bio | website | math.tamu.edu/~branimir |
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location | College Station, TX | |
age | 28 | |
visits | member for | 4 years, 9 months |
seen | 2 hours ago | |
stats | profile views | 457 |
I'm a noncommutative geometer doing a postdoc at Texas A&M, having just completed my PhD at Caltech.
Mar 22 |
awarded | Announcer |
Feb 27 |
awarded | Announcer |
Jan 4 |
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Correspondences as generalized morphism between $C^*$-algebras
See, for instance, mathoverflow.net/questions/82871/… and similar discussions on MathOverflow and Math.SE. I suppose the point is that naive $\ast$-homomorphisms $A \to B$ aren't necessarily all that natural in the nonunital case, when seen through the guiding lens of Gelfand–Naimark. |
Jan 4 |
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Correspondences as generalized morphism between $C^*$-algebras
I think that the heart of the matter here is that Gelfand–Naimark duality for locally compact Hausdorff spaces is a subtle business. One way to cut the Gordian knot is Woronowicz's approach, which identifies a continuous map $f: X \to Y$ as inducing a nondegenerate $\ast$-homomorphism $f^t : C_0(Y) \to M(C_0(X))$, where $M(C_0(X))$ denotes the multiplier algebra of $C_0(X)$; indeed, if $\phi : A \to M(B)$ is a nondegenerate $\ast$-homomorphism, so that $\phi(A)B$ is dense in $B$, then the induced bimodule ${}_A B_B$ satisfies both nondegeneracy and fullness. |
Jan 4 |
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Correspondences as generalized morphism between $C^*$-algebras
I completely forgot; thanks for reminding me. There is a corresponding condition on the right, which says that $(E,E)$ should be dense in $A$; again, it holds for ${}_A A_A$ precisely because $A$ admits an approximate unit. |
Jan 3 |
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Correspondences as generalized morphism between $C^*$-algebras
In particular, observe that the “trivial line bundle” ${}_A A_A$ satisfies this condition precisely because $A$ admits an approximate unit. |
Jan 3 |
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Correspondences as generalized morphism between $C^*$-algebras
In the nonunital case, one often imposes the additional condition on ${}_A E_B$ that $A E$ be dense in $E$; this should take care of the issue in your edit. |
Jan 3 |
awarded | Announcer |
Jan 1 |
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Commutative spectral triples
...firmly from the perspective of mathematical physics. |
Jan 1 |
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Commutative spectral triples
...and hence, given a choice of spin structure, a canonical unbounded representative for the so-called fundamental $K$-homology class of your manifold. On the other hand, spin manifolds turn out to be natural to consider in the context of physics. So, I suspect it's a mixture of spectral triples' historical origins in $K$-homology and of Connes's interest in physics, already in the early and mid '90s, that resulted in a cultural association of spectral triples with noncommutative spin manifolds. This was certainly cemented by the account in Gracia-Bondia–Varilly–Figueroa, which was written... |
Jan 1 |
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Commutative spectral triples
Now to your latest question. As late as Connes's book Noncommutative Geometry (1994), spectral triples were still known by the older name of unbounded $K$-cycles, since the definition is exactly what you need to get a reasonable unbounded (in the sense of Baaj–Julg) representative for a $K$-homology class, and indeed, in $K$-homology, it turns out to be most natural to consider spin$^\mathbb{C}$ manifolds; in particular, the moment you impose a spin structure on a compact spin manifold, you automatically get an orientation, a Riemannian metric, and the spin Dirac operator... |
Jan 1 |
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Commutative spectral triples
Not just sufficient but genuinely necessary, given the proof that we have, which really uses every last condition in an essential manner; for instance, it's the orientability condition that gives you the candidate atlas. |
Jan 1 |
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Commutative spectral triples
@truebaran No, it's $A$ commutative plus a whole laundry list of highly non-trivial conditions concerning every part of the spectral triple; when I say that those spectral triples fail to be commutative, it's precisely because they fail some of those conditions. Please take a look at the introduction to Connes's paper for the full definition. |
Dec 30 |
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Commutative spectral triples not coming from manifolds
Another very interesting example, coming from a purely differential-geometric context, is Hasselmann's recent work on constructing spectral triples for Carnot manifolds: arxiv.org/abs/1404.5494. |
Dec 30 |
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Commutative spectral triples
@truebaran I've added a bit more about your Question 3; there's all sorts of things you can do to break the conditions for a commutative spectral triple, with the upshot that no meaningful characterisation is possible. Let me know if anything's unclear. |
Dec 30 |
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Commutative spectral triples
Added details about natural spectral triples for manifolds that aren't commutative spectral triples. |
Dec 30 |
revised |
Commutative spectral triples
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Dec 30 |
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Commutative spectral triples not coming from manifolds
In particular, you might also take a look at this paper, arxiv.org/abs/1112.6401, where the authors construct a family of spectral triples for the Sierpinski gasket, which, for instance, can be made to have spectral dimension the Hausdorff dimension of the Sierpinski gasket. |
Dec 30 |
revised |
Commutative spectral triples
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Dec 30 |
answered | Commutative spectral triples |