First fundamental form question.

Question

The question I posted;

$6.1.2\quad$ Show that and apply an isometry of $\Bbb R^3$ to a surface does not change its first fund. form. What is the effect of a dilation (i.e., a map $\Bbb R^3\to \Bbb R^3$ of the form $v\mapsto av$ for some constant $a\ne 0$.

And I posted its answer as below photo. I cannot generally understand its answer.

$6.1.2\,$ Applying a translation to a surface patch $\boldsymbol{\sigma}$ does not change $\boldsymbol{\sigma}_u$ or $\boldsymbol{\sigma}_v$. If $P$ is a $3\times3$ orthogonal matrix, $P(\boldsymbol{\sigma})_u=P(\boldsymbol{\sigma}_u)$, $P(\boldsymbol{\sigma})_v=P(\boldsymbol{\sigma}_v)$, and $P$ preserves dot products $(P(\mathbf p)\cdot P(\mathbf q)=\mathbf p\cdot\mathbf q$ for all vectors $\mathbf p$, $\mathbf q\in\Bbb R^3)$. Applying the dilation $(x,y,z)\mapsto a(x,y,z)$, where $a$ is a non-zero constant, multiplies $\boldsymbol{\sigma}$ by $a$ and hence the first fundamental form by $a^2$.

Please explain that more clearly and explicitly. Thanks a lot.

Answer 1

Let $\sigma: U\subseteq \mathbb R^2\rightarrow \mathbb R^3$ $(u,v)\mapsto \sigma(u,v)=(x(u,v),y(u,v),z(u,v))$ parametrize a 2 dimensional surface in $\mathbb R^3$ by coordinates $(u,v)$. With

$$I(X,Y):=\langle X,Y\rangle:=\sum_{i,j=1}^2 a_ib_j\langle \sigma_i,\sigma_j\rangle $$

we denote its first fundamental form, computed for all tangent vectors

$$X=a_1\sigma_1+a_2\sigma_2, $$

$$Y=b_1\sigma_1+b_2\sigma_2, $$

at any given point of the surface. We introduced the notation $\sigma_1:=\sigma_u=\frac{\partial \sigma}{\partial u}$ and $\sigma_2:=\sigma_v=\frac{\partial \sigma}{\partial v}$ for sake of clarity.

• Translations

A translation

$$\sigma'(u,v):=\sigma(u,v)+b, $$

with $b\in\mathbb R$ has no effect on the tangent vectors $\sigma_u$ and $\sigma_v$, i.e.

$$\sigma'_u=\sigma_u, $$ $$\sigma'_v=\sigma_v, $$

and the first invariant form is unaffected by translations.

This follows immediatly by computing $\frac{\partial \sigma'}{\partial u}$ and $\frac{\partial \sigma'}{\partial v}$.

• Orthogonal transformations

By definition of orthogonal transformation (represented by a matrix $O$, once we choose a basis in $\mathbb R^3$)

$$I(OX,OY):=\langle OX,OY\rangle=\langle O^TOX,Y\rangle=\langle X,Y\rangle=I(X,Y),$$

as $O^TO=I$. The first invariant form is then invariant under orthogonal transformations. In the reference presented in the OP this fact is summarized under "...and $P$ preserves the dot product".

• Dilations

A dilation

$$\sigma'(u,v):=\alpha\sigma(u,v), $$

with $\alpha\in\mathbb R^{+}$ is such that

$$\sigma'_u=\alpha\sigma_u, $$ $$\sigma'_v=\alpha\sigma_v; $$

we arrive at

$$I(X',Y')=\sum_{i,j=1}^2 a_ib_j\langle \sigma'_i,\sigma'_j\rangle =\sum_{i,j=1}^2 a_ib_j\langle \alpha\sigma_i,\alpha\sigma_j\rangle=a^2I(X,Y),$$

i.e. applying a dilation by $\alpha$ the first fundamental form is multiplied by $\alpha^2$.