Timeline for A question about flat connection
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2016 at 8:01 | comment | added | Ho Man-Ho | Let us continue this discussion in chat. | |
| Feb 21, 2016 at 19:25 | comment | added | Sebastian Goette | @indextheory Actually, the Bismut-Lott torsion form depends on the flat connection, see Astérisque 275, Thm 3.45. Feel free to contact me per email if you need more details. | |
| Feb 21, 2016 at 18:20 | comment | added | Ho Man-Ho | I mean if we look at the statement of the RRG for two flat connections, then the difference between $\displaystyle\textrm{Im(CCS)}(H(Z, F|_Z), \nabla_k^{H(Z, F|_Z)})-\int_{X/B}e(TZ, \nabla^{TZ})\cup\textrm{Im(CCS)}(F, \nabla_k^F)$ for $k=0, 1$ is the exterior differential of the same $T$. And I don't know whether it makes sense to say that the path joining two flat connections is not necessarily flat makes RGG theorem to be harder to prove than other secondary index theorems. In the above comment I meant $\tilde{\eta}$ also depends on $\nabla^E$, although it doesn't have to be flat. | |
| Feb 21, 2016 at 18:13 | comment | added | Ho Man-Ho | Thanks, I see. Actually I really need $\tilde{\nabla}$ to be a flat connection. The real question behind this is the Riemann-Roch-Grothendieck theorem for complex flat vector bundle by Bismut-Lott. I know that you also have some work on it. In the imaginary part of this theorem, it is very interesting to me that the real analytic torsion form does not depend on the flat connection on $F\to X$, in contrast to the Bismut-Cheeger eta form which depends on more data. | |
| Feb 21, 2016 at 17:41 | comment | added | Sebastian Goette | It is an instructive exercise to compute the number $\tilde{\mathrm{ch}}(\nabla^0,\nabla^1)[S^1]$ explicitly for two flat connections on $S^1\times\mathbb C\to S^1$. | |
| Feb 21, 2016 at 17:33 | comment | added | Sebastian Goette | Yes, you can, if $\tilde\nabla$ itself is not flat. Easiest example: take $X=S^1$, then all connections are flat. But you can distinguish them by their holonomy, which can be any element in $GL(r,\mathbb C)$, where $r$ is the rank. If the holonomies of $\nabla_0$ and $\nabla_1$ are not conjugate, the flat connections are not isomorphic. Higher dimensional examples can be constructed using paths in the representation variety of $\pi_1(X)$. If $\tilde\nabla$ is flat, then all $\nabla_t$ are isomorphic (use parallel translation in $t$-direction to see this). | |
| Feb 21, 2016 at 17:17 | history | edited | Sebastian Goette | CC BY-SA 3.0 |
Typo corrected
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| Feb 21, 2016 at 17:14 | comment | added | Ho Man-Ho | First of all thank you for your answers. Let me change my question a little bit. Let say I have only one flat connection $\nabla$ on $E\to X$. Can I construct a flat connection $\tilde{\nabla}$ on $p^*E\to X\times[0, 1]$ such that $\tilde{\nabla}|_{X\times 0}=\nabla$ and $\tilde{\nabla}|_{X\times t}$ is any arbitrary flat connection for $0<t\leq 1$ and $\tilde{\nabla}|_{X\times 1}$ is not $\nabla$? That means I just want to specify one flat connection. | |
| Feb 21, 2016 at 16:48 | history | answered | Sebastian Goette | CC BY-SA 3.0 |