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What Is a Tensor?
Continue reading the Original PF Insights Post.
What Is a Tensor?
Continue reading the Original PF Insights Post.
If I interpret correctly this sentence says the underlying set of U⊗V is U×V. This is not correct. This works for the direct sum U⊕V but the tensor product is a "bigger" set than U×V. One usual way to encode it is to quotient ##\mathbb{R}^{(U \times V)}## (the set of finitely-supported functions from U×V to ##\mathbb{R}##) by the appropriate equivalence relation.Definition: A tensor product of vector spaces U⊗V is a vector space structure on the Cartesian product U×V that satisfies ...
The tensor product of two 1 dimensional vector spaces is 1 dimensional so it is smaller not bigger than the direct sum. The tensor product of two 2 dimensional vector spaces is 4 dimensional so this is the the same size as the direct sum not bigger.If I interpret correctly this sentence says the underlying set of U⊗V is U×V. This is not correct. This works for the direct sum U⊕V but the tensor product is a "bigger" set than U×V. One usual way to encode it is to quotient ##\mathbb{R}^{(U \times V)}## (the set of finitely-supported functions from U×V to ##\mathbb{R}##) by the appropriate equivalence relation.
This is correct but missing the relevant point: that the presentation contains a false statement. The fact that you can indeed find counter examples where the direct sum is bigger than the tensor product does not makes the insight presentation any more correct.The tensor product of two 1 dimensional vector spaces is 1 dimensional so it is smaller not bigger than the direct sum. The tensor product tof two 2 dimensional vector spaces is 4 dimensional so this is the the same size as the direct sum not bigger.
No, this interpretation was of course not intended, rather a quotient of the free linear span of the set ##U \times V##.If I interpret correctly this sentence says the underlying set of U⊗V is U×V. This is not correct. This works for the direct sum U⊕V but the tensor product is a "bigger" set than U×V. One usual way to encode it is to quotient ##\mathbb{R}^{(U \times V)}## (the set of finitely-supported functions from U×V to ##\mathbb{R}##) by the appropriate equivalence relation.
Thank youNo, this interpretation was of course not intended, rather a quotient of the free linear span of the set ##U \times V##.
I added an explanation to close this trapdoor. Thank you.
I know, or at least assumed all this. And I was tempted to explain a lot of these aspects. However, as I recognized that this would lead to at least three or four parts, I concentrated on my initial purpose again, which was to explain what kind of object tensors are, rather than to cover all aspects of their applications. It was meant to answer this basic question which occasionally comes up on PF and I got bored retyping the same stuff over and over again. That's why I've chosen Strassen's algorithm as an example, because it uses linear functionals as well as vectors to form a tensor product on a very basic level, which could easily be followed.- Students of General Relativity learn about tensors ...
- Students of Quantum Mechanics learn about tensors ...
- If one wants to discuss tensor products purely mathematically ...
You can consider every matrix as a tensor (defining the matrix rank by tensors) or write a tensor in columns, as it is a vector (element of a vector space) in the end. Personally I like to view a tensor product as the solution of a couniversal mapping problem. As I said, I was tempted to write more about the aspect of "How to use a tensor" instead of "What is a tensor" but this would have led to several chapters and the problem "Where to draw the line" were still an open one. Therefore I simply wanted to take away the fears of the term and answer what it is, as I did before in a few threads, where the basic question was about multilinearity and linear algebra and the constituencies of tensors. The intro with the numbers should show that the degree of complexity depends on the complexity of purpose. I simply wanted to shortcut future answers to threads rather than write a book about tensor calculus. That was the main reason for the examples, which can be understood on a very basic level. Otherwise I would have written about the Ricci tensor and tensor fields which I find far more exciting. And I would have started with rings and modules and not with vector spaces. Thus I only mentioned them, because I wanted to keep it short and to keep it easy: an answer for a thread. Nobody on an "A" and probably as well on an "I" level reads a text about what a tensor is.Sure, you can represent a tensor by a multidimensional array, but this does not mean that a tensor is a multidimensional array or that a multidimensional array is a tensor. Let us take the case of tensors in ##V\otimes V## for definiteness. A basis change in ##V## can be described by a matrix that will tell you how the tensor components transform, but in itself this matrix is not a tensor.
This must mean I am B-level.Nobody on an "A" and probably as well on an "I" level reads a text about what a tensor is.
Yes, you are right. My goal was really to say "Hey look, a tensor is nothing to be afraid of." and that's why I wrotePerhaps I was mislead regarding the intended audience from the beginning. I am pretty sure most engineering students will not remember what a homomorphism is without looking it up. Certainly a person at B-level cannot be expected to know this?
In the end, I suspect we would give different answers to the question in the title based on our backgrounds and the expected audience. My students would (generally) not prefer me to give them the mathematical explanation, but instead the physical application and interpretation, more to the effect of how I think you would interpret "how can you use tensors in physics?" or "how do I interpret the meaning of a tensor?"
