According to Elementary Differential Geometry by A N Pressley:

$\Large\textbf{5.3. Conformal Mappings of Surfaces}$

Now that we understand how to measure lengths of curves on surfaces, it is natural to ask about angles. Suppose that two curves $\boldsymbol{\gamma}$ and $\boldsymbol{\tilde\gamma}$ on a surface $\cal S$ intersect at a point $P$ that lies in a surface patch $\boldsymbol{\sigma}$ of $\cal S$. Then, $\boldsymbol\gamma(t)=\boldsymbol\sigma(u(t),v(t))$ and $\boldsymbol{\tilde\gamma}(t)=\boldsymbol\sigma(\tilde u(t),\tilde v(t))$ for some smooth function $u,v,\tilde u$ and $\tilde v$, and for some parameter values $t_0$ and $\tilde {t_0}$, we have $\boldsymbol\sigma(u(t_0),v(t_0))$ $=$ $P$ $=$ $\boldsymbol\sigma(\tilde u(\tilde{t_0}),\tilde v(t_0))$.

The angle $\theta$ of intersection of $\boldsymbol{\gamma}$ and $\boldsymbol{\tilde\gamma}$ at $P$ is defined to be the angle between the tangent vectors $\dot{\boldsymbol{\gamma}}$ and $\dot{\boldsymbol{\tilde\gamma}}$ (evaluated at $t\mathop =t_0$ and $t\mathop=\tilde{t_0}$, respectively). Using the dot product formula for the angle between vectors, we see that $\theta$ is given by $$\cos\theta=\frac{\dot{\boldsymbol{\gamma}}\cdot\dot{\boldsymbol{\tilde\gamma}}}{\|\dot{\boldsymbol{\gamma}}\|\|\dot{\boldsymbol{\tilde\gamma}}\|}.$$ By the chain rule, $$\dot{\boldsymbol{\gamma}}=\boldsymbol\sigma_u\dot{u}+\boldsymbol\sigma_v\dot{v},\quad\dot{\boldsymbol{\tilde\gamma}}=\boldsymbol\sigma_u\dot{\tilde u}+\boldsymbol\sigma_v\dot{\tilde v},$$ so $$\begin{align}\dot{\boldsymbol{\gamma}}\cdot\dot{\boldsymbol{\tilde\gamma}}&=(\boldsymbol\sigma_u\dot{u}+\boldsymbol\sigma_v\dot{v})\cdot(\boldsymbol\sigma_u\dot{\tilde u}+\boldsymbol\sigma_v\dot{\tilde v})\\&=(\boldsymbol\sigma_u\cdot\boldsymbol\sigma_u)\dot{u}\dot{\tilde u}+(\boldsymbol\sigma_u\cdot\boldsymbol\sigma_v)(\dot{u}\dot{\tilde v}+\dot{\tilde u}\dot{v})+(\boldsymbol\sigma_v\cdot\boldsymbol\sigma_v)\dot{v}\dot{\tilde v}\\&=E\dot{u}\dot{\tilde u}+F(\dot{u}\dot{\tilde v}+\dot{\tilde u}\dot{v})+G\dot{v}\dot{\tilde v}.\end{align}$$ Replacing $\boldsymbol{\tilde\gamma}$ by $\boldsymbol\gamma$ (resp. $\boldsymbol\gamma$ by $\boldsymbol{\tilde\gamma}$) gives similar expressions for $\|\dot{\boldsymbol{\gamma}}\|^2=\dot{\boldsymbol{\gamma}}\cdot\dot{\boldsymbol{\gamma}}$ (resp. $\|\dot{\boldsymbol{\tilde\gamma}}\|^2$), which finally give the formula $$\cos\theta=\frac{E\dot{u}\dot{\tilde u}+F(\dot{u}\dot{\tilde v}+\dot{\tilde u}\dot{v})+G\dot{v}\dot{\tilde v}}{(E\dot{u}^2+2F\dot{u}\dot{v}+G\dot{v}^2)^{1/2}(E\dot{\tilde u}^2+2F\dot{\tilde u}\dot{\tilde v}+G\dot{\tilde v}^2)^{1/2}}.\tag5$$

(${\bf \sigma}_u=:\dfrac{\partial \sigma}{\partial u}$).

The above text several times assuming that $\dfrac{\partial \sigma}{\partial u}=\dfrac{\partial \sigma}{\partial \bar{u}}$ and $\dfrac{\partial \sigma}{\partial v}=\dfrac{\partial \sigma}{\partial \bar{v}}$. Why these assumptions are correct?


1 Answer 1


It is only a matter of notation. The parametrization $\sigma:U\to\mathbb{R}^3$ has its two directional derivatives, in the directions of the standard basis vectors. These directional derivatives are denoted by $\sigma_u$ and $\sigma_v$. Together, these directional derivatives give the differential of $\sigma$. Then we have different paths in $U$, but whenever calculating a tangent vector, we use the chain rule and plug in the differential of $\sigma$. This differential does not depend on the path in $U$.

Edit: More explicitly: We have two curves - $\gamma$ and $\tilde{\gamma}$. Instead of using the confusing notation used in your text book, let us write$$\gamma=\sigma\circ\delta,\quad\tilde{\gamma}=\sigma\circ\tilde{\delta},$$where $\delta$ and $\tilde{\delta}$ are paths in $U$. By the chain rule, we have$$\left.\frac{d}{dt}\right|_{t=0}\gamma(t)=d\sigma_{\delta(0)}\left(\left.\frac{d}{dt}\right|_{t=0}\delta(t)\right),\qquad\left.\frac{d}{dt}\right|_{t=0}\tilde{\gamma}(t)=d\sigma_{\tilde{\delta}(0)}\left(\left.\frac{d}{dt}\right|_{t=0}\tilde{\delta}(t)\right).$$Assumnig $\delta(0)=\tilde{\delta}(0)$ (otherwise, there isn't any angle between the two paths), we see that the differential $d\sigma$ is the same in both expressions.

This is all there is to it.

  • 2
    $\begingroup$ $u$ and $\bar {u}$ are different in general, so $\dfrac{\partial \sigma}{\partial u}=\dfrac{\partial \sigma}{\partial \bar{u}}$ may not hold in general. $\endgroup$
    – user200918
    Aug 28, 2015 at 15:23
  • 1
    $\begingroup$ @LuckyGuy Really, it's just notation. Anyway, please see the edit in my answer. $\endgroup$ Aug 28, 2015 at 20:52

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