bio | website | math.tamu.edu/~branimir |
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location | College Station, TX | |
age | 28 | |
visits | member for | 4 years, 2 months |
seen | 6 hours ago | |
stats | profile views | 387 |
I'm a noncommutative geometer doing a postdoc at Texas A&M, having just completed my PhD at Caltech.
Jul 29 |
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Reference request for instantons
If you don't mind a bit of noncommutative differential geometry, then you can take a look at this expository article by Landi and Van Suijlekom for an account of instantons on noncommutative (real) $4$-tori: arxiv.org/abs/hep-th/0603053 The heart of the matter is that if you apply Rieffel's strict deformation quantisation to a compact $\mathbb{T}^N$-manifold $X$, then any $\mathbb{T}^N$-equivariant object over $X$ (e.g., an equivariant vector bundle) can be deformed to an analogous noncommutative-geometric object over the deformation of $X$. |
Mar 12 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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. |