# Is the space of trace class operators finitely representable in an $L^1$-space?

I am interested in knowing whether the space of trace class operators is (crudely) finitely representable in an $L^1$-space. I suspect that the answer is negative but I am unable to find any argument confirming my intuition.

As for motivation, I am working on matrix-valued versions of some inequalities coming from harmonic analysis, and I would like to know if the generalisation I seek is non-trivial, if true.

Definition: A Banach space $X$ is said to be crudely finitely representable in a Banach space $Y$ if there exists a constant $C>0$ such that every finite-dimensional subspace $V$ of $X$ is $C$-isomorphic to a subspace of $Y$, i.e. there exists an isomorphism $T: V \to T(V) \subset Y$ satisfying $\|T\|\cdot \|T^{-1}\| \leqslant C$.

• crudely finitely represented in an $L^1$-space - is there a definition? Dec 13, 2013 at 16:03
• books.google.de/… Dec 13, 2013 at 16:06
• Thanks for the comment, Marc; I've added the definition to the body of the question. Dec 13, 2013 at 16:13
• The answer is no, but right now I don't recall a proof. Dec 13, 2013 at 16:14
• BTW: A separable Banach space $X$ is crudely finitely representable in an $L_1$ space iff $X$ is isomorphic to a subspace of $L_1(0,1)$. This is proved in the Lindenstrauss-Pelczynski "Absolutely summing operators" paper. (Use ultra products to embed into an abstract $L_1$ space and quote Kakutani's representation theorem.) Dec 13, 2013 at 16:18

You can deduce that $S_1$ is not finitely crudely representable in an $L_1$ space from the paper
• Thanks a lot for this answer; it crossed my mind that local unconditional structure might be relevant. To be more specific, the result follows from theorem 2.1, because $L^1$ has local unconditional structure (as na $\mathcal{L}^1$-space) and finite cotype. Dec 14, 2013 at 10:24
• On a tangential note, do we know what are the reflexive subspaces of $S_1$? At least do we know if they are super-reflexive? Mar 15, 2014 at 20:22
The question was answered long ago. But my favorite answer (to myself) is: The trace class $S_1$ has not Analytic UMD property while commutative $L^1$ has this property. Since Analytic UMD property is a local property, it follows immediately that $S_1$ is not finitely representable in $L^1$.