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Suppose $ds^2 = Edu^2 + 2Fdudv + Gdv^2$ is the first fundamental form of some regular surface patch, show $EG - F^2 > 0$ for each point on the surface patch.

So technically I could put $E$, $F$, and $G$ into the metric tensor and define a surface patch $\sigma$ to find the determinant and show it is not $0$.

But the problem is that I want to find a way to show this is indeed strictly positive without refering to the surface patch $\sigma$. Is it possible to do this in terms only in $E$ , $F$, and $G$?

Lemon
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  • It is impossible because $E$, $F$ and $G$ are the components of the first fundamental form in a coordinate patch $(u,v)$ – Yuri Vyatkin Apr 17 '13 at 06:51
  • Impossible as in, it is not possible to prove it without referring it to the surface patch $\sigam$? – Lemon Apr 17 '13 at 07:04
  • To be honest, your question is not clear to me. Do you want to prove that the first fundamental form is positive definite in a coordinate-free manner? When I said "impossible" I meant that in the form as the problem is stated this is, probably, not expected. Of course, one can show this in an invariant way, but using a different notation... – Yuri Vyatkin Apr 17 '13 at 08:22
  • Yes. I want to prove that the first fundamental form is positive definite in a coordinate free manner – Lemon Apr 17 '13 at 18:56

1 Answers1

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Even though the answer is in a sense completely obvious, I have to admit that the question is good, because it checks the understanding.

Answer. The first fundamental form (a.k.a. the induced metric) is positive-definite because the metric in the background space is assumed to be positive-definite, and on the vectors tangential to the surface both metrics have the same value by definition.

Explanations. I decided to add a few lines to exhibit the context. So, let $f \colon U \to \Bbb R^3$ be a parametrization of a regular surface $S_f$. Regularity means that $\mathrm{d}f$ is injective. Denote by $G$ the inner product in $\Bbb R^3$ (usually $G$ is taken to be the Euclidean metric). The first fundamental form of the surface $S_f$ is defined by its action on the vectors $X$ and $Y$ tangent to the surface as $$ g(X,Y) := G(\mathrm{d}f(X), \mathrm{d}f(Y)) $$ Looking at this definition one can immediately convince themselves that the first fundamental form is positive definite.

Yuri Vyatkin
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