tr(ab)=tr(ba), part 2. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:34:58Z http://mathoverflow.net/feeds/question/76604 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76604/trabtrba-part-2 tr(ab)=tr(ba), part 2. Bill Johnson 2011-09-28T06:14:08Z 2011-10-24T12:59:30Z <p>This is a Banach space version of Andre Henriques' question</p> <p><A HREF="http://mathoverflow.net/questions/76386/trab-trba" rel="nofollow"> Trace Question </A></p> <p>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 <code>$X^*$</code> and <code>$Y^*$</code> have the bounded or even metric approximation property), so that the trace of $ab$ and of $ba$ are well defined. Then must $tr(ab)=tr(ba)$?</p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/76604/trabtrba-part-2/78920#78920 Answer by Bill Johnson for tr(ab)=tr(ba), part 2. Bill Johnson 2011-10-23T19:24:38Z 2011-10-24T12:59:30Z <p>My question has a negative answer.</p> <p><strong>Lemma.</strong> 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. $TX\subset Y$, <code>$T^2=0$</code>, and $tr(T)=1$.</p> <p>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 $ab=0$.</p> <p>Experts will see immediately that you can realize the situation in the previous paragraph by letting $Z$ be a James-Lindenstrauss space s.t. <code>$Z^{**}/Z$</code> fails the AP while <code>$Z^{**}$</code> and $Z$ have Schauder bases. More remarkable is that you can even have $X=\ell_p$ with <code>$1&lt;p&lt;2$</code> and $Y$ isomorphic to $\ell_p$. This was proved by A. Szankowski a couple of years ago.</p> <p>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 <code>$(X/Y)^*$</code> 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 </p> <p><code>$Tx = \sum Q^*f_n(x) x_n$</code>.</p> <p>$QT=0$ because $\sum \langle f_n, z \rangle z_n=0$ for all $z\in X/Y$ and hence $TX \subset Y$.</p> <p>$T_{|Y} =$ because <code>$Q^*$</code> ranges in the annihilator of $Y$ in <code>$X^*$</code> and hence <code>$T^2=0$</code>.</p> <p>Finally, <code>$tr(T)= \sum \langle Q^*f_n, x_n \rangle =\sum \langle f_n, z_n \rangle =0$</code>.</p> <p>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). </p> <p>ADDED 10/24/11: The paper of Szankowski I mentioned is</p> <p>Szankowski A (2009)<br> Three-space problems for the approximation property.<br> J. Eur. Math. Soc., 11(2): 273-282.</p> <p>Although obvious, I should have mentioned that from the negative answer to the question for $X=Y=\ell_p$, <code>$1&lt;p&lt;2$</code>, by duality you also get a negative answer for $X=Y=\ell_p$, <code>$2&lt;p&lt;\infty$</code>.</p>