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There are many different ways to define tensors. Actually it seems that the word "tensor" is applicable to many various concepts/objects.

In any case, it also seems that when we use the multilinear map definition (tensors are multilinear forms from $V^* \times V^* \times \dots \times V^* \times V \times \dots \times V$ to the associated field $\mathbb{F}$) and we apply that to imply, for instance, that vectors are $(1,0)$-tensors i.e. linear forms from $V^*$ to $\mathbb{F}$, $l \to l(v)$, we need that $V^{**}$ be isomorphic to $V$. And this seems to imply that $V$ has finite dimension. Why? And more importantly, does this mean that this definition (tensors as multilinear forms) is not applicable when $V$ has infinite dimension ?

Carla
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  • https://math.stackexchange.com/questions/58548/why-are-vector-spaces-not-isomorphic-to-their-duals – Asaf Karagila May 17 '19 at 15:16
  • You are confusing two different things. (1) The natural map $V \to V^{}$ is an isomorphism; (2) the spaces $V$ and $V^{}$ are isomorphic. In fact, (1) is what you want for most purposes; and is equivalent to finite dimension. But (2) is in your title. – GEdgar May 17 '19 at 15:24
  • I agree. I tried to be too concise in the title. But my question really is in the text below the title. :) – Carla May 17 '19 at 17:46

2 Answers2

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Let $V$ be a vector space over a field $\mathbb{F}$. There is always an injective map $\Psi : V \to (V^*)^*$ given by $v \mapsto \operatorname{ev}_v$ where $\operatorname{ev}_v : V^* \to \mathbb{F}$ is given by $\varphi \mapsto \varphi(v)$. If $V$ is finite-dimensional, then $\Psi$ is an isomorphism, while if $V$ is infinite-dimensional, then $\Psi$ is not an isomorphism; see this answer.

Whether $V$ is finite-dimensional or not, given a vector $v$, we have $\Psi(v) \in (V^*)^*$. That is, $v$ corresponds to a linear map $\operatorname{ev}_v : V^* \to \mathbb{F}$, i.e. a $(1, 0)$-tensor. However, given an arbitrary $(1, 0)$-tensor, we can only state that this corresponds to a vector if $V$ is finite-dimensional.

Michael Albanese
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  • You did not explain the "why?" part. – GEdgar May 17 '19 at 15:18
  • "given an arbitrary (1,0)-tensor, we can only state that this corresponds to a vector if V is finite-dimensional" - this depends on the chosen definition of a "tensor". What you said is true if "tensor" means "multilinear map", but if an (a,b)-tensor means "element of the tensor product $V^{\otimes a}\otimes (V^*)^{\otimes b}$", then a (1,0)-tensor is the same as a vector. – lisyarus May 17 '19 at 15:20
  • @GEdgar: I will get back to this, I have to go to class. – Michael Albanese May 17 '19 at 15:24
  • @lisyarus: I am aware of that, but the question is specifically about the definition via a multilinear map. – Michael Albanese May 17 '19 at 15:24
  • @Michael Albanese OK. Thanks for the link. But to deserve the green check you have to answer the important question: does that mean that using the multilinear form definition works only for finite-dimension V spaces (as opposed to the "element of tensor product" definition) ? – Carla May 17 '19 at 17:55
  • @Carla: It works perfectly fine, it just doesn't coincide with an alternative definition which is an element of $V^a\otimes (V^*)^b$ as lisyarus alludes to. – Michael Albanese May 17 '19 at 18:05
  • @MichaelAlbanese: sorry but just to make sure I understand you correctly: are you are saying that the multilinear map definition also works for infinite-dimension vector spaces ? But then how do you show that a vector is a [1,0] tensor in that infinite-dimension space ? – Carla May 17 '19 at 18:19
  • @Carla: Yes I am. A vector $v$ always defines a $(1, 0)$-tensor, namely $\operatorname{ev}_v : V^* \to \mathbb{F}$ where $\operatorname{ev}_v(\varphi) = \varphi(v)$. – Michael Albanese May 17 '19 at 18:29
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    @MichaelAlbanese Thank you for your time and patience. Got it! A [1,0] tensor is not always a vector if V is infinite-dimensional. As soon as I get enough credit I will upvote your answer (I gave an answer for another question but the post is too old ha ha ha!) – Carla May 18 '19 at 21:18
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To give a very concrete example, suppose that $V$ is a vector space admitting a numerable basis, $$ V=\bigoplus_{n\in\Bbb N}\Bbb Fe_n. $$ Then the map $\phi:V\rightarrow V^{*}={\rm Hom}(V,\Bbb F)$ which associates to each basis element $V$ its dual, namely $$ \phi(v)(e_i)=e_i^*,\qquad\text{where $e_i^*(e_j)= \left\{\begin{array}{cl}1&\text{if $i=j$}\\ 0 & \text{if $i\neq j$}\end{array}\right.$}, $$ is not surjective because the image consists of the elements in $V^*$ which are finite sums of the $e_i^*$'s missing many linear forms on $V$ like, for instance, $$ \lambda\in V^*\quad\text{such that}\quad \lambda(e_i)=\left\{\begin{array}{cl}1&\text{if $i$ is odd}\\ 0 & \text{if $i$ is even}\end{array}\right. $$

Andrea Mori
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  • Oh, so the answer is kind of "because direct sum doesn't equal direct product for infinite index sets"? Interesting, I would not have thought to connect these two facts, +1. – Chill2Macht Feb 26 '20 at 01:23
  • Why can't another map establish an isomorphism? $\phi$ doesn't work but this doesn't mean there isn't a $\tilde{\phi}$ which does the trick. No? – Kadmos Nov 20 '24 at 16:24
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    @Kadmos : you run into a problem of incompatible cardinalities. Given any subset $S\subset\mathbb{N}$ an isomorphism must have in its image the linear functional $e_S^$ that has value $1$ on each $e_i$ with $i\in S$ and has value $0$ on each $e_i$ such that $i\notin S$. Although the $e_S^$ are not linearly independent, the fact that the cardinality of the set $\mathrm{P}(\mathbb{N})$ is strictly bigger than numerability implies--after some non-obvious work--that it is impossible to obtain all of the $e_S^*$ from a numerable basis. – Andrea Mori Nov 21 '24 at 14:05