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I am following some lecture notes on Ricci flow and reached the section where we linearize the Ricci tensor and obtain the principal symbol for the resulting operator. We have $T \in \: \Gamma(Sym^2 T^{*}M)$ smooth, fixed and positive definite and then compute the time derivative for the divergence of $G(T)$:

$\bigg(\frac{\partial}{\partial t}\delta G(T) \bigg)Z = -T \bigg((\delta G(h))^{\#},Z\bigg) + \bigg<h,\nabla T(.,.,Z) - \frac{1}{2}\nabla_{Z}T \bigg>, $

where $\frac{\partial g}{\partial t}=h$, $Z$ an arbitrary vector field and $G(T)=T-\frac{1}{2}(tr \: T)h$.

The notes then say that this implies

$\frac{\partial}{\partial t}T^{-1}\delta G(T) = -\delta G(h)\: + \:\: ...$

where the dots indicate terms which don't have derivatives of $h$. I see that you can apply $T^{-1}$ as it doesn't depend on $t$ and then lose the $Z$ as it is arbitrary, but not sure exactly what happens to the right-hand side such that we end up with $-\delta G(h)$ or where the sharp goes.

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  • $\begingroup$ I gotta admit that, despite my preference for nice abstract notation, I only like what I invent myself. So when faced with deciphering someone else's abstract notation like this, I start by converting it all to local coordinates and write the tensors with indices. Here, it's particularly confusing, at least for me, with the dependence on $t$. $\endgroup$
    – Deane Yang
    Commented Feb 15, 2019 at 16:23
  • $\begingroup$ Yes, this is why I am confused, I have studied other treatments of the Ricci-DeTurck flow in Chow and others and it is completely explicit and done in indices, so I find this treatment difficult to decipher. $\endgroup$
    – Hollis Williams
    Commented Feb 15, 2019 at 18:05
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    $\begingroup$ I don't see why the sharp has been dropped, should I just convert to indices, do the calculation and then convert back to the notation of the lecture notes at the end? Not sure what else to do. $\endgroup$
    – Hollis Williams
    Commented Feb 15, 2019 at 18:18
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    $\begingroup$ I think there is a typo in the definition of $G(T)$. It should be $G(T) = T -\frac 12 tr(T)\cdot \mathbf g$. $\endgroup$
    – Panagiotis Konstantis
    Commented Feb 15, 2019 at 21:01
  • $\begingroup$ I think in Peter Topping, Lecture notes on Ricci flow, he refers to do computations in a coordinate-free way. Some people refer to this way. This avoids local coordinate sickness. Of course, one of the drawbacks is "a heavier notation" $\endgroup$
    – Truong
    Commented Feb 16, 2019 at 3:05

1 Answer 1

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It looks like this is from the Lecture notes on Ricci flow from Peter Topping. He mentions that he uses the symbol $T$ as the symmetric, positive definite bilinear form and also for the map $\Gamma(TM^\ast) \to \Gamma(TM^\ast)$ induced by $T$ and the metric $g$ in the following way $$ T(\alpha)(Z)=T(\alpha^\#,Z) $$ where $\alpha^\#$ is the dual vector field to the $1$-form $\alpha$ with respect to $g$. Thus looking at \begin{eqnarray*} \bigg(\frac{\partial}{\partial t}\delta G(T) \bigg)Z &=& -T \bigg((\delta G(h))^{\#},Z\bigg) + \bigg<h,\nabla T(.,.,Z) - \frac{1}{2}\nabla_{Z}T \bigg> \\ &=& -T(\delta G(h))(Z) + \ldots \end{eqnarray*} we just apply to the equation the inverse of $T$ to obtain your desired expression (note that $\delta G(T)$ is a $1$-form)

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  • $\begingroup$ Hi this is correct, I was working from Topping's lecture notes on Ricci flow. Also, once we have the desired expression he says that this implies that $\frac{\partial}{\partial t} L_{(T^{-1} \delta G(T))^{\#}}g = -L_{(\delta G(T))^{\#}} + ... $ where $L$ is the Lie derivative. So, it has to be a sharp as it needs to be a vector field in the direction of which the Lie derivative acts, but why is it $\delta G(T)$ and not $\delta G(h)$? $\endgroup$
    – Hollis Williams
    Commented Feb 16, 2019 at 15:26
  • $\begingroup$ Also, in this expression with the Lie derivatives the term on the right-hand side involves a second derivative of $h$ and the dots indicate terms which contain $h$ and its first derivative. $\endgroup$
    – Hollis Williams
    Commented Feb 16, 2019 at 15:35
  • $\begingroup$ I've thought about it and I think that the fact that Topping has it with a $T$ rather than a $h$ must be a typo, as it changes back to $h$ when he linearizes the operator, then the implication comes from definition of the Lie derivative. $\endgroup$
    – Hollis Williams
    Commented Feb 17, 2019 at 16:41
  • $\begingroup$ That would have been my guess too. $\endgroup$
    – Panagiotis Konstantis
    Commented Feb 17, 2019 at 18:43

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