A proof of the fundamental theorem of algebra using homotopy

fundamental theorem of algebra는 보통 Liouville’s theorem로 증명하는 것이 가장 일반적이지만 그래도 쫌 다르게 (폼나게?) 증명하는 증명법을 소개한다.

이 내용은 모두 C.T.C Wall책[1]에 있는 내용임.

 


1. notation >

X,Y are topological spaces.
I is the unit interval on \mathbb{R} ( = [0,1] ).
S^1 is the unit circle on complex plane.
H^0 (X) is the set of continuous map X \to \mathbb{Z}. (every subset of \mathbb{Z} is open.)
\pi_0(X) is path components of X.
F(X) is free abelian group of X.
H_0(X) := F(\pi_0(X))
[Y,X] is set of homotopy classes of maps f : Y \to X
H^1(X) := [X,S^1]
e : \mathbb{R} \to S^1, e(t) := \exp(2\pi it)

2. definition > lifting maps from S^1 up to \mathbb{R}

If f : X \to S^1 is continuous map, does there exist \hat{f} ?

If there is, \hat{f} is lifting map of f

3. theorem >

Any continuous map f : I \to S^1 has a lift \hat{f} : I \to \mathbb{R} which is unique up to translation by integer.
증명 생략

4. definition > degree of map

for any continuous map f : S^1 \to S^1,

there exist g : I \to \mathbb{R}
e(g(1)) = f(e(1)) = f(e(0)) = e(g(0)) (∵ e(0) = e(1) = 1)
g(1)-g(0) is an integer
\deg f := g(1)-g(0)
사실 degree는 S^1 에서 한 바퀴 돌 동안에 f 의 image에서 돈 횟수이다.

5. a horde of theorems >

  1. \deg(f\circ g) = \deg f + \deg g
  2. degree 자체가 function이므로 \deg : H^1(S^1) \to \mathbb{Z}
    The degree map is group isomorphism.
  3. The two homotopic maps have same degree.
  4. the following conditions are equivalent
    i) f is nullhomotopic
    ii) \deg f = 0
    iii) f has continuous lift \hat{f} : S^1 \to \mathbb{R}
  5. If the map f is not surjective, \deg f = 0. (i.e f is nullhomotopic)

3번만 빼면 쉽다.

6. theorem > Fundamental theorem of algebra

proof >
We will suppose not, and find a contradiction.
Write the polynomial as

\displaystyle P(z) = \sum_{i=0}^n a_i z^i

Without loss of generality, we may suppose a_n=1
We define a map F : S^1 \times \mathbb{R}^+ \to S^1 by F(z,r) = P(rz)/|P(rz)|
If we write f_r(z) = F(z,r), then F provides a homotopy between the maps f_r. Now f_0 is constant, so has zero degree. We will show that for r large enough, f_r has degree n. This will give the required contradiction.

Choose

\displaystyle R > \max\left( \sum_{i=0}^{n-1} |a_i| , 1\right)

Then for |z| = 1,

\displaystyle \left| \sum_{i=0}^{n-1} a_i (Rz)^i \right| \le \sum_{i=0}^{n-1} |a_i|R^i \le R^{n-1} \sum_{i=0}^{n-1} |a_i| < R^n = |(Rz)^n|

So

\displaystyle \left| \frac{\sum_{i=0}^{n-1}a_i(Rz)^i}{(Rz)^n}\right| < 1

It follows that P(Rz)/(Rz)^n has a positive real part. Hence so has

\displaystyle \frac{P(Rz)}{(Rz)^n} \left|\frac{(Rz)^n}{P(Rz)}\right| = \frac{f_R(z)}{z^n}

So, the map f_R(z)/z^n has zero degree. Hence \deg f_R equals the degree of the map z^n.

The theorem now follows

 


[1] C. T. C. Wall, A geometric introduction to topology. Addison-Wesley 1972, Dover 1993.

 


2014.7.10

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