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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 $\varepsilon:TM\rightarrow E$.

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.

Authors say it is always possible to find such an isotropic splitting but I don't see why. Could someone can explain this fact?

Thank you.

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1 Answer 1

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This answer, due to Roytenberg, is taken from http://arxiv.org/abs/math/9910078 page 48 (below Corollary 3.8.4)

"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."

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Thank you! Finally it is not as easy as I thought... –  Benjamin Oct 4 '12 at 5:31

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