Let $X$ be a Banach space. Let $B(X)$ be the space of all bounded linear operators on $X$. Does $B(X)$ have an empty character space for any $X$? I know the proof of the fact that $M_n(\mathbb{C})$ has no characters and also $B(H)$, where $H$ is a Hilbert space.
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4 Answers
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I guess that you mean that $B(H)$ has no character (=continuous unital algebra homomorphism into $\mathbf{C}$) if $H$ has dimension $\neq 1$ (idem for $M_n(\mathbf{C})$ for $n\neq 1$), and thus that your question assumes $\dim(X)\ge 2$ (and hence $=\infty$).
Argyros and Haydon (Acta Math, 2011: arXiv, Project Euclid unrestricted access) constructed a Banach space $X$ of infinite dimension in which every bounded self-operator is scalar+compact. Hence for such a space, modding out by the ideal of compact self-operators yields a character.
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$\begingroup$ Maybe I'm just reading this too early, but surely it's not true that an at-least-$2$-dimensional Banach space is always infinite dimensional? Or did you just mean that we can, so we might as well, construct an infinite dimensional example? $\endgroup$– LSpiceCommented Apr 3, 2018 at 12:54
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2$\begingroup$ @LSpice thanks: in the second case I needed to say infinite dimension. In the remaining occurrence what is meant is that any Banach space $X$ of dimension $\ge 2$ such that $B(X)$ has a character, has infinite dimension. $\endgroup$– YCorCommented Apr 3, 2018 at 13:43
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4$\begingroup$ The AH-space is a good example, and it may be worth noting (for people reading these answers) that the methods were pushed further in the thesis of M. Tarbard to produce $X$ such that $B(X)/K(X)$ is isomorphic to the convolution algebra $\ell^1({\bf Z}_+)$, hence $B(X)$ has a character space that is "big" and "nice" $\endgroup$ Commented Apr 3, 2018 at 20:07
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2$\begingroup$ @YemonChoi Thanks, I didn't know this. Arxiv link to Tarbard's paper is arxiv.org/abs/1011.4776 $\endgroup$– YCorCommented Apr 3, 2018 at 20:21
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Examples were known before the Argyros-Haydon space mentioned in Yves Cornulier's answer. For instance, if $J$ denotes the James space, then the image of the canonical map $J\to J^{**}$ has codimension $1$, and from this one can show that the closed 2-sided ideal $W(J)$ of all weakly compact operators on $J$ has codimension one in $B(J)$, thus giving us a character.
I am not sure where this was first observed: I learned of it from this paper of Loy and Willis:
MR1044280 (91f:46069) R. J. Loy, G. A Willis. Continuity of derivations on B(E) for certain Banach spaces E. J. London Math. Soc. (2) 40 (1989), no. 2, 327–346.
In the same paper, they exhibit some other examples of $E$ for which $B(E)$ admits a closed ideal of codimension $1$. For instance, one can take $K$ to be the scattered compact space obtained from the interval $[0,\omega_1]$ in the order topology, and then $E=C(K)$ has this property. (An alternative construction of the character for $B(C(K))$ can also be found in the 2014 paper of Kania, Koszmider and Lauststen, "A weak∗-topological dichotomy with applications in operator theory".) They also construct a James-type example of an $E$ such that $B(E)$ quotients on $\ell^\infty$, and hence has a lot of characters. All these examples are of a different flavour to the spaces that have been constructed more recently using the Argyros-Haydon machinery and its descendants.
At the other extreme, it is worth noting that if $X=\ell_p$ ($1\leq p\leq\infty$) or $X=L_p[0,1]$ ($1\leq p \leq\infty$) then $B(X)$ has no characters.
answered Apr 3, 2018 at 20:05
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2$\begingroup$ Trying to prove your last statement as an exercise ($\ell^p$ or $L^p$ has no character), it seems that whenever $X\simeq Y\oplus Y$ for some $Y$ then $X$ has no character [let $u,v$ be the corresponding projections: then $u,v$ are conjugate, so have the same image in any character, and since they have image 0 or 1, it follows they have image zero, but since $u+v=1$ we get a contradiction]. More generally such an argument applies if $X\simeq Y^{\oplus n}$ for some $n\ge 2$. $\endgroup$– YCorCommented Apr 3, 2018 at 20:28
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$\begingroup$ @YCor Thanks - that's a nice way of looking at it. $\endgroup$ Commented Apr 3, 2018 at 21:06
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$\begingroup$ @YCor Hey Here do you mean that if $X\simeq Y^{\bigoplus n}$ for some $n\geq 2$ then $B(X)$ has no characters? $\endgroup$ Commented Apr 4, 2018 at 7:10
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2$\begingroup$ @YCor, you are completely right. There is a more general condition preventing the existence of characters, it is called the continued bisection of the identity. You will find the definition in [Loy-Willis]. $\endgroup$ Commented Apr 4, 2018 at 12:13
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In order to complement answers given by Ycor and Yemon Choi, let me mention the space $X_M$ constructed by Mankiewicz (Isreael J. Math., 1989), which has the following remarkable properties:
- $X_M$ is separable and super-reflexive,
- $B(X_M)$ has a continuous homomorphism onto $\ell_\infty$.
Since $\ell_\infty$ has $2^{2^{\aleph_0}}$ characters, so has $B(X_M)$.
Another example of a space $X$ for which $B(X)$ has a character is G. Edgar's long James space. Recently, many spectacular separable examples of spaces whose algebras of operators admit characters have been constructed so the list goes on.
answered Apr 4, 2018 at 12:09
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I think that this question is a suitable reason to remind everyone about a very interesting (but not so well known) unpublished paper C.J. Read, "Different forms of the approximation property", Unpublished manuscript, approximate date: 1989. In this paper Read shows the existence of Banach spaces with "approximate characters". The techniques used by Read is of the same type as the techniques of Mankiewicz in the paper mentioned in the answer of Tomek Kania. The techniques was introduced by Efim Gluskin in his papers in 1981.
(I posted Read's paper on my web page because one of my papers relies heavily on Read's construction, which is not available anywhere.)
answered Apr 4, 2018 at 20:07
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$\begingroup$ Many thanks for this - I only heard about these approximate characters last year, from one of my colleagues. Does this concept of approximate characters appear also in Gluskin's work, and if so, how is it used? $\endgroup$ Commented Apr 4, 2018 at 20:59
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2$\begingroup$ I do not remember the details now. One of the ideas of Gluskin was to make finite-dimensional spaces in which every "mixing" operator has "large" norm. I think that approximate characters appeared only in later work of Mankiewicz and Read. You can look at the survey of Mankiewicz and Tomczak-Jaegermann in "Handbook of the Geometry of Banach spaces", volume 2. My favorite sources on this topic are papers of Szarek in "Acta Math." (1983, 1987), and his paper in "Transactions" (1986). $\endgroup$ Commented Apr 4, 2018 at 21:11