extending bilinear form with tensor product Let $V$ and $W$ be vector spaces over a field $K$ (or possibly modules over a commutative ring), and suppose I’m given a bilinear form, $B:V \times V \rightarrow K$.  Is there a natural way to extend this to a bilinear map on the tensor product:
$B_W: (V \otimes_K W) \times (V \otimes_K W) \rightarrow W$?
One way to try to do this is to pick a basis, $e_i$, of $W$, and define:
$$ B_W( a^i \otimes e_i, b^i \otimes e_i ) \equiv B(a^i,b^i) e_i $$
and extend by linearity.  However, this seems like it will behave badly under change of basis, so I’m guessing the answer to my question is no.
On the other hand, suppose I’m only interested in finding pairs of elements that map to zero.  One can check that this will be true in the above definition in any basis provided we impose:
$$ B(a^i, b^j) = 0 , \forall i,j $$
Is there a more natural way of phrasing this condition, ie, without having to pick a basis?
 A: You're probably aware that a bilinear form $f : V \times V \rightarrow K$ can also be viewed as a map $f : V \otimes_K V \rightarrow K$.
Write $U = V \otimes_K W$. Supplying a map $B_W : U \otimes_K U \rightarrow W$ is quite similar to supplying a map $W \otimes_K W \rightarrow W$ (i.e. a non-associative non-unitial $K$-algebra structure on $W$). If there is a map $\mu: W \otimes_K W \rightarrow W$, then we get a map 
$$\phi : U \otimes_K U \rightarrow (V \otimes_K V) \otimes (W \otimes_K W) \stackrel{B \otimes_K \mu}{\rightarrow} K \otimes_K W \cong W$$
The way you are trying seems to be to define $\phi : (V \otimes_K W) \otimes_K (V \otimes_K W) \rightarrow W$ by setting $(e_i \otimes f_j \otimes e_{i'} \otimes f_{j} ) = B_V(e_i \otimes e_{i'}) f_j$. But note that $f_j$ were the same here and not every tensor in $W \otimes_K W$ can be written as a sum of simple tensors of the form $f_j \otimes f_j$. So we could not extend that linearly to the whole space. If we have $\mu : W \otimes_K W \rightarrow W$. Then we can set  $(e_i \otimes f_j \otimes e_{i'} \otimes f_{j'} ) = B_V(e_i \otimes e_{i'}) \mu(f_j \otimes f_{j'})$. That's equivalent to the above.
This construction is called extension of scalars. We turned a bilinear form on the $K$-vector space $V$ into a bilinear form on the $W$-vector space $V \otimes_K W$.

Continuing, one might wonder if there is a way to put a commutative (unitial, associative, distributive) algebra structure on $W$. There is no such canonical way, unless we count the map $0 : W \otimes_K W \rightarrow W$. However, after choosing a basis $\{ w_i \}_{i \in I}$, we may define $w_i \cdot w_j = \delta_{ij} w_i = \delta_{ij} w_j$. Moreover, we can recover the basis $\{ w_i \}_{i = 1}^n$ of $W$ from the idempotents of this algebra. The upshot is that we must have chosen a basis to define this algebra structure.

Here's something similar:

Definition: Let $K$ be a field and let $V$ and $W$ be finite dimensional $K$-vector spaces. Let $\beta$ and $\gamma$ be bilinear forms on $V$ and $W$ respectively, and let $f : V \otimes_K V \rightarrow K$ and $g : W \otimes_K W \rightarrow K$ be the corresponding $K$-linear maps. Then we have a map $f \otimes g : (V \otimes_K V) \otimes_K (W \otimes_K W) \rightarrow K \otimes_K K$. Write $U = V \otimes_K W$. We set $\delta : U \otimes_K U \rightarrow K$ to be the composition
  $$U \otimes_K U \stackrel{\cong}{\rightarrow} (V \otimes_K V) \otimes_K (W \otimes_K W) \stackrel{f \otimes g}{\rightarrow} K \otimes_K K \stackrel{\cong}{\rightarrow} K $$

Let's calculate the matrix $M$ corresponding to this bilinear form $\delta$. Seems like it should be the Kronecker product, but let's see. Take $e_i \otimes f_j$ and $e_{i'} \otimes f_{j'}$ in $U = V \otimes_K W$, where $\{ e_i \}_{i = 1}^n$ is some designated basis of $V$ and $\{ f_j \}_{j = 1}^m$ is some designated basis of $M$. For a matrix to make sense, we have to order the basis $e_i \otimes f_j$, but it doesn't really matter how we do that, so I'll just refer to the entries of $M$ as $M_{(i, j), (i', j')}$. Anyways, 
$$M_{(i, j), (i', j')} = \delta(e_i \otimes e_j \otimes e_{i'} \otimes e_{j'}) = \beta(e_i \otimes e_{i'}) \gamma (e_j \otimes e_{j'})$$
If we write $B_{ii'} = \beta(e_i \otimes e_{i'})$ for the matrix corresponding to $\beta$ and $C_{jj'} = \gamma(f_j \otimes f_{j'})$ for the matrix corresponding to $\gamma$, then $M_{(i, j), (i', j')} = B_{ii'} C_{j j'}$. This is the Kronecker product.
