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Apart from the function $(LN-M^2)$ composed of purely second fundamental form coefficients ( or with derivatives) what are some other examples of such scalars which are isometrically invariant?

EDIT1:

Without basis and as a pure guess I imagined inclusion of lesser coefficients $(g,f,c)$ in some higher invariant :) as determinant,

\begin{bmatrix} L & M & g\\[0.3em] M & N & f\\ g & f & c \end{bmatrix}

or as..

\begin{bmatrix} L & M & E\\[0.3em] M & N & G\\ E & G & F \end{bmatrix}

EDIT2:

i.e., if each of $(g,f,c)=1,$ whether by any chance $ (2 M - N - L)$ is also an isometric invariant?

EDIT 3:

In other words, what qualifies $f(L,M,N)=0$ for isometric invariance?

Narasimham
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  • This is a very good question. Are you only considering the case of surfaces in $\mathbb{R}^3$? In that case, I believe that (functions of) the Gauss curvature $\det(\text{II}) = LN - M^2$ is the only intrinsic isometric scalar-valued invariant derivable from the second fundamental form $\text{II}$, but I don't have a proof. – Jesse Madnick Sep 03 '16 at 22:03
  • I do not know. For a time I thought L+M may be... anyway please mention anything you consider interesting. It seemed after Gauss proved the part in Egregium thm no one entered such an ambit. – Narasimham Sep 03 '16 at 22:13
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    $L+N$ certainly is not. The flat plane and a rolled up toilet roll are isometric, but one is minimal and the other is not. ($L+M$ is not even invariant under local changes of coordinates.) And since the only two scalar invariants associated to a two dimensional symmetric matrix are its trace and determinant, this justify Jesse's claim. – Willie Wong Sep 03 '16 at 22:17
  • There might be more possibilities if you allow derivatives, but the obvious ones (e.g. $|\nabla K|$) need the first fundamental form too. – Anthony Carapetis Sep 03 '16 at 22:20
  • @Willie Wong Yes i know, I was just imagining in line of invariants $I_1,I_2,I_3...$ when there is nothing at all. – Narasimham Sep 03 '16 at 22:37

0 Answers0