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Abstract definition

I'm giving here the definition of Wikipedia trough universal property.

A tensor product of $V$ and $W$ is a pair $(T,h)$ where $T$ is a vector space and $h:V\times W\to T$ is a bilinear map such that for every bilinear map $f: V\times W\to Z$, there exists a unique linear map $\tilde f:T\to Z$ such that $f=\tilde f\circ h$.

One can show that every tensor product of $V$ and $W$ are isomorphic.

Usual realization

Most of people define the tensor product $V^*\otimes V^*$ as the set of bilinear maps over $V$. Example from here:

A function $T:V^2\to \mathbb{R}$ is a tensor if it is bilinear.

My question

If I understand correctly, the space of bilinear forms over $V$ is a tensor product.

QUESTION: what is the function $h$ such that $(L_2(V,\mathbb{R}),h)$ is a tensor product of $V^*$ and $V^*$ ?

I made a guess here: Universal property of multilinear maps (how to make them a tensor product) but I'm not sure it works.

Laurent Claessens
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1 Answers1

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Let $\{\alpha_i\}$ be a basis of $V^*$ and $\{\beta_j\}$ be a basis of $W^*$.

We define $h:V^*\times W^*\to L_2(V\times W,\mathbb{R})$ by $$ h(\xi,\sigma)(v,w)=\xi(v)\sigma(w). $$ It has the property that the elements $h(\alpha_i,\beta_j)$ form a basis of $L_2(V\times W,\mathbb{R})$.

Now we prove that the couple $\big( L_2(V\times W,\mathbb{R},h) \big)$ is a tensor product for $V^*$ and $W^*$.

If $f:V^*\times W^*\to Z$ is a bilinear map, it has its "matrix" given by the element $f_{ij}=f(\alpha_i,\beta_j)\in Z$.

The trick is to consider the map $\tilde f:V^*\times W^*\to Z$ defined by $$ \tilde f\Big( \sum_{ij}a_{ij}h(\alpha_i,\beta_j) \Big)=\sum_{ij}a_{ij}f_{ij}. $$

It is not so complicated to check that $\tilde f\circ h=f$.

Laurent Claessens
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