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I want to show $$ (\delta S)Z = \delta (S(\cdot, Z))+\frac{1}{2}\langle S, \mathcal L_Zg\rangle \tag{1} $$
where $\delta= -tr_{12}\nabla$ is divergence operator, $\mathcal L$ is Lie derivative, $S\in\Gamma(Sym^2 T^*M)$

What I try: I use abstract index notation. First $$ (\delta S)Z =-(\nabla^a S_{aj})Z = -(\nabla^a S_{aj})Z^j $$ Besides, $$ \delta (S(\cdot,Z)) =-\nabla^a[S(\cdot, Z)]_a $$ since $S(\cdot, Z)= S_{ij}Z^j$, I think $[S(\cdot, Z)]_a=S_{aj}Z^j$. Therefore $$ \delta(S(\cdot, Z)) =-\nabla^a(S_{aj}Z^j)=-(\nabla ^a S_{aj})Z^j - S_{aj}(\nabla^a Z^j) =(\delta S)Z- S_{aj}(\nabla^a Z^j) \tag{2} $$ On the other hand, when $\delta$ is restricted on $\Gamma (Sym^2 T^*M)$, its formal adjoint is $\omega\rightarrow \frac{1}{2}\mathcal L_{\omega^\sharp}g$. Therefore, I have $$ \frac{1}{2}\langle S, \mathcal L_Z g \rangle = \langle \delta(S), Z^\flat \rangle = \langle -\nabla^a S_{aj} , Z^i {g_{ik}} \rangle = -(\nabla^a S_{aj}) Z^j \tag{3} $$ However, I can't get (1) from (2) and (3).

If the answer is complex, a picture of handwritten draft is enough for me. Thanks very much.

Enhao Lan
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1 Answers1

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It's good that you are trying to use the abstract index notation, but you need one more piece for you puzzle, the formula for the Lie derivative of the metric: $$ (\mathcal L_Z g)_{a b} = \nabla_a Z_b + \nabla_b Z_a = 2 \nabla_{(a} Z_{b)} $$ where $Z^a$ is a vector field, and $Z_a = g_{a b} Z^b$ is the corresponding $1$-form. Of course, $\nabla$ is the Levi-Civita connection, corresponding to the Riemannian metric $g_{a b}$, and we use the latter along with its inverse $g^{a b}$ to raise and lower the indices without mention (the Ricci calculus).

With that in mind, the calculation will be straightforward: $$ \delta (S(\cdot, Z)) = - \nabla^a (S_{a b} Z^b) = - (\nabla^a S_{a b})Z^b - S_{a b} \nabla^a Z^b = \\ (\delta S) Z - S_{a b} \nabla^{(a} Z^{b)} = (\delta S) Z - \tfrac{1}{2} S_{a b} (\mathcal L_Z g)^{a b} = (\delta S) Z - \tfrac{1}{2} \langle S, \mathcal L_Z g \rangle $$

Here I use the notation for the symmetric part $t_{(a b)} := \tfrac{1}{2}(t_{a b} + t_{b a}) $ of a $2$-tensor $t_{a b}$ and the fact that if a tensor $s_{a b}$ is symmetric ($s_{a b} = s_{b a} = s_{(a b)} $), then for any tensor $t_{a b}$ we have the identity: $$ s_{a b} t^{a b} = s_{(a b)} t^{a b} = s_{a b} t^{(a b)} $$

Yuri Vyatkin
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