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I am looking at the following exercise of the book of A.Pressley:

Let $$Edu^2+2F dudv+Gdv^2$$ be the first fundamental form of a surface patch $\sigma(u, v)$ of a surface $S$.

Show that, if $p$ is a point in the image of $\sigma$ and $v, w \in T_pS$, then $$\langle v, w \rangle = E d u (\textbf{v})d u (\textbf{w}) + F(du(\textbf{v})dv(\textbf{w}) + du(\textbf{w})dv(\textbf{v})) + Gdv(\textbf{w})dv(\textbf{w})$$

$$$$

Can you give an idea?

I don't really have a clue what to do.

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    $\begingroup$ how do you understand $du^2, dudv$ and $dv^2$? $\endgroup$ – janmarqz Dec 22 '15 at 1:25
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    $\begingroup$ and how do you understand $T_p S$, what is a vector $v \in T_p S$? $\endgroup$ – Curiosity Dec 22 '15 at 8:05
  • $\begingroup$ $T_pS$ is the tangent space and a vector $v\in T_pS$ is a tangent vector. Correct? @LeVanTu $\endgroup$ – user175343 Dec 22 '15 at 14:53
  • $\begingroup$ The First fundamental form is a dot product. Correct? @janmarqz $du$ and $dv$ are linear maps. Or not? $\endgroup$ – user175343 Dec 22 '15 at 14:56
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    $\begingroup$ the point of view in A.Pressley is a little difference from the one I known in Do Carmo. And I think the Do Carmo "Differential geometry of Curves and Surface" would give you more information to understand the notion "first fundamental form". $\endgroup$ – Curiosity Dec 22 '15 at 15:08
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Ok, if you understand that the $du$ and $dv$ are linear functionals, then you must realise that the symbolisms $du^2$, $dudv$ and $dv^2$ are abbreviations for three bilinear maps $T_pS\times T_pS\to\Bbb R$ which are defined as soon as one understand that there is a basis for $T_pS$. And in fact, they should be written as $du\otimes du$, $du\otimes dv$ and $dv\otimes dv$.

But the basis for $T_pS$ is $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ where let us briefly rename as $\partial_u=\frac{\partial}{\partial u}$ and $\partial_v=\frac{\partial}{\partial v}$.

So, because $du(\partial_u)=1$, $du(\partial_v)=0$, $dv(\partial_v=0$ and $dv(\partial_v)=1$. Then $$du^2(\partial_u,\partial_u)=du(\partial_u)du(\partial_u)=1$$ $$du^2(\partial_u,\partial_v)=du(\partial_u)du(\partial_v)=0$$ $$du^2(\partial_v,\partial_u)=du(\partial_v)du(\partial_u)=0$$ $$du^2(\partial_v,\partial_v)=du(\partial_v)du(\partial_v)=0$$

Similar for $dudv$ and $dv^2$.

Now, if you are compelled to have a inner product in $T_pS$ then, the natural way to pair a couple of tangent vectors $X,Y$, one must consider them as a linear combinations $$X=X^u\partial_u+X^v\partial_v,$$ $$Y=Y^u\partial_u+Y^v\partial_v,$$ and one can define first the metric coefficients $E=\langle \partial_u,\partial_u\rangle$, $F=\langle \partial_u,\partial_v\rangle$ and $G=\langle \partial_v,\partial_v\rangle$.

Hence \begin{eqnarray*} \langle X,Y\rangle&=&(Edu^2+2Fdudv+Gdv^2)(X,Y)\\ &=&Edu^2(X,Y)+2Fdudv(X,Y)+Gdv^2(X,Y)\\ &=&EX^uY^u+2FX^uY^v+GX^vY^v\\ \end{eqnarray*} gives us a scalar product in $T_pS$.

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