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.



Isotropic splitting for exact Courant algebroids
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."


