# tr(ab)=tr(ba), part 2.

This is a Banach space version of Andre Henriques' question

Trace Question

for Hilbert spaces. Let $a:X\to Y$ and $b:Y\to X$ be bounded linear operators between Banach spaces s.t. $ba$ and $ab$ are both nuclear. Assume whatever approximation properties on $X$ and $Y$ that you want (say, assume that both $X^*$ and $Y^*$ have the bounded or even metric approximation property), so that the trace of $ab$ and of $ba$ are well defined. Then must $tr(ab)=tr(ba)$?

When $X$ and $Y$ are Hilbert spaces, you can find three correct proofs and one interesting but incomplete proof at the above link. None of these generalize immediately to the Banach space setting.

Caveat: I have not done a literature search or thought much about this problem, but it is natural to consider it after reading Andre's question.

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@Mark Meckes: As a student I found the result as well as the fact that the trace is well defined mysterious. Other students laughed at me, but later, after reading Grothendieck's Memoirs, I realized that I was right. As for a simple proof in finite dimensions, the flawed proof in the link works and is not too bad. –  Bill Johnson Sep 29 '11 at 2:48
I thought about similar things in my thesis-- the only reference I could find was M. Grosser, "The trace of certain commutators", Rev. Roumaine Math. Pures Appl. 34 (1989), no. 5, 413–418. ams.org/mathscinet-getitem?mr=1021948 This imposes strong conditions on $a$ and $b$ (but the proof is much too complicated, I give an easy, half-page proof in my thesis). So I suspect that the general case might be hard... –  Matthew Daws Sep 29 '11 at 12:41
@Bill: the flawed proof in the link is of course the elementary one I had in mind for finite dimensions. But I agree with you that the trace is more mysterious than is often recognized. Are you familiar with the textbook "Linear Algebra Done Right" by Axler? He defines the trace of an operator on a finite-dimensional space to be the sum of its eigenvalues, and only after that proves that it can be calculated by summing diagonal matrix entries. –  Mark Meckes Sep 29 '11 at 13:47
@Bill, where in Grothendieck's Memoirs were your suspicions of mysteriousness of the trace confirmed? –  Mariano Suárez-Alvarez Oct 6 '11 at 18:15
That the trace of the zero operator is not well defined on a space that fails the approximation property. That is, if $X$ fails the approximation property, then there are a sequence $f_n$ in $X^*$ and $x_n$ in $X$ s.t. $\sup \|f_n\| <\infty$, $\sum \|x_n\| = 1$, $\sum f_n(x_n) = 1$, yet $\sum f_n(x) x_n = 0$ for all $x$ in $X$. –  Bill Johnson Oct 6 '11 at 18:44

My question has a negative answer.

Lemma. Suppose $X$ has the approximation property (AP), $Y$ is a subspace of $X$, and $X/Y$ fails the AP. Then there is a nuclear operator $T$ on $X$ s.t. $TX\subset Y$, $T^2=0$, and $tr(T)=1$.

Suppose you have $X$, $Y$, $T$ as in the lemma and $Y$ has the AP. Define $a:X\to Y$ to be $T$ considered as an operator into $Y$ and let $b:Y\to X$ be the inclusion map. Then $ba=T$ has trace one but $ab=0$.

Experts will see immediately that you can realize the situation in the previous paragraph by letting $Z$ be a James-Lindenstrauss space s.t. $Z^{**}/Z$ fails the AP while $Z^{**}$ and $Z$ have Schauder bases. More remarkable is that you can even have $X=\ell_p$ with $1<p<2$ and $Y$ isomorphic to $\ell_p$. This was proved by A. Szankowski a couple of years ago.

The lemma is easy: Since $X/Y$ fails the AP, by Grothendieck's classical characterization of the AP there is an absolutely summable sequence $f_n$ in $(X/Y)^*$ and a sequence $z_n$ in the open unit ball of $X/Y$ s.t. for all $z\in X/Y$, $\sum \langle f_n, z \rangle z_n=0$ but $\sum \langle f_n, z_n \rangle =1$ (that is, the trace of the zero operator on $X/Y$ is not well defined). Let $Q$ be the quotient mapping from $X$ onto $X/Y$ and get $x_n$ in the unit ball of $X$ s.t. $Qx_n=z_n$. Define a nuclear operator $T$ on $X$ by

$Tx = \sum Q^*f_n(x) x_n$.

$QT=0$ because $\sum \langle f_n, z \rangle z_n=0$ for all $z\in X/Y$ and hence $TX \subset Y$.

$T_{|Y} =$ because $Q^*$ ranges in the annihilator of $Y$ in $X^*$ and hence $T^2=0$.

Finally, $tr(T)= \sum \langle Q^*f_n, x_n \rangle =\sum \langle f_n, z_n \rangle =0$.

This construction raises more questions than it answers. For what Banach spaces $X$ and $Y$ is there an affirmative answer to the trace question? The only positive result I see is when one of the spaces is a Hilbert space and the other one is a weak Hilbert space in the sense of Pisier. The affirmative answer follows because Pisier proved that the Lidskii trace formula is valid for nuclear operators on a weak Hilbert space whose eigenvalues are absolutely summable (an old result due to Konig, Maurey, Retherford and me says that on any Banach space that is not isomorphic to a Hilbert space, there is a nuclear operator whose eigenvalues are not summable, so it is not clear that the trace question has an affirmative answer when $X$ and $Y$ are both weak Hilbert spaces).

ADDED 10/24/11: The paper of Szankowski I mentioned is

Szankowski A (2009)
Three-space problems for the approximation property.
J. Eur. Math. Soc., 11(2): 273-282.

Although obvious, I should have mentioned that from the negative answer to the question for $X=Y=\ell_p$, $1<p<2$, by duality you also get a negative answer for $X=Y=\ell_p$, $2<p<\infty$.

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