# Traces of operators in nuclear spaces

I am currently reading up on nuclear spaces in Jarchow, "Locally Convex Spaces", but I got confused and don't seem to find my mistake. In said book, theorem 21.5.9 states:

Let $F$ be a nuclear Frechet space. Then, $F'_\beta \otimes_{\pi} F = L_\beta (F,F)$. i.e. the projective tensor product of $F$ with its strong dual is homeomorphic to the operators on $F$ with the strong topology.

Now, that means every operator on $F$ has a trace, right?

So take the the space $X=\Pi_\mathbb{N} \mathbb{R}$. This is a nuclear Frechet space as it is a countable infinite product of those. Let $id_k : \mathbb{R} \rightarrow \mathbb{R}$ denote the identity on the k'th component of $X$. We have $id_X = \Sigma_k id_k$. Now, by above theorem $id_X$ and all $id_k$ must be trace class, but clearly, the sum can not converge to $id_X$ as otherwise the trace must be infinite.

If I read correctly, the strong topology has a 0-basis of $L(B,U)$, maps from some bounded set into an open set. The partial sums $f_n=\Sigma_{1≤k≤n} id_k$ however satisfy that $f_n-id_X$ is zero on the first $n$ components and thus maps any set into a chosen open set for n big enough. Thus, the partial sum converges and there is a contradiction.

I simply can't spot my mistake, perhaps you can help me?

• In the stereotype world, en.wikipedia.org/wiki/Stereotype_space, this equality is not true: ${\mathbb R}_{\mathbb N}\circledast {\mathbb R}^{\mathbb N}\ne {\mathcal L}({\mathbb R}^{\mathbb N})$. – Sergei Akbarov Oct 4 '15 at 15:34
• That might be, but for Nuclear Frechet Spaces and the projective tensor product, it holds. – J.L.R. Oct 4 '15 at 16:11
• Something is strange for me here... For a locally convex space $X$ let us denote by $X^{\mathbb N}$ and $X_{\mathbb N}$ the direct product and the locally convex sum of contable copies of $X$. Then I would think that $({\mathbb R}^{\mathbb N})'_{\beta}\otimes_{\pi}{\mathbb R}^{\mathbb N}=({\mathbb R}^{\mathbb N})_{\mathbb N}\ne({\mathbb R}_{\mathbb N})^{\mathbb N}=L_{\beta}({\mathbb R}^{\mathbb N})$. – Sergei Akbarov Oct 4 '15 at 16:23
• The first equality does not hold as projective tensor products do not commute with direct sums. – J.L.R. Oct 4 '15 at 16:31
• The author of the book you mention is Hans Jarchow. – Jochen Wengenroth Oct 5 '15 at 20:17

It is indeed true that for a nuclear Frechet space $F$, the complete projective tensor product $F'_\beta \tilde{\otimes}_\pi F$ is isomorphic to $L_\beta(F,F)$. The mistake is your claim that the complete projective tensor product $E\tilde{\otimes}_\pi F$ consists of convergent series $\sum\limits_{n=1}^\infty \lambda_n e_n \otimes f_n$. This is true if $E$ and $F$ are both Frechet spaces or both strong duals of Frechet spaces. In the mixed case as in your situation the complete inductive tensor product can be described by such series but it is much smaller than the projective tensor product.

EDIT: I meant the inductive tensor product (the injective coincides indeed with the projective one because of nuclearity).

Your answer is wrong (unfortunately, I can't comment, so here is another answer instead). Now, I'm quoting from "Notes on locally convex vector spaces" by J.L. Taylor.

Since $F$ is nuclear, the (completed, you are right, I meant that) injective and projective tensor products agree, so "is smaller than" doesn't apply here.

Furthermore, theorem 21.5.9 in Yarchow explicitely states that the maps are homeomorphisms, and not just an isomorphism. We have:

$F'_\beta \hat{\otimes}_\pi F=F'_\beta \hat{\otimes}_\epsilon F=L_\beta(F,F)$

(corrolary 5.19 in the aforemention lecture notes). Thus the injective tensor product is just as big, "is smaller than" does not make much sense here.

Also, we have (proposition 3.5):

Any bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R}$ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$.

Which brings me to the real answer to the problem, which I just found myself:

Now, this would be fine if the trace was indeed continuous. But: Let $U$ be an open set in $\mathbb{R}^\mathbb{N}$, $V$ be an open set in $\mathbb{R}_\mathbb{N}$. Then, we always have: $U(V)=\mathbb{R}$, as can always be seen by choosing a sufficiently small open subset. Hence, the trace is not a bilinear continous map, hence can not be extended! Please correct me if you see any mistake, but I think that is the answer, no?

• J.L.R: you seem to have created two different accounts. One for the question, and one for the answer... (you can see that the icons next to "J.L.R" are different). – André Henriques Oct 6 '15 at 22:19
• Yes, sorry for that. The registration process was a little confusing to me. – J.L.R. Oct 6 '15 at 22:21
• You could try requesting the moderators to merge the accounts --- please visit MO.meta e.g., meta.mathoverflow.net/search?q=merge to figure out how! – Suvrit Oct 6 '15 at 22:30
• Nah, I got it working now, I am fine, but thank you for the suggestion. – J.L.R. Oct 6 '15 at 22:31
• See the edit to my answer. For any non-normed locally convex space $F$ the evalution $F'_\beta \times F \to \mathbb C$, $(\phi,x)\mapsto \phi(x)$ is separately continuous but discontinuous. This shows that the inductive tensor product $F'_\beta \otimes_\iota F$ (defined by the universal property, that all separately continuous bilinear maps factorize continuously) has a much finer topology than the projective tensor product. – Jochen Wengenroth Oct 7 '15 at 6:29