I've encountered several definitions for conformal maps, and I was wondering whether they are equivalent or not. My goal is understanding how the concepts of conformality and holomorphy relate to each other.
One of the definitions says that conformal maps are the ones which preserve angles at a certain point (both orientation and size). Thus, for a map to be conformal in an open set U, it must be conformal at each point in U.
On the other hand, a conformal map can be defined as a holomorphic function with non-vanishing derivative at all points in some open set U.
Under the second definition, it is easy to prove that conformality and holomorphy are equivalent. Under the first definition, it is clear that holomorphy still implies conformality, but it isn't as obvious that conformality implies holomorphy.
Could anyone shed some light on this topic? Thanks in advance.
Best Answer
I assume you are talking about a two-dimensional set $U$. For simplicity, $U\subset\mathbb R^2$, $f\colon U\to \mathbb R^2$ to avoid geometric issues.
Let's say that in the first definition you assume that the function is $C^1$. Take a conformal map $f$ as in the first definition. The fact that $f$ preserves angles can be rewritten as follows (this should be proved of course, but I will skip this):
If you also assume the orientation is preserved, the jacobian of $f$ at every point is of the form $$ Jf(x,y)=\left[\begin{aligned}a\,\,&b\\-b\,\,&a \end{aligned}\right],\qquad\exists a,b\in\mathbb R.$$ That is, $f$ satisfies the Cauchy-Riemann equations, that is, if $f(x,y)=(u(x,y),v(x,y))$, $$ \frac{du}{dx}=\frac{dv}{dy},\,\,\,\,\frac{dv}{dx}=-\frac{du}{dy}. $$ This is in fact equivalent to the function $f$ being holomorphic if $f$ is $C^1$.
Edit. So, I guess that the Jacobian is invertible is necessary to keep curves smooth when you transport them through the map $f$ (think about the holomorphic function $z\mapsto z^2$, which maps lines that pass through the origin to angle-shaped figures).
Once your jacobian is invertible, you essentially want that the cosine of two tangent vectors is preserved through the Jacobian. If you impose this to hold for arbitrary two vectors in the tangent plane, you should end up saying that the Jacobian is indeed an orthogonal transfomation (i.e., that preserves the scalar product) possibly multiplied by a non-zero constant: this follows from the fact that orthogonal pairs of vectors are mapped to orthogonal pairs of vectors. This is an exercise of linear algebra, I don't know how a reference for this sorry (it works in any dimension by the way).
Once you have that, in the two dimensional case it is easy to classify all orthogonal matrices that preserve orientation: $$ A=\left[\begin{aligned}\cos(\theta)\,\,&\sin(\theta)\\-\sin(\theta)\,\,&\cos(\theta) \end{aligned}\right],\qquad\exists \theta\in\mathbb R.$$ So when you allow for multiplication by a constant, you end up with the above ansatz.