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So this is part of a proof in Peter Topping's text, Lectures on Ricci Flow that I don't understand. Let $\delta A=-\text{tr}_{12}\nabla A$ and $\omega$ be a time dependent one-form. Also we use the notation $h=\partial_tg$.We know that

$$\delta(f\alpha)=-\langle df,\alpha \rangle+f(\delta\omega)$$

By the divergence theorem we know that

$$\int \delta(f\omega)dV=0$$

Peter then concludes that

$$\int\partial_t(\delta\omega)fdV=-\int\langle h,\omega\rangle dV+\int\langle df,\partial_t\omega\rangle dV-\int[(\delta\omega)f-\langle df,\omega\rangle]\frac{1}{2}(\text{tr}h)dV$$

Ii don't see how he gets the extra terms

$$-\int[(\delta\omega)f-\langle df,\omega\rangle]\frac{1}{2}(\text{tr}h)dV$$

Can someone explain to me where these terms come from. Thanks.

Edit So I have figured out the reasoning, and I'll record it here for future reference. Note that

$$\partial_t\int(\delta\omega)fdV=\int\partial_t(\delta\omega)fdV+\int(\delta\omega)f\partial_tdV$$

Since $\partial_tdV=\frac{1}{2}(\text{tr}h)dV$ we get that the above equals

$$\partial_t\int(\delta\omega)fdV=\int\partial_t(\delta\omega)fdV+\frac{1}{2}\int(\delta\omega)f(\text{tr}h)\partial_tdV$$ Now we compute the left hand side a different way by noting that

$$(\delta f\omega)=-\langle df,\omega\rangle+(\delta\omega)f$$

Noting that the integral of the second term vanishes by divergence theorem we have $$\begin{align} \partial_t\int(\delta\omega)fdV&=\partial_t\int\langle df,\omega\rangle dV\\ &=\int\langle df,\partial_t\omega\rangle dV+\frac{1}{2}\int\langle df,\omega\rangle(\text{tr}h)V \end{align}$$

Combining these results and rearranging we get the desired identity.

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