# Insights What Is a Tensor? - Comments

#### fresh_42

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2018 Award
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#### burakumin

Definition: A tensor product of vector spaces U⊗V is a vector space structure on the Cartesian product U×V that satisfies ...
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.

#### lavinia

Gold Member
- Students of General Relativity learn about tensors from their transformation properties. Tensors are arrays of number assigned to each coordinate system that transform according to certain rules. Arrays that do not transform according to these rules are not tensors. I think that it would be helpful to connect this General Relativity approach to the mathematical approach that you have explained in this Insight.

Also these students need to understand how a metric allows one to pass back and forth between covariant and contra-variant tensors. One might show how this is the same as passing between a vector space and its dual.

- Students of Quantum Mechanics learn about tensors to describe the states of several particles e.g. two entangled electrons. In this case, the mathematical definition is more like the Quantum Mechanics definition but for the Quantum Mechanics student it is also important to understand how linear operators act on tensor products of vector spaces.

- If one wants to discuss tensor products purely mathematically, then one might show how they are defined when the scalars are not in a field but in a commutative ring - or even a non-commutative ring. The formal properties do not depend on a field per se.

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#### lavinia

Gold Member
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.
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.

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#### burakumin

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.
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.

#### fresh_42

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2018 Award
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.
No, 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.

#### burakumin

No, 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.
Thank you

#### fresh_42

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2018 Award
- Students of General Relativity learn about tensors ...

- Students of Quantum Mechanics learn about tensors ...

- If one wants to discuss tensor products purely mathematically ...
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.

#### Orodruin

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Let me first say that I think that the Insight is well written in general. However, I must say that I have had a lot of experience with students not grasping what tensors are based on them being introduced as multidimensional arrays. 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.

Furthermore, you can represent a tensor of any rank with a row or column vector - or (in the case of rank > 1) a matrix for that matter (just choose suitable bases). This may even be more natural if you consider tensors as multilinear maps. An example of a rank 4 tensor being used in solid mechanics is the compliance/stiffness tensors that give a linear relation between the stress tensor and the strain tensor (both symmetric rank 2 tensors). This is often represented as a 6x6 matrix using the basis $\vec e_1 \otimes \vec e_1$, $\vec e_2 \otimes \vec e_2$, $\vec e_3 \otimes \vec e_3$, $\vec e_{\{1} \otimes \vec e_{2\}}$, $\vec e_{\{1} \otimes \vec e_{3\}}$, $\vec e_{\{2} \otimes \vec e_{3\}}$ for the symmetric rank 2 tensors. In the same language, the stress and strain tensors are described as column matrices with 6 elements.

#### fresh_42

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2018 Award
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.

#### Orodruin

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Perhaps 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?"

Nobody on an "A" and probably as well on an "I" level reads a text about what a tensor is.
This must mean I am B-level.

#### fresh_42

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2018 Award
Perhaps 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?"
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 wrote
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.
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.
This must mean I am B-level.
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 is the skeleton I originally planned:

\subsection*{Covariance and Contravariance}
\subsection*{To Rise and to Lower Indices}
\subsection*{Natural Isomorphisms and Representations}
\subsection*{Tensor Algebra}
\section*{Stress Energy Tensor}
\section*{Cauchy Stress Tensor}
\section*{Metric Tensor}
\section*{Curvature Tensor}
\section*{The Co-Universal Property}
\subsection*{Graßmann Algebras}
\subsection*{Clifford Algebras}
\subsection*{Lie Algebras}
\section*{Tensor Fields}

#### fresh_42

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I am pretty sure most engineering students will not remember what a homomorphism is without looking it up.
Corrected. Thanks.

#### WWGD

Gold Member
I thought it would be nice to have a good understanding of what a singleton $a \otimes b$ represents in a tensor product. It is one of these things that I have understood and then forgotten many times over.

#### WWGD

Gold Member
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.
I think this is done before the moding out and arranging into equivalence classes is done.

#### fresh_42

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2018 Award
I think this is done before the moding out and arranging into equivalence classes is done.
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.

#### WWGD

Gold Member
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 was replying to someone else's post.

#### fresh_42

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2018 Award
I was replying to someone else's post.
Sorry, was a bit in "defensive mode".

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#### stevendaryl

Staff Emeritus
The way that tensors are manipulated implicitly assumes isomorphisms between certain spaces.

If $A$ is a vector space, then $A^*$ is the set of linear functions of type $A \rightarrow S$ (where $S$ means "scalar", which can mean real numbers or complex numbers or maybe something else depending on the setting).

The first isomorphism is $A^{**}$ is isomorphic to $A$.

The second isomorphism is $A^* \otimes B^*$ is isomorphic to those function of type $(A \times B) \rightarrow S$ that are linear in both arguments.

So this means that a tensor of type $T^p_q$ can be thought of as a linear function that takes $q$ vectors and $p$ covectors and returns a scalar, or as a function that takes $q$ vectors and returns an element of $V \otimes V \otimes ... \otimes V$ ($p$ of them), or as a function that takes $p$ covectors and returns an element of $V^* \otimes ... \otimes V^*$ ($p$ of them), etc.

#### WWGD

Gold Member
The way that tensors are manipulated implicitly assumes isomorphisms between certain spaces.

If $A$ is a vector space, then $A^*$ is the set of linear functions of type $A \rightarrow S$ (where $S$ means "scalar", which can mean real numbers or complex numbers or maybe something else depending on the setting).

The first isomorphism is $A^{**}$ is isomorphic to $A$.

The second isomorphism is $A^* \otimes B^*$ is isomorphic to those function of type $(A \times B) \rightarrow S$ that are linear in both arguments.

So this means that a tensor of type $T^p_q$ can be thought of as a linear function that takes $q$ vectors and $p$ covectors and returns a scalar, or as a function that takes $q$ vectors and returns an element of $V \otimes V \otimes ... \otimes V$ ($p$ of them), or as a function that takes $p$ covectors and returns an element of $V^* \otimes ... \otimes V^*$ ($p$ of them), etc.
Good point; same is the case with Tensor Contraction, i.e., it assumes/makes use of , an isomorphism.

#### lavinia

Gold Member
Is there a natural way how the $V's$ come up contravariant and the $V^{*'}s$ covariant? Without coordinate transformations?
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.

One might guess that this is the reason for the terms covariant and contravariant tensor though I do not know the history.

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#### fresh_42

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2018 Award
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.

One might guess that this is the reason for the terms covariant and contra-variant tensor though I do not know the history.
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.

#### lavinia

Gold Member
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.
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.

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"What Is a Tensor? - Comments"

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