Note: Quadratics for Grown Ups

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We study the quadratic formula $\left(-b\pm\sqrt{b^2-4ac}\right)/2a$ geometrically, as a map from the space of coefficients to the space of roots for polynomials of degree $\leq 2$. Our main goal is the following theorem: restricted to real polynomials with complex roots, the quadratic formula realizes an isometry from the projective model to the conformal model of the hyperbolic plane.

Geometry of the map Polynomials $\longrightarrow$ Roots

Our goal is to understand in detail the geometry of the quadratic formula as a map from coefficients to roots, and from this understanding extract a cool fact: namely that restricted to real coefficients, the relevant geometry includes both a projective and a conformal model of the hyperbolic plane. Here we start the road to viewing the relationship between a polynomial and its roots as a geometric problem by first casting it as a topological one. As our interest lies in the quadratics everything is quite explicit: however just here at the beginning I will talk about general $n$, as it’s my hope that some of these ideas will (with much more work) continue to tell a good story in higher degree.

The fundamental theorem of algebra guarantees every complex nonconstant polynomial has at least one root, or equivalently every degree $n$ polynomial has exactly $n$ roots (counting multiplicity). One recasting of this as a topological statement about the map $\Roots:Polynomials\mapsto Their;Roots$, is as follows. Let $\Pol_n\cong\C^\times\times\C^n$ denote the space of degree $n$ polynomials over $\C$ identified with their coefficients, and for each $f\in\Pol_n$ let $\Roots(f)$ be its multiset of roots. To describe the space of these multi-sets, its useful recall the notion of \emph{symmetric power}: the $n^{th}$ symmetric power of a space $X$ is the collection of all unordered $n$-tuples, $\SP^n(X)=X^n/\Sym(n)$. The map $\Roots\colon\C^\times\C^n\to\SP^n(\C)$ is continuous, and factors through projectivization ($f$ and $cf$ have the same roots) to a map sending a monic polynomial to its roots. The result is a continuous bijection; and thus homeomorphism, between the space of polynomials viewed-as-their-coefficients, and the space of polynomials-viewed-as-their-roots.

This familiar setting is not quite where our story will take place however. Instead of projectivizing the space $\C^\times\times\C^n$ of degree $n$ polynomials onto their monic representatives, we will instead \emph{include back in} the missing polynomials of lower degree, and study the space $\Pol_{\leq n}\cong\C^{n+1}$. This at first seems a difficult task, as the roots map no longer lands in a single space but rather in the union of the symmetric powers $\SP^n(\C)\cup \SP^{n-1}(\C)\cup\cdots\cup \SP^1(\C)\cup\SP^0(\C)$. However, it is natural to topologize the union of the first $n$ symmetric powers of a space $X$ by using the one-point compactification $X\cup{\infty}$: if $k<n$ we represent the unordered $k$-tuple ${x_1,\ldots, x_k}$ by the unordered $n$-tuple ${x_1,\ldots, x_k, \infty,\ldots, \infty}$, essentially filling in the correct number of empty slots with the `filler point' $\infty$.

For polynomials, this amounts to (for example) saying that, viewed in relation to cubic polynomials, a linear polynomial has \emph{two roots at infinity}. This is actually quite natural, as when we write down families of cubics which in the limit become linear, two of their roots escape all bounded sets in $\C$. This leads to the following observation: when we topologize the union $\SP^n(\C)\cup\SP^{n-1}(\C)\cup\cdots$ via the aforementioned identification with $\SP^n(\C\cup{\infty})$, the natural extension of the roots map $\Roots\colon \Pol_{\leq n}\to\SP^n(\C\cup{\infty})$ remains continuous. This map again factors through projectivization of the domain to a continuous bijection, and thus homeomorphism from $\CP^{n}$ onto $\SP^n(\CP^1)$.

