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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 $\|T\|\cdot \|T^{-1}\| \leqslant C$.

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  • $\begingroup$ crudely finitely represented in an $L^1$-space - is there a definition? $\endgroup$
    – Marc Palm
    Dec 13, 2013 at 16:03
  • $\begingroup$ books.google.de/… $\endgroup$
    – Marc Palm
    Dec 13, 2013 at 16:06
  • $\begingroup$ Thanks for the comment, Marc; I've added the definition to the body of the question. $\endgroup$
    – Mateusz Wasilewski
    Dec 13, 2013 at 16:13
  • $\begingroup$ The answer is no, but right now I don't recall a proof. $\endgroup$
    – Bill Johnson
    Dec 13, 2013 at 16:14
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    $\begingroup$ 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.) $\endgroup$
    – Bill Johnson
    Dec 13, 2013 at 16:18

2 Answers 2

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You can deduce that $S_1$ is not finitely crudely representable in an $L_1$ space from the paper

Pisier, Gilles Some results on Banach spaces without local unconditional structure. Compositio Math. 37 (1978), no. 1, 3–19.

However, I think that the result you want might have been known earlier. Maybe it follows from Kwapien and Pelczynski's "Main triangle projection" paper, which I do not now have at hand.

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  • $\begingroup$ 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. $\endgroup$
    – Mateusz Wasilewski
    Dec 14, 2013 at 10:24
  • $\begingroup$ 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? $\endgroup$
    – Tomasz Kania
    Mar 15, 2014 at 20:22
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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 $L^1$.

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