Given a Hilbert space $H$, let $S_1(H)$ denote the space of trace-class operators on $H$, with the trace-class norm or Schatten 1-norm. That is
$$ \Vert T \Vert_1 = \sum_{j\geq 1} |s_j| $$
where $(s_1,s_2,\dots)$ is the sequence of eigenvalues of the operator $|T|=(T^*T)^{1/2})$ written in any order.
Let $H_1$ and $H_2$ be Hilbert spaces of dimention $n$ and $m$ respectively. Let $H_1\otimes_2 H_2$ denote their Hilbert-space tensor product (hence, an inner product space of dimension $nm$).
Consider the map $$\varphi:S_1(H_1)\hat{\otimes}S_1(H_2)\to S_1(H_1\otimes_2 H_2),$$ which is defined by
$$\varphi( A \otimes B) (\xi_1\otimes\xi_2) = (A\xi_1 )\otimes (B\xi_2) \quad(\xi_1\in H_1, \xi_2\in H_2) $$
Is it the case that $$\Vert\varphi^{-1}\Vert=\min\lbrace{m, n}\rbrace ?$$
Here, $\hat{\otimes}$ denotes the projective tensor product of Banach spaces.
Best Answer
Revised answer:
(Notation from p 61ff of the reference below). Note that for trace class norm we have $\mathcal{B}(H_1) = H_1 \hat\otimes H_1' = H_1\hat \otimes H_1$. For operator norm we have $\mathcal{B}(H_1) = H_1\hat{\hat\otimes} H_1'= H_1\hat{\hat\otimes} H_1$ where the tensor product is now the inductive one (also called $\epsilon$-tensor product). Using this and finite dimensionality we then can write
$\varphi:\mathcal{B}(H_1)\hat{\otimes}\mathcal{B}(H_2)\to\mathcal{B}(H_1\hat{\otimes}H_2)$ as the natural isomorphism $$ (H_1 \hat\otimes H_1)\, \hat\otimes\, (H_2 \hat\otimes H_2) \to (H_1 \hat\otimes H_2)\hat\otimes (H_1 \hat\otimes H_2)' = (H_1 \hat\otimes H_2)\hat\otimes (H_1 \hat{\hat\otimes} H_2) $$ which is $Id\,\hat\otimes\,j$ where $j:H_1 \hat\otimes H_2\to H_1 \hat{\hat\otimes} H_2$ is the embedding (here iso) of trace class norm into operator norm. The norm of $j$ is indeed $min\{\dim(H_1,\dim H_2)$, and the norm of $j^{-1}$ is its inverse, which can be seen by fixing bases and then writing any operator $A:H_1\to H_2$ as $U_2\circ D\circ U_1$ for isometries $U_i$ and a diagonal operator $D$.
Ah,
I see now that the question changed. Here is the version for the new question: $$ (H_1 \hat\otimes H_1)\, \hat\otimes\, (H_2 \hat\otimes H_2) = (H_1 \hat\otimes H_2)\, \hat\otimes\, (H_1 \hat\otimes H_2) \to (H_1 \otimes_2 H_2)\hat\otimes (H_1 \otimes_2 H_2)' = (H_1 \otimes_2 H_2)\hat\otimes (H_1 \otimes_2 H_2) $$ which is $i\hat\otimes i$ where $i: H_1 \hat\otimes H_2 \to H_1 \otimes_2 H_2$ is the natural iso. Its norm is the norm of the embedding $\ell^1\to \ell^2$ for the s-numbers.