Timeline for A question about the first eigenvalue for two Kahler metrics
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 18, 2014 at 11:52 | answer | added | username | timeline score: 1 | |
| Mar 18, 2014 at 11:44 | comment | added | username | I transformed it into an answer, feel free to accept it. | |
| Mar 18, 2014 at 10:24 | comment | added | Daniel | Wonderful! Thank you for your helpful idea. | |
| Mar 18, 2014 at 9:27 | comment | added | username | Let $$ J_{\omega}(v)=\frac{\int_{M}\left|\nabla v\right|\omega^{n}}{\int_{M}v^{2}\omega^{n}-\left(\int_{M}v\omega^{n}\right)^{2}}. $$ Since $\int_{M}\omega^{n}=1$, we have \begin{eqnarray*} J_{\omega}(v) & = & \frac{\int_{M}\left|\nabla v\right|\omega^{n}}{\int_{M}\left(v-\int_{M}v\omega^{n}\right)^{2}\omega^{n}}\\ & = & J_{\omega}(v+C) \end{eqnarray*} for any constant $C$. So the constant term does not matter, as you can always define $w=v+C$ such that $\int_{M}w\omega^{\prime n}=\lambda\int_M v \omega^n$ for any convenient $\lambda$. | |
| Mar 18, 2014 at 1:19 | comment | added | Daniel | It do not work. If $\inf_M v\omega^n=0$, we may not have $\inf_M v\omega'^n=0$. | |
| Mar 17, 2014 at 19:05 | comment | added | username | Does it help if you write the denominator $\int_M (v-\int_M v \omega^n)^2\omega^n$, and then change the infinum to be over functions of zero average ? | |
| Mar 17, 2014 at 16:52 | comment | added | Daniel | Sorry, I forgot it. And I add the normalize condition of the metrics. | |
| Mar 17, 2014 at 16:50 | comment | added | Deane Yang | My immediate instinct, without going through your calculation, is to insert a factor $\int \omega^n$ to the first term in the denominator. | |
| Mar 17, 2014 at 16:48 | history | edited | Daniel | CC BY-SA 3.0 |
added 118 characters in body
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| Mar 17, 2014 at 16:43 | history | asked | Daniel | CC BY-SA 3.0 |