# Why does a space vector $V$ must have finite dimension to be isomorphic to its bidual $V^{**}$?

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 ?

• math.stackexchange.com/questions/58548/… May 17, 2019 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. May 17, 2019 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. :) May 17, 2019 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.

• You did not explain the "why?" part. May 17, 2019 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. May 17, 2019 at 15:20
• @GEdgar: I will get back to this, I have to go to class. May 17, 2019 at 15:24
• @lisyarus: I am aware of that, but the question is specifically about the definition via a multilinear map. May 17, 2019 at 15:24
• @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!) 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, $$\lambda\in V^*\quad\text{such that}\quad \lambda(e_i)=\left\{\begin{array}{cl}1&\text{if i is odd}\\ 0 & \text{if i is even}\end{array}\right.$$

• 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. Feb 26, 2020 at 1:23