Tensor products are rather useful gadgets that are ubiquitous
in applications, for example in quantum mechanics
Let V,W be vector spaces over a field k.
A bilinear function f:VxW-->L to a vector
space L is called a tensor product of V and W
if it satisfies the following universal property:
Whenever g:VxW-->M is a bilinear function to a vector space
then there exists a unique linear transformation F:L-->M such that
At this point we don't know that there can be any such tensor
products. But we will show that if there is a tensor product
then it is unique. So suppose that f:VxW-->L and
g:VxW-->M both satisfy the universal property.
By the universal property for f (with the other bilinear map
being g) there exists
a linear transformation F:L-->M such that
Ff=g. Likewise using the universal property for
g (with f the other bilinear map) there exists
a linear transformation G:M-->L such that
Gg=f. From these two equations it follows that FGg=g.
But now think about the universal property for g when the other
bilinear map is just g. Clearly a solution to the universal property
is the the identity function 1M:M-->M. But
as FGg=g another solution is FG. By the uniqueness in
the universal property we deduce that FG=1M. By
a symmetric argument we see that GF=1M. So we see that
F is an isomorphism of vector spaces. By the uniqueness of the
universal property it is the unique isomorphism with Ff=g.
Putting it all together.
If a tensor product exists then it is unique up to unique isomorphism.
Notation Thus we can speak of the
tensor product of V and W. It is usual to abuse notation
(as I did in the theorem) and refer to the vector space as the tensor product
with the bilinear map being implicit. The notation is
VxW (actually usually the x is denoted by
a circle with a x in it).
That just leaves one tiny detail. Proving that there is a tensor product.
I'll do the finite dimensional case to keep the notation simpler but the same idea works in general.
Fix ei : 1<=i<=n and
fj : 1<=j<=m bases
for V and W. Now let L be some mn-dimensional
vector space with basis gi,j for 1<=i<=n
and 1<=j<=m. Define a map f:VxW-->L by
f(a1e1+...+ anen, b1f1+...+ bmem)
= Sum(1<=i<=n,(1<=j<=m) aibjgi,j.
It is routine to check that f
satisfies the universal property
and so is the tensor product of V
Note that the above shows:
Proposition If V,W are finite dimensional
vector spaces then dim VxW = (dim V).(dim W)
More generally, if M and N are modules for a commutative
ring R then
a tensor product of M and N is
a bilinear map f:MxN-->L to an R-module
that is universal. That is, whenever g:MxN-->K is a bilinear
map to a module there is a unique R-module homomorphism F:L-->K
such that Ff=g. Exactly as above one shows that such a
tensor product is unique up to unique isomorphism. The construction is
a little more involved though.