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Question 1

Let $\Sigma$ a surface parametrized by $X(u,v)$. Than at $p=(u,v)\in \Sigma$, the tangent plan $T_p\Sigma$ is generated by $\{X_u,X_v\}$. So any vector in $T_p\Sigma$ can be written as $aX_u+bX_v$. In particular, the scalar product on $T_p\Sigma\times T_p\Sigma$ has matrix $(g_{ij})=\begin{pmatrix}E&F\\F&G\end{pmatrix},$ where $g_{ij}=\left<X_i,X_j\right>$. Now, it's written on my course that if $ds$ is an infinitesimal length of a curve on $\Sigma$ at the neighborhood of $p$, then $$ds^2=Edu^2+2Fdudv+Gdv^2,$$ and I don't really understand where it come from.

Attempt

I think that $$ds=\frac{\partial s}{\partial u}du+\frac{\partial s}{\partial v}dv,$$ and thus $$ds^2=\left(\frac{\partial s}{\partial u}\right)^2du^2+2\frac{\partial s}{\partial u}\frac{\partial s}{\partial v}dudv+\left(\frac{\partial s}{\partial v}\right)^2dv^2,$$ but why $E=\left(\frac{\partial s}{\partial u}\right)^2$, $F=\frac{\partial s}{\partial u}\frac{\partial s}{\partial v}$ and $G=\left(\frac{\partial s}{\partial v}\right)^2$ ?


Question 2

suppose that $\Sigma$ is a graph, i.e. $\{f(x,y)\mid x,y\in D\}$. Let $z=f(x,y)$. and suppose that the plane $z=0$ is tangent to the surface. Then, we can write $$z=L\frac{x^2}{2}+Mxy+N\frac{y^2}{2},$$ and the second fundamental form at the origin in the coordinate $x,y$ is the quadric form $$Ldx^2+2Mdxdy+Ndy^2,$$ with $$L=\frac{\partial ^2f}{\partial x^2}\cdot n, \quad M=\frac{\partial^2 f}{dxdy}\cdot n,\quad N=\frac{\partial ^2f}{\partial y^2}\cdot n,$$ where $n$ is normal to the plane $z=0$. What is the motivation for such a definition ? Does it give a new metric to the space ? I don't really see what it gives.

user386627
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  • Question 1: Write down what is means for $g_{ij}$ to be the matrix determining the inner product, and compute the length of the vector $ds=(du,dv)$. – DKS Nov 03 '17 at 16:47

1 Answers1

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For the first question: $ds^2$ is just another notation for the inner product, as in $ds^2(\vec{v},\vec{w}) = \langle \vec{v},\vec{w}\rangle$. Then $$\begin{align} ds^2(\vec{v},\vec{v}) &= \langle \vec{v},\vec{v}\rangle \\ &= \langle aX_u+b X_v, aX_u+bX_v\rangle \\ &= a^2 \langle X_u,X_u\rangle + 2ab\langle X_u,X_v\rangle + b^2\langle X_v,X_v\rangle \\ &= E a^2 + 2Fab + Gb^2 \\ &= E \;du(\vec{v})^2 + 2F\,du(\vec{v})dv(\vec{v}) + Gdv(\vec{v})^2 \\ &= (E\,du^2 + 2Fdu\,dv+G\,dv^2)(\vec{v},\vec{v}),\end{align}$$and so $ds^2 = E\,du^2 + 2Fdu\,dv + G\,dv^2$.

For the second question: you can think that $X_u$ and $X_v$ are always tangent to the surface, but the second derivatives $X_{uu}$, $X_{uv}$ and $X_{vv}$ need not be tangent. The Second Fundamental Form is a quadratic form which sort of measures how these second derivatives deviate from being tangent. In general, you write $$X_{ij} = \sum_k \Gamma_{ij}^k X_k + h_{ij}N,$$where $\Gamma_{ij}^k$ are certain coefficients (called the Christoffel Symbols of $X$, you'll meet them in due time), and $N$ is the unit normal of the surface. Taking the product with $N$ we obtain $h_{ij} = \langle X_{ij},N\rangle$. The Second Fundamental Form is not another metric in the surface, in general, because the quadratic form defined by it need not be definite (as in positive-definite or negative-definite). The quantity which will tell you if this is the case is called "Gaussian curvature" (and it has a whole lot of other interesting applications).

The text looks at the particular case where $\Sigma$ is a graph because this will tell you exactly how the Second Fundamental Form measures how $\Sigma$ looks like near an arbitrary point (and you do not lose any generality because any surface can be written locally as a graph over its tangent plane, and applying an isometry you can assume that said tangent plane is just $z=0$).

Being more precise, if the Second Fundamental Form at $p \in \Sigma$ is definite, $\Sigma$ looks like an elliptic paraboloid near $p$ (being entirely contained in one of the half-spaces determined by $T_p\Sigma$ - $p$ is said to be an elliptic point in this case). If the quadratic form is indefinite, it'll look like an hyperbolic paraboloid, and $p$ is said to be an hyperbolic point. You can look at pages $122$ and $123$ of Christian Bär's Elementary Differential Geometry (for example) for pictures and more details.

Ivo Terek
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  • Why do we denote $ds^2(u,v)=<u,v>$ ? By the way, thank you for all your explanation. I'll stud it more in detail tomorrow. – user386627 Nov 03 '17 at 18:03
  • Think of $\Bbb R^2$. We have $ds = \sqrt{dx^2 + dy^2}$ (Pythagorean theorem), so if $\vec{v} = (a,b)$, we have $$ds^2(\vec{v},\vec{v}) = dx(\vec{v})^2 + dy(\vec{v})^2 = a^2+b^2 = \langle \vec{v},\vec{v}\rangle.$$ – Ivo Terek Nov 03 '17 at 18:18