# What is this Tensor mapping to?

• B
When I'm going from a smooth manifold to R with V* X V -> R what does the R scalar stand for. Is it some length in the manifold? and Does this have to do with the way V* and V are defined, since one is a contra-variant and one is a co-variant, are they related in the way the Pythagoras formula is related ? c^2 = a^2 + b^2

fresh_42
Mentor
You shouldn't confuse the manifold with its tangent space. The main difficulty is, that differentiation can be viewed in so many ways, that any answer depends on the perspective taken. Just to illustrate what I mean, have a look at the 10 point list about a derivative at the beginning of
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?

WWGD
Gold Member
When I'm going from a smooth manifold to R with V* X V -> R what does the R scalar stand for. Is it some length in the manifold? and Does this have to do with the way V* and V are defined, since one is a contra-variant and one is a co-variant, are they related in the way the Pythagoras formula is related ? c^2 = a^2 + b^2
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.

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

WWGD
Gold Member
I would recommend a set of YouTube videos by Eigenchris. They ( videos) lay things out pretty clearly.

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

fresh_42
Mentor
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?
##(V^*\otimes V)^*=\operatorname{Hom}_\mathbb{R}(V^*\otimes V,\mathbb{R})## is what I said.

Well, I didn't see that statement of yours.

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

WWGD
Gold Member
Well, I didn't see that statement of yours.

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
Metric space? Where are we using a metric here?

fresh_42
fresh_42
Mentor
Well, I didn't see that statement of yours.

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
It is generally true. The Hom functor neither requires a metric nor a finite generation nor dimension.
Are we still in the math forums?

WWGD
Infrared
Gold Member
You do need ##V## to be finite dimensional for ##(V^*\otimes V)^*=V\otimes V^*## to hold (otherwise ##V^{**}## is bigger than ##V##), but if ##V## is the tangent space of a Riemannian manifold...

WWGD
Gold Member
You do need ##V## to be finite dimensional for ##(V^*\otimes V)^*=V\otimes V^*## to hold (otherwise ##V^{**}## is bigger than ##V##), but if ##V## is the tangent space of a Riemannian manifold...
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).

Infrared
Gold Member
Algebraic. In the continuous case for inner product spaces, it is true for Hilbert spaces (Riesz representation theorem). You definitely need completeness though (otherwise if ##v_n## is a non-convergent, Cauchy sequence, then ##\lim_{n\to\infty}\langle v_n,\cdot\rangle## is a dual vector not in the image of ##V\to V^*##)

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

WWGD
Gold Member
Ultimately, it is a legitimate general question. We have a bilinear map with inputs in ##V, V^*## (or a linear map on some element ##v \otimes v^*##)and output in the Reals. There are many options, many choices for these elements. The output value in ##\mathbb R## will depend on both the choice of map as well as the specific inputs. It would be nice if someone had a specific example, at the moment, I don't, will look for one.

fresh_42
Mentor
What do you mean by example? How about Strassen's algorithm for matrix multiplication?

WWGD
Gold Member
What do you mean by example? How about Strassen's algorithm for matrix multiplication?
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).

Infrared
Gold Member
Are you asking for examples of endomorphisms of the tangent bundle of a manifold? If so, a (almost) complex structure or curvature are typical examples.

fresh_42
fresh_42
Mentor
I have a problem with manifold and linear in one sentence. Curvature is of course a tensor like tangents are linear maps, but they are derived concepts from the manifold. And something like ##v\otimes v^*## has at prior nothing to do with any manifold, except those which are vector spaces. I find Strassen's algorithm a good example for bilinear maps outside physics. Reading the physical questions on PF one could get the impression that tensors et al. require a manifold, and this is simply not the case.

WWGD
Gold Member
I have a problem with manifold and linear in one sentence. Curvature is of course a tensor like tangents are linear maps, but they are derived concepts from the manifold. And something like ##v\otimes v^*## has at prior nothing to do with any manifold, except those which are vector spaces. I find Strassen's algorithm a good example for bilinear maps outside physics. Reading the physical questions on PF one could get the impression that tensors et al. require a manifold, and this is simply not the case.
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.

fresh_42
Mentor
Op is asking for examples in Riemannian manifolds
and says, quote:
When I'm going from a smooth manifold to R with V* X V -> R what does the R scalar stand for.
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.

WWGD
Gold Member
Do you really think that at an intro level someone is considering general rings R and not the Reals? That they're thinking of generalized vector spaces and not the tangent/cotangent spaces? My take is s/he read something about manifolds and tensors and wants an idea for motivating the use or existence of tensors. I dont see the need to illustrate things in a perfectly bottom-up format, it seems giving an example would be helpful itself.

fresh_42
Mentor
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.
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.

_______________
##^*)## rhetorical exaggeration

Hi, yeah I looked at video 9 of eigenchris and I think it kind of answered my question.

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!

WWGD