Timeline for A survey for various $K$-homology theories and their relationship
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2015 at 2:54 | comment | added | user2015 | @Paul Siegel Could you give me a reference on this product ? i do not have much background on index theory or operator algebra. | |
| Oct 9, 2015 at 2:16 | comment | added | Paul Siegel | @user2015 Yes - this is a special case of the Kasparov product in KK-theory. In geometric K-homology, the Kasparov product is just the product of Dirac operators. | |
| Oct 8, 2015 at 21:45 | comment | added | user2015 | @Paul Siegel Is there an "external" product for $K$ homology? say, if i have $u\in K_n(X)$ and $v\in K_m(Y)$,is there a product "$\times$",s.t. $u\times v\in K_{m+n}(X\times Y)$ | |
| Oct 8, 2015 at 1:26 | vote | accept | user2015 | ||
| Apr 27, 2015 at 19:44 | comment | added | Paul Siegel | @user2015 If you're referring to the Baum-Connes assembly map, you can formulate the map in either the analytic or geometric model of K-homology, though I suppose the analytic model is typically used. Some of the various different approaches are formulated in this paper: arxiv.org/pdf/0907.2066v1.pdf, a follow-up to the paper I linked to in my answer. (It is also useful to formulate the assembly map using the E-theory picture of K-homology; I believe it is this model that was used to prove the Baum-Connes conjecture for A-T-menable groups.) | |
| Apr 27, 2015 at 19:35 | comment | added | Paul Siegel | @user2015 The relationship between analytic K-homology and the Bott spectrum comes down to the fact that the space of Fredholm operators on a separable Hilbert space represents the K-theory functor. One can then build a model for the K-theory spectrum by looking at spaces of Fredholm operators which anti-commute with Clifford generators (in the spirit of the multigrading structure in Kasparov's theory). This was worked out by Atiyah and Singer in this paper: maths.ed.ac.uk/~aar/papers/askew.pdf | |
| Apr 27, 2015 at 15:40 | comment | added | user2015 | @ Paul Siegel Could you make it more explicit the relation between the analytic $K$ homology and the $K$ homology defined using Bott spectrum? For countable group $\pi$,there is an assembly map by Kasparov $K_\ast(B\pi)\to K_\ast(C_r^{\ast}(\pi))$.What's the $K$-homology on the left hand side? | |
| Apr 26, 2015 at 19:47 | comment | added | Branimir Ćaćić | A possible first step towards this might be found in Hilsum's notion of bordism of unbounded KK-cycles---nothing to do with cobordism of bounded cycles---which has recently been developed further by Deeley--Goffeng--Mesland: arxiv.org/abs/1503.07398 | |
| Apr 26, 2015 at 19:35 | comment | added | Paul Siegel | (In fact it would be more elegant to build a version of the theory in which the generators are unbounded operators since Dirac operators are supposed to give the fundamental class in K-homology, but I don't think anybody knows what the relations would look like. For instance two unbounded operators could give the same K-homology class even if their domains intersect trivially.) | |
| Apr 26, 2015 at 19:32 | comment | added | Paul Siegel | @მამუკაჯიბლაძე The idea is to replace the unbounded Fredholm operator $D$ with a bounded operator which has the same index, and any operator of the form $f(D)$ where $f(t)$ is a smooth odd function on $\mathbb{R}$ which satisfies $f(t) > 0$ for $t > 0$ and $f(t) \to 1$ as $t \to \infty$ will do the job. This is explained in detail in chapter 10 of Higson-Roe. | |
| Apr 26, 2015 at 18:23 | comment | added | მამუკა ჯიბლაძე | Could you please hint at why does $F$ have this particular form? In your reference at that place they refer to another source (book by Higson and Roe). | |
| Apr 26, 2015 at 15:48 | history | answered | Paul Siegel | CC BY-SA 3.0 |