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According to many sources I've looked through, part of the definition of trace-class operators is that they be compact. What's the need for this caveat? Why not just look at all those operators $T$ for which $\sum_{i \in I} \langle T e_i,e_i \rangle$ converges absolutely for some orthonormal basis $\{e_i\}_{i \in I}$? This is the definition of trace-operator with the compactness assumption dropped, right?

As an aside, is there an example of an operator $T$ that is not compact yet $\sum_{i \in I} \langle T e_i,e_i \rangle$ converges absolutely for some orthonormal basis $\{e_i\}_{i \in I}$

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  • $\begingroup$ @qbert Oh, so being trace-class implies that the operator is compact? $\endgroup$
    – user193319
    Nov 5, 2019 at 0:35
  • $\begingroup$ @qbert So, do we lose anything if we drop the compactness assumption? That is, is the set of operators $T$ such that $\sum_{i \in I} \langle Te_i,e_i \rangle$ absolutely converges for some ONB $\{e_i\}_{i \in I}$ still of any interest, or does the set of trace-class operators so defined lose nice properties? $\endgroup$
    – user193319
    Nov 5, 2019 at 1:06

2 Answers 2

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Let $\pi$ be any permutation of positive integers such that $\pi (i) \neq i$ for any $i$. Then $T(e_i)=e_{\pi (i)}$ gives you an isometric isomorphism with $\langle Te_i , e_i \rangle =0$ for all $i$. This $T$ is surely not compact.

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Your definition is not correct. It would be correct (but it is not trivial) if you require that $\sum_j|\langle Te_j,e_j\rangle|<\infty$ for all orthonormal bases $\{e_j\}$. Here is an example of a compact, not trace-class operator such that all terms in your sum are zero for a certain orthonormal basis.

The usual definition is that $T$ is trace class if $\sum_j\langle |T|e_j,e_j\rangle < \infty$ for some (and then all!) orthonormal basis $\{e_j\}$. Here $|T|=(T^*T)^{1/2}$. In particular $T$ is trace-class if and only if $|T|$ is. As $|T|$ is positive, it is easy to see that if it is trace-class then it is compact. Then the polar decomposition and the fact that the trace-class operators form an ideal gives you that $T$ is compact if it is trace-class.

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  • $\begingroup$ Ah, so compactness $T$ is a consequence of of $T$ being trace-class! Very nice! $\endgroup$
    – user193319
    Nov 5, 2019 at 2:00

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