There is always an injection between $V^* \otimes V^*$ and $(V \otimes V)^*$ given by $$ f(v^* \otimes w^*)(x \otimes y)=v^*(x)w^*(y), $$ where $x,y \in V$. I've been given to understand that in infinite dimension it is not surjective. Does anybody can explain me why it is the case? Does anybody have a concrete and simple example where $V^* \otimes V^*$ is not isomorphic to $(V \otimes V)^*$

Edit. This problem is involved in basic theory of Hopf Algebra since the fact that in infinite dimension $V^* \otimes V^*$ is only a proper subset of $(V \otimes V)^*$ is the key point for which the dual of a co-algebra is an algebra, but in general the dual of an algebra is not a co-algebra.

Second Edit The answer given here doesn't seem to answer my question as the answer of Mariano does. So even if the question may seem a duplicate, the answer is not. And as Hardmath notes "the proposed duplicate does not restrict to the tensor product of a vector space with itself, and indeed the answer given there involves one factor being the dual of the other factor. It doesn't address the problem here."

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    $\begingroup$ Assuming the axiom of choice we can show that $V^* \otimes V^*$ and $(V\otimes V)^*$ will always be isomorphic, since $\kappa^2=\kappa$ for infinite cardinals and $\dim(V\otimes V)=\dim(V)^2$. However, this doesn't make your particular choice of injection an isomorphism. From the top of my head I don't know an easy counterexample to this. $\endgroup$ – Christoph Feb 22 '18 at 12:38
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    $\begingroup$ @Christoph I'm quite sure that in infinite dimension $V^* \otimes V^*$ is only a proper subset of $(V \otimes V)^*$ this is a key problem for which the dual of a co-algebra is an algebra, but in general the dual of an algebra is not a co-algebra. $\endgroup$ – Dac0 Feb 22 '18 at 15:30
  • $\begingroup$ I agree with @Dac0, infinite dimensional spaces are not isomorphic to their dual generally. $\endgroup$ – Chickenmancer Feb 22 '18 at 15:44
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    $\begingroup$ @chicken, in fact, an infinite dimensional vector space is never isomorphic to its dual. You can find several arguments for that here. $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '18 at 6:17
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    $\begingroup$ I did not claim that $V\cong V^*$. I'm just saying $\dim(V^*\otimes V^*) = \dim(V^*)^2 = \dim(V^*)$ and since $\dim(V)=\dim(V)^2=\dim(V\otimes V)$ we also have $\dim((V\otimes V)^*) = \dim(V^*)$. So the dimensions agree and the vector spaces are isomorphic. $\endgroup$ – Christoph Feb 23 '18 at 10:01

The question that makes sense is:

if $V$ is infinite dimensional, is the canonical map $T:V^*\otimes V^*\to (V\otimes V)^*$ surjective?

and the answer is no. To see this we have to find something about the elements of the image of $T$ which is special.

Suppose that $\phi\in(V\otimes V)^*$ and consider the set $U(\phi)$ of all vectors $u\in V$ such that $\phi(u\otimes v)=0$ for all $v\in V$. This is a subspace of $V$.

If $a\in V^*\otimes V^*$, so that there are $n\geq0$ and $\phi_1,\dots,\phi_n$, $\psi_1,\dots,\psi_n\in V^*$ such that $a=\sum_{i=1}^n\phi_i\otimes\psi_i$, then the space $K(a)=\bigcap_{i=1}^n\ker\phi_i$ has finite codimension in $V$ and is contained in $U(T(a))$. This tells us that

if $\phi\in(V\otimes V)^*$ is in the image of the map $T$, then $U(\phi)$ is a subspace of $V$ of finite codimension.

Now, to show that $T$ is not surjective it is enough that we exhibit a $\phi\in (V\otimes V)^*$ such that $U(\phi)$ is not of finite codimension!

Let $\{e_i\}_{i\in I}$ be a basis for $V$. There is a unique $\phi\in(V\otimes V)^*$ such that $$\phi(e_i\otimes e_j)=\begin{cases}1, & \text{if $i=j$;} \\0, & \text{if not.}\end{cases}$$ Let $u\in V$. As we have a basis, there are scalars $a_i$, one for each $i\in I$ and almost all of which are zero, such that $u=\sum_{i\in I}a_ie_i$. If $j\in I$ is such that $a_i\neq0$, then $\phi(u\otimes e_i)=a_i\neq0$, and we see that $u\not\in U(\phi)$ unless $u=0$. In other words, we have $U(\phi)=0$ and certainly the zero subspace of $V$ does not have finite codimension in $V$. The element $\phi$ is therefore not in the image of the map $T$.

In fact, we can reverse this. Suppose that $\lambda\in(V\otimes V)^*$ is such that $U(\lambda)$ has finite codimension in $V$. Let $n$ be that codimension, let $\{u_1,\dots,u_n\}$ be a basis of $V/U(\lambda)$ and let $\{\bar\phi_1,\dots,\bar\phi_n\}$ be the corresponding dual basis for the dual space of $V/U(\lambda)$. Let $p:V\to V/U(\lambda)$ be the canonical map and for each $i\in\{1,\dots,n\}$ let $\phi_i=\bar\phi_i\circ p\in V^*$. As $\lambda$ vanishes on the subspace $U(\phi)\otimes V$ of $V\otimes V$, it induces a map $\Lambda:(V/U(\lambda))\otimes V\to k$ (here $k$ is the field) For each $i\in\{1,\dots,n\}$ we consider the map $\psi_i\in V^*$ such that $\psi_i(v)=\Lambda(u_i\otimes v)$. After all this setup, one can easily check now that $\lambda=T(\sum_{i=1}^n\phi_i\otimes\psi_i)$.

We have this proved:

The image of the canonical map $V^*\otimes V^*\to(V\otimes V)^*$ is precisely the subset of all $\lambda\in(V\otimes V)^*$ such that $U(\lambda)$ has finite codimension in $V$.

  • $\begingroup$ This is similar and related to the fact that the image of the canonical map $V\otimes V^*\to\operatorname{End}(V)$ is the subspace of all endomorphisms of finite rank. $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '18 at 7:07
  • $\begingroup$ Thank You! I upvoted and if nobody come up with a better answer I will consider it as solved, but is there a direct way to show that the injection I gave is not surjective? $\endgroup$ – Dac0 Feb 23 '18 at 8:08
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    $\begingroup$ @Dac0 (to Mariano) It should be $u\in V$ in the part "(...) consider the set $U(\phi)$ of all vectors $v\in V$ such (...)". (to Dac0) But Mariano proved it directely. 1. He proved that $\phi\in Im(T)$ implies that $\phi$ has a finite codimension; 2. He constructed $\phi$ with infinite codimension. Thus, $T$ in not surjective. I doubt you can get a better answer. $\endgroup$ – Fallen Apart Feb 23 '18 at 13:32
  • $\begingroup$ @Dac0 PS. I should have written $U(\phi)$ has a finite/an infinite codimension. $\endgroup$ – Fallen Apart Feb 23 '18 at 13:54

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