As each space of lower degree polynomials is naturally built into the construction, we may take the direct limit of this over inclusions for increasing $n$ to give a third and final topological incarnation of the fundamental theorem:

What does this tell us? Firstly, that there is a uniform, natural way to consider polynomials of lower degree as degenerations of higher degree polynomials, allowing us to union to domains, codomains and root maps for each lower degree into a single morphism $\CP^n\to\SP^n(\CP^1)$. Using this, from the viewpoint of coefficients the geometry of polynomials is naturally contained in the geometry of complex projective space, but from the viewpoint of roots the geometry of polynomials is naturally contained in the geometry of symmetric powers of the sphere. The roots map provides a way of passing between these geometries: if $X\subset \Pol_{\leq n}$ is a collection of poylnomials, we may consider both a \emph{projective model} $\mathsf{P}(\mathsf{Coefs}(X))$ and a \emph{symmetric powers model} $\mathsf{Roots}(X)$.

Quadratics

Restricting to (at most) quadratic polynomials, gives the map $\Roots\colon \CP^2\to\SP^2(\CP^1)$.
We will write the projectivized coefficients of a polynomial as $[a:b:c]$ and the roots as ${z,w}$. The discriminant of the polynomial $f(z)=az^2+bz+c$ will be denoted $\delta(f)=b^2-4ac$.

In this low degree, an explicit description of the second symmetric power of $\CP^1$ is useful: $\SP^2(\CP^1)=(\CP^1\times \CP^1)/\Z_2$ where $\Z_2$ acts by switching the first and second coordinates. We may even draw a useful picture of this, after identifying $\CP^1$ with the unit sphere in $\R^3$: let $\mathsf{pr}\colon\S^2\to I$ be the map sending a point on the sphere to its height; its $z$ coordinate in $I=[-1,1]$. Then the map $\mathsf{pr}\times\mathsf{pr}\colon\S^2\times\S^2\to I\times I$ gives us a kind of degenerate fibration: the preimage of $(x,y)\in I^2$ is generically a torus, consisting of the latitude circle at height $x$ times the latitude at height $y$.

The diagonal embedding $\Delta\colon \S^2\inject\S^2\times\S^2$ is the preimage of the diagonal in $I\times I$, and the $\Z_2$ action permuting coordinates acts freely off of here. Along this diagonal the action is trivial, and so geometrically $\Delta(\S^2)$ is a locus cone singularities in the quotient $\SP^2(\S^2)\cong\CP^2$. Subsets of $\SP^2(\S^2)$ can be profitably understood by `folding' a corresponding subset of $\S^2\times\S^2$ along the diagonal $\S^2$.

Forming the symmetric product $\mathsf{SP}^2(\CP^1)$ as a quotient of $\CP^1\times\CP^1$ by the map exchanging coordinates. (Note that in many of these cartoon drawings it will appear that $\Delta(\CP^1)$ is a boundary of $\SP^2(\CP^1)$. This is of course not true; in reality it is a codimension-2 sphere shaped singular locus.)

It’s instructive to view some important subsets of polynomials in both the roots and coefficient spaces. In the coefficients model, the at-most-linear polynomials show up as the $\CP^1$ at infinity with respect to the affine patch centered at $[1:0:0]\simeq z^2$. In the roots model, this same set identifies with the $\CP^1$ worth of points ${\infty, z}$ in $\SP^2(\CP^1)$. This $\CP^1$ intersects another important $\CP^1$ in the space of quadrics - the set of polynomials with a double root - in a single point (the constant polynomial, which by our convention has a double root at $\infty$). In the roots model, the quadratics with a double root are easy to describe, they are the image of the diagonal embedding $\Delta(\CP^1)$; the singular locus of the orbifold structure on $\SP^2(\CP^1)$.

In the coefficient model this set is more difficult to spell out: it is the projective variety cut out by the discriminant $V(\delta)=V(b^2-4ac)$. To see this is a 2-sphere, we may change variables so that $b^2-4ac$ is diagonal, $x^2+y^2+z^2$ (all non-degenerate quadratic forms are equivalent over $\C$), and note that in the affine patch $z=-1$ we have $x^2+y^2=1$ in $\C^2$, whose solution set is a cylinder; together with two points (the north and south poles) at infinity with respect to this patch.

