Does there exists polynomial map of degree m sending the S^n to itself?

polynomial-sphere-map.md

Polynomial Maps from $S^n$ to $S^n$

Motivated by https://x.com/gabrielpeyre/status/1837156819799577034

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Understanding Polynomial Maps on Spheres

Let's clarify what we mean by a polynomial map from $S^n$ to $S^n$:

• $S^n$ is the $n$-dimensional unit sphere in $ℝ^{n+1}$, defined as:

$$ S^n = \lbrace x \in \mathbb{R}^{n+1} \mid |x| = 1 \rbrace $$

• A polynomial map is a function $F: S^n \rightarrow S^n$ where each component of $F(x)$ is a polynomial in the coordinates of $x$.
• The goal is to construct such maps where $F(x)$ remains on $S^n$ for all $x \in S^n$. Caveat: $F(x)$ is not allowed to be a linear transformation and must have degree $m$.

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Quadratic Maps: The Starting Point

Intuition: in $S^1$, the unit circle, think of doubling angles:

• If $x = (\cos \theta, \sin \theta)$, then:

$$ F(x) = (\cos 2\theta, \sin 2\theta) $$

Notice the double-angle identities:

$$ \cos 2\theta = 2\cos^2 \theta - 1, \quad \sin 2\theta = 2\sin \theta \cos \theta $$

So, $F(x_0, x_1) = (2x_0^2 - 1, 2x_0 x_1)$ is a quadratic map such that $F(S^1) \subset S^1$.

In general, for a point $x = (x_0, x_1, \dots, x_n) \in S^n$, consider the map:

$$ F(x) = \left( 2x_0^2 - 1, 2x_0 x_1, 2x_0 x_2, \dots, 2x_0 x_n \right) $$

Verification

$$ \begin{align*} |F(x)|^2 &= \left( 2x_0^2 - 1 \right)^2 + \sum_{i=1}^n \left( 2x_0 x_i \right)^2 \\ &= \left( 4x_0^4 - 4x_0^2 + 1 \right) + 4x_0^2 \sum_{i=1}^n x_i^2 \\ &= \left( 4x_0^4 - 4x_0^2 + 1 \right) + 4x_0^2 (1 - x_0^2) \quad (\text{since } |x|^2 = 1) \\ &= 4x_0^4 - 4x_0^2 + 1 + 4x_0^2 - 4x_0^4 \\ &= 1 \end{align*} $$

Thus, $F(x)$ maps $S^n$ onto itself.

Generalizing to Higher-Degree Maps

The Big Idea

Can we construct a degree- $m$ polynomial map $F: S^n \rightarrow S^n$ for any positive integer $m$? Yes! The key lies in leveraging Chebyshev polynomials and trigonometric identities. We see what we did above is a special case of this, because we are now not doubling angles, but multiplying them by $m$.

Chebyshev Polynomials Connection

• Chebyshev Polynomials of the First Kind ($T_m$):

$$ T_m(x) = \cos(m \arccos x) $$

• Properties:
  • $T_m(\cos \theta) = \cos m\theta$
  • They are polynomials of degree $m$.

Now for $x = (x_0, x_1, \dots, x_n) \in S^n$, define:

$$ F(x) = \left( T_m(x_0), U_{m-1}(x_0) x_1, U_{m-1}(x_0) x_2, \dots, U_{m-1}(x_0) x_n \right) $$

where $U_{m-1}$ is the Chebyshev polynomial of the second kind.

Why Does This Work?

• Degree $m$: Both $T_m$ and $U_{m-1}$ are polynomials of degrees $m$ and $m-1$, respectively.
• Maps $S^n$ to $S^n$: Using trigonometric identities, we can show that $|F(x)| = 1$. For $x_0 = \cos \theta$, and $1 - x_0^2 = \sin ^2 \theta$, the map essentially multiplies the angle $\theta$ by $m$.

Verification

$$ |F(x)|^2 = T_m(x_0)^2 + \left( U_{m-1}(x_0) \right)^2 \sum_{i=1}^n x_i^2 = \cos^2 m\theta + \sin^2 m\theta = 1 $$