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I'm self-study Mean Curvature Flow and I'm stuck on item $(ii)$ of the lemma below

enter image description here

My doubts are referent the equalities marked with a red rectangle

  1. Why $|A|^2 = \langle h_{ij}, h_{ij} \rangle$ ? I know that $|A|^2 = g^{ij}g^{kl}h_{ik}h_{jl}$, but I can't see how $|A|^2 = \langle h_{ij}, h_{ij} \rangle$

    1. What is the definition of $\nabla A$? The closer I got to the definition was on the proof of Lemma 2.2 of this article by Huisken. I think the second equality marked is just by definition, but I would like to know what is the definition of $\nabla A$.

    2. Why $\text{tr} (A^3) = \langle h_{ij}, h_{ik} h^k_j \rangle$?

Thanks in advance!

$\textbf{EDIT:}$ I finally understood why $\text{tr} (A^3) = \langle h_{ij}, h_{ik} h^k_j \rangle$. I will post how to develop $\langle h_{ij}, h_{ik} h^k_j \rangle$ to arrive at this.

By the definition of inner product of tensors (see this topic as well as the comments that I did in Lee's answer for the definition and for a motivation of it),

$\langle h_{ij}, h_{ik} h^k_j \rangle = g^{ii} g^{jj} h_{ij} h_{ik} h^k_j = g^{ii} g^{jj} g^{lk} h_{ij} h_{ik} h_{lj}$

and, as pointed by Anthony on his answer, I can consider an orthonormal frame (it's just assume a local chart with normal coordinates), then

$\langle h_{ij}, h_{ik} h^k_j \rangle = g^{ii} g^{jj} g^{lk} h_{ij} h_{ik} h_{lj} = h_{ij} h_{ik} h_{kj} = h_{ij} h_{jk} h_{ki} = \text{tr} (A^3)$.

George
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1 Answers1

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I'm not sure what the source of this screenshot is, but it uses (very slightly) odd notation: the quantity $\langle h_{ij},h_{ij} \rangle$ would often instead be written as one of the following:

  • $|A|^2,$ which is by definition
  • $\langle A, A \rangle;$
  • As you did, $g^{ij}g^{kl}h_{ik}h_{jl},$
  • or raise indices to write this as $h_{ij}h^{ij};$
  • or assume an orthonormal frame and write $h_{ij} h_{ij}.$

Here, $A$ is the second fundamental form, whose components are called $h_{ij}.$ (I've always found this to be a strange convention, but it's nearly universal in MCF.)

Once you understand the notation, the answers to your questions are very simple:

  1. By definition we have $|A|^2 = h_{ij}h_{ij},$ so of course $\Delta |A|^2 = \Delta (h_{ij}h_{ij}).$
  2. The second fundamental form $A$ is a 2-tensor field defined on the hypersurface, and $\nabla A$ is simply the 3-tensor field obtained from $A$ by taking the covariant derivative.
  3. In an orthonormal frame, $A^3$ has components $(A^3)_{il} = h_{ij}h_{jk}h_{jl},$ so taking the trace we find $\mathrm{tr}(A^3) = h_{ij}h_{jk}h_{ki}.$
Anthony Carapetis
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