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

  • 2
    $\begingroup$ Do you mean $\nabla^* \circ \nabla$ in the final equation? For example, acting on one forms, the "equation" that you wrote is not true (you can check it on $\mathbb{R}^n$ to see there it is only true on closed one-forms, a fact which also uses that $\mathbb{R}^n$ is Ricci-flat). $\endgroup$ Jun 27, 2016 at 14:40
  • $\begingroup$ @Willie. Thanks for your point. I edited it. $\endgroup$
    – C.F.G
    Jun 27, 2016 at 18:25
  • 3
    $\begingroup$ What happens if you then just plug and chug? (Using, say, the top definition, which is a special case of the second one.) $\endgroup$ Jun 27, 2016 at 18:30
  • $\begingroup$ But How i prove it? $\endgroup$
    – C.F.G
    Jun 27, 2016 at 19:24
  • 2
    $\begingroup$ Can you prove it for scalar functions? On $\mathbb{R}^n$? Show your work and say where you are stuck. If you want help, you need to tell us what you need help with. $\endgroup$ Jun 27, 2016 at 20:48

1 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

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

  • $\begingroup$ I do not understand your question. Can you be more precise? $\endgroup$ Jul 25, 2016 at 13:53
  • $\begingroup$ Ok. What is the role of vector fields $X_1,\cdots, X_n$ used in the first definition . can you re define $D^*T(X_1,\cdots,X_r)=-tr \{(X,Y)\to D_XT(Y,X_1,\cdots,X_r)\}$ without using $X_i$'s? $\endgroup$
    – C.F.G
    Jul 25, 2016 at 14:14
  • $\begingroup$ The $X_1, \ldots, X_r$ are just arbitrary place holders; they are included to make the "trace" operation concrete (so you know which two slots to trace). If you use the index notation the definition is $$(D^*T)_{i_1, \ldots, i_r}^{j_1, \ldots, j_s} = g^{kl} D_k T_{l, i_1, \ldots, i_r}^{j_1, \ldots, j_s} $$ without resorting in explicitly specifying the place holder vector fields. Alternatively you can also use Penrose's diagrammatic notation and it will also not require the place holders. // The role of $X_1,\ldots$ is no different from the role of $x$ in the function definition $f(x) = x^2$. $\endgroup$ Jul 25, 2016 at 14:24
  • $\begingroup$ Thanks. Can you answer my another question? Is $\nabla\circ\nabla _{X,Y}T=\nabla_{\nabla_XY}T-\nabla_X\nabla_Y T$? $\endgroup$
    – C.F.G
    Jul 25, 2016 at 14:38
  • $\begingroup$ (1) That's a matter of notation and should be introduced by whatever thing you are reading. (2) But almost certainly not, the right hand side has the wrong sign. It is more likely $\nabla_X\nabla_Y - \nabla_{\nabla_X Y}$. $\endgroup$ Jul 25, 2016 at 14:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .