Expression for codifferential in terms of interior product

Question

Let $(M^n,g)$ be a Riemannian manifold with local orthonormal frame $\{e_1,\ldots,e_n\}$ with dual basis $\{e^1,\ldots,e^n\}$ and with Levi-Civita connection $\nabla$. It can be checked on basis that $d \colon \Omega^k M \to \Omega^{k+1} M$ can be represented in the form $$ d = \sum_{i=1}^n e^i \wedge \nabla_{e_i}. $$ I have to show that the coddiferential $\delta \colon \Omega^{k+1} M \to \Omega^k M$ can be represented in the form $$ \delta = -\sum_{i=1}^n e_i \, \lrcorner \, \nabla_{e_i}. $$ We have the following formulas: $$ \langle X \, \lrcorner \, \omega, \tau \rangle = \langle \omega, X^\flat \wedge \tau \rangle, \quad X \in TM, \; \omega \in \Lambda^k M, \; \tau \in \Lambda^{k-1} M, $$ and $$ e^i \wedge \nabla_{e_i} \tau = \nabla_{e_i} ( e^i \wedge \tau) - (\nabla_{e_i} e^i) \wedge \tau. $$ Using these formulas we find that $$ \langle e^i \wedge \nabla_{e_i} \tau, \omega \rangle = \langle \nabla_{e_i} (e^i \wedge \tau), \omega \rangle - \langle (\nabla_{e_i} e^i) \wedge \tau, \omega \rangle \\ = - \langle \tau, e_i \, \lrcorner \, \nabla_{e_i} \omega \rangle - \langle \tau, (\nabla_{e_i} e^i)^\sharp \, \lrcorner \, \omega \rangle. $$ Hence, the last scalar product must always vanish, but I'm not sure that it is always true. Am I missing something?

Answer 1

The problem is in the last line of your computation. You seem to be using $$ \langle \nabla_{e_i} (e^i \wedge \tau), \omega \rangle = - \langle e^i \wedge \tau, \nabla_{e_i} \omega \rangle . $$ All of your earlier formulas involving $e^i$ or $e_i$ are local formulas, which make sense only in the domain of the orthonormal frame $(e_i)$. But this one is not true locally; in order to move the $\nabla_{e_i}$ from one side of the inner product to the other, you need to invoke integration by parts (or, more precisely, the divergence theorem). But since these expressions are not globally defined, this doesn't work. The problem is that neither $e^i\wedge \tau$ nor $\nabla_{e_i}\omega$ has any invariant global meaning independent of the frame.

You can try to make sense of this globally by defining a global vector field $$ T = \langle e^i\wedge \tau,\, \omega\rangle e_i. $$ (You can check that transforming $(e_i)$ to another orthonormal frame $\widetilde e_j = A^i_j e_i$ leaves this expression unchanged.) The divergence theorem shows that $\int_M(\operatorname{div}T)\,dV_g=0$. But the divergence of $T$ is not equal to $\nabla_{e_i}\langle e^i\wedge \tau,\, \omega\rangle$ in general: there's an additional term involving the connection coefficients (Christoffel symbols) for the frame $(e_i)$.

However, the following formula is true pointwise: $$ \operatorname{div} T = \langle e^i \wedge \nabla _{e_i}\tau,\, \omega\rangle + \langle \tau,\, e_i \, \lrcorner \, \nabla_{e_i}\omega\rangle,\tag{1} $$ and when you integrate this, you get the formula you want.

To prove $(1)$, note first that each of the two terms on the right-hand side is well-defined globally, independently of the choice of frame (again, by computing the transformation law under a change of frame). Thus we can prove the identity at a given point $p\in M$ by choosing any convenient frame in a neighborhood of $p$. It's always possible to choose an orthonormal frame such that $\nabla_{e_i} e_j=0$ at $p$ for each $i$ and $j$, and thus the connection coefficients all vanish at $p$. (Start with any orthonormal frame at $p$, and extend it to be parallel along radial geodesics starting at $p$.) In this frame, you can do the computation you were trying to do, and the troublesome terms are all zero.