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.