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An answer to a request for an example of a tensor here is as follows:

Here is the concrete example: concrete. When a structural material is stressed, the state of stress is described by a tensor. It’s not a scalar and it’s not a vector. Any tiny spherical element is deformed in general into an ellipsoid. It takes five numbers to specify that ellipsoid: three for the stretchings of the principal axes, and two for the orientation of the long axis. That’s a tensor. (There are other ways to specify the five parameters but they come to the same thing.)

This makes a lot of sense; however, I "understand" a tensor according to the definition:

A $(p,q)$ tensor, $T$ is a MULTILINEAR MAP that takes $p$ copies of $V^*$ and $q$ copies of $V$ and maps multilinearly (linear in each entry) to $k:$

$$T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times\times \cdots V\times V}} \overset{\sim}\rightarrow K\tag 1$$

According to this definition, the "end product" of a tensor would be a real number, not $5$.

This misunderstanding may also be at the root of my confusion about the result of this example of a tensor product.

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Antoni Parellada
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  • I wouldn't try and understand what a tensor is based on your quote... (and the stress tensor in $\mathbb{R}^3$ should depend on $6$ parameters, not $5$). It really mixes things up in a bad way. It's not that the "end product" of the stress tensor is six numbers, but that the stress tensor (considered as a map $T \colon \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ has nine components with respect to an orthonormal basis and it is symmetric so it has $6$ free parameters. – levap Feb 24 '17 at 16:37
  • @levap I almost follow... You introduce the "stress" qualifier, which throws me off. Are you saying that the definition with the critical $\rightarrow \mathbb R$ is correct, but there can be a number of components within a tensor, sort of like a Russian doll... I know it's not making sense, but I'm going for the intuition... – Antoni Parellada Feb 24 '17 at 16:40
  • More or less. A vector in $\mathbb{R}^3$ will have three components with respect to some given basis $(v_1,v_2,v_3)$ so in this sense, it is described by three parameters. A linear map $T \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ can be represented by a $3 \times 3$ matrix with respect to some given basis so in this sense, it is described by nine components. A bilinear form $T \colon \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ can also be represented by a $3 \times 3$ matrix with respect to some basis so it depends on $9$ parameters. If $T$ is symmetric, the matrix will be – levap Feb 24 '17 at 16:45
  • symmetric and so there are only $6$ "free parameters". And you can ignore the "stress" qualifier, just treat $T$ as a symmetric bilinear form. – levap Feb 24 '17 at 16:45
  • Thank you! I get the "...can also be represented by a $3×3$ matrix with respect to some basis so it depends on $9$ parameters." (emphasis mine). Yet the output is still one single number... in $\mathbb R$. – Antoni Parellada Feb 24 '17 at 16:49
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    Well, the output of a bilinear form (an $(0,2)$-tensor in your notation) is indeed a single number. However, the output of a linear map is a vector and a linear map can be considered as a $(1,1)$-tensor. There is a natural identification between bilinear maps $T \colon V^{*} \times V \rightarrow \mathbb{R}$ and linear maps $S \colon V \rightarrow V$ so you can think of certain tensors as outputing (certain) vectors. – levap Feb 24 '17 at 17:52
  • Very useful! Thank you. You should consider compiling an answer with these comments... – Antoni Parellada Feb 24 '17 at 17:54
  • I don't feel like I answered your original question, just gave you some information regarding "how tensors works" and what kind of objects they are... – levap Feb 24 '17 at 17:56

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The "definition" you give of a tensor is bad, and this is a great example of why: tensors can be interpreted as functions in many different ways, and the "definition" only describes one of them. A tensor of type $(p, q)$ can be interpreted as a function in $2^{p+q}$ ways. For example, a tensor of type $(1, 1)$ can be interpreted as a function in the following four ways:

  • A linear function $V \to V$,
  • A linear function $k \to V^{\ast} \otimes V$, or equivalently an element of $V^{\ast} \otimes V$.
  • A linear function $V \otimes V^{\ast} \to k$, or equivalently a bilinear function $V \times V^{\ast} \to k$,
  • A linear function $V^{\ast} \to V^{\ast}$.

The stress tensor apparently has type $(0, 2)$, so it can also be interpreted as a function in four ways.

Qiaochu Yuan
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  • You hinted at it yesterday. What would be a good definition? – Antoni Parellada Feb 24 '17 at 20:53
  • @Antoni: there are a few different considerations you might want to optimize for in a definition. One is to remove any explicit mention of duals: in that case, a $(p, q)$ tensor is a linear function from $V^{\otimes q}$ to $V^{\otimes p}$. One (which is what your definition accomplishes) is to remove any explicit mention of tensor products; personally I think both of these goals are silly because duals and tensor products are really important and you should understand them if you want to understand tensors. So my preferred definition does neither: a $(p, q)$ tensor is an element of... – Qiaochu Yuan Feb 24 '17 at 20:57
  • ...the tensor product $V^{\otimes p} \otimes (V^{\ast})^{\otimes q}$. Now, once you've absorbed some useful facts about how to manipulate tensor products and duals, which this definition encourages you to do in order to even understand it, you can not only see why all of these definitions are equivalent, you can also see all $2^{p+q}$ ways of interpreting such a tensor as a function, by shifting around the copies of $V$ back and forth between the source and target. – Qiaochu Yuan Feb 24 '17 at 20:58
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    There is a "2-dimensional" notation for doing all of this which is much clearer than the usual notation, and it goes back in some form to Penrose. See https://qchu.wordpress.com/2012/11/05/introduction-to-string-diagrams/ and https://qchu.wordpress.com/2012/11/06/string-diagrams-duality-and-trace/. In this notation the different ways of interpreting a tensor as a function arise naturally from bending some wires back and forth. – Qiaochu Yuan Feb 24 '17 at 21:01
  • @QiaochuYuan: Would you be able to show me a concrete example of how all four of those "functions" are equivalent (similar to what was done here: https://math.stackexchange.com/questions/2158892/working-out-a-concrete-example-of-tensor-product)? I'm very interested in understanding their equivalence because I'm trying to understand how a tensor can produce anything other than a real number (based on what the "standard" definition states). I'm also confused by your statement: " shifting around the copies of V back and forth between the source and target". – RobbieFresh Aug 08 '17 at 14:31