This question is a follow-up to this recent question and related to that one.

Is there an easy example of an (infinite-dimensional) Banach space $X$ and a non-empty compact set $K \subset \mathbb{C}$ that can't be the spectrum of a bounded operator $A: X \to X$?

Of course, if $X$ contains an infinite-dimensional Hilbert space as a norm one complemented subspace and $\emptyset \neq K \subset \mathbb{C}$ is an arbitrary compact set then we can produce an operator $A: X \to X$ such that $\sigma(A) = K$.

As Jonas Meyer pointed out in his answer, the recent breakthrough by Argyros and Haydon settling the long-standing scalar-plus-compact problem (see Gowers's blog entry for some background) shows that there is a space with the property that the only possbile spectra of bounded operators are the countable and compact subsets of $\mathbb{C}$ with at most one accumulation point. (Update: Jonas Meyer has added further information and pointers to the literature to his answer, I'll refrain from repeating this information here since I couldn't add anything of interest.) But these examples are definitely far more involved than I would like them to be. If it turns out that the example has to be so difficult for some reason that eludes me, I'd like to hear about that, too.

More optimistically, one might ask:

Are there known classes of infinite-dimensional Banach spaces for which there is a characterization of the compact subsets of $\mathbb{C}$ that may arise as spectra of bounded operators?

A nice answer to this optimistic question would be: The class of such-and-such Banach spaces has the property that only/precisely the, say, totally disconnected compact subsets of $\mathbb{C}$ arise as spectra of bounded operators. Update: In view of the question in the title, I'm of course most interested in answers that exclude certain compact subsets of $\mathbb{C}$.

Update 2: I asked a slightly updated version of this question on MathOverflow.

  • $\begingroup$ To avoid trivialities, you want to specify that $X$ is infinite-dimensional. $\endgroup$ – Robert Israel Apr 28 '11 at 16:56
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    $\begingroup$ A sufficient condition for all nonempty compact $K$ to be spectra of operators is that $X$ has an unconditional basis, because if $\{\lambda_n\}$ is a sequence dense in $K$ and $\{x_n\}$ an unconditional basis, you can define $A$ by $A \sum_n c_n x_n = \sum_n \lambda_n c_n x_n$. $\endgroup$ – Robert Israel Apr 28 '11 at 17:19
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    $\begingroup$ Hereditarily indecomposable spaces also have the property that operators have countable spectrum with at most one accumulation point, and these were known before Argyros and Haydon's result. This doesn't meet the easiness criterion, so instead of answering here I just mentioned it (with references) in an addendum to my answer to Jack's question. $\endgroup$ – Jonas Meyer Apr 28 '11 at 20:08
  • $\begingroup$ @Robert: Thanks a lot for your comments. The first one is of course just a carelessness on my part. The second point is a good one, I should have looked at Lindenstrauss-Tsafriri more closely before posting. $\endgroup$ – t.b. Apr 29 '11 at 0:03
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    $\begingroup$ @Jonas: I now asked the question on MO. (The link is the first one in the Update 2 of the question) $\endgroup$ – t.b. May 29 '11 at 8:38

Jonas asked me to add an answer to avoid letting the bounty wither away. Unfortunately, the question on MO did not lead to a conclusive answer so far. However, Bill Johnson — a world renowned expert on the theory of Banach spaces — left the following two comments (I quote):

  1. I am not aware of an example before Gowers-Maurey of a Banach space such that not every compact subset of the plane is the spectrum of an operator on the space.
  2. To realize every compact subset of the plane as the spectrum of some operator on a space, it is enough that the space has a complemented subspace with an unconditional basis. Before Gowers-Maurey, there were known to exist separable spaces (such as the Kalton-Peck twisted sum of two Hilbert spaces) such that no complemented subspace has an unconditional basis.

From comment 1. I feel that it is safe to conclude:

The question does not seem to have a simple positive answer. If it should, rather surprisingly, happen to have one, it probably hasn't been found or noticed so far.

Let me quickly indicate my thoughts on comment 2.

First of all, Bill is referring to the twisted sum construction introduced in the paper

MR0542869 (82g:46021) Kalton, N. J.; Peck, N. T., Twisted sums of sequence spaces and the three space problem. Trans. Amer. Math. Soc. 255 (1979), 1–30.

Now, as Bill notes, a necessary condition for $X$ to possibly exclude certain compact sets as spectra is the absence of any complemented infinite-dimensional subspace admitting an unconditional basis. One of the corollaries of Kalton-Peck's work (precisely their corollary 6.9) is that the twisted sum $Z_{p}$ of two $\ell^{p}$-spaces for $1 \lt p \lt \infty$ does not admit such a complemented subspace. As the paper by Kalton-Peck is well writtan and it's getting late here, I'm not entering into details now.

Thus, $Z_{p}$ would be a possible candidate. However, it is not obvious to me at all to decide whether this actually yields the desired example or not. Let me mention that the Kalton-Peck construction has a very nice homological algebra interpretation in terms of non-triviality of Yoneda-Exts of Banach spaces and explicit constructions using what they call quasi-linear maps. I will elaborate in the next few days as I still need to check some details.

I'm accepting this answer in order to close this question here and because I have little doubt that this question will remain unanswered for quite a bit. If you should happen to know anything of interest, please leave a comment here or answer the question on MO. Any input will be very very welcome, of course.

  • $\begingroup$ Jonas: As it's getting late here, I simply posted a draft with some thoughts I had over the last few days. I'll elaborate in the next few days. Thank you for your interest and I hope you're not too disappointed in what I produced now. $\endgroup$ – t.b. Jun 2 '11 at 23:34
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    $\begingroup$ Theo: Thanks! This is more than worthy of the bounty. $\endgroup$ – Jonas Meyer Jun 2 '11 at 23:38
  • $\begingroup$ @Jonas: Thank you! As I said, I have a few ideas that I would like to explain in somewhat more detail, but most certainly they won't lead to an answer of this question. I'll notify you as soon as this is ready. Best wishes, $\endgroup$ – t.b. Jun 2 '11 at 23:50

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