What's a local angle?

Question

When I was trying to understand the definition of conformal map I got confused. A conformal map is a function $f: U \to \mathbb C$ where $U \subset \mathbb C$ such that $f$ preserves local angles.

But what is the definition of a local angle?

I had no trouble imagining the meaning of oriented angle. But the meaning of local angle... I cannot guess what it is. I tried to find a definition on the internet but the search results were all unhelpful.

Also... in the article it seems that conformal and biholomorphic (=holomorphic and bijective) are used as synonyms.

Is a map conformal if and only if it is bijective and holomorphic?

Answer 1

Lets say you have two straight lines in the complex plane, $\gamma_1(t) = a_1 + d_1 t$ and $\gamma_2(t) = a_2 + d_2 t$, and lets further stipulate that there are $t_1$ and $t_2$ such that $\gamma_1(t_1) = \gamma_2(t_2)$. Now, $d_1$ and $d_2$ are complex numbers, so we can see that the angle between the lines where they meet is $\arg(d_2) - \arg(d_1).$

If $f:\mathbb{C} \to \mathbb{C}$ is a general function, then $g_1 = f\circ\gamma_1$ and $g_2 = f\circ\gamma_1$ might not be straight lines, but we want to find a way to talk about the angle between $g_1$ and $g_2$. When they were straight lines it was pretty easy, but now, if $f$ is differentiable as a transformation of $\mathbb{R}^2,$ they are only curves. To find their direction at any moment, we define $$g_1'(t_0) = \frac{dg_1}{dt}(t_0) = \lim_{h\to 0} \frac{g_1(t+h) - g(t)}{h}$$ as you would expect, and you can see that this is a complex number if we think of $\mathbb{C}$ as being the codomain of $g_1$, or $\mathbb{R}^2$, if we think of it as being vector valued.

So, we know that $g_1(t_1) = g_2(t_2)$ and we can calculate $\arg(g_2'(t_2)) - \arg(g_1'(t_1))$ to find the angle between these momentary directions. Its this quantity that they are calling the "local angle". Its the right generalisation for the angle between lines to the angle between curves.

It turns out that, if $f:\mathbb{R}^2\to \mathbb{R}^2$ and $f(a, b) = (u(a, b), v(a, b))$ then we can define $$f' = \begin{pmatrix} \frac{ \partial u }{\partial x} & \frac{ \partial u }{\partial y} \\ \frac{ \partial v }{\partial x} & \frac{ \partial v }{\partial y} \end{pmatrix}$$ and if a curve $c : \mathbb{R} \to \mathbb{R}^2$ is differentiable at $a$ then $(f\circ c)' = f' c'$ where the operation between the terms on the RHS is matrix multiplication. Now, if you think about what the Cauchy-Riemann equations say about complex differentiable functions, then either $f' = 0$ or $f'$ is a rotation and a dilation. So, complex differentiability implies that the derivative of the function is $0$ or it will rotate directions of curves at the same point by the same amount, and then change their length by a scaling factor. This implies you cannot have changed the angle between curves, and so $f$ is conformal if its complex differentiable and its derivative is non zero.

Regarding your second question, $exp$ is conformal and holomorphic, but its not a bijection.

Hope that helps.

Answer 2

This means that the derivative of $f$ is a linear map that preserves angles (so it's a combination of scaling and rotation). Maybe it's helpful to look at this picture from Wikipedia: infinitesimally, the grid blocks are being rotated and scaled.

Holomorphic functions with nonzero derivative always have this property; if you write $f'(z_0) = r e^{i \theta}$, then locally at $z_0$, $f$ is a scaling by $r$ and rotation by $\theta$.

No, a conformal map, in the usual definition, is only a local biholomorphism in general - in other words, at any $z_0$, there's an open neighborhood $U$ such that $f : U \rightarrow f(U)$ is a biholomorphism. A counterexample is $f(z) = e^z$ on $\mathbb{C}$, which has nonzero derivative everywhere (so it's conformal) but it's not injective.