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 ?

  • $\begingroup$ math.stackexchange.com/questions/58548/… $\endgroup$
    – Asaf Karagila
    May 17, 2019 at 15:16
  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    May 17, 2019 at 15:24
  • $\begingroup$ I agree. I tried to be too concise in the title. But my question really is in the text below the title. :) $\endgroup$
    – Carla
    May 17, 2019 at 17:46

2 Answers 2


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.

  • $\begingroup$ You did not explain the "why?" part. $\endgroup$
    – GEdgar
    May 17, 2019 at 15:18
  • $\begingroup$ "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. $\endgroup$
    – lisyarus
    May 17, 2019 at 15:20
  • $\begingroup$ @GEdgar: I will get back to this, I have to go to class. $\endgroup$ May 17, 2019 at 15:24
  • $\begingroup$ @lisyarus: I am aware of that, but the question is specifically about the definition via a multilinear map. $\endgroup$ May 17, 2019 at 15:24
  • 1
    $\begingroup$ @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!) $\endgroup$
    – Carla
    May 18, 2019 at 21:18

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, $$ \lambda\in V^*\quad\text{such that}\quad \lambda(e_i)=\left\{\begin{array}{cl}1&\text{if $i$ is odd}\\ 0 & \text{if $i$ is even}\end{array}\right. $$

  • $\begingroup$ 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. $\endgroup$ Feb 26, 2020 at 1:23

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