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 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 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 at 17:46

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 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 at 15:20
  • $\begingroup$ @GEdgar: I will get back to this, I have to go to class. $\endgroup$ – Michael Albanese May 17 at 15:24
  • $\begingroup$ @lisyarus: I am aware of that, but the question is specifically about the definition via a multilinear map. $\endgroup$ – Michael Albanese May 17 at 15:24
  • $\begingroup$ @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) ? $\endgroup$ – Carla May 17 at 17:55

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.