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This is the Theorema Egregium of Gauss.

Gaussian curvature (something defined from derivative of the normal unit vector) depends only from coefficients of the first fundamental form, so it is intrinsic. My question is "Why"?

The first fundamental form is given by these three coefficients $E=(\partial_1,\partial_1), F,G$ defined similarly.

Now, we have that $I(v_1\partial_1+v_2\partial_2)=Ev_1^2+2Fv_1v_2 + Gv_2^2$ is independent from the local chart, but its coefficients change. So, ok that $I$ is intrinsic, but why we call also its coefficients intrinsic if they are actually defined from a local chart and depend from the local chart? Why can I call everything which uses only $E,F,G$ to be intrinsic? Why is $E,F,G$ themselves intrinsic?

HaroldF
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  • Might be helpful: understanding of the first fundamental form –  Jul 05 '17 at 21:36
  • See also: Does intrinsic mean existing regardless of some bigger space? –  Jul 05 '17 at 21:39
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    The word "intrinsic" means "having to do only with $I$ and not with the unit normal vector." The coefficients $E$, $F$, $G$ are intrinsic, yes, but also depend on the local chart. There is no contradiction in that. Meanwhile, the definition of Gauss curvature $K$ appears to be extrinsic, yet it is independent of the chart used to compute it. Once $K$ is expressed in terms of $E$, $F$, $G$ (and their derivatives) alone, we see that $K$ is actually intrinsic. The formula for $K$ in terms of $E, F, G$ is independent of the local chart, because $K$ is. – Jesse Madnick Jul 05 '17 at 23:29
  • The problem is probably with the meaning (to me) of the word intrinsic in this particular subject. To me (and my textbook) the word intrinsic stands for independence from "how I see the surface in the ambient space $\mathbb{R}^3$. For $I$ there is a good theorem which says that the tangent plane is canonically isomorphic to the derivation space of an equivalence class of smooth functions. So my questions are: Is the word intrinsic has nothing to do with chart independence? Is this the real reason why $I$ is intrinsic? – HaroldF Jul 06 '17 at 09:11

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