Does equality of Laplacians imply Kähler? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:46:13Z http://mathoverflow.net/feeds/question/100530 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100530/does-equality-of-laplacians-imply-kahler Does equality of Laplacians imply Kähler? Michael Albanese 2012-06-24T16:27:15Z 2012-12-12T05:39:03Z <p>This question follows on from <a href="http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible" rel="nofollow">this one</a>. </p> <p>Let $(X, \omega)$ be a Hermitian manifold and define the Laplacians $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$ and $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$. </p> <p>If $(X, \omega)$ is a K&auml;hler manifold, that is $d\omega = 0$ (or equivalently $\partial\omega = 0$ or $\bar{\partial}\omega = 0$), we have $\Delta_{\bar{\partial}} = \Delta_{\partial}$. </p> <p>More generally, on any Hermitian manifold we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]$ where:</p> <ul> <li> $[\bullet, \bullet]$ is the graded commutator;</li> <li> $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are the adjoints of wedging with the forms $\partial\omega$ and $\bar{\partial}\omega$ respectively; and</li> <li> $L$ is the Lefschetz operator, that is, wedging with $\omega$.</li> </ul> <p>It is clear how the additional terms relating the Laplacians in the Hermitian case vanish if the metric is K&auml;hler ($\partial\omega = 0$ and $\bar{\partial}\omega = 0$, so $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are both zero). What about the converse? That is:</p> <blockquote> <p>If $\Delta_{\bar{\partial}} = \Delta_{\partial}$ on a Hermitian manifold $(X, \omega)$, is it necessarily K&auml;hler?</p> </blockquote> <hr> <p>The accepted answer in the linked question refers to <i>balanced</i> manifolds. These are manifolds with the property that $\Delta_{\bar{\partial}}f = \Delta_{\partial}f$ for any smooth <i>function</i> $f$. Not all such manifolds are K&auml;hler. The above question is stronger as it requires equality for all smooth <i>forms</i>.</p> http://mathoverflow.net/questions/100530/does-equality-of-laplacians-imply-kahler/100551#100551 Answer by YangMills for Does equality of Laplacians imply Kähler? YangMills 2012-06-25T00:05:12Z 2012-06-25T00:05:12Z <p>The answer is yes, as proved by Y. Ogawa in <a href="http://projecteuclid.org/euclid.jdg/1214429279" rel="nofollow">this paper,</a> see Theorem 3.10. Apparently equality on functions and $1$-forms is enough to conclude that the metric is K&auml;hler.</p> <p>There is also a <a href="http://dx.doi.org/10.1090/S0002-9947-1966-0189067-0" rel="nofollow">related paper by C.C.Hsiung</a>, see Theorem 4.2.</p> http://mathoverflow.net/questions/100530/does-equality-of-laplacians-imply-kahler/116146#116146 Answer by Michael Albanese for Does equality of Laplacians imply Kähler? Michael Albanese 2012-12-12T05:39:03Z 2012-12-12T05:39:03Z <p>In addition to the papers mentioned by YangMills, there is also the <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=0205205&amp;loc=fromreflist" rel="nofollow">earlier paper</a> by A. W. Adler which shows that if $\Delta = 2\Delta_{\bar{\partial}}$ on a hermitian manifold, then it is Kähler.</p>