Suppose that $X$ is a Banach space. How to prove that the center of the algebra $B(X)$ (the algebra of bounded operators on $X$) consists only of operators of the form $aI$, where $a$ is scalar and $I$ is identity operator?
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3 Answers
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Maybe this is hw ? Anyway, you always have rank one operators $x\mapsto L(x)v$, $v\in X$, $L\in X^*$, and any operator which commutes with all of them is easily seen to be scalar.
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The argument of BS works also in the case where $X$ is a Hausdorff locally convex space since the topological dual still separates points (by Hahn-Banach). This is enough to show that the (continuous) finite rank operators act transitively on non-zero vectors from which it follows that their center is already trivial. But then all bounded operators have at most the center of the finite rank ones and the multiples of the identity.
So if you want to go beyond trivial center you have to consider topological vector spaces which are not lc, pretty weird ones ;) I don't know the answer for $L^p$ spaces with $p < 1$...?
I guess the more interesting problem is whether there are more bounded operators than the continuous finite rank operators and the multiples of the identity...
answered Mar 17, 2011 at 14:51
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1$\begingroup$ @Stefan: There are subspaces of $L_p$, $p<1$, on which every continuous linear operator is a multiple of the identity. See MR0617557 (82j:46039) Kalton, N. J.; Roberts, James W. A rigid subspace of $L_{0}$L0. Trans. Amer. Math. Soc. 266 (1981), no. 2, 645–654. 46E30 (46A22) $\endgroup$ Commented Mar 17, 2011 at 15:42
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$\begingroup$ @Stefan : On a Banach space you have at least the operators $\lambda I + N$ with $N$ nuclear, i.e. the sum of an (operator-norm) absolutely convergent series of rank one operators. According to Gowers, it is not known if this can be all of the bounded operators, but Argyros and Haydon have constructed a space with all operators scalar + compact. See gowers.wordpress.com/2009/02/07/… for this interesting story. $\endgroup$– BS.Commented Mar 17, 2011 at 16:10
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$\begingroup$ @BS: Thanks also for that reference. Interesting indeed. Yeah, of course you can still complete the finite rank ones, so my above statement is sort of stupid :) $\endgroup$ Commented Mar 17, 2011 at 16:12
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2$\begingroup$ Take rigid $X$ and $Y$ s.t. the only operator from $X$ to $Y$ is zero and similarly from $Y$ to $X$. Then the center of $B(X \oplus Y)$ is itself and is two dimensional. Unfortunately, Kalton is no longer with us to prove that such a pair of spaces exists... $\endgroup$ Commented Mar 17, 2011 at 16:33
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$\begingroup$ Such spaces DO exist. This follows from the proof of Corollary 4.3 in the Kalton-Peck paper. Following the notation there, if $S: X/N_1 \to X/N_2$ is non zero, we see from the proof that $N_1$ must be a subspace of $N_2$. Take $N_1$ and $N_2$ both of dimension one but not equal. $\endgroup$ Commented Mar 17, 2011 at 17:09
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If I see it correctly, then take a Hamel basis $(v_i)$ with $i \in I$ of the Banach space $X$. Then the coordinate switch $T_{i,j}:v_k \mapsto \delta_{i,k}v_j$ is bounded, because finite dimensional subspaces have a complement in a Banach space. Writing down any $S$ in the centre of $\mathscr{B}(X)$ as a matrix with rows and columns indexed by $I$, commutativity of $S$ and $T_{i,i}$ shows that is must be diagonal and commutativity of $S$ and $T_{i,j}$ shows that all diagonal entries must be the same. Hence $S$ is a scalar multiple of the identity.
The essence of the argument is that finite dimensional subspaces are complemented in Banach spaces.
