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I'm studying by myself Mean Curvature Flow and I'm trying understand the definition of $\ast$-operation given by Richard Hamilton in the beginning of the section $13$ (page $40$) of his paper "Three-manifolds with positive Ricci curvature":

"If $A$ and $B$ are two tensors we write $A \ast B$ for any linear combination of tensors formed by contraction on $A_{i \cdots j}B_{k \cdots l}$ using the $g^{ik}$".

I'm trying understand this definition in order to understand how Huisken obtained the evolution equation of Chistoffel symbols in the beginning of the section $7$ (final of the page $19$) on his paper "Flow by mean curvature of convex surfaces into spheres":

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

I found a master thesis which do the computation of this evolution equation as follows

enter image description here

I'm stuck in how to develop these two equalities where appears $\ast$ operation. I will put what I did in order to understand $\ast$ operation and develop these two equalities.

$\textbf{My attempt:}$

I don't know if I was clear the first time, but my doubt is what is the $\ast$-operation, because by the Hamilton's explanation on his paper lead me to think that $\ast$-operation is such that all terms of the linear combination must have a contraction of at least one of the tensors involved on $\ast$-operation, so $T \ast S$ would be one of the following forms: $\sum_{p = 1}^{p = r} \sum_{q = 1}^{q = s} \text{tr}_g \ T \ S_{i_pj_q}$, $\sum_{p = 1}^{p = r} \sum_{q = 1}^{q = s} T_{i_pj_q} \ \text{tr}_g \ S$ or $\text{tr}_g \ T \text{tr}_g \ S$, where $1 \leq i_1 < i_2 < \cdots < i_p \leq n$, $1 \leq j_1 < j_2 < \cdots < j_q \leq n$ and $p,q \in \{ 1, \cdots, n \}$ (I'm considering the tensors $T$ and $S$ are $(0,2)$-tensors which has $n^2$ components). According this interpretation, we have in normal coordinates that

$\begin{align} (\circ) - H(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk}) &=& - H \nabla_j h^i_k - H \nabla_k h^i_j + H \nabla^i h_{jk}\\ &=& \text{tr}_g A (\nabla_j h^i_k) + \text{tr}_g A (-\nabla_k h^i_j) + \text{tr}_g A \nabla^i h_{jk}\\ &=& \text{tr}_g A (\nabla_j g^{ir}h_{rk}) + \text{tr}_g A (-\nabla_k g^{ir}h_{rj}) + \text{tr}_g A \nabla^i h_{jk}\\ &=& g^{ir} \text{tr}_g A (\nabla_j h_{rk}) + g^{ir} \text{tr}_g A (-\nabla_k h_{rj}) + \text{tr}_g A \nabla^i h_{jk} = A \ast \nabla A, \end{align}$

where the last equality is valid according my interpretation because all the terms of the linear combination has a product among the contraction of the tensor $A$ and a component of the tensor $\nabla A$.

Furthermore, we have

$\begin{align} (\circ \circ) -h^i_k \nabla_j H -h^i_j \nabla_k H + h_{jk} \nabla^i H &=& -h^i_k \nabla_j (g^{rs}h_{rs}) -h^i_j \nabla_k (g^{rs}h_{rs}) + h_{jk} \nabla^i (g^{rs}h_{rs})\\ &=& -h^i_k \left( g^{rs} \nabla_j h_{rs} \right) -h^i_j \left( g^{rs} \nabla_k h_{rs} \right) + h_{jk} \left( g^{rs} \nabla^i h_{rs} \right)\\ &=& -h^i_k ( \text{tr}_g \nabla_j A ) -h^i_j ( \text{tr}_g \nabla_k A ) + h_{jk} ( \text{tr}_g \nabla^i A )\\ &=& -(g^{ir}h_{rk}) ( \text{tr}_g \nabla_j A ) -(g^{ir}h_{rj}) ( \text{tr}_g \nabla_k A ) + h_{jk} ( \text{tr}_g \nabla^i A )\\ &=& A \ast \nabla_j A + A \ast \nabla_k A + A \ast \nabla^i A, \end{align}$

where the last equality is valid according my interpretation because all the terms of the linear combination has a product among the contraction of the covariant derivative of the tensor $A$ and a component of the tensor $A$.

I'm stuck here, because I don't know how to show that $(\circ) + (\circ \circ) = \nabla H \ast A + H \ast \nabla A$. I think I misunderstood the definition because I cannot conclude what I want, so my question it's how to write the linear combination $T \ast S$ in order to understand how $\ast$-operation is defined.

$\textbf{EDIT:}$

$\begin{align} -h^i_k \nabla_j H -h^i_j \nabla_k H + h_{jk} \nabla^i H - H(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk}) &=& - \nabla_j (h^i_k H) - \nabla_k (h^i_j H) - \nabla^i(h_{jk} H)\\ &=& - \nabla_j (g^{ir} h_{rk} \text{tr}_g (A)) - \nabla_k (g^{ir} h_{rj} \text{tr}_g (A)) - \nabla^i (h_{jk} \text{tr}_g (A))\\ &=& -g^{ir}(\nabla_j (h_{rk})\text{tr}_g(A) + h_{rk} \text{tr}_g(\nabla_j A)) -g^{ir}(\nabla_k (h_{rj}) \text{tr}_g(A) + h_{rj} \text{tr}_g (\nabla_k A))\\ &-& (\nabla^i (h_{jk}) \text{tr}_g(A) - h_{jk} \text{tr}_g(\nabla^i A))\\ &=& \left( -g^{ir} \nabla_j (h_{rk}) -g^{ir} \nabla_k (h_{rj}) - \nabla^i (h_{jk}) \right) \text{tr}_g(A)\\ &-& g^{ir} h_{rk} \text{tr}_g(\nabla_j A) - g^{ir} h_{rj} \text{tr}_g (\nabla_k A) - h_{jk} \text{tr}_g(\nabla^i A)\\ &=& \left( -g^{ir} \nabla_j (h_{rk}) -g^{ir} \nabla_k (h_{rj}) - \nabla^i (h_{jk}) \right) \text{tr}_g (A)\\ &-& g^{ir} h_{rk} \nabla_j (\text{tr}_g A) - g^{ir} h_{rj} \nabla_k(\text{tr}_g A) - h_{jk} \nabla^i(\text{tr}_g A)\\ &=& \text{tr}_g^{45} (\nabla A \otimes A) + \text{tr}_g^{45} (A \otimes \nabla A) = \nabla A \ast A. \end{align}$

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  • $\begingroup$ I'm confused. Ultimately, all you need is $\nabla A * A$, and you already know how to do that, right? $\endgroup$ – Deane Yang 11 hours ago
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I am not sure if I understood your question correctly: Note that in Hamilton's notation the term $\text{tr}_g T$ can be written as

$$ \text{tr}_g T =T_{il}g^{il}=T_{ik}\delta_l^kg^{il}=T_{ik}g_{lj}g^{jk}g^{il}=T\ast g. $$ And I would say that in this notation one suppress the metric $g$. This means that in $T\ast S$ there can be linear combination involving the traces, but you need only the tensors $T$ and $S$ as well as $g$.

Edit: Note that $g^{ij}T_{jl}$ is seen in this definition also as a "contraction" which is not precisely the definition of contration (see e.g. Wikipedia). Then $$ h^i_k\nabla_j H = g^{il}h_{lk}\nabla_j H = A \ast \nabla H =A \ast \nabla A. $$ The last equation is due to the fact $$ g^{il}h_{lk}\nabla_j H = g^{il}g^{pq}h_{lk}\nabla_j h_{pq}. $$

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  • $\begingroup$ Firstly, I would like to thanks to answer my question, but can you explain with more details what did you mean when you said "you need only the tensors $T$ and $S$ as well as $g$."? I edited my topic and tried give more details about how I understood the $\ast$-operation by the Hamilton's explanation. $\endgroup$ – George Jan 13 at 23:48
  • $\begingroup$ In the definition of Hamilton's $\ast$-operator you can also take contractions with one index of $T$ and one of $S$ which is not of the forms you mentioned in your edit. $\endgroup$ – Panagiotis Konstantis Jan 15 at 8:32
  • $\begingroup$ Maybe it's an stupid question, but I never did a course on Riemannian geometry, I just study by my self it. A metric contraction it's an operator which takes a tensor $T$ of type $(k,l)$ and returns a tensor $\text{tr}_g T := g^{ij}T_{ij}$ of type $(k,l-2)$, but I never heard or read about contraction of one index like you said. Can you give more details how can I take contractions with one index of $T$ and one of $S$ please? I tried think what this would be, but I couldn't understand what is this kind of contraction. $\endgroup$ – George Jan 17 at 11:01
  • $\begingroup$ I think that I understood when you said take contractions with one index of $T$, when you said this, you want to say that I can have terms like $\nabla_i (\text{tr}_g T)$? Because if the tensor $T$ has type $(k,l)$, the tensor $\nabla_i (\text{tr}_g T)$ have type $(k,l-1)$. This would able me derive $\nabla H \ast A + H \ast \nabla A$, is this right? $\endgroup$ – George Jan 17 at 12:09
  • $\begingroup$ No, contraction needs always two indices. If you contract an index of $T$ with an index of $S$ then you contracted again two indices. For example you can contract the tensor product $T \otimes S$ with one entry in $T$ and one in $S$. Actually Hamilton's definition is nothing else than a multiple contraction of $T \otimes S$. In other words (for $(0,2)$-tensors the tensor product is represented in index notation by $T_{ij}S_{lk}$ and $T_{ij}S_{lk}g^{jl}$ is an example of what I am speaking about. You can also understand it as $\text{tr}_g^{23} T\otimes S$ (where you contract the 2. with the 3.) $\endgroup$ – Panagiotis Konstantis Jan 17 at 13:15

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