Timeline for Hodge theory, conformal manifolds and Fredholm modules-understanding the proof of one Lemma
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2019 at 0:51 | comment | added | truebaran | ... $Sh'$ is of the form $\frac{1+\gamma}{2}d\alpha$ thus $h'+Sh'=d\alpha$ and $\omega+S\omega=h+d\alpha$. I don't see how to get rid off this harmonic piece $h$. Maybe in the formulation there should be $S$ restricted to the forms orthogonal to harmonic forms instead of the full domain? | |
| Jan 21, 2019 at 0:48 | comment | added | truebaran | Thank you very much for your answer, now I understand it much better. However still I don't see why the image of the exterior derivative is the graph of $S$. To be more precise: let $\omega=h+h'$, $\omega \in \mathcal{H}_0^{-}$ where $h$ is harmonic and $h'$ is orthogonal to harmonic forms. Then $S\omega \in \mathcal{H}_0^{+}$ thus the graph of $S$ being $\{(\omega,S\omega): \omega \in \mathcal{H}_0^{-}\}$ may be identified with $\{\omega+S\omega: \omega \in \mathcal{H}_0^{-}\}$ and $\omega+S\omega=h+h'+Sh+Sh'=h+h'+Sh'$. Now $h'$ is of the form $\frac{1-\gamma}{2} d \alpha$ while... | |
| Jan 14, 2019 at 1:09 | history | bounty ended | truebaran | ||
| Jan 14, 2019 at 1:08 | vote | accept | truebaran | ||
| Jan 13, 2019 at 14:31 | history | answered | ubik | CC BY-SA 4.0 |