2
$\begingroup$

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?

I find some question on this case in Formal adjoint of the covariant derivative but do not help.

Thanks in advance

$\endgroup$
  • 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). $\endgroup$ – Willie Wong Jun 27 '16 at 14:40
  • $\begingroup$ @Willie. Thanks for your point. I edited it. $\endgroup$ – C.F.G Jun 27 '16 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.) $\endgroup$ – Willie Wong Jun 27 '16 at 18:30
  • $\begingroup$ But How i prove it? $\endgroup$ – C.F.G Jun 27 '16 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. $\endgroup$ – Willie Wong Jun 27 '16 at 20:48
4
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Excuse me. why in the first definition used $X_i$ for tensor field $T$? $\endgroup$ – C.F.G Jul 24 '16 at 16:57
  • $\begingroup$ I do not understand your question. Can you be more precise? $\endgroup$ – Willie Wong Jul 25 '16 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 '16 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 $f(x) = x^2$. $\endgroup$ – Willie Wong Jul 25 '16 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 '16 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.