Tensor products are rather useful gadgets that are ubiquitous in applications, for example in quantum mechanics.

Definition 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 Ff=g.

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.

Theorem 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 and W. 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.

The above is certainly formidable, but would not scare most applied mathematicians. They are most happy in defining the product space in terms of bases:
If U has basis {u1,u2,...,um} and V has basis {v1,v2,...,vn} then simply define U×V to be a space with basis {u1×v1,u1×v2,...,u1×vn, u2×v1,u2×v2,...,u2×vn,... um×v1,um×v2,...,um×vn}
while the basis independent definition is more general and makes no assumption of such a space existing, this definition is sometimes easier to stomach.

An example of this use of this in Quantum Mechanics is used for dealing with states of particles with spin: if |m> is a spin state and |x> is the usual position ket, then define product kets as |x,m> = |x> × |m>. This can then be used to define the complete wave function of the particle, as detailed in Angular Momentum in Quantum Mechanics.
Also, if some operator acts on U, its action naturally be specified on U×V by U(u × v) = (U(u)) × v ie. it acts as the identity on V.

Log in or register to write something here or to contact authors.