Definition of formal adjoint of covariant derivative

Question

I read in Einstein Manifolds, L. Besse that the covariant derivative $D: \mathcal{J}^{(r,s)}(M)\to \Omega^1(M)\otimes\mathcal{J}^{(r,s)}(M)$ admit an formal adjoint $D^*:\Omega^1(M)\otimes\mathcal{J}^{(r,s)}(M) \to \mathcal{J}^{(r,s)}(M)$ as follows: $$D^*T(X_1,\cdots,X_r)=-tr \{(X,Y)\to D_XT(Y,X_1,\cdots,X_r)\}$$ where $X_i$s are vector field and $T\in \Omega^1(M)\otimes\mathcal{J}^{(r,s)}(M)$.

In general, for every linear differential operator(LDO) $D:\Gamma(E)\to\Gamma(F)$ where $E$ and $F$ are vector bundle over $M$, admit a formal adjoint operator $D^*:\Gamma(F)\to\Gamma(E)$ defined by: $$\int_M\left<D\xi,\eta\right>dvol_g=\int_M\left<\xi,D^*\eta\right>dvol_g$$ where $\xi\in\Gamma(F), \eta\in\Gamma(E) $ with compact support. (Difinition 2.2)

Question: Which of the above difinition is related to the following equation: $$\nabla^*\circ \nabla=-tr\nabla^2$$ and how can I prove this equation?

Related: Formal adjoint of the covariant derivative

Answer 1

How to obtain the first definition from the second

Let $E$ be the bundle $\mathcal{J}^{r,s}$ and $F$ be the bundle $\Omega^1 \times \mathcal{J}^{r,s}$. The precise definition of $\mathcal{J}^{r,s}$ doesn't matter that much here, but for concreteness let's say it is the set of rank $(r,s)$ tensor fields over $M$ on which the metric inner product extends as $\langle\cdot,\cdot\rangle$. More precisely, if $\{e_i\}$ is a frame of orthonormal vectors and $\{f_i\}$ a frame of orthonormal covectors, we have, for $W,Z\in \Gamma(\mathcal{J}^{r,s})$ $$ \langle W, Z\rangle = \sum_{i_1, \ldots, i_r, j_1, \ldots, j_s = 1}^{\dim M} W(e_{i_1}, \ldots, e_{i_r}, f_{j_1} , \ldots, f_{j_s}) Z(e_{i_1}, \ldots, e_{i_r}, f_{j_1} , \ldots, f_{j_s}) $$

The inner product can be defined on $F$ (aka $\Omega^1 \times \mathcal{J}^{r,s}$) analogously.

Now, let $W \in \Gamma(\mathcal{J}^{r,s})$ and $Z\in \Gamma(\Omega^1 \times\mathcal{J}^{r,s})$, the definition requires $\nabla^*$ satisfies

$$ \int \langle \nabla W, Z \rangle ~\mathrm{dvol}_g = \int\langle W, \nabla^* Z\rangle ~\mathrm{dvol}_g \tag{1}$$

for any choice of $W$ and $Z$. Now, a direct computation using the metric compatibility of the inner product shows you that, if we write $P\in \Gamma(\Omega^1)$ as

$$ P(X) = \sum_{i_1, \ldots, i_r, j_1, \ldots, j_s = 1}^{\dim M} W(e_{i_1}, \ldots, e_{i_r}, f_{j_1} , \ldots, f_{j_s}) Z(X, e_{i_1}, \ldots, e_{i_r}, f_{j_1} , \ldots, f_{j_s}) $$

we have, by Leibniz rule,

$$ \sum_{e_i} (\nabla_{e_i} P)(e_i) = \langle \nabla W, Z\rangle - \langle W, \nabla^* Z \rangle \tag{2}\label{lb}$$

where $\nabla^*$ is defined as in your first definition. To conclude if suffices to observe that if $W, Z$ have compact support, the left hand side of \eqref{lb} integrates to 0 by the divergence theorem.

How to obtain the formula for the Laplacian

Directly compute using the first definition you have

$$ \nabla^* \circ \nabla W (\cdots)= - \mathrm{tr} \{ ((X,Y) \mapsto \nabla_X (\nabla W)(Y, \cdots) = \nabla^2_{X,Y} W(\cdots)\} $$