# In which sense is composition a tensor product

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.

• Do you mean composition? If so, the composition of two linear maps is linear again, not bilinear. If not, can you add a definition? A Google search gives no meaningful results. – Marc Paul Mar 14 '16 at 8:57
• I do mean composition. See edit. I do not mean, however, that the composition of two linear operators is bilinear, but that the composition map itself, taking two linear operators and giving another lienar operator is bilinear. – Bananach Mar 14 '16 at 9:05
• Ah yes, thanks for clearing up my confusion. It seems that the space you are looking for is just $\mathcal L(U,V) \otimes \mathcal L(V, W)$, right? – Marc Paul Mar 14 '16 at 9:30
• If I understand Gerard's answer right, then yes, under the identification of your space with $\mathcal{L}(U,W)$, where this identification happens through contraction – Bananach Mar 14 '16 at 10:25

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.

%$$\require{AMScd} %\begin{CD} %\Gamma(X,\mathcal O_X) @>>> \mathcal O_{X,x}\\ %@AAA @AAA \\ \Gamma(Y,\mathcal O_Y) @>>> \mathcal O_{Y,f(x)} %\end{CD}$$

• Thanks a lot, despite not knowing much about tensors, I can mostly understand what you're saying. Does that mean that the identification of $\circ$ with $\otimes$ depends on some choice of basis for $V$? Also, in my particular application, I have bounded operators between Banach (Hilbert if that helps) spaces. What then? A reference would be great – Bananach Mar 14 '16 at 10:07
• Actually, it would also be great if you could expand on how exactly the contraction works. If I have two rank-1 operators $u^*\otimes v\in U^*\otimes V$ and $v^*\otimes w\in V^*\otimes W$, then I can interpret their tensor product as an operator from $U$ to $W$ as follows: Apply $u^*$ to the input. Apply $v^*$ to the result times $v$. Output $w$ times the result. How do I interpret in the same way the contraction $u^*\otimes (v\otimes v^*) \otimes w$? Apply $u^*$ to the input, ...? – Bananach Mar 14 '16 at 10:27
• @Bananach O.K. I'll put details this evening. The contraction is indeed the "trace operator". – Duchamp Gérard H. E. Mar 14 '16 at 11:20
• @Bananach [Output w times the result. How do I interpret in the same way the contraction $u^*\otimes (v\otimes v^*) \otimes w$ ? Apply $u^∗$ to the input]---> you were right, see details and do not hesitate. – Duchamp Gérard H. E. Mar 14 '16 at 22:25
• Je sais que on n'est pas censé dit ca ici, mais merci beaucoup pour les explications extraordinaires – Bananach Mar 15 '16 at 12:49