Notes:Tensor product

Snippets

Page 47 of notebook, example is [ilmath]\mathbb{R}^m\otimes\mathbb{R}^n[/ilmath]

Definitions

There are two. First of all is an arbitrary (finite?) operation [ilmath]\otimes[/ilmath] where we define:

• [math]v_1\otimes w_1+\cdots+v_k\otimes w_k\in U\otimes W[/math] for arbitrary [ilmath]k\in\mathbb{N} [/ilmath]

Second is via this property:

• [math] \begin{xy} \xymatrix{ {U\times V} \ar@2{->}[r]^t \ar@2{->}[dr]_f & {U\otimes V} \ar@{.>}[d]^{\tilde{f}}\\ & X } \end{xy} [/math] (note that the [ilmath]\implies[/ilmath] arrows are bilinear, and the single arrows linear maps. THIS IS NON-STANDARD

This diagram says:

• There exists a bilinear map, [ilmath]t[/ilmath] such that whenever [ilmath]f[/ilmath] is a bilinear map, there exists a linear map, [ilmath]\tilde{f} [/ilmath], such that the diagram commutes

Notes

Right now (9/6/2015 @ 0827) I have two "definitions" of tensor products. One as detailed on Notes:ToMond and another as the book(s) I have read.

Bilinear function

A function, [ilmath]f:U\times V\rightarrow W[/ilmath] is bilinear if:

• It is linear in both variables, that is:
  • [math]f(\alpha u_1+\beta u_2,v)=\alpha f(u_1,v) + \beta f(u_2,v)[/math] and
  • [math]f(u,\alpha v_1+\beta v_2)=\alpha f(u,v_1) + \beta f(u,v_2)[/math]

As can be seen on Bilinear map (which is a page in need of cleanup!)

Scalar multiplication

Note that:

• [math]f(\lambda u,v)=\lambda f(u,v)[/math] and
• [math]f(u,\lambda v)=\lambda f(u,v)[/math]

So we can conclude that:

• [math]\lambda f(u,v)=f(\lambda u,v)=f(u,\lambda v)[/math]

Tensor product

The tensor product of the vector spaces is [ilmath]U\otimes V[/ilmath] and the elements are [ilmath]u\otimes v[/ilmath] for a bilinear function: [ilmath]\otimes:U\times V\rightarrow W[/ilmath]

Questions

What is the 0 tensor

I have been told that the [ilmath]0[/ilmath] of [ilmath]U\otimes V[/ilmath] is [ilmath]0_U\otimes 0_V[/ilmath] however I am not convinced of this yet. What I do know that the [ilmath]0[/ilmath] vector is given by the [ilmath]0[/ilmath] scalar multiplied by any vector, so I know:

• [math]0(u\otimes v)=[/math]
  • [math](0u)\otimes v=0_u\otimes v[/math]
  • [math]u\otimes(0v) = u\otimes 0_v[/math]

I am convinced that [ilmath]0_u\otimes v=u\otimes 0_v[/ilmath] but I am not yet convinced that we must therefor have [ilmath]=0_v\otimes 0_u[/ilmath]

Solved

Page 46 of notebook, gist is to show that (u,v) is in the kernel when u=0 or v=0 (done on bilinear page) then suppose that [math]u\ne 0\wedge v\ne 0[/math] and construct bilinear map with $f(u,v)\ne 0$ then by the characteristic property of the tensor product the tensor product is non zero.