Timeline for A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$
Current License: CC BY-SA 4.0
12 events
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| May 6 at 12:48 | comment | added | Jeffrey Case | @Nikolai Yes, his argument is incorrect, though it is true that if $\xi$ is harmonic, then $\xi \wedge \omega^{m-1}$ is harmonic (Proof: $\xi \wedge \omega^{m-1} = L^{m-1}\xi$ and $\Delta$ commutes with $L$). For an example of two harmonic forms whose product is not harmonic, consider a nonzero harmonic $(0,1)$-form $\xi$ on a compact Riemann surface $\Sigma_g$ of genus $g \geq 2$. Then $\star\xi$ is also harmonic, but $\xi \wedge \star\xi$ is necessarily nonzero but vanishes somewhere, and hence (being nonconstant) cannot be harmonic. | |
| May 6 at 11:15 | comment | added | Nikolai | I see, so there is a flaw in Siu's argument on $\xi \wedge \omega^{m-1}$ being harmonic? @JeffreyCase | |
| May 6 at 10:40 | history | edited | Jeffrey Case | CC BY-SA 4.0 |
added 459 characters in body
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| May 6 at 10:32 | comment | added | Jeffrey Case | @Nikolai Yes, that was what I intended to write in my previous comment. In Siu’s lecture notes he assumes that the complex manifold is compact. In this setting, harmonic is equivalent to being $\bar\partial$- and $\smash{\bar\partial}^*$-closed, and so one gets the statement “if $\eta$ is a closed $(1,1)$-form, then $\eta$ is harmonic if and only if $\Lambda\eta$ is locally constant”. The converse of your original statement is false without compactness: On $\mathbb{C}$, take $\eta = zL1$. Then $z=\Lambda\eta$ is harmonic but not constant, yet $\eta$ is harmonic. | |
| May 5 at 19:11 | comment | added | Nikolai | Not sure I follow anymore. Did you meant to say that "The wedge product of two harmonic forms need not be harmonic, even if one of the factors is parallel"? Also the argument I gave relies on compactness and connectedness, but I suppose the other direction which you gave did not, even in the first place. There should be a way to prove the converse without resorting to compactness and connectedness and I thought this came from parallelity of $\omega$, but guess not? @JeffreyCase | |
| May 5 at 10:36 | vote | accept | Nikolai | ||
| May 4 at 23:28 | comment | added | Jeffrey Case | Sorry, I made a mistake in my last comment. The wedge product of two harmonic forms need not be harmonic, even if one of the factors is harmonic. I’ve also edited my answer to address this and to make sure my answer isn’t implicitly assuming $X$ is compact and connected, as in David Speyer’s comment to your question. | |
| May 4 at 23:24 | history | edited | Jeffrey Case | CC BY-SA 4.0 |
Corrected mistakes
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| May 4 at 21:34 | comment | added | Jeffrey Case | No, the wedge product of harmonic forms need not be harmonic. But it is if one of those forms is parallel. | |
| May 4 at 20:04 | comment | added | Nikolai | Thank you for this. About your last comment, in general is the wedge product of a harmonic form with a parallel form harmonic again? @jeffrey-case | |
| May 4 at 14:39 | comment | added | LSpice |
TeX note: You probably want the first, rather than the second, of $\smash{\bar\partial}^* \bar\partial^*$ \smash{\bar\partial}^* \bar\partial^* (note the height of the $*$), but I didn't edit in case you prefer it the latter way.
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| May 4 at 11:37 | history | answered | Jeffrey Case | CC BY-SA 4.0 |