4
$\begingroup$

I'm working on a problem from Hatcher's Algebraic Topology, and I want to show that if we take polynomial $f(x)$ definded on the Riemann sphere mapping to the Riemann sphere, the local degree of the map (when viewed as a map $S^2 \to S^2$) at a root is equal to the multiplicity of the root.

Intuitively this seems obvious because if we look at a neighborhood of some root with multiplicity $n$ and translate the function to the origin, we get a map that kind of looks like the map $z \to z^n$, given a sufficiently small neighborhood, which has degree $n$ as desired. My problem is making my claim rigorous because the polynomial does not map perfectly from $z \to z^n$ in the neighborhood. I'm not sure how to compute the local degree directly. I know it means looking at the induced map $H_2(U,U-x_0) \to H_2(f(U), f(U)-0)$ where $x_0$ is a root and $U$ is a neighborhood of $x_0$.

$\endgroup$

2 Answers 2

4
$\begingroup$

I believe I figured it out.

We can use the argument principle, which states the number of zeros minus the number of poles around a small contour around the root $x$ is the change in the argument of $f(z)$ as $z$ travels around the contour, divided by $2 \pi$. This means the multiplicity of $x$ is number of times $f(z)$ travels around $0$ when $z$ travels around a sufficiently small contour surrounding $x$. In other words, this means the local degree of $f$ at the root is the multiplicity of the root as desired.

$\endgroup$
1
$\begingroup$

Here's a way to do this without complex analysis, which is based on the idea that $f$ "looks like" (i.e., is homotopic in some sense to) the map $z \mapsto z^n$ near the root.

If $f$ has a root at $a \in \mathbb{C}$, we can write $f(z) = \lambda (z - a)^k(z - b_1) \cdots (z - b_m)$ by the fundamental theorem of algebra. For simplicity, let's assume $a = 0$ and $\lambda = 1$. We have a homotopy $$ h_t(z) = z^k(tz - b_1) \cdots (tz - b_m) $$ from the map $g(z) = \beta z^k$ (where $\beta = b_1 \cdots b_m$) to $f$. Observe that if $z$ is close to $0$, each term $tz - b_j$ is close to $b_j$, and in particular is nonzero. Therefore, for a sufficiently small neighborhood $U$ of $0$, the homotopy $h_t$ restricts to a homotopy $h_t : (U, U - 0) \to (\mathbb{C}, \mathbb{C} - 0)$. It follows that the induced map $f_* : H_*(U, U - 0) \to (\mathbb{C}, \mathbb{C} - 0)$ is the same as the induced map $g_* : H_*(U, U - 0) \to (\mathbb{C}, \mathbb{C} - 0)$. Note that these are the maps involved in the computation of the local degrees of $f$ and $g$, respectively. Thus, the local degree of $f$ at $0$ is the same as the local degree of $g$ at $0$, which we know is $k$.

Note that it is crucial that the maps $h_t$ send $U - 0$ to $\mathbb{C} - 0$, since that allows us to invoke the homotopy invariance of singular homology. Thus, we are using a fact that is stronger than $f$ and $g$ merely being homotopic (which is rather trivial since $\mathbb{C}$ is contractible).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .