# Link between the hairy ball theorem and the fundamental theorem of algebra

I read in the book "Concepts of modern mathematics" by Ian Stewart that it was possible to proof the fundamental theorem of algebra using the hairy ball theorem (complete reference to the page is in the following quora question : https://www.quora.com/How-can-the-hairy-ball-theorem-be-used-to-prove-the-fundamental-theorem-of-algebra). I am not aware of such a proof. It is not obvious to me (compactification does not seem to lead to a satisfactory answer) and I would be delighted if someone could help me with that.

• I see the idea, but I am not sure that it works that easy. It is not obvious to me that we can extend a polynomial into a continuous function on its one-point compactification. For instance P(z) = z tends to +\infty when $z$ real tends to $+\infty$ or $-\infty$ if $z$ tends to $-\infty$, and thus the function on the compactified space would not be continuous, isn't it ? Also, take a constant polynomial, and the approach seems to be contradictory. maybe this is the right approach, but there are some technical difficulties that I do not manage to overcome. – RandomTopics May 2 '16 at 10:22
• non-constant Polynomials are proper hence extend to the one point compactification – Thomas Rot May 2 '16 at 11:49
• Right, I didn't read your comment properly. The 'point' of the one-point compactification is that all infinities ($+\infty, -\infty, i \infty$ etc) are the same point (often just denoted $\infty$) so that the extension to the sphere IS continuous (as Thomas also says). But this extension takes values in the sphere rather than $\mathbb{C}$ as, for instance, $f(\infty) = \infty$ – Vincent May 2 '16 at 11:53
• Both theorems amount to $\pi_1(S^1)=\mathbb{Z}$, so in this sense they are equivalent. – Alex Degtyarev May 2 '16 at 11:58
• You probably want the slightly more sophisticated version of the Hairy Ball theorem which says that the sum of the indices of the zeros equals 2. – HJRW May 2 '16 at 17:50

Maybe this is an idea. It suffices to prove that any polynomial $f\in \mathbf C[z]$ of even degree $2d\geq2$ has a zero in $\mathbf C$. Assume that it has no zero. Then the rational section $$s=f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d}$$ of the $d$-th tensor power $T^{\otimes d}$ of the holomorphic tangent bundle on the complex projective line $\mathbf P^1(\mathbf C)$ is a nowhere vanishing global section. This is so since the holomorphic vector field $\partial/\partial z$ does not vanish on $\mathbf C$, has a zero of order $2$ at $\infty$, the polynomial $f$ does not vanish on $\mathbf C$, and has a pole of order $2d$ at $\infty$. Consider the section $s$ as a topological surface in the complex line bundle $T^{\otimes d}$ over $\mathbf P^1(\mathbf C)$. As image of the sphere $S^2$, it is homeomorphic to $S^2$ and does not have any nontrivial topological coverings. Its inverse image under the power-$d$-map $$T\rightarrow T^{\otimes d}$$ from the complex line bundle $T$ into the complex line bundle $T^{\otimes d}$ is, therefore, a disjoint union of $d$ disjoint copies of the sphere $S^2$ in the complement of the zero section of the tangent bundle of $S^2$. Each would define a nowhere vanishing vector field on $S^2$. Contradiction by the hairy ball theorem.