Geometric interpretation of a quintic's roots as a pentagon?

Solution 1:

Polynomials aside, this question is about how arbitrary sets of $n$ points on a line (namely, the $x$-axis) relate to vertices of regular $n$-gons. Even with the polynomial stuff, it's worthwhile to generalize to arbitrary sets of points in the plane (identified with the complex plane for root-of-polynomial purposes).

The fact is: Any set of $n$ points in the plane can be obtained from a "sum" of vertices of $n$ regular $n$-gons.

We can derive the desired relationship in two steps:

1. Show that any polygonal cycle of $n$ points in the plane can be written as an appropriately-defined "sum" of $1+\lfloor n/2 \rfloor$ affinely-regular $\{n/m\}$-gons (that is, images of regular $\{n/m\}$-gons under affine ---in fact, merely linear--- transformations).

The figure shows a polygon (upper-right) with coordinate matrix $$\left[\begin{matrix} 3 & 1 & 4 & 15 & 9 \\ 2 & 6 & 5 & 3 & 5 \end{matrix} \right]$$ and its three affinely-regular components: a convex $5$-gon, a "starry" $\{5/2\}$-gon, and a black $\{5/0\}$-gon (with all vertices coinciding at the centroid of the polygon). The tinted segments indicate how the vertices of the polygon are "vector sums" of the centroid and corresponding vertices of the other affine components.

2. Show that any affinely-regular polygon (that isn't already regular) is the "sum" of two actually-regular polygons, so that, ultimately, the original polygonal cycle is the "sum" of $n$ regular polygons.

The next figure decomposes our polygon's affinely-regular components into regular components. The tinted paths emanating from the centroid are updated to show the vector sums:

In what follows, we take $P = \{p_0, p_1, \dots, p_{n-1}\}$ to be an ordered list of points (the vertices of a $n$-gonal cycle, or simply an $n$-gon) in the plane, and we write $[P]$ for the matrix whose $k$-th column is the coordinate vector of $p_k$ in $\mathbb{R}^2$. Given a linear transformation $L$, we write $[L]$ for its matrix. And we let matrix addition induce a "sum" on polygons and on transformations: $$[P_1 \oplus P_2] := [P_1] + [P_2] \qquad\qquad [L_1 \oplus L_2] := [L_1] + [L_2]$$

Also, I'm using the Schlafli polygon symbol $\{n/m\}$ in the Grunbaum sense for not-relatively-prime $m$ and $n$. For instance, I consider the $\{6/2\}$-gon to be, not a compound of two triangles on six points, but a doubly-traced triangle on three points. (See the definition of $R_m$ below.)

For Step 1, consider these coordinate matrices of regular $\{n/m\}$-gons for $m = 0, 1, 2, \dots, \lfloor n/2 \rfloor$:

$$R_m := \begin{cases} \sqrt{\frac{1}{n}} \;\left[\;\matrix{1 & \phantom{-}1 & 1 & \cdots & \phantom{-}1} \;\right] & m = 0 \\[6pt] \sqrt{\frac{1}{n}} \;\left[\;\matrix{1 & -1 & 1 & \cdots & -1} \;\right] & m = n/2 \\[6pt] \sqrt{\frac{2}{n}} \;\left[\;\matrix{1 & \cos\frac{2\pi m}{n} & \cos\frac{4\pi m}{n} & \cdots & \cos\frac{2(n-1)\pi m}{n} \\ 0 & \sin\frac{2\pi m}{n} & \sin\frac{4\pi m}{n} & \cdots & \sin\frac{2(n-1)\pi m}{n}} \;\right] & \text{otherwise} \\ \; \end{cases}$$ (See that, for $m=6$, $n=3$, the corresponding hexagon wraps-around twice on the vertices of an equilateral triangle. This is why we diverge from the formal definition of the Schlafli symbol.) A bit of trig, along with the convenient scale factors, ensure this key relation: $$\sum_{m=0}^{\lfloor n/2 \rfloor} R_m^\top R_m = I \tag{2}$$

Now, for a given $n$-gon $P$ in the plane, consider the family of $n$-gons $P_m$ defined by $$[P_m] := [P] R_m^\top R_m$$

By $(2)$, $$\sum_{m=0}^{\lfloor n/2 \rfloor} [P_m] = [P]\;\sum_{m=0}^{\lfloor n/2 \rfloor} R_m^\top R_m = [P]I = [P] \qquad\to\qquad\bigoplus_{m=0}^{\lfloor n/2 \rfloor} P_m = P \tag{3}$$ so that $P$ is the "sum" of the $P_m$s, each of which is an affinely-regular $\{n/m\}$-gon. This completes Step 1. $\square$

Step 2 amounts to a simple observation about real $2$-by-$2$ matrices. $$M = U S V^\top = U\left[\matrix{ s_1 & 0 \\ 0 & s_2 } \right] V^\top = U \left( Q_1 + Q_2 \right) V^\top = U Q_1 V^\top + U Q_2 V^\top \tag{1}$$ where $USV^\top$ is the singular value decomposition of $M$, and $$Q_1 := \frac{s_1+s_2}{2}\left[\matrix{ 1 & 0 \\ 0 & 1 } \right] \qquad\qquad Q_2 := \frac{s_1-s_2}{2}\left[\matrix{1 & \phantom{-}0 \\ 0 & -1 } \right]$$ are scaled-orthogonal matrices. Since $U$ and $V$ are necessarily orthogonal themselves, $(1)$ describes a decomposition of any $2$-by-$2$ matrix into the sum of two scaled-orthogonal matrices. (If $M$ is already orthogonal, then $s_1 = s_2 = 1$, so that $Q_2$ vanishes.)

