On page 16 of Prof. Jones' online von Neumann algebras notes(you can find the book here: von Neumann algebras notes

There is an exercise 3.3.9(iv) (labeled as a harder one),

Suppose $H$ is a separable Hilbert space, show that there is no nonzero linear map $tr: B(H)\rightarrow \mathbb{C}$ satisfying $~tr(ab)=tr(ba)$.

I can only figure out that if such $tr$ exists, it must be zero on any finite rank operator by using the right shift operator $S$, but how to process from finite dimensional operators to all operators in $B(H)$ by just using the property $tr(ab)=tr(ba)$, can any one give some hint?

  • $\begingroup$ Have you done parts ii and iii? $\endgroup$ – Kevin Carlson Oct 31 '12 at 3:39
  • $\begingroup$ for part ii, just use $Tp_k\rightarrow T$, but $tr(Tp_k)=0$, where $p_k$ is the orthognormal projection on the n-dimensional subspace; while for iii, my method only works if we assume $tr(x^*x)>0$ for all nonzero operators. $\endgroup$ – ougao Oct 31 '12 at 4:07

The same idea that you use to show that $tr$ is zero on finite-rank operators, can be used to see that it is zero on any projection.

Consider a projection $q$ with infinite dimension and codimension. Consider an orthonormal basis $\{\xi_n\}$ such that $\{\xi_{2n}\}$ spans the range of $1-q$. Let $\{e_{kj}\}$ be the matrix units associated with our orthonormal basis, and let $$ x=\sum_je_{j,2j}. $$ Then $$ x^*x=\sum_je_{2j,2j}=1-q,\ \ xx^*=\sum_je_{j,j}=1. $$ So $tr(1)=tr(xx^*)=tr(x^*x)=tr(1-q)$. Then $tr(q)=0$. As $q$ was any projection with infinite dimension and co-dimension, we also have $tr(1-q)=0$. Thus $$ tr(1)=tr(q+(1-q))=tr(q)+tr(1-q)=0+0=0. $$ Now if $p$ is a projection with finite co-dimension, then $tr(1-p)=0$ (as $1-p$ is finite-rank), but then $tr(p)=0$ since $tr(1)=0$.

We have shown that $tr(p)=0$ for all projections in $B(H)$. It is well-known (see here for references) that every operator in $B(H)$ is a finite linear combination of projections. So $tr=0$.

(in part iii one can avoid this last argument: the fact that $tr$ is positive allows one to use Cauchy-Schwarz, and this together with $tr(1)=0$ can be used to see that $tr=0$: indeed, in that case for any $x$ $$ |tr(x)|=|tr(1x)|\leq tr(1)^{1/2}tr(x^*x)^{1/2}=0. $$ so $tr=0$)

  • $\begingroup$ But the key point is that why can you use $tr(\sum_{i=1}^{\infty}a_i)=\sum_{i=1}^{\infty}tr(a_i)$, to use it, you have assumed that $tr$ is weakly continouous. $\endgroup$ – ougao Oct 31 '12 at 11:18
  • $\begingroup$ I'm not using that identity, and I'm not assuming any continuity. Where do you think I'm using it? $\endgroup$ – Martin Argerami Oct 31 '12 at 14:39
  • $\begingroup$ Sorry for my careless reading of your proof and thanks to the answer. I know that I must have be unaware of some points, now I know that I have to use the matrix units in infinite dimensional, thanks for your help! $\endgroup$ – ougao Oct 31 '12 at 15:47
  • $\begingroup$ I left my first comment because I did not check the reference you gave me, since before I check the reference, my knowledge is that every operator in $B(H)$ is generated by projections, but not necessarily finite, since I only know to use the spectrum theorem to find this linear composition. Now I got it. Thanks again! $\endgroup$ – ougao Oct 31 '12 at 16:17
  • $\begingroup$ No problem. I edited the answer at the end to clarify how to solve iii without using the result about the linear combinations of projections). $\endgroup$ – Martin Argerami Oct 31 '12 at 16:27

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