# Induced inner product

1. Jul 7, 2005

### Palindrom

Hi,
Say I have two inner product spaces, V and W.
What is the definition of their tensor product?
Is this product naturally always an inner product space?
Thank!

2. Jul 7, 2005

### mathwonk

the tensor product of V and W is bydefinition any vector space V(tens)W equipped with a bilinear map VxW-->V(tens)W such that every bilinear map VxW-->U into any vector spave U, occurs uniquely as a composition VxW-->V(tens)W-->U, for some unique lienar map V(tens)W-->U.

if the spaces are finite dimensional it can be constructed as the dual of the space of bilinear functions from VxW to the scalar field k. if v1,...vn, and w1,...wm are bases of V and W, then the bilinear functions taking (vi,wj) to 1 and the other pairs to zero, form a basis of the dual of the tensor product. hence the tensor product has dimension equal to the product of the dimensions of V and W.

now ask yourself how to use the inner products on V and W to define one on V(tens)W.

i.e. how would you define <vtensw, v'tensw'>. is your definition bilinear? positive definite?

Last edited: Jul 7, 2005
3. Jul 7, 2005

### Palindrom

If V(tens)W is the dual of the space of bilinear functions from VxW to K, then it is a space of functionals that operate on bilinear functions, isn't it? So how come the bilinear functions you mentioned belong to it? They operate on VxW directly, don't they?

And I'm still thinking about that inner product.

4. Jul 7, 2005

### mathwonk

they do not belong to it, i said they belong to its dual.

i.e. if v,w Re elements of V,W then v(tens)w is an element dual to the space of bilinear functions on VxW.

e.g. if F( , ) is a bilinear function on VxW, then v(tens)w acts on F by sending F to

F(v,w), a number.

5. Jul 7, 2005

### Palindrom

Oohps, sorry, I hadn't noticed you wrote "dual of the tensor product".

Is the map you mentioned (that is, (vi,wj) |----> f(vk,wl)=delta(k,i)*delta(j,l)) defined linearly or bilinearly?
That is, if f is assigned to (vi,wj) and g to (vk,wl), then the functional assigned to (vi+vk,wj+wl) is f+g, or f+g+h+t, when h and t are assigned to the two remaining pairs?

I'm sorry for the annoying questions, I'm just trying to understand the nature of the connection between VxW and V(tens)W.

6. Jul 7, 2005

### mathwonk

f is not assigned to anything. F is a bilinear map. it takes a pair of vectors to a number. it defiens a bilinear map F:VxW-->k.

as such it induces a linear map G:V(tens)W-->k such that G(v(tens)w) = F(v,w).

VxW is a vector space of dimension dimV + dimW. V(tens)W has dimension dim(V).dim(W).

If v1,...vn, w1,...,wm are bases of V and W, then (v1,0),...,(vn,0),(0,w1),...(0,wm) is a basis of VxW.

while (v1(tens)w1),.....,(vn(tens)wm)), is a basis of V(tens)W.

there is a bilinear map from VxW to V(tens)W, that takes (v,w) to v(tens)w.

but not everything in V(tens)W has that form. for instance v1(tens)w1 + v2(tens)w2 is also an element of V(tens)W. I.e. typical elements of V(tens)W are sums of things like v(tens)w.

7. Jul 7, 2005

### Palindrom

O.K.
So in order to define the inner product on V(tens)W, let's take two guys from there, a and b. They can each be broken into sums of v(tens)w. We'll define the inner product of v(tens)w with u(tens)x as <v,u><w,x>. This induces an inner product on all of V(tens)W, since the inner product must be bilinear.
I'm just trying to see if it is, indeed, an inner product. Am I on the right track?

8. Jul 7, 2005

### mathwonk

well that was my idea. does it have the right properties?

9. Jul 7, 2005

### Palindrom

It's the definite positivity that I'm worried about. But I'll check tommorow, right now I'm going to bed. I'm beat.
Thanks a lot, you don't know how much you've helped me. It's been bothering me for weeks.

10. Jul 7, 2005

### mathwonk

try choosing bases.

i.e. maybe if e1,...,en, and f1,...,fm are orthonormal bases for V,W, then {ei(tens)fj} is an orthonormal basis for V(tens)W.

that should do it. ???

Last edited: Jul 7, 2005
11. Jul 14, 2005

### Palindrom

Sorry, I've been under pressure lately. I think it's ok.

I've run into this problem in a class in Quantum Physics, of all things.