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 VW(VW) which sends ifigiVW to the unique functional in (VW) sending vwifi(v)gi(w) for all (v,w)V×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 (VW) 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(XY)=dimXdimY holds. It is clear that |X|=|k|dimX 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 dimX|k| then in fact |X|=dimX 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|dimXdimY it seems we have
dim(XY)=|k|dimXdimY=|k|dimY=dimY=dimXdimY=dim(XY)

which implies that in fact (XY) and XY 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 (XY)XY?

Answer

The map is not an isomorphism because an element in XY is a finite
sum of functionals of the form xy, where xX and yY. However, when X and Y are infinite dimensional, not every functional on XY 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 (XY), namely the evaluation map
which takes a tensor xy 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 XY.


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
XY embeds into End(X)
(the space of linear operators from X to itself) as the space of finite rank linear operators (i.e. those whose image is finite dimensional). Denote
this image by FREnd(X)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 XY to the ground field
given by evaluation of the functionals in Y on the vectors in X.
(This is precisely the functional on XY that we considered above,
reexpressed in the language of operators.)

Replacing X by X in the preceding paragraph, we see that
XY=FREnd(X). Note that there is a natural map
FREnd(X)FREnd(X) given by mapping an endomorphism ϕ to its
transpose ϕt. (In terms of tensor products, this is the natural
map XY=XXXXXX=XY, the isomorphism being the canonical one which switches the
two factors.)

The map XY(XY)
can then be reintrepreted as the pairing between
FREnd(X) and FREnd(X)
defined as follows:
for ϕFREnd(X) and ψFREnd(X),
ϕ,ψ:=trace(ϕψt).

And now we see why this map is not surjective: for example, if
ϕ is any endomorphism of X, i.e. any element of End(X),
then the composite ϕψt has finite rank (since ψt does),
and so trace(ϕψt) 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 XY(XY) can be reinterpreted as the embedding
FREnd(X)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.

Attribution
Source : Link , Question Author : Mike F , Answer Author : jimmij

Leave a Comment