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I am teaching myself Riemannian Geometry in order to studying Mean Curvature flow. I was reading Lecture Notes on Mean Curvature Flow by Carlo Mantegazza and I'm trying understand the following definition:

The metric $g$ of $M$ extended to tensors is given by

$$g(T,S) = g_{i_1s_1} \cdots g_{i_ks_k}g^{j_1z_1} \cdots g^{j_lz_l} T^{i_1 \cdots i_k}_{j_1 \cdots j_l} S^{s_1 \cdots s_k}_{z_1 \cdots z_l},$$

where $g_{ij}$ is the matrix of coefficients of $g$ in local coordinates and $g^{ij}$ is its inverse. Clearly, the norm of a tensor is

$$|T| = \sqrt{g(T,T)}.$$

My doubt is why make sense define $g(T,S)$ as defined? I would like to know too if my thoughts below are lead me to the definition of $g(T,S)$ and how can I conclude my thoughts.

$\textbf{My attempt in order to understand the definition:}$

Firstly, I know that the squared norm of the second fundamental form is

$$|A|^2 = g^{mn}g^{st}h_{ms}h_{nt}$$

by this lecture notes and I know that the second fundamental form $A$ is a $(0,2)$- tensor.

This lead me to think that I would be able to understand the definition given by Mantegazza if I understand how define $g(T,S)$ when $T$ and $S$ are $(0,2)$- tensors, because if $T$ and $S$ are $(k,l)$- tensors, then I can see them as $(0,2)-$ tensors just fixing the $k$ coordinates and the $l - 2$ coordinates.

I know that there is an isomorphism between the space of endomorphisms of a finite-dimensional vector space $V$ and the space of $(1,1)-$ tensors defined on $V$, then I thought to raise an index of the tensor $A = (h_{ij})$ in order to obtain a $(1,1)-$ tensor $(g^{ik}h_{kj})$ and I thought define the squared norm of $A$ using the operator norm of the endomorphism associated to $(g^{ik}h_{kj})$ by the isomorphism quoted previously.

I'm stuck here in understand how use the operator norm in order to define the squared norm of $A$. Is it the way to understand the definition of $g(T,S)$? If so how can I proceed in order to conclude that $|A|^2 = g^{mn}g^{st}h_{ms}h_{nt}$?

Thanks in advance!

Arctic Char
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George
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1 Answers1

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It's not the operator norm -- it's the Frobenius norm.

Jack Lee
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  • I understand now how define the squared norm of $A$, thanks for this, but I'm having difficult now to understand how generalize the norm of $A$ to a norm of a tensor $T$ of type $(k,l)$. I see that $g_{i_1s_1} \cdots g_{i_ks_k}g^{j_1z_1} \cdots g^{j_lz_l} T^{i_1 \cdots i_k}{j_1 \cdots j_l} S^{s_1 \cdots s_k}{z_1 \cdots z_l} = T^{i_1 \cdots i_k}{j_1 \cdots j_l} S^{j_1 \cdots j_l}{i_1 \cdots i_k}$ so that this led me to think that I need to define something like a "generalized Frobenius norm" and use the fact that the inner product of tensors(a bilinear form) can be recovered from – George Aug 05 '18 at 23:10
  • a quadratic form $Q(v) := B(v,v,)$ noticing that $B(u,v) = \frac{1}{2} \left[ Q(u+v) - Q(u) - Q(v) \right]$, where $B$ is the inner product of tensors, but I'm having difficult about how define this "generalized Frobenius norm" in order to understand how define $g(T,S)$ for tensors in general case. Can you point how to do this please? Thanks in advance! – George Aug 05 '18 at 23:15
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    @George: One way to do it is just to take the formula you wrote as a definition, and check that it yields the same result when you change to a different basis. (All of the transition matrices cancel out.) Another way is to note that every tensor can be written as a sum of decomposable tensors (those that can be written as tensor products of $1$-tensors), and define the inner product of two such tensors by $\langle \phi_1\otimes\dots\otimes\phi_k,\ \psi_1\otimes\dots\otimes \psi_k\rangle = \langle \phi_1,\psi_1\rangle \cdot \dots\cdot \langle \phi_k, \psi_k\rangle$, then extend by bilinearity. – Jack Lee Aug 05 '18 at 23:25
  • I understood now, thanks a lot! – George Aug 06 '18 at 12:16