39,108 reputation
32974
bio website uregina.ca/science/mathstat/…
location Regina, Canada
age 46
visits member for 2 years, 7 months
seen 3 mins ago

I'm an Operator Algebraist (which interestingly makes me an analyst!)


Jul
9
answered non-faithful KMS states
Jul
9
comment No trace on $B(H)$ if $H$ is infinite dimensional
Yes, of course. I work at the University of Regina. And thanks because I had not noticed that the link was not working; I have replaced it with the good one.
Jul
9
comment No trace on $B(H)$ if $H$ is infinite dimensional
Please see the edit.
Jul
9
revised No trace on $B(H)$ if $H$ is infinite dimensional
added 48 characters in body
Jul
9
comment Ultra weakly continuous trace on a von Neumann Algebra
You can probably do with strongly continuous on bounded sets. This equivalent to ultraweak continuity. You can find a proof in chapter 7 of Kadison-Ringrose (most likely section 7.2).
Jul
9
answered No trace on $B(H)$ if $H$ is infinite dimensional
Jul
8
answered Suppose that $R$ is a commutative ring and $|R|=30$. If $I$ is an ideal of $R$ and $|I|=10$, prove that $I$ is maximal ideal
Jul
7
answered A question on a lemma about the product map
Jul
7
answered Is $\mathbb{C}$ equal to $\mathbb{R}^2$?
Jul
6
revised Volterra Operator is compact but has no eigenvalue
added 25 characters in body
Jul
6
comment Examples of hyperstonean space
What you want to do is not intuitive at all. The $Y$ you are looking for is called the hyperstonean cover of $X$. There are a few papers about it.
Jul
5
answered Examples of hyperstonean space
Jul
5
comment Unique trace on a type $II_1$ von Neumann Algebra
You are welcome! I love this stuff :)
Jul
5
answered Lifting a unitary to a partial isometry
Jul
5
answered Unique trace on a type $II_1$ von Neumann Algebra
Jul
4
answered Minimal projections and Type II von Neumann Algebras.
Jul
4
answered Where is commutativity of $b$ needed?
Jul
3
comment Ultra weakly closed *-subalgebra of B(H)
I see your point now. If you look at the first part of my answer, it shows that $M=p_KMp_K$, so you can consider $M\subset B(K)$ where it is a von Neumann algebra. Then it contains $I_{B(K)=p_K$.
Jul
3
comment 2-norm of a canonical Jordan form and spectral radius
Computations are easier if you take $A=\begin{bmatrix}0&1\\0&0\end{bmatrix} $. Then the spectral radius is zero and the norm is one.
Jul
2
comment Ultra weakly closed *-subalgebra of B(H)
Yes, if it is non-degenerate. In that case, the largest projection you are looking for is the identity. In general, it will be the identity of $M$, which may or may not agree with the identity of $B(H)$.