Next we aim to impose some notions of geometry on the space of quadratics. The coefficient model, $\CP^2$, has a natural notion of geometry, with automorphism group $\PSL(3;\C)$. Note that this is different than what we would get considering only monic degree 2-polynomials (which would be the affine group for $\C^2$), and it allows the mixing of linear with quadratic polynomials by sending roots to $\infty$. To understand the geometry of $\SP^2(\CP^1)$, we first look to its orbifold universal cover $\CP^1\times\CP^1$. This has automorphism group given by $\Aut(\CP^1)$ on each factor extended by swapping the factors, or $\Z_2\ltimes \PSL(2;\C)\times\PSL(2;\C)$. The automorphisms of $\SP^2(\CP^1)$ are those elements above which are well defined (and nontrivial) on the quotient: that is, the diagonal elements $\Delta(\PSL(2;\C))$. This has a nice geometric interpretation: automorphisms of $\SP^2(\CP^1)$ preserve the singular locus which is itself a copy of $\CP^1$, and any automorphism of this $\CP^1$ is admissable. Equivalently, all automorphisms of $\SP^2(\CP^1)$ are induced from automorphisms of the underlying extended complex plane, and then applied to multi-sets of cardinality 2.

Automorphisms of $\mathsf{SP}^2(\mathbb{CP}^1)$ must preserve the singular locus, and thus all arise from automorphisms of the underlying space $\CP^1$

Finally, it will be useful to return to the coefficients model, and compute the restricted group of symmetries which `play nicely' with the roots map. As symmetries of $\SP^2(\CP^1)$ must preserve the singular locus, the diagonal $\CP^1$ representing double roots, their realization in the coefficients model must preserve the discriminant locus $V(\delta)$. As $\delta$ is a quadratic form on $\C^3$, the automorphisms of $\CP^2$ preserving $V(\delta)$ form the complex orthogonal group $\PO(\delta,\C)$. This is just an incarnation of the irreducible representation $\SL(2,\C)\to\SO(3;\C)$, taking the $\SL(2,\C)$ action on the extended complex plane, applying it to unordered 2-tuples, then interpreting those as roots of a polynomial and viewing the action on coefficients. The fact that all orthogonal groups are isomorphic over $\C$ tells us that $\SO(\delta,\C)$ contains subgroups isomorphic to $\SO(3;\R)$ and $\SO(2,1;\R)$, whose actions we can understand geometrically. With $\SO(\delta,\C)\cong\SL(2;\C)$ acting as conformal transformations of the sphere $V(\delta)$, as expected an $\SO(3)$ subgroup acts as rigid rotations of this sphere, and $\SO(2,1)$ subgroups act as M"obius transformations preserving some circle on $V(\delta)$. A particular one of these, namely the real points $\SO(\delta;\R)$ will be important below.

In root space, the automorphisms of $\SP^2(\CP^1)$ are the diagonal embedding of $\PSL(2;\mathbb{C})$ as we saw above.
In coefficient space, this group appears instead as a complex orthogonal group, $\SO(\delta,\C)\subset\PSL(3;\C)$.

It is interesting to note that from this perspective, it is still natural to allow automorphisms mixing quadratic with linear polynomials, but no longer mixing polynomials with distinct roots with those having a double root.

Real Quadratics

If we restrict to polynomials with real coefficients, $\Pol_{\leq 2}(\C)$ becomes $\Pol_{\leq 2}(\R)$ and correspondingly the space of projectivized coefficients changes from $\CP^2$ to $\RP^2$. It is more work to describe the image under the roots map. As $\mathsf{Roots}\colon\CP^2\to\SP^2(\CP^1)$ is a homeomorphism, that restriction is an embedding of $\RP^2$ into $\SP^2(\CP^1)$ is clear.
To understand this embedding, we will realize this $\RP^2$ as a certain 2-complex in $\CP^1\times\CP^1$, `folded over' the diagonal $\Delta(\CP^1)$.