And to be honest, I'm bad at basis changes, i.e. frame changes and this whole rising and lowering indices is mathematically completely boring stuff. I first wanted to touch all these questions but I saw, that would need a lot of more space. So I decided to write a simple answer and leave the "several parts" article about tensors for the future. Do you want to know where I gave it up? I tried to get my head around the covariant and contravariant parts. Of course I know what this means in general, but what does it mean here? How is it related? Is there a natural way how the ##V's## come up contravariant and the ## V^{*'}s## covariant? Without coordinate transformations? In a categorial sense, it is again a different situation. And as I've found a source where it was just the other way around, I labeled it "deliberate". Which makes sense, as you can always switch between a vector space and its dual - mathematically. I guess it depends on whether one considers ##\operatorname{Hom}(V,V^*)## or ##\operatorname{Hom}(V^*,V)##. But if you know a good answer, I really like to hear it.Depending on whom you ask, how many room and time there is for an answer, where the emphases lie or what you want to use them for, the answers may vary significantly.
Well, your motivation can't have been to learn what a tensor is. That's for sure. Maybe you have been curious about another point of view. As I started, I found there are so many of them, that it would be carrying me away more and more (and thus couldn't be used as a short answer anymore). It is as if you start an article "What is a matrix?" by the sentence: "The Killing form is used to classify all simple Lie Groups, which are classical matrix groups. There is nothing special about it, all we need is the natural representation and traces ... etc." Could be done this way, why not.This must mean I am B-level.
Corrected. Thanks.I am pretty sure most engineering students will not remember what a homomorphism is without looking it up.
I think this is done before the moding out and arranging into equivalence classes is done.If I interpret correctly this sentence says the underlying set of U⊗V is U×V. This is not correct. This works for the direct sum U⊕V but the tensor product is a "bigger" set than U×V. One usual way to encode it is to quotient ##\mathbb{R}^{(U \times V)}## (the set of finitely-supported functions from U×V to ##\mathbb{R}##) by the appropriate equivalence relation.
It's the freely generated vector space (module) on the set ##U \times V##. The factorization indeed guarantees the multilinearity and the finiteness of sums which could as well be formulated as conditions to hold.I think this is done before the moding out and arranging into equivalence classes is done.
I was replying to someone else's post.It's the freely generated vector space (module) on the set ##U \times V##. The factorization indeed guarantees the multilinearity and the finiteness of sums which could as well be formulated as conditions to hold.
Sorry, was a bit in "defensive mode".I was replying to someone else's post.
No problem.Sorry, was a bit in "defensive mode".
Good point; same is the case with Tensor Contraction, i.e., it assumes/makes use of , an isomorphism.The way that tensors are manipulated implicitly assumes isomorphisms between certain spaces.
If [itex]A[/itex] is a vector space, then [itex]A^*[/itex] is the set of linear functions of type [itex]A \rightarrow S[/itex] (where [itex]S[/itex] means "scalar", which can mean real numbers or complex numbers or maybe something else depending on the setting).
The first isomorphism is [itex]A^{**}[/itex] is isomorphic to [itex]A[/itex].
The second isomorphism is [itex]A^* \otimes B^*[/itex] is isomorphic to those function of type [itex](A \times B) \rightarrow S[/itex] that are linear in both arguments.
So this means that a tensor of type [itex]T^p_q[/itex] can be thought of as a linear function that takes [itex]q[/itex] vectors and [itex]p[/itex] covectors and returns a scalar, or as a function that takes [itex]q[/itex] vectors and returns an element of [itex]V \otimes V \otimes ... \otimes V[/itex] ([itex]p[/itex] of them), or as a function that takes [itex]p[/itex] covectors and returns an element of [itex]V^* \otimes ... \otimes V^*[/itex] ([itex]p[/itex] of them), etc.
Given a linear map between two vector spaces ##L:V →W## then ##L## determines a map of the algebra of tensor products of vectors in ##V## to the algebra of tensor products of vectors in ##W##. This is correspondence is a covariant functor. ##L## also determines a map of the algebra of tensor products of dual vectors in ##W## to the algebra of tensor products of dual vectors in ##V##. This correspondence is a contravariant functor.Is there a natural way how the ##V's## come up contravariant and the ## V^{*'}s## covariant? Without coordinate transformations?
Yes, but one could as well say ##T_q^p(V) = \underbrace{V \otimes \ldots \otimes V}_{p-times} \otimes \underbrace{V^* \otimes \ldots \otimes V^*}_{q-times}## has ##p## covariant factors ##V## and ##q## contravariant factors ##V^*## and in this sourceGiven a linear map between two vector spaces ##L:V →W## then ##L## determines a map of the algebra of tensor products of vectors in ##V## to the algebra of tensor products of vectors in ##W##. This is correspondence is a covariant functor. ##L## also determines a map of the algebra of tensor products of dual vectors in ##W## to the algebra of tensor products of dual vectors in ##V##. This correspondence is a contravariant functor.
One might guess that this is the reason for the terms covariant and contra-variant tensor though I do not know the history.
I think I said the same thing. The covariant factors are the tensor products of the vectors, the contravariant are the tensors of the dual vectors.Yes, but one could as well say ##T_q^p(V) = \underbrace{V \otimes \ldots \otimes V}_{p-times} \otimes \underbrace{V^* \otimes \ldots \otimes V^*}_{q-times}## has ##p## covariant factors ##V## and ##q## contravariant factors ##V^*## and in this source
http://www.math.tu-dresden.de/~timmerma/texte/tensoren2.pdf (see beginning of section 2.1)
it is done. So what are the reasons for one or the other? The fact which are noted first? Are the first ones always considered contravariant? As someone who tends to confuse left and right I was looking for some possibility to remember a convention, one or the other. So I'm still looking for a kind of natural, or if not possible, at least a canonical deduction.