# Branimir Ćaćić

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bio website math.tamu.edu/~branimir location College Station, TX age 27 member for 3 years, 8 months seen 22 hours ago profile views 358

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

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 Feb16 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 $C_0(\mathbb{R})^+ \cong C_0((0,1))^+ \cong C(S^1)$. Feb4 awarded Nice Answer Feb2 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? Jan8 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? Dec30 awarded Announcer Sep11 awarded Announcer Aug5 reviewed No Action Needed Axiom of dependent choice (up to $\omega_1$) and group rank Aug2 revised unique continuation property for overdetermined elliptic PDE LaTeX and grammar cleanup Aug2 suggested suggested edit on unique continuation property for overdetermined elliptic PDE Jul26 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. Jul20 reviewed No Action Needed Level sets of Hamiltonians of S^1 actions Jul20 awarded Necromancer Jul19 awarded Custodian Jul19 reviewed No Action Needed Kernel of perturbation of biharmonic operator Jul17 answered (Preferably rare) Audio/Video recordings of famous mathematicians? Jul16 answered Uppercase Point Labels in High-School Diagrams: from Euclid? Jul15 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. Jul15 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 $S^1$. Jul4 comment Why is this operator compact? Would you happen to know, by any chance, when $0$ can be an isolated point in the spectrum of the Dirac operator on a complete but non-compact spin manifold? Because then one could still use the spectral triples convention of replacing $g(t) = |t|^{-n}$ with some $\tilde{g} \in C_0(\mathbb{R})$ with $\tilde{g}(t) = g(t) = |t|^{-n}$ for $|t|>\epsilon$ and $\tilde{g}(0) = 0$, where $\sigma(D) \cap [-\epsilon,\epsilon] = \{0\}$. Otherwise, I suppose one would have no choice but too use, rather, something like $g(t) = (1+|t|)^{-n}$ to make things work? Jul4 comment Why is this operator compact? In the literature on spectral triples, where one usually considers symmetric Dirac-type operators on compact manifolds, $|D|^{-1}$ is defined as $0$ on $\operatorname{ker}(D)$ and as $|D|^{-1}$ on $\operatorname{ker}(D)^\perp$, whilst one often finds $\langle D \rangle := \sqrt{1 + D^2}$ precisely to avoid having to fuss with this.