Integration of Ricci curvature on a compact manifold.

Question

$\mathbf {The \ Problem \ is}:$ Let, $(M,g)$ be a compact Riemannian manifold and $X\in \chi(M).$ Show that $\int Ric_g(X,X)\mu_g=\int ((tr(\nabla_.X))^2-tr(\nabla_.X\circ \nabla_.X))\mu_g.$

$\mathbf {My \ approach}:$ There's a hint: to use $\int_M \Delta f \mu_g=0$ where $\Delta f$ is Laplacian of $f.$

I started with $\nabla$ is torsion-free . Define $\operatorname{T}(v)=\nabla_vX$ then $\operatorname{T}:T_pM\to T_pM$ is linear for each $p\in M.$

Let, $p\in M$ and $(U,\{e_i\}_{i=1}^n)$ be a normal neighbourhood around $p.$

Now,$\nabla_X(\nabla_{e_i}X)=\nabla_{\nabla_{e_i}X}X+[X,\nabla_{e_i}X] \implies \nabla_{e_i}(\nabla_XX)-R(e_i,X)X-\nabla_{[e_i,X]}X=\nabla_{\nabla_{e_i}X}X+[X,\nabla_{e_i}X] \implies R(e_i,X)X=\nabla_{e_i}(\nabla_XX)-\nabla_{[e_i,X]}X-\nabla_{\nabla_{e_i}X}X-[X,\nabla_{e_i}X] \implies R(e_i,X)X=\nabla_{e_i}(\nabla_XX)-\nabla_X(\nabla_{e_i}X)-\operatorname{T}\circ \operatorname{T}(e_i)$ where we used $\nabla_{e_i}X=[e_i,X]$ as $\nabla$ is torsion-free and again we use torsion-freeness on two vector fields $X$ and $[e_i,X].$ Then taking trace of the left side and of $\operatorname{T}\circ \operatorname{T},$ we obtain 2 terms in the expression but I can't proceed after this .

I don't know how to bring $(tr(\nabla_.X))^2.$

Do we need to start with $f(p)=\frac{1}{2}\langle X,X\rangle(p) ?$

Thanks in advance for any help .

Answer 1

Using the abstract index notation, we can rewrite the identity in question as $$ \int_M Ric_{a b}X^a X^b \mu_g = \int_M \big( (\nabla_a X^a)^2 - (\nabla_a X^b) (\nabla_b X^a) \big) \mu_g $$

To show that this holds on a closed (compact, without boundary) manifold $M$, equipped with a Riemannian metric $g$, and $\nabla$ being the Levi-Civita connection for $g$, we can start from the definition of the Riemann curvature operator: $$ Riem_{a b}{}^{c}{}_{d}X^d = \nabla_a \nabla_b X^c - \nabla_b \nabla_a X^c $$

Taking into account that $Ric_{b d} := Riem_{a b}{}^{a}{}_{d}$ by definition, we get: $$ Ric_{b d} X^b X^d = X^b \nabla_a \nabla_b X^a - X^b \nabla_b \nabla_a X^a $$

Renaming the dummy indices of the first term on the right-hand side, we see that $$ Ric_{b d} X^b X^d = X^a \nabla_b \nabla_a X^b - X^b \nabla_b \nabla_a X^a \tag{1} $$

By the Leibniz rule, we have the identities $$ \nabla_b (X^a \nabla_a X^b) = (\nabla_b X^a)(\nabla_a X^b) + X^a \nabla_b \nabla_a X^b \tag{2} $$ and $$ \nabla_b (X^b \nabla_a X^a) = (\nabla_b X^b) (\nabla_a X^a) + X^b \nabla_b \nabla_a X^a \tag{3} $$ which we can use to express the terms on the right-hand side of $(1)$ as follows: $$ Ric_{b d} X^b X^d = \nabla_b(X^a \nabla_a X^b) - (\nabla_b X^a)(\nabla_a X^b) - \nabla_b(X^b \nabla_a X^a) + (\nabla_b X^b) (\nabla_a X^a) $$

The left-hand sides of both $(2)$ and $(3)$ are in the divergence form, so the corresponding terms (1-st and 3-rd) integrate to zero over a closed manifold.