703 reputation
813
bio website math.tamu.edu/~branimir
location College Station, TX
age 27
visits member for 4 years, 1 month
seen 2 days ago

I'm a noncommutative geometer doing a postdoc at Texas A&M, having just completed my PhD at Caltech.


Mar
12
comment Strange (?) definition of the spectrum
In any event, the fact that any reasonable notion of spectrum can act very strangely for noncommutative $C^\ast$-algebras should rather be seen as an indication that "noncommutative topology" contains fundamentally new phenomena that necessitate new tools.
Mar
12
comment Strange (?) definition of the spectrum
The spectrum is made for the representation theory, not representation theory for the spectrum, and from the standpoint of the representation theory of $C^\ast$-algebras, unitarily equivalent representations really are equal in every meaningful way. In other areas of mathematics, moduli spaces can be extraordinarily difficult to handle, but this doesn't make them any less meaningful or interesting.
Feb
16
comment What is the character that compactifies $\mathbb{R}$ through the Gelfand transform?
Maybe I'm misreading you, but the Stone-Čech compactification of $\mathbb{R}$ can't possibly be the one-point compactification, since $C_b(\mathbb{R})$ isn't isomorphic to $C_0(\mathbb{R})^+ \cong C_0((0,1))^+ \cong C(S^1)$.
Feb
4
awarded  Nice Answer
Feb
2
comment Computing noncommutative geometries
What you're probably looking for is deformation quantization, for which there are several methods appearing in the literature. In a specifically operator-algebraic context, what you might want to use is Rieffel's strict deformation quantisation: see, for instance, these survey articles by Rieffel himself: math.berkeley.edu/~rieffel/papers/deformation.pdf math.berkeley.edu/~rieffel/papers/quantization.pdf For instance, the noncommutative torus can be very nicely obtained from the usual torus through strict deformation quantisation. What context are you working in, anyway?
Jan
8
comment When is the dual module isomorphic to conjugate module of a *-algebra
Is $\mathcal{M}$ and $A$-bimodule, and if so, is $\mathcal{M}^\ast = \operatorname{Hom}_{A \otimes A^o}(M, A)$ (i.e., simultaneously left and right $A$-linear functionals), or what?
Dec
30
awarded  Announcer
Sep
11
awarded  Announcer
Aug
5
reviewed No Action Needed Axiom of dependent choice (up to $\omega_1$) and group rank
Aug
2
revised unique continuation property for overdetermined elliptic PDE
LaTeX and grammar cleanup
Aug
2
suggested suggested edit on unique continuation property for overdetermined elliptic PDE
Jul
26
comment Structure theorem for finite dimensional $C^*$-algebras and their representations
I have Farenick's book, and it definitely constructs the theory of finite-dimensional operator algebras in complete detail; if I recall correctly, he even puts the real case on equal footing with the complex case.
Jul
20
reviewed No Action Needed Level sets of Hamiltonians of S^1 actions
Jul
20
awarded  Necromancer
Jul
19
awarded  Custodian
Jul
19
reviewed No Action Needed Kernel of perturbation of biharmonic operator
Jul
17
answered (Preferably rare) Audio/Video recordings of famous mathematicians?
Jul
16
answered Uppercase Point Labels in High-School Diagrams: from Euclid?
Jul
15
comment Is a circle action on M_n necessarily inner?
The obstruction to this lift being a group homomorphism, then, will be precisely a $U(1)$-valued $2$-cocycle on $S^1 (=U(1))$, so if you can figure out the group cohomology $H^2(S^1,U(1))$, you should have your answer one way or another. I'm afraid, though, that I know nothing about the computation of group cohomology.
Jul
15
comment Is a circle action on M_n necessarily inner?
If I'm not mistaken, the datum of your continuous action $\alpha : S^1 \to \operatorname{Aut}(M_n(\mathbb{C}))$ should be equivalent to the datum of a continuous projective unitary representation $\beta : S^1 \to PU(n)$, via $\alpha(\theta)(T) = \beta(\theta)T\beta(-\theta)$; your action $\alpha$ is inner if and only if $\beta$ lifts to a continuous unitary representation $\tilde{\beta} : S^1 \to U(n)$ of $S^1$ on $\mathbb{C}^n$. I don't know, though, off the top of my head, whether or not the obstruction to such a lift necessarily vanishes for $S^1$.