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I'm reading a master thesis about Mean Curvature Flow and I'm trying understand how was developed the following equation:

$$\frac{\partial \Gamma^i_{jk}}{\partial t} = \nabla A \ast A.$$

It's how the equation above was developed:

enter image description here

I didn't understand how the terms highlighted appears. I know that $A \ast \nabla A$ means that I'm considering the linear combination of the contraction of $A$ and $\nabla A$ with respect to the metric $g$ (as you can read on the beginning of the section $13$ on page $40$ of Hamilton's article), but I can't see why the line above $(3)$ it's a linear combination of contractions of $\nabla H, A, H$ and $\nabla A$ and why $\nabla H \ast A + H \ast \nabla A = \nabla A \ast A$. I think if I have more details about the operator $\ast$, then I will be able to understand $(3)$.

Thanks in advance!

$\textbf{EDIT:}$

I understood how obtain $(1)$ and (2) and I have an idea for answer my doubt about $(3)$, but I couldn't understand how obtain $(3)$ yet.

$(1)$: using the fact that $\frac{\partial g^{il}}{\partial t} = 2H h^{il}$, we observe that

$$\frac{\partial g^{il}}{\partial t} = \frac{\partial (g^{is}g_{sz}g^{zl})}{\partial t} = \frac{\partial (g^{is}g^{zl})}{\partial t} g_{sz} + (g^{is}g^{zl}) \frac{\partial g_{sz}}{\partial t}$$

$$ = \left( \frac{\partial g^{is}}{\partial t} g^{zl} + g^{is} \frac{\partial g^{zl}}{\partial t} \right) g_{sz} + (g^{is}g^{zl}) \frac{\partial g_{sz}}{\partial t}$$

$$ = \left( 2H h^{is} g^{zl} + g^{is} 2H h^{zl} \right) g_{sz} + (g^{is}g^{zl}) \frac{\partial g_{sz}}{\partial t}$$

$$= 2H h^{is} g^l_s + g^i_z 2H h^{zl} + (g^{is}g^{zl}) \frac{\partial g_{sz}}{\partial t}$$

$$= 2H h^{il} + 2H h^{il} + (g^{is}g^{zl}) \frac{\partial g_{sz}}{\partial t}$$

$$=2 (2H h^{il}) + g^{is} \frac{\partial (g_{sz})}{\partial t} g^{zl}$$

$$=2 (\frac{\partial g^{il}}{\partial t}) + g^{is} \frac{\partial (g_{sz})}{\partial t} g^{zl}$$

$$\Longrightarrow \frac{\partial g^{il}}{\partial t} = - g^{is} \frac{\partial (g_{sz})}{\partial t} g^{zl}$$

$(2)$: it's just observe that

$$\Gamma^z_{jk} = g^{zl} \left( \frac{\partial g_{kl}}{\partial x_j} + \frac{\partial g_{jl}}{\partial x_k} - \frac{\partial g_{jk}}{\partial x_l} \right)$$

$(3)$: Computing in normal coordinates, we observe that

$(\circ) - H(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk}) = - g^{ij}h_{ij}(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk})$

$(\circ \circ) -h^i_k \nabla_j H -h^i_j \nabla_k H + h_{jk} \nabla^i H = -h^i_k \nabla_j (g^{rs}h_{rs}) -h^i_j \nabla_k (g^{rs}h_{rs}) + h_{jk} \nabla^i (g^{rs}h_{rs}) = -h^i_k \left( g^{rs} \nabla_j h_{rs} \right) -h^i_j \left( g^{rs} \nabla_k h_{rs} \right) + h_{jk} \left( g^{rs} \nabla^i h_{rs} \right)$,

I don't be sure if I'm using the definition of $\ast$ correctly, but if I'm not wrong, then $(\circ)$ it's a linear combination which terms involve the trace of the tensor $ A$ and the components of the tensor $\nabla A$ (it's the components of the tensor $\nabla A$ if we rewrite $\nabla_j h^i_k = g^{is} \nabla_j h_{sk}$ and $\nabla_k h^i_j = g^{is} \nabla_k h_{sj}$), while $(\circ \circ)$ it's a linear combination which terms involve the trace of the tensor $\nabla A$ and the components of the tensor $A$ (again, it's the components of the tensor $A$ if we rewrite $h^i_k = g^{is}h_{sk}$ and $h^i_j = g^{is}h_{sj}$) nad this justify why $(\circ) + (\circ \circ) = \nabla A \ast A$, but this not answer why $(\circ) + (\circ \circ) = \nabla H \ast A + H \ast \nabla A$, which lead us to my question:

let be $T$ and $S$ $(0,2)$-tensors (just for simplicity) with components $T_{ij}$ and $S_{kl}$ and denote by $\text{tr}_g T$ the trace of the tensor $T$ with respect to the metric $g$. Is it $T \ast S$ a linear combination with terms like $\text{tr}_g T \ S_{ij}$, $T_{ij} \ \text{tr}_g S$, $\text{tr}_g T \ \text{tr}_g S$ or it's a linear combination with other kind of terms?

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1 Answer 1

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I think you can prove (1) a bit faster by observing that $g_{sz}g^{zl}=\delta_s^l$ is constant, so \begin{align*} 0 \,&=\, g^{is}\,\frac{\partial}{\partial t}\left(g_{sz}\,g^{zl}\right) \\ &=\, g^{is}\,\frac{\partial}{\partial t}g_{sz}\,g^{zl}\,+\,g^{is}\,g_{sz}\,\frac{\partial}{\partial t}g^{zl} \\ &=\, g^{is}\,\frac{\partial}{\partial t}g_{sz}\,g^{zl}\,+\,\delta_z^i\,\frac{\partial}{\partial t}g^{zl} \\ &=\, g^{is}\,\frac{\partial}{\partial t}g_{sz}\,g^{zl}\,+\,\frac{\partial}{\partial t}g^{il}\,. \end{align*} Concerning (3), I am not familiar with the notation $\ast$ used here (so take everything I'm writing with a grain of salt...) but I would say that what you wrote is correct: $\ast$ is not really a well defined operator but it just says that the result is a linear combination of contractions. Since $H$ is a contraction of $A$, in particular both $\nabla H\ast A$ and $H\ast\nabla A$ are linear combinations of contractions of $A$ and $\nabla A$, hence they both can be written as $A\ast\nabla A$. As much as this may seem a more elegant and compact way of writing the result, it seems to me that the equality $\nabla H\ast A+H\ast\nabla A=\nabla A\ast A$ is a bit misleading, because you are actually losing some information. While every element of the form $\nabla H\ast A+H\ast\nabla A$ can be written in the form $\nabla A\ast A$, the converse seems to be false. For instance, the tensor $$ g^{ik}\,A_{ij}\,\nabla_k A_{lm} $$ stays in $\nabla A\ast A$ but not in $\nabla H\ast A+H\ast\nabla A$.

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  • $\begingroup$ Thanks for the answer, I have forgotten this topic! I discussed on the comments here about the point $(3)$ and I think that discussion solves my problem. I put the link if you are interested to see the discussion. I think that the answer that I receive there is more suitable, because you see there that the $\ast$-operator give us a tensor, which is reasonable if you have in mind the property that $|A \ast B| \leq C |A| |B|$ for some constant $C > 0$, where $| \cdot |$ is the tensor's norm $\endgroup$
    – George
    Apr 12, 2019 at 22:47

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