $X,Y,V,W$ be finite dimensional vector spaces. Denote with a prime their dual spaces.

Then it is well known that $L(V,W) = V^\prime \otimes W $ (where $L(V,W)$ denotes the vector space of linear maps $V\to W$) and in the same way $L(X,Y) = X^\prime \otimes Y$.

Furthermore $X^\prime \otimes Y^\prime = (X \otimes Y)^\prime $ for any two finite dimensional vector spaces.

And so using the usual rules for the tensor product: $$L(V,W) \otimes L(X,Y) = (V^\prime \otimes W) \otimes (X^\prime \otimes Y)= ( V \otimes X)^\prime \otimes ( W \otimes Y) = L( V \otimes X, W \otimes Y) $$ with the isomorphism defined for $T \in L(V,W) $ and $S \in L(X,Y)$ by the relation $(T \otimes S) (v \otimes x) = (Tv) \otimes (Sx)$ for all $x \in X, v \in V$.

Therefore $ L( V \otimes X, W \otimes Y)$ together with the bilinear tensor product of linear maps map (as defined in the first paragraph) also has the universal property.

I am not sure if a similar result also holds in a more general setting. I also found this post which offers a more direct proof.

jd27
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  • Where does $(V^\prime \otimes W) \otimes (X^\prime \otimes Y)= ( V \otimes X)^\prime \otimes ( W \otimes Y)$ come from? – Jens Wagemaker Sep 08 '23 at 21:33
  • Using associativity and the fact that the order can be exchanged and then finishing with $V^\prime \otimes X^\prime = (V\otimes X)^\prime$ – jd27 Sep 09 '23 at 06:53