# Topological degree of a complex valued map defined over a circle

Given a continuous map $f \colon S^n \to S^n$, it induces a map $f_{*} \colon \tilde{H}_n(S^n) \to \tilde{H}_n(S^n)$ of the form $f_{*}(z)=k*z$, where $k$ is an integer. Define the degree of $f$ as $k$.



Let $$p(z) = a_nz^n+a_{n-1}z^{n-1} + \dots +a_0$$ a complex valued polynomial function. Let $\gamma : C \to S^1$ $$\gamma(z) = \dfrac{p(z)}{\mid p(z)\mid}$$ where $C$ is a circle contained in $\mathbb{C}$ (the same for $S^1$). Obviously $C$ doesn't touch any roots, so $\gamma$ is "well defined".

1) I have to prove that if the circle contains only one root of $p$ (call it $\alpha$), then the topological degree is $m$, the multiplicity of such root of $p$.

My Attempt. let $p(z)=(z-\alpha)^m\tilde{p}(z)$. Then consider the homotopy $$\Gamma(s,z) := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m}{\mid \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m \mid}$$ where $s$ ranges in $[0,1]$ and then I have to prove that $\dfrac{(z-\alpha)^m}{\mid (z-\alpha)^m \mid}$ has degree $m$.

2) Second task is to prove that if the circle contains two roots $\alpha, \beta$ then the degree of the map $\gamma$ is the sum of the multiplicities ($m,n$) of such roots. Here is my attempt I'm unsure about the correctness of such homotopy. Can someone help me pointing out why it is correct (or incorrect?)

consider the homotopy $$H(z,t) := := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)\beta+sa])^n}{\mid \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)\beta+sa])^n \mid}$$

The doubts arise because it's not very clear to me when an homotopy is valid and when not. Here I think it is right, because it is obviously continuous, but the fact that my polynomial is defined over a circle (so there is a hole in between) doesn't let me to be sure of such function.

• If you write a+\dots b, you see $a+\dots b$, but if you write a+\dots+b, you see $a+\dots+b$, with the dots centered. I take the latter to be standard usage and I changed it in the first line of this question (i.e. I added the seecond "+"). – Michael Hardy May 21 '14 at 15:56
• The problem is that there are so many equivalent definitions of the degree that it is hard to tell what you suppose is known and what not. The intuition why the degree of $z^n: S^1\to S^1$ is $n$ is that the image of the circle winds around the target circle $n$ times (the angle is $n\theta$). But if you choose a circle $C$ that doesn't contain $0$ in the interior, then $c\in C\mapsto z^n/|z^n|$ has degree zero, because it can be extended to a map $z^n/|z^n|$ from the interior of $C$ to $S^1$. The best written short book on this topic it Milnor's Topology from the diferentiable viewpoint. – Peter Franek May 21 '14 at 16:06
• @MikeMiller edited. I'm interested in the case $S^1$ obviously. – Riccardo May 21 '14 at 22:19
• The polynomial is defined on all of $\mathbb{C}$. So the only problem with your intended homotopies can be that the denominator vanishes for some $s\in [0,1]$ and $z\in C$. The assumptions on $p$ resp. $\tilde{p}$ guarantee that that doesn't happen here. So the homotopies are okay, and the only thing left is that $z\mapsto z^n$ has degree $n$. – Daniel Fischer May 22 '14 at 20:58
• @DanielFischer perfectly clear. Is it a problem if I'm citing this comment and make an answer so the question is "complete" and useful for other users? – Riccardo May 22 '14 at 21:01

First of all, the fact that $(z-\alpha)^n$ has degree $n$ over $C$ is equivalent to prove that $z^n$ has degree $n$ over a small circle around the origin. Intuitively $z^n$ wraps around the circle $n$ times. This can be formalized using the concepts of local degree and the formula for calculating the degree of a map from its local degrees (see for example Hatcher vol.1 page 136-137)
Then for the second task, we can consider the homotopy $$H(z,t) := := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)b+sa])^n}{\mid \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)b+sa])^n \mid}$$
Which is well defined thanks to the assumption on $p$ and $\tilde{p}$. (in particular the denominator does not vanish (the other roots are outside the circle and so $\tilde{p}(s\alpha + (1-s)z) \neq 0$ for every $s \in [0,1]$. To conclude, just note that the degree of $(z-\alpha)^{m+n}$ is obviously $m+n$