bio | website | math.tamu.edu/~branimir |
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location | College Station, TX | |
age | 28 | |
visits | member for | 4 years, 7 months |
seen | 2 hours ago | |
stats | profile views | 439 |
I'm a noncommutative geometer doing a postdoc at Texas A&M, having just completed my PhD at Caltech.
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
added 744 characters in body |
Dec 30 |
answered | Commutative spectral triples |
Dec 3 |
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Geometrical interpretation of a Schrödinger operator
That is the correct form. From a purely mathematical perspective, your operator is a sort of truncation of the Bochner Laplacian, so that the difference, from one perspective, is just the price you pay for the truncation. On the other hand, if you really take $g$ seriously as a Riemannian metric, and if that difference really is a scalar function, then I guess, if nothing else, you can think of $H = \nabla^\ast \nabla + V$, where the difference $V$ is treated as a scalar potential? I must admit I'm only culturally semi-literate in these particular matters, so take this with a grain of salt. |
Dec 2 |
awarded | Yearling |