In which sense is composition a tensor product

Question

Let $\Phi\colon U\to V$ and $\Psi\colon V \to W$ be linear operators, and consider their composition $$ \Psi\circ \Phi $$

The operation, $$\circ:\mathcal{L}(U,V)\times\mathcal{L}(V,W)\to \mathcal{L}(U,W)\\ (\Phi,\Psi)\mapsto \Psi\circ \Phi $$ is bilinear. So I expect that we can understand $\Psi$ and $\Phi$ identified (by $\iota$) within a tensor space $T$ such that $$ \Psi\circ\Phi= \iota(\Psi \otimes \Phi)\ . $$ However, I cannot figure out what $T$ should be.

Answer 1

Yes, in fact the tensor product $U^*\otimes V$ can be identified to $\mathcal{L}^{fin}(U,V)$, the space of operators with finite rank. Now, with a suitable topology (the pointwise convergence, $V$ being endowed with the discrete topology, to put it explicit) one has $$ \mathcal{L}(U,V)\cong U^*\hat\otimes V $$
then, at the level of finite rank operators, your composition, $$ \circ : \mathcal{L}^{fin}(U,V)\otimes \mathcal{L}^{fin}(V,W)\rightarrow \mathcal{L}^{fin}(U,W) $$ reads as the trace contraction of factors 2-3 as follows $$ U^*\otimes (V \otimes V^*)\otimes W\ . $$ This passes to the completion.

So, this was the general scheme. Let us now go into details

The isomorphism $\mathcal{L}^{fin}(U,V)\cong U^*\otimes V$
One has a natural arrow $j_{U,V} : U^*\otimes V\rightarrow \mathcal{L}(U,V)$ given by $j_{U,V}(f\otimes v)[x]=f(x)v$ as you remarked in the comments. Its image is $\mathcal{L}^{fin}(U,V)$ as it is easy to see that $Im(j_{U,V})\subset \mathcal{L}^{fin}(U,V)$. For each $T\subset V$ of finite dimension, one constructs an approximate section of $j_{U,V}$ by means of a finite basis ${t_j}_{j\in J}$ of $T$ to $\phi\in \mathcal{L}^{fin}(U,V)$ with image in $T$, we set $$ s_T(\phi)=\sum_{j\in J}(t_j^*\circ \phi)\otimes t_j $$
where $t_j^*$ is the coordinate family (i.e. $t_j^*(t_i)=\delta_{ij}$). It can be shown that $s_T$ does not depend on the chosen basis and that the $s_T$ extend each other (inductive system). So setting $s=\lim_{T\rightarrow V}s_T$, we get a section of $j_{U,V}$ which is, in fact, the inverse isomorphism.

Topology on $\mathcal{L}(U,V)$. Endowing $V$ with the discrete topology and $\mathcal{L}(U,V)$ with the pointwise convergence, we get the criterium that $(f_\alpha)_{\alpha\in A}$ ($A$ is a sup-directed set) tends to zero iff $$ (\forall u\in U)(\exists B\in A)(\alpha\geq B\Longrightarrow f_\alpha(u)=0) $$ likewise, one has the summability criterium $(f_i)_{i\in I}$ is summable iff $$ (\forall u\in U)(\exists F\subset_{finite} I)(i\notin F\Longrightarrow f_i(u)=0) $$ one can see at once that we can consider $\sum_{i\in F}f_i(u)$ as the limit and check that the map $u\rightarrow \sum_{i\in F_u}f_i(u)$ ($F$ depends on $u$) is linear. Let us call $l$ this map. We can prove easily that $l=lim_{F\rightarrow_{finite} I}$ and we have $l=\sum_{i\in I}f_i$.

Representation of $\mathcal{L}(U,V)$ as $U^*\hat\otimes V$

With the preceding topology it can be shown that

1. $\mathcal{L}(U,V)$ is complete
2. $\mathcal{L}^{fin}(U,V)$ is dense in $\mathcal{L}(U,V)$ still calling $j_{U,V}$ the embedding $U^*\otimes V\rightarrow \mathcal{L}(U,V)$, one gets a topology on the tensor product and its completion gives the isomorphism $$ j_{U,V}: U^*\hat\otimes V\cong \mathcal{L}(U,V) $$ (by a little abuse of language, we still note it $j_{U,V}$).

Concrete computations We can give two expressions for the inverse of $j_{U,V}$ (representation of linear maps). Let $(u_i)_{i\in I}$ (resp. $(v_j)_{j\in J}$) be a basis of $U$ (resp. $V$), then, for any $\phi\in \mathcal{L}(U,V)$, the families $$ \Big(u_i^*\otimes \phi(u_i)\Big)_{i\in I}\ ;\ \Big((v_j^*\circ \phi)\otimes v_j)\Big)_{j\in J} $$
are summable and their sums are $\phi$, hence $$ \phi=\sum_{i\in I}u_i^*\otimes \phi(u_i)=\sum_{j\in J}(v_j^*\circ \phi)\otimes v_j \qquad (2) $$
(where, for the sake of expressiveness, by a little abuse of language, we identified the tensors to their image through $j_{U,V}$)

Continuity of the composition

For the aforementioned topologies, the composition $$ \circ : \mathcal{L}^{fin}(U,V)\otimes \mathcal{L}^{fin}(V,W)\rightarrow \mathcal{L}^{fin}(U,W) $$ is continuous (means, separately continuous and jointly continuous at $(0,0)$), it extends to the completions as the usual $\circ$. This proves by isomorphisms that the usual trace operator (between second and third factors) $$ tr_{23}:(U^*\otimes V)\otimes (V^*\otimes W)\rightarrow U^*\otimes W $$ extends as $$ \hat{tr}_{23}:(U^*\hat\otimes V)\otimes (V^*\hat\otimes W)\rightarrow U^*\hat\otimes W $$ giving the interpretation of the composition as a tensor contraction. To figure out concretely $\hat{t_{23}}$, the best is to take a basis of $V$ and use the second representation for $\phi$ and the first for $\psi$, one gets $$ \hat{t_{23}}\Big(\sum_{(i,j)\in I^2}(v_i^*\circ \phi)\otimes v_i)(v_j^*\otimes \psi(v_j)\Big)=\sum_{i\in I}(v_i^*\circ \phi)\otimes \psi(v_i) $$ which represents $\psi\circ\phi$.

Nota I gave the scheme (a lot of piled notions, but not very difficult each), all can be unfolded on request.

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