Timeline for index of a family of Dirac operators in $K^1$
Current License: CC BY-SA 2.5
8 events
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| Dec 20, 2018 at 17:24 | comment | added | Ho Man-Ho | What can we say when the dimension of the family of kernels of the self-adjoint elliptic operator is not a constant? It's well known that in the even dimensional fiber case, if the dimension of the kernels of the $\mathbb{Z}_2$-graded Dirac operator is not constant we can perturb the Dirac operator so that its index can be represented by the kernel bundle of the perturbed operator minus a trivial bundle. I wonder if there is any analog of this "perturbation method" to the odd dimensional fiber case. | |
| Dec 20, 2018 at 17:23 | comment | added | Ho Man-Ho | Sorry to add another question on this post. I am recently considering a question on the index of a family of self-adjoint Dirac operators in the setting of (local) family index theorem, where the dimension of the fibers of the underlying fibration is odd. Theorem 4.1 in your paper "A vanishing theorem for characteristic classes of odd-dimensional manifold bundles" is very helpful to me. My question concerns the case that when the dimension of the family of kernels of the self-adjoint Dirac operators is not constant. Since my question is a bit long, I type it in the next comment. | |
| Nov 22, 2010 at 13:10 | comment | added | Johannes Ebert | I see; on the fundamental group, you can only detect a $Z^3$-factor of $K^1 (T^3)$. For the last one, you need some information on $H^3$ indeed, which is not captured by the spectral flow. Did you try to compute the Chern character of the index, using the index formula for selfadjoint operators? | |
| Nov 22, 2010 at 11:57 | comment | added | J Fabian Meier | Sorry, I should have explained my question in more detail: I mean, you explained above what to do in the case of a 2-torus. But in the case of the 3-torus, the image of an index element in $K^1(B)$ splits into a part in $H^1(B)$ and a part in $H^3(B)$. If I understood the situation correctly I can find out about the $H^1(B)$-part by calculating three appropriate spectral flows. But what about the image of the index in $H^3(B)$? | |
| Nov 22, 2010 at 10:13 | comment | added | Johannes Ebert | Can't you do the following? By the Künneth formula (proven by Atiyah, of course), $K^* (T^3) = K^* (S^1)\otimes K^* (S^1) \otimes K^* (S^1)$. The K-theory of the circle is $K^0 (S^1)=Z$ and $K^1(S^1)=Z$. | |
| Nov 22, 2010 at 9:59 | comment | added | J Fabian Meier | Thank you very much so far! An additional question: If B happens to be a 3-torus, what might I try to calculate the "$H^3$-part" of $K^1(B)$ ? | |
| Oct 28, 2010 at 10:37 | vote | accept | J Fabian Meier | ||
| Oct 22, 2010 at 15:36 | history | answered | Johannes Ebert | CC BY-SA 2.5 |