KindiakMath

Surprising Theorems with Short Proofs

While our implementation of the fundamental group in algebraic topology is highly nontrivial—and rather technical—the fruit of our labor is surprisingly far-reaching.

Our first result is called Brouwer’s fixed point theorem (Theorem 1), which states conditions for which there exists x such that f(x) = x. This theorem was used by John Nash to develop the foundations of modern game theory, which is itself a crucial model in economic theory.

Theorem 1. Define \mathbb D := \{ z \in \mathbb C : |z| < 1\}. We remark that \bar{\mathbb D} = \mathbb D \cup S^1. For any continuous map g : \bar{\mathbb D} \to \bar{\mathbb D}, there exists z \in \bar{\mathbb D} such that g(z) = z.

Proof. Suppose for a contradiction that for any z \in \bar{\mathbb D}, g(z) \neq z. Define the half-open ray

R = \{ g(z) + t (z - g(z)) : t > 0\}.

Then there exists a unique r(z) \in S^1 such that

|g(z) + r(z) (z - g(z))| = 1.

Therefore, we obtain a continuous map r : \bar{\mathbb D} \to S^1. Furthermore, if z \in \bar{\mathbb D}, then r(z) = z. Hence, denoting the (continuous) inclusion map i(z) = z by i : S^1 \to \mathbb D, we obtain a continuous map

(r \circ i)(z) = r(i(z)) = r(z) = z\quad \Rightarrow \quad r \circ i = \mathrm{id}_{\bar{\mathbb D}}.

Since \pi_1(S^1, 1) = \mathbb Z and \pi_1(\mathbb D, 1) = 0, we obtain the composite map r_* \circ i_* = \mathrm{id}_{ \pi_1(S^1, 1) }, so that 0 = \pi_1(\mathbb D, 1) = \pi_1(S^1, 1) = \mathbb Z as group isomorphisms, a blatant contradiction.

Another famed theorem is the fundamental theorem of algebra (Theorem 2), which asserts the existence of at least one complex root for any complex-valued polynomial with complex coefficients.

Theorem 2. Given n \in \mathbb N^+ and constants a_0,a_1,\dots,a_{n-1} \in \mathbb C, define the polynomial p : \mathbb C \to \mathbb C by

\displaystyle p(z) := \sum_{k=0}^{n-1} a_k z^k + z^n.

Then there exists some z \in \mathbb C such that p(z) = 0.

Proof. Suppose for a contradiction that p \neq 0. In particular, p(z) \neq 0 for any z \in S^1. Define \hat p := p/|p| : S^1 \to S^1, which is nonzero by hypothesis.

In the case |z| < 1, define the homotopy H_1 : S^1 \times [0, 1] \to S^1 from e_{\hat p(0)} to \hat f by

\displaystyle H_1(z, t) = \frac{p(tz)}{|p(tz)|},

so that f_0 \simeq e_{\hat p(0)} \simeq \hat p, and j([\hat p]) = j([f_0]) = 0.

In the case |z| > 1, define the homotopy H_2 : S^1 \times [0, 1] \to S^1 from f_n to \hat p by

\displaystyle H_2(z, t) = \frac{t^n p(z/t)}{|t^n p(z/t)|},

so that f_n \simeq \hat p, and j([\hat p]) = j([f_n]) = n.

Hence, n = 0, a blatant contradiction.

Corollary 1. Given n \in \mathbb N^+, there exists some z_1,\dots,z_n \in \mathbb C such that p(z_k) = 0.

Proof. Induction using Theorem 2.

As discussed, there is much more to be explored in algebraic topology. But for one final application of topology, we will switch gears and explore topological vector spaces, which turns out the be the basis for modern Fourier analysis, which is itself a foundation for many engineering use-cases. We will only look at the first bit though—topological vector spaces.

—Joel Kindiak, 18 Jun 25, 1345H

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