Tensor Product.

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  • #1
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Hello all.

Most modern treatments of the tensor product use equivalence classes to define a quotient space in order to define the tensor product. However in Tensor Analysis on Manifolds, Bishop and Goldberg are much less complicated. I have attached a near word for word copy of their definition.

Why is their's so simple. Why do the other treatments approach it using a more complicated method. Is it just more rigorous.

Matheinste.
 

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  • #2
matt grime
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That doesn't define the tensor product of two spaces. It defines a symbol [itex]f\otimes g[/itex] of two things in the dual spaces of two vector spaces. It doesn't say anything about what the symbol is or does.
 
  • #3
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Hello matt grime.

Thanks for pointing that out.

Matheinste.
 
  • #4
Hurkyl
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Any (good) treatment of tensors of vector spaces will tell you that the tensor product of two linear functionals is naturally equivalent to the bilinear functional given in that PDF.

And the book you're studying had better also contain the "more complicated method" -- it is fairly important to know that [itex]V^* \otimes W^*[/itex] is spanned by the set of all such [itex]\tau \otimes \theta[/itex], and to know the algebraic relations satisfied by the tensor product.

But one big picture thing you're missing: the tensor product is defined for any pair of vector spaces, not just for vector spaces of linear functionals.
 
  • #5
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In that book, they define [tex]T^r_s = V \otimes ... \otimes V \otimes V^* \otimes ... \otimes V^*[/tex] to be the set of all multilinear functions on [tex]V^* \times ... \times V^* \times V \times ... \times V[/tex] into the reals. Then they verify that elements of this space satisfy the usual tensor relations (associativity and distributivity). And say, as you point out, that it is spaned by elements on the form you say.

I guess this is one way to define the tensor product (maybe how many people doing analysis would do it???), but as you talk about there is a completely algebraic way of defining the tensor product. Don't know which is better, as far as I know you can show that the two spaces are isomorphic.
 
  • #6
matt grime
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That definition would only appear to make sense for finite dimensional vector spaces (i.e. ones where taking double duals yields the thing you first thought of. There is no harm in defining it that way, and I've no idea why you think it's analytic.

The more conventional way to define it is via some universal property, but this doesn't actually show such a thing exists! One then needs to go the extra mile to construct it.
 
  • #7
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only thought it was analytical because iv'e seen it in lectures on analysis, and the other way seems to be more algebraic. You are right this is for finite dimensions, because they only want to use it on the tangent spaces.

Could you elabrorate on why it only makes sense in the finite case? Never thought about that. Is it because the isomorphism only exist in that case? I think I've seen it constructed where you need a basis for the vectorspace V.
 
  • #8
matt grime
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I said why it appears to make sense only for f.d. spaces: it seems that it wants to treat V as the dual space to V*. That is not true for infinite dimensional things.
 
  • #9
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Hello all.

Further to my last post I am slowly getting a grip on tensor products using the modern approach but wish to check my reasoning a step at a time. Would someone please tell me if the following is correct.

In all cases the vector spaces will be considered as being over the reals.

As a first step we generate the “free” vector space F(U x V) over the Cartesian product U x V . We do this by using some rule of vector addition between all elements (u,v) of U x V and some rule of scalar multiplication of all elements of U x V, so that we have a vector space. The set of all members of the set U x V can be considered as a basis for this vector space.

Matheinste.
 
  • #10
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just on the side, what texts cover the infinite dimensional vector spaces? We only ever covered finite dimensional cases in my classes. It would be interesting to see how things become different....
 
  • #11
Try any book on Functional Analysis.
 
  • #12
Hello all.

Most modern treatments of the tensor product use equivalence classes to define a quotient space in order to define the tensor product. However in Tensor Analysis on Manifolds, Bishop and Goldberg are much less complicated. I have attached a near word for word copy of their definition.

Why is their's so simple. Why do the other treatments approach it using a more complicated method. Is it just more rigorous.

Matheinste.


hola hermano me gustaria que me des mas informaciones de analisis tensorial por favor estare muy agradecido
 

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