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!


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?


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 $\nabla H\ast A+H\ast\nabla A$.

  • $\begingroup$ 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 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.