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I want to show $$ \operatorname{tr} \nabla^2_{X,~\cdot} h(\cdot, W)= -(\nabla \delta h)(X,W) $$ where $X,W$ are tangent vector fields on Riemannian manifold. And $$ \nabla^2_{X,Y} = \nabla _X \nabla_Y - \nabla _{\nabla_XY} \\ \delta:\Gamma(\otimes^k T^* M) \rightarrow \Gamma(\otimes^{k-1} T^* M),~~~~ \delta(T)= -\operatorname{tr}_{12} (\nabla T). $$

What I try: $$ \operatorname{tr} \nabla^2_{X,~\cdot} h(\cdot, W) = g^{ij}\nabla ^2_{X, \partial_i} h(\partial_j, W) \tag{1} $$ Besides \begin{align} -(\nabla\delta h)(X,W)&= [\nabla_X(\operatorname{tr}_{12}(\nabla h))] W \\ &=[\nabla_X (g^{ij} \nabla h(\partial_i, \partial_j, \cdot))]W \\ &=g^{ij} \nabla_X[\nabla h(\partial_i, \partial_j, \cdot)] W \\ &= g^{ij}[\nabla_X\nabla_{\partial_i } h(\partial_j, \cdot)- \nabla_{\nabla_X\partial_i } h(\partial_j, \cdot) - \nabla_{\partial_i } h(\nabla_X\partial_j, \cdot) ] W\\ &=g^{ij}\nabla^2_{X, \partial_i} h(\partial_j, W) - \nabla_{\partial_i } h(\nabla_X\partial_j, W) \tag{2} \end{align} Obviously, (2) is not equal to (1). How should I do?

Arctic Char
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Enhao Lan
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1 Answers1

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For those who wonder how this can be shown using the abstract index notation, here is a brief summary.

Rewrite both sides of the sought identity: $$ \mathrm{tr} \nabla^2_{X,\cdot} h(\cdot, W) = g^{b c} X^a W^d \nabla_a \nabla_b h_{c d} = X^a W^d \nabla_a \nabla^c h_{c d} = X^a W^c \nabla_a \nabla^b h_{b c} $$ and $$ -(\nabla \delta h) (X, W) = X^a W^c \nabla_a \nabla^b h_{b c} $$ so that the equality is simply read off from expanding the notation in abstract indices!

Here $g_{a b}$ is a Riemannian metric, $\nabla$ is the corresponding Levi-Civita connection, $\delta$ is the codifferential, which on a (tensor-bundle-valued) $k$-form $\omega_{a_1 \dots a_k}$ can be computed by the formula $$ (\delta \omega)_{a_2 \dots a_k} = - \nabla^{a_1} \omega_{a_1 a_2 \dots a_k} $$

We treat the $2$-tensor $h_{a b}$ as a $1$-form $h_a$ with values in the contagent bundle (with index $b$) in order to properly apply the above formula.

See similar calculations in this and this answers.

Yuri Vyatkin
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