Real quadratics come in two flavors: those with real roots, and those with complex conjugate roots.
In the $\CP^1\times\CP^1$ cover, polynomials with complex conjugate roots correspond to the collection $(z,\overline{z})$ as $z$ ranges in $\C\cup\infty$, forming a sphere. This sphere intersects the sphere $\Delta(\CP^1)$ along the great circle ${(x,x)\mid x\in\R\cup\infty}$, and so in the quotient gets folded along this great circle into a disk. The polynomials with real roots correspond to the collection ${(x,y)\mid x,y\in\R\cup\infty}$, which forms a torus in $\CP^1\times\CP^1$. This torus also contains the real diagonal ${(x,x)\mid x\in\R\cup \infty}$ as a $(1,1)$ curve and is folded along it onto a M"obius band in the quotient. Thus, in $\CP^1\times\CP^1$ the relevant 2-complex is the union of a sphere and a torus, glued along a circle as in the cartoon below.
In the quotient, the sphere becomes a disk and the torus a M"obius band, glued along their common circle of intersection: this is the $\RP^2$ embedded by the roots map.

The preimage of the $\RP^2$ of real polynomials in $\SP^2(\CP^1)$, viewed in the orbifold cover $\CP^1\times\CP^1$ and its quotient, a creased $\RP^2$.

Note that this $\RP^2$ contains part of the singular locus of $\SP^2(\CP^1)$. It is constructed from $\S^2\cup_{\S^1}T^2$ folded in half along their common $\S^1$, which we saw above to consist of real double roots. To remember this, I will draw a crease along this curve when a picture denotes the image of $\RP^2$ under the roots map.

The real quadratic polynomials form an $\RP^2$ inside of both $\CP^2\cong \SP^2(\CP^1)$. In the root-space, this $\RP^2$ intersects the singular locus in the circle of polynomials with a double real root.

To calculate the automorphisms of this subset in each the coefficient and roots viewpoints, we find the subgroup which fixes the real points setwise. In the coefficient view this is easy: the smoothly embedded $\RP^2\subset\CP^2$ has automorphism group $\PSL(3;\R)<\PSL(3;\C)$. On the roots side, any automorphism of this creased $\RP^2$ must send the crease to itself (to see this, note that all automorphisms of $\SP^2(\CP^1)$ preserve the singular locus, and so the subgroup preserving real polynomials must preserve the intersection of this singular set with the real points, which is precisely the creased circle). This circle represents the real quadratics with a double root $ \Delta(\R\cup\infty)\subset\Delta(\CP^1)$ and the automorphisms preserving it setwise are none other than the diagonal embedding of $\PSL(2;\R)$. Viewed back on the other side, the automorphisms of $\RP^2\subset\CP^2$ which preserve the division into polynomials with a double root and those with distinct roots are the intersection of $\Aut(\RP^2)$ with $\Aut(V(\delta))$; that is $\SO(\delta;\R)=\PGL(3;\R)\cap\SO(\delta;\C)$. Over $\R$, the discriminant $\delta=b^2-4ac$ has signature $(2,1)$, so this is a conjugate of $\SO(2,1)$ in $\PGL(3;\R)$ preserving the divison of $\RP^2$ into a disk, M"obius band by the discriminant.

The automorphism group of real quadrics in coefficient space is all of $\PGL(3;\R)$. In root space, any automorphism must preserve the singular locus, so the symmetries are reduced to a copy of $\PSL(2;\R)$. Back in coefficient space, this subgroup is realized as $\SO(\delta;\R)<\PSL(3;\R)$.

