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

490 Views Asked by At

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 ?

2

There are 2 best solutions below

9
On BEST ANSWER

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.

1
On

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