Isotropic splitting for exact Courant algebroids - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T15:50:39Z http://mathoverflow.net/feeds/question/108712 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108712/isotropic-splitting-for-exact-courant-algebroids Isotropic splitting for exact Courant algebroids Benjamin 2012-10-03T15:10:14Z 2012-10-03T17:09:15Z <p>An exact Courant algebroid $E$ is one such that the sequence $0\to T^\star M\xrightarrow{\rho^\star} E^\star\simeq E\xrightarrow{\rho} TM\to 0$ is exact. Here $\rho$ is the anchor of the algebroid. Since the sequence is exact, we have a splitting $\varepsilon:TM\rightarrow E$. </p> <p>However I would like to show that we can always choose an isotropic splitting, that is, such that $\langle\varepsilon(X),\varepsilon(Y)\rangle=0$ for all $X$ and $Y$ in $TM$, where $\langle\cdot,\cdot\rangle$ is the pairing of the algebroid. </p> <p>Authors say it is always possible to find such an isotropic splitting but I don't see why. Could someone can explain this fact?</p> <p>Thank you.</p> http://mathoverflow.net/questions/108712/isotropic-splitting-for-exact-courant-algebroids/108720#108720 Answer by yael fregier for Isotropic splitting for exact Courant algebroids yael fregier 2012-10-03T16:39:44Z 2012-10-03T17:09:15Z <p>This answer, due to Roytenberg, is taken from <a href="http://arxiv.org/abs/math/9910078" rel="nofollow">http://arxiv.org/abs/math/9910078</a> page 48 (below Corollary 3.8.4)</p> <p>"First, remark that $\rho^{\star}T^{\star}M$ is isotropic. Once we have one isotropic subbundle $T^{*}M$, transversal isotropic subbundles are sections of a bundle over $M$ whose fiber is an open cell in the Grassmanian of isotropic subspaces of half dimension in a pseudo-Euclidean space of signature zero; the fiber is contractible (it is diffeomorphic to the linear space of skew-symmetric matrices), so sections always exist."</p>