# Evolution equation of Christoffel symbols under Mean Curvature Flow

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: 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 $$(3)$$.

$$\textbf{EDIT:}$$

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?

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 $$\nabla H\ast A+H\ast\nabla A$$.
• 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 – George Apr 12 at 22:47