# Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism?

Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the wikipedia article on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.

Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$\dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$

which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?

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The map is not an isomorphism because an element in $X^*\otimes Y^*$ is a finite sum of functionals of the form $x*\otimes y*$, where $x^* \in X^*$ and $y^* \in Y^*$. However, when $X$ and $Y$ are infinite dimensional, not every functional on $X\otimes Y$ will be of this form.

One case to consider, which will make this clear, is the case when $Y = X^*$. There is one obvious element of $(X\otimes Y)^*$, namely the evaluation map which takes a tensor $x\otimes y$ to the value of the functional $y$ on the vector $x$. Now one can check that this element is not in the image of the map from $X^*\otimes Y^*$.

I am going to take a little time to rewrite the preceding example in a different language, because I think that it helps illustrate what is going on.

First note that if $Y = X^*$, then $X\otimes Y$ embeds into $End(X)$ (the space of linear operators from $X$ to istelf) as the space of finite rank linear operators (i.e. those whose image is finite dimensional). Denote this image by $FREnd(X)\subset End(X)$. Note that any element of $FREnd(X)$ has a well-defined trace (because even though the domain is infinite dimensional, the range is finite dimensional); in the tensor product description, this is just the natural map from $X\otimes Y$ to the ground field given by evaluation of the functionals in $Y$ on the vectors in $X$. (This is precisely the functional on $X\otimes Y$ that we considered above, reexpressed in the language of operators.)

Replacing $X$ by $X^*$ in the preceding paragraph, we see that $X^*\otimes Y^* = FREnd(X^*)$. Note that there is a natural map $FREnd(X) \to FREnd(X^*)$ given by mapping an endomorphism $\phi$ to its transpose $\phi^t$. (In terms of tensor products, this is the natural map $$X \otimes Y = X\otimes X^* \to X^{**}\otimes X^* \cong X^*\otimes X^{**} =X^*\otimes Y^*,$$ the isomorphism being the canonical one which switches the two factors.)

The map $X^*\otimes Y^*\to (X\otimes Y)^*$ can then be reintrepreted as the pairing between $FREnd(X^*)$ and $FREnd(X)^*$ defined as follows: for $\phi\in FREnd(X^*)$ and $\psi \in FREnd(X),$ $$\langle \phi,\psi\rangle := trace(\phi\circ \psi^t).$$

And now we see why this map is not surjective: for example, if $\phi$ is any endomorphism of $X^*$, i.e. any element of $End(X^*)$, then the composite $\phi\circ \psi^t$ has finite rank (since $\psi^t$ does), and so $trace(\phi\circ\psi^t)$ is defined.

Thus we in fact have an embedding of all of $End(X^*)$ into $FREnd(X)^*$. With a little more work you can check that this latter embedding is an isomorphism. The conclusion in this case is that the embedding $X^*\otimes Y^* \to (X\otimes Y)^*$ can be reinterpreted as the embedding $$FREnd(X^*) \to End(X^*),$$ which is not surjective when $X^*$ (or equivalently, $X$) is infinite dimensional, since it does not contain the identity map (for example).

Note that under the identification of $End(X^*)$ with $FREnd(X)^*$, the identity map is identified precisely with the trace on $FREnd(X)$, and so we get a reinterpretation of our original example, and see more clearly what is going on: the point is that the identity endomorphism of an infinite dimensional vector space does not have finite rank.

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Maybe you should fix your ambiguous texts in line 6 and 7? @Matt E – awllower Apr 6 '11 at 4:04
Thank you so much for this fantastic answer. I've carefully checked the details of all the maps involved and found the process VERY educational. For instance: even checking an easy little thing like "the transpose of a finite rank map is finite rank" required a little flip of perspective which I don't think I've ever made properly before. – Mike F Apr 6 '11 at 23:42
@Mike: Dear Mike, I'm glad that this answer helped. I agree that it's enjoyable and enlightening to work through all the details. (I had fun doing exactly that while writing the answer!) Best wishes, – Matt E Apr 7 '11 at 0:09