Thus, if $R$ is any "point-set" (in particular, vertices of a regular $n$-gon), and $L$ any linear transformation, then we have $$[L][R] = [T_1\oplus T_2] [R] = [T_1][R] + [T_2][R] \qquad\to\qquad L(R) = T_1(R) \oplus T_2(R)$$
where the $[T_i]$ are scaled-orthogonal matrices, so that the $T_i$ are origin-fixing similarities (ie, isometries composed with dilations). This says exactly that the linear image of any planar point-set is the "sum" of two similar images of that set; if the point-set consists of the vertices of a regular polygon, then so do the similar images. This completes Step 2, and shows that an arbitrary $n$-gon in the plane can be expressed as the "sum" of regular $n$-gons. (To see that there are $n$ components in total, note that we needn't decompose an affinely-regular $\{n/0\}$-gon or an $\{n/(n/2)\}$-gon, as these are inherently regular; for all other $m$, the affinely-regular $\{n/m\}$-gonal component gives rise to two regular components (one of which might be "zero"). $\square$

So, any set of points in the plane can be derived from combinations of vertices of regular polygons. If we restrict those points to a line, then Step 1 above still applies. Step 2, however, simplifies according to the observation that the projection of any affine image of a regular polygon is equivalent to the projection of an isometric image of that polygon; thus, there's no reason to "decompose" the projecting affine polygon into two regular components, since one component will do.

Some comments:

1. Except for $m=0$, all of the regular components have their centers at the origin. The $m=0$ component (all of whose vertices coincide) effectively accounts for the translation of the sum into position.

2. We imposed an order on the points to get a particular polygon $P$ and derive its regular components. Obviously, for $n > 3$, we can impose a different order, get a different polygon, and thus derive a new set of regular components. I have not investigated how these various sets of regular components relate (except, of course, for having a common set of points as their "sum").

3. Polygon-izing the points is just one approach. Another is to consider our $n$ points as vertices of an $(n-1)$-simplex. Or, more generally, as the vertices of any combinatorial graph we care to construct by connecting the points with edges. Step 1's analysis, as-is, no longer applies; however, we can replace the $R_m$s with coordinate matrices of other "regular" figures (say, the skeletons of Platonic polyhedra), as described in my note "Spectral Realizations of Graphs". A straightforward generalization of Step 2 ensures that any linear image of a $d$-dimensional figure decomposes into $d$ similar figures, as outlined in "An Extension of a Theorem of Barlotti to Multiple Dimensions".

4. The process described readily generates coordinate matrices for various regular components from the coordinates of the original points themselves. (The Singular Value Decomposition is the only tricky part.) Thus, all metric properties of those components are computable. However, if $F$ is a polynomial having those points as roots, it's unclear what aspects of the components relate to properties of the polynomial. Given that the polynomial doesn't know or care what order (or other structure) we might impose on the roots, it's not even clear what relations to suspect. (Analysis in this area bears on comment (2).)

Edit to add... As it turns out, the specific nature of our matrix products allows us to be more explicit about the decomposition in Step 2, avoiding the Singular Value Decomposition altogether, and providing insight into the geometry.

Let $P$ be a coordinate matrix ... $$P := \left[\begin{matrix} x_0 & x_1 & x_2 & \cdots & x_{n-1} \\ y_0 & y_1 & y_2 & \cdots & y_{n-1} \end{matrix}\right]$$ ... and consider $R_m$ as defined above, with $m \neq 0$ and $m \neq n/2$ (for simplicity). Then, we have $$M := P R_m^\top = \sqrt{\frac{2}{n}}\;\left[\begin{matrix} \sum x_k \cos\frac{2\pi k m}{n} & \sum x_k \sin\frac{2\pi k m}{n} \\[6pt] \sum y_k \cos\frac{2\pi k m}{n} & \sum y_k \sin\frac{2\pi k m}{n} \end{matrix}\right]$$ with summations taken over $k = 0$ to $n-1$. With a quick detour into the complex domain to define $$z_k := x_k + i y_k \qquad\qquad \omega := \exp(2\pi i/n)$$ $$\rho_m := r_m \exp i\theta_m = \frac{1}{n} \sum_{k=0}^{n-1}\;z_k \omega^{km} \qquad \sigma_m := s_m \exp i\phi_m = \frac{1}{n} \sum_{k=0}^{n-1}\;z_k \overline{\omega}^{km}= \sum_{k=0}^{n-1}\;z_k \omega^{-km}$$ (Thus, $\rho_m$ is the centroid of the polygon in which the $k$th vertex has been rotated by $2\pi km/n$; likewise for $\sigma_m$, but with oppositely-oriented rotations. Interestingly, the coordinates of these centroids are invariant under translation of the original polygon (why?), so that $r_m$ and $s_m$ measure the same distance from the center of vertex rotation, no matter where that center happens to be.) These allow us to write

$$x_k = \frac{1}{2}(z_k + \overline{z}_k) \qquad y_k = \frac{1}{2i}(z_k - \overline{z}_k)$$ $$\cos\frac{2\pi k m}{n} = \frac{1}{2}( \omega^{km} + \overline{\omega}^{km} ) \qquad \sin\frac{2\pi k m}{n} = \frac{1}{2i}( \omega^{km} - \overline{\omega}^{km} )$$ and therefore $$\begin{align} M &=\sqrt{\frac{n}{2}} \left[ \begin{matrix} \frac{1}{2}(\rho_k + \overline{\rho}_m + \sigma_m + \overline{\sigma}_m) & \phantom{-}\frac{1}{2i}(\rho_m - \overline{\rho}_m - \sigma_m + \overline{\sigma}_m) \\ \frac{1}{2i}(\rho_m - \overline{\rho}_m + \sigma_m - \overline{\sigma}_m) & -\frac{1}{2}(\rho_m + \overline{\rho}_m - \sigma_m - \overline{\sigma}_m) \end{matrix} \right] \\[6pt] &= \sqrt{\frac{n}{2}}\left[ \begin{matrix} r_m \cos\theta_m + s_m \cos\phi_m & \phantom{-}r_m\sin\theta_m - s_m \sin\phi_m \\ r_m \sin\theta_m + s_m \sin\phi_m & -r_m \cos\theta_m + s_m \cos\phi_m \end{matrix}\right] \\[6pt] &= r_m \sqrt{\frac{n}{2}}\left[ \begin{matrix} \cos\theta_m & \phantom{-}\sin\theta_m \\ \sin\theta_m & -\cos\theta_m \end{matrix}\right] + s_m \sqrt{\frac{n}{2}}\left[ \begin{matrix} \cos\phi_m & - \sin\phi_m \\ \sin\phi_m & \phantom{-}\cos\phi_m \end{matrix}\right] \end{align}$$ which is our decomposition of $M$ into (real!) scaled-orthogonal matrices.

Consequently, the regular $\{n/m\}$-gonal components of polygon with coordinate matrix $P$ have coordinate matrices found in this sum: $$PR_m^\top R_m = M R_m = r_m \left[ \begin{matrix} \cdots & \cos\left(\theta_m - \frac{2\pi k m}{n} \right) & \cdots \\ \cdots & \sin\left(\theta_m - \frac{2\pi k m}{n} \right) & \cdots \end{matrix} \right] + s_m \left[ \begin{matrix} \cdots & \cos\left(\phi_m + \frac{2\pi k m}{n} \right) & \cdots \\ \cdots & \sin\left(\phi_m + \frac{2\pi k m}{n} \right) & \cdots \end{matrix} \right]$$ It's worth noting that the component polygons are traced in opposite orientations. More importantly, the parameters $r_m$, $s_m$, $\theta_m$, and $\phi_m$, which have geometric significance relative to those polygons with rotated vertices, provide the radii and "starting angles" for $P$'s regular components.

Solution 2:

I'll use a separate answer to address OP's request for handling the specific case where the points are the roots of this quintic: $$x^5 + x^4 - 4 x^3 - 3 x^2+3 x+1 = 0$$

We'll polygon-ize by imposing an order on the roots from least to greatest; that is we take this to be the coordinate matrix: $$P := \left[\begin{matrix} -1.918\dots & -1.309\dots & -0.284\dots & 0.830\dots & 1.682\dots \\ 0 & 0 & 0 & 0 & 0\end{matrix}\right]$$

Here's the graph of the polynomial, showing the (tinted) roots and the (black) $\{5/0\}$ component at the roots' centroid (aka, their average).

Following the process outlined in my other answer, we get an affinely-regular origin-centered $5$-gon and $\{5/2\}$-gon; each of these decomposes into two actually-regular components, as shown:

You'll notice that in this special case, where $P$'s vertices are collinear, the regular components of each type are mirror images of each other, in order that their $y$-coordinates in the vector sums cancel. If we choose to introduce projection to zero-out $y$-coordinates, then we can eliminate one of each of the regular components, and double the size of the remaining one. Here's one possibility:

In the above, I took the opportunity to re-center the regular components with the centroid. The extra dots floating around are the vector sums of corresponding vertices of the components; projections of these onto the $x$-axis give the polynomial's roots.