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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.

Amir Sagiv
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RandomTopics
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  • 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 02 '16 at 10:22
  • I guess what helps is to ASSUME your function $f$ is non-constant and nowhere zero from the start (hoping to find a contradiction) and then work with the function $1/f$. But then we still are stuck with a continuous function from the sphere to $\mathbb{R}^2$ without a canonical way to identify this $\mathbb{R}^2$ with the tangent space to the sphere in each point. – Vincent May 02 '16 at 11:01
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    non-constant Polynomials are proper hence extend to the one point compactification – Thomas Rot May 02 '16 at 11:49
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    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 02 '16 at 11:53
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    Both theorems amount to $\pi_1(S^1)=\mathbb{Z}$, so in this sense they are equivalent. – Alex Degtyarev May 02 '16 at 11:58
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    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 02 '16 at 17:50
  • don't directl answer your question but here http://mathoverflow.net/questions/234777/gauss-proof-of-fundamental-theorem-of-algebra and http://mathoverflow.net/questions/132036/has-the-fundamental-theorem-of-algebra-been-proved-using-just-fixed-point-theory – john mangual May 02 '16 at 18:27
  • Perhaps one should have a look through the 52 answers at http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra (and maybe add a 53rd). – Gerry Myerson May 03 '16 at 01:00
  • I did not see any proof using fundamental theorem of algebra among these. – RandomTopics May 03 '16 at 15:15

2 Answers2

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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.

Johannes Huisman
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This is a (very) partial answer: Suppose the polynomial is real and of odd degree. Then this polynomial is the characteristic polynomial of some matrix acting in an odd number of dimensions. Restricting this matrix to the sphere and then projecting onto the tangent space of the sphere defines a vector field on an even dimensional sphere, hence by the Hairy Ball Theorem it vanishes somewhere. The place where it vanishes is an eigenvector of the matrix, hence there is a corresponding eigenvalue, hence the polynomial has a root.

I'm not sure if something similar can be done for arbitrary polynomials though.

Josh Lackman
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    Note that in the real odd degree case the polynomial takes different signs for large positive and negative arguments, so you can get by with a topologically simpler connectedness and continuity argument. – Noah Stein May 02 '16 at 18:20
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    @NoahStein Yes of course, that's the standard proof, but it gives no connection of the result with the hairy ball theorem which is what the question is about. – Josh Lackman May 02 '16 at 22:09