This action of $\SO(2,1)$ is well known - it acts as the isometries of both the Klein (projective) model of the hyperbolic plane when restricted to the disk, and the projective model of de Sitter 1+1 space when restricted to the M"obius band exterior (or anti de Sitter 1+1 space, as the two are isomorphic in this dimension). The $\SL(2;\R)$ action on the real polynomials in $\SP^2(\CP^1)$ is maybe not as well known, but comes from gluing together two natural actions on the universal cover $\CP^1\times\CP^1$. Up here, recall that the $\RP^2$ of real polynomials is double covered by a more interesting object; the union of a sphere and torus along a circle. The sphere consists of pairs $(z,\overline{z})$ for $z\in\CP^1$ and the torus of points $(x,y)$ for each coordinate in the extended reals. The subgroup of $\Aut(\CP^1\times\CP^1)$ fixing this sphere a copy of $\PSL(2,\C)$ embedded in $\Aut(\CP^1\times\CP^1)$ as $A\mapsto (A,\overline{A})$ and the subgroup fixing the torus is the real points, $\PSL(2;\R)\times\PSL(2;\R)$.

Automorphisms of the branched cover of $\RP^2_\mathsf{Roots}$ in $\CP^1\times\CP^1$.

Most of the automorphsims of either component individually do not descend to the quotient $\SP^2(\CP^1)$; to do so an automorphism must preserve the singular locus, which intersects the picture here in the same $\S^1$ common to the sphere and torus. On the sphere, this selects out only conformal automorphisms preserving the equator, which divides $\CP^1$ into two disks, each of which when equipped with this $\PSL(2;\R)$ action is a conformal model of the hyperbolic plane (either the Poincare disk or upper half plane, depending if the stereographic projection to $\C$ has projection point on, or off the equatorial circle). Thinking of the original $\CP^1$ as the boundary of hyperbolic $3$-space when equipped with the $\PSL(2,\C)$ action, we can see fixing a circle in the boundary as restricting the possible isometries to those that preserve that hyperbolic plane (which we can view conformally by then projecting onto the upper, lower hemispheres of the ideal boundary if we would like). On the torus, the corresponding picture is similar, but for anti-de Sitter space. Much like the sphere is the ideal boundary of $\mathbb{H}^3$, the torus is the ideal boundary of $\mathsf{AdS}^3$. Fixing certain curves (in our case the (1,1) curve with respect to the decomposition of isometries as $\PSL(2;\R)\times\PSL(2;\R)$ above) correspond to preserving a lower-dimensional anti de Sitter space $\mathsf{AdS}^2$, which is isomorphic to de Sitter 2-space as a coincidence of low dimensions and has model an open M"obius band. The $\PSL(2;\R)$ action on the creased $\RP^2$ in $\SP^2(\CP^1)$ is what you get from gluing these two actions together along the common ideal boundary of their spaces, where they agree.

The $\RP^2$ of real polynomials is a union of two natural geometries along their ideal boundary.

The Quadratic Formula

We’ve done all the hard work now; having described the domain and codomain of the roots map assigning coefficients in $\CP^2$ to their roots in $\SP^2(\CP^1)$ as well as the symmetries relevant to each side.

The complex roots of a real quadratic come in conjguate pairs, and so are represented in root space by the disk ${z,\overline{z}}$ in $\SP^2(\CP^1)$. In coefficient space, the real polynomials with complex roots are the negative cone of the discriminant, represented by the disk $\P{\delta<0}$ in $\RP^2\subset\CP^2$. The first of these spaces is naturally a conformal model of the hyperbolic plane (looking in the double cover $\CP^1\times\CP^1$, it is a hemisphere of $\CP^1$ together with the M"obius transformations preserving it), and the second is naturally a projective model (the action of $\SO(\delta,\R)\cong\SO(2,1)$ is by projective transformations preserving the metric given by the cross ratio). Restricting the root map to this disk gives the map $\Roots\colon \D_\mathsf{Coefs}^2\to\D_{\mathsf{Roots}}^2$, which is equivariant with respect to the action of hyperbolic isometries on each side (that is, if $g\in\Isom(\mathbb{H}^2)$ then $\Roots(g.f)=g.\Roots(f)$ where in the first case $g$ acts as an element of $\SO(\delta;\R)$ and in the second as an element of $\Delta(\PSL(2;\R))$.)

The roots map is an intertwiner for the action of isometries on both sides, and is itself an isometry from the coefficients model to the roots model.

It only remains to show that $\Roots$ is an isometry. But this is no work at all, thanks to the symmetry we have established along the way. Since $\Roots$ is $\Isom(\Hyp^2)$ equivariant and this action is transitive, $\Roots$ is completely determined by its value at any point. Consider $f=z^2+1$ with $\Roots(f)={i,-i}$.
Now let $\Phi$ be any isometry between these two models of hyperbolic space equivariant with respect to the given group actions. Without loss of generality we may assume that $\Phi(z^2+1)={i,-i}$ (if not, pre- and post-compose by isometries of the domain / codomain to make it so). But now $\Roots, \Phi$ are $\Isom(\Hyp^2)$-equivariant maps from $\D^2_{\mathsf{Coefs}}$ to $\D^2_{\mathsf{Roots}}$ agreeing on the point $z^2+1$, and so they are equal.

To finish it off, we will write this map down in coordinates, as an explicit map from a disk in $\RP^2$ to a disk in $\CP^1$. This also constitutes a second proof of the statement above, in which you are free to ignore everything up to here in this writeup and just directly compare the formula to the standard conversion from the Klein disk to the Upper Half Plane, for instance as found on Wikiepdia. Starting with a quadratic $f=az^2+bz+c$, we represent it in the space of projectivized coefficients as $[a:b:c]$, and get at its roots via the familiar quadratic formula. $$\Roots:[a:b:c]\mapsto \left{ \frac{-b\pm\sqrt{b^2-4ac}}{2a}\right}$$

To get useful coordinates on the codomain, note that as the roots are always a complex conjugate pair, one must be in the upper half plane and the other in the lower. Thus we may unambigiously select the root in the upper half plane and represent our pair of roots by a single number $x+iy$ for $y>0$. Noting that $\delta<0$, we may write this as follows, using the convention that $\sqrt{-}$ means `the unique positive square root of' when applied to real numbers. $$\Roots\colon[a:b:c]\mapsto \frac{-b}{2a}+i\frac{\sqrt{4ac-b^2}}{2a}$$ This provides us with useful coordinates on the codomain, and so our next move is to do something similar for the domain. The first obvious choice is the reduction to monic quadratics using the affine patch $a=1$, which lets us think of the domain as the set of points in $\R^2={(b,c)}$ with $b^2<4c$:

$$\Roots\colon (b,c)\mapsto \frac{-b}{2}+i\frac{\sqrt{4c-b^2}}{2}$$

The quadratic formula as usually written is a map from a projective model of} $\mathbb{H}^2$ \textbf{as a paraboloid onto the upper half plane.

This is nothing more than the quadratic formula as learned in grade school, but we know from the above that we may interpret this as an isometry between two copies of the hyperbolic plane! The copy in the codomain is familiar; by selecting coordinates given by the root with positive imaginary part we have naturally landed in the upper half plane. But what is the model in the domain? This `paraboloid' model is in fact the Klein model in disguise - we have just chosen the wrong affine patch; one that runs parallel to the lightcone instead of transverse to it.

projective model of the hyperbolic plane is a paraboloid, hyperboloid disk depending on choice of affine patch.

To reconstruct the more familiar picture, we need to change the affine patch. The following transformation does the job:

$$ \pmat{ a\b\c }= \pmat{ \tfrac{w+u}{2}\v\\frac{w-u}{2} }$$

In these coordinates, the discriminant is $\delta=u^2+v^2-w^2$ and the quadratic formula is the map taking the polynomial $(w+u)z^2+2vz+(w-u)=0$ to its roots:

$$\Roots\