0
$\begingroup$

Let $X$ be a $\mathbb R$-Banach space and $X'\:\hat\otimes_\pi\:X$ denote the completion of the tensor product of $X'$ and $X$ with respect to the projective norm. The trace functional $\operatorname{tr}:X'\:\hat\otimes_\pi\:X\to\mathbb R$ is the unique linearization of the bounded bilinear form $$X'\times X\to\mathbb R\;,\;\;\;(\varphi,x)\mapsto\varphi(x)\tag1.$$ Now, I've seen that there are two related notions:

  1. If $E$, $F$ and $G$ are $\mathbb R$-vector spaces, $$C_G:(E\otimes G')\otimes(G\otimes F)\to E\otimes F\;,\;\;\;(x\otimes u')\otimes(u\otimes y)\mapsto\langle u',u\rangle_{G',\:G}x\otimes y\tag2$$ is called tensor contraction or vector-valued trace
  2. If $H$ and $K$ are separable $\mathbb R$-Hilbert spaces and $T$ is a trace-class operator on the Hilbert-Schmidt tensor product $H\:\hat\otimes_2\:K$, then there is a unique trace-class operator $\operatorname{tr}_K(T)$ on $H$, called partial trace, with $$\operatorname{tr}\left(\operatorname{tr}_K(T)B\right)=\operatorname{tr}\left(T\left(B\otimes_2\operatorname{id}_H\right)\right)\tag3$$ for all $B\in\mathfrak L(H)$

(I guess that in 2. there is a similar construction possible for the projective tensor product; but I haven't found any reference for that.)

Now, if I'm not missing anything, there should be a third thing we can construct: If $Y$ is another $\mathbb R$-Banach space, we could define $\operatorname{tr}_{X,\:Y}:\mathfrak L(X,Y)\:\hat\otimes_\pi\:X\to Y$ as the unique linearization of the bounded bilinear operator $$\mathfrak L(X,Y)\times X\to Y\;,\;\;\;(L,x)\mapsto Lx\tag4.$$ How are these notions (especially the tensor contraction and $\operatorname{tr}_{X,\:Y}$) related?

$\endgroup$
3
  • $\begingroup$ I think 2) can be understood as a version of 1) using the projective tensor product of operator spaces, but I have not sat down to check the details $\endgroup$
    – Yemon Choi
    Commented May 21, 2020 at 6:44
  • $\begingroup$ 1) admits a version for the projective tensor product of Banach spaces, which then looks very like your (4), but I admit that right now I don't see a direct deduction of (4) from the Banach-space version of 1) $\endgroup$
    – Yemon Choi
    Commented May 21, 2020 at 6:46
  • 1
    $\begingroup$ The basic principle is that composition of two operators can be linearized to a contractive linear map ${\mathcal L}(E,F) \hat{\otimes}_\pi {\mathcal L}(F,G) \to {\mathcal L}(E,G)$. This seems to imply both 1) and 4) $\endgroup$
    – Yemon Choi
    Commented May 21, 2020 at 6:48

0

You must log in to answer this question.