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- Thread starter sqljunkey
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fresh_42

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https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

##V^*\otimes V## are all linear functions ##V \longrightarrow V##. They form again a vector space and the functions ##V^*\otimes V \longrightarrow \mathbb{R}## are its dual vector space. In case of manifolds, ##V## is usually the tangent space of the manifold.

To have a look at what tensors are, see

https://www.physicsforums.com/insights/what-is-a-tensor/If you check the example at the end, the matrix multiplication, you will see, that there doesn't have to be a manifold. And in case the vector space is indeed a tangent space of a manifold, then the question is: what do you want to express, or calculate, and which perspective are you taken?

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A tensor here in your case is a linear map that contains precisely the information needed, no more, no less, to define a multilinear map so that the multilinear map is replaced by a linear one. I assume instead of x here you mean ##V\otimes V*##. Then a bilinear map in those two arguments is replaced by a linear one. The meaning of the expression will depend on the context . Often in cases of ##V \otimes V*## you do a pairing ( an index contraction) and evaluate to v*(v) and this gives you a number.

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I was referring to the Riemannian manifold. But I did enjoy the answers.

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This is trivially true - ##V^*\otimes V:V \times V^* \to \mathbb{R}## is the general form, but yes, if ##V## is a vector space, then for sure so is ##V \times V^*##. Likewise, since ##\mathbb{R}## is a field it is equally a vector space.##V^*\otimes V## are all linear functions ##V \longrightarrow V##.

How can ##V^*\otimes V## be its own dual?They form again a vector space and the functions ##V^*\otimes V \longrightarrow \mathbb{R}## are its dual vector space.

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##(V^*\otimes V)^*=\operatorname{Hom}_\mathbb{R}(V^*\otimes V,\mathbb{R})## is what I said.This is trivially true - ##V^*\otimes V:V \times V^* \to \mathbb{R}## is the general form, but yes, if ##V## is a vector space, then for sure so is ##V \times V^*##. Likewise, since ##\mathbb{R}## is a field it is equally a vector space.

How can ##V^*\otimes V## be its own dual?

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And yet.... ##(V^* \otimes V)^* = V \otimes V^*## is only true in a metric space on finite-dimensional vector spaces, so that the hom-set (your RHS term ) of all Real maps from the dual space to its "parent" is equal to the tensor product of a double dual space and its "parent" space?

This is not generally true

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Metric space? Where are we using a metric here?

And yet.... ##(V^* \otimes V)^* = V \otimes V^*## is only true in a metric space on finite-dimensional vector spaces, so that the hom-set (your RHS term ) of all Real maps from the dual space to its "parent" is equal to the tensor product of a double dual space and its "parent" space?

This is not generally true

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fresh_42

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It is generally true. The Hom functor neither requires a metric nor a finite generation nor dimension.

And yet.... ##(V^* \otimes V)^* = V \otimes V^*## is only true in a metric space on finite-dimensional vector spaces, so that the hom-set (your RHS term ) of all Real maps from the dual space to its "parent" is equal to the tensor product of a double dual space and its "parent" space?

This is not generally true

Are we still in the math forums?

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WWGD

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But then are we considering the continuous or algebraic dual? For algebraic answer is always no but I believe it us yes in some cases for the continuous dual ( also depending on our choice of topology).

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Anyway, this is drifting rather far away from anything to do with smooth manifolds.

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What do you mean by example? How about Strassen's algorithm for matrix multiplication?

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I mean a map into the Reals defined on a pair ##(v,v^*)## or ## v \otimes v^* ##on a manifold . And if possible an interpretation of the output, e.g., the curvature tensor ( I am rusty on this, just to illustrate).What do you mean by example? How about Strassen's algorithm for matrix multiplication?

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Op is asking for examples in Riemannian manifolds( see post 3) And, of course, I am referring to maps defined on tangent/cotangent spaces. An example of why we are interested in pairing a vector with its dual to produce a number.

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and says, quote:Op is asking for examples in Riemannian manifolds

which is underdetermined, to say the least. If ##M## is this smooth manifold, then we make silently the following assumptions: ##R = \mathbb{R}## and ##V=T_pM##. But nobody said this. The entire process from ##M## to ##V## seems to be unclear to the OP. Hence the entire discussion above is pretty much off topic. That's why I start earlier: There is no ##V## for ##M## unless specified. One is linear and flat, including all parts of its tensor algebras and dual spaces, the other one is curved. There is no connection between flat and curved, except we clarify what is meant. And I want the OP to understand that the manifold is the ball and the vector space a plate lying on the ball: two objects, not one.When I'm going from a smooth manifold to R with V* X V -> R what does the R scalar stand for.

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Yes, me, too. And that's why I say: ##V## and ##M## do have nothing in common. The next step would be to answer: Why is a tangent##\,^*)## a linear map? Then why is a linear map an element of ##V^*\otimes V## and sometime later curvature.Do you really think that at an intro level someone is considering general rings R and not the Reals? My take is s/he read something about manifolds and tensors and wants an idea for motivating the use or existence of tensors.

_______________

##^*)## rhetorical exaggeration

- #23

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I failed to specify initially that the manifold had a tangent space, and was equipped with a metric tensor. I kinda assumed that V* X V -> R was all metric tensors.

And yeah I'm hoping to create a system of meshes like the CFD people do to simulate Navier Stokes solutions.

I did enjoy all the answers though.

Happy Turkey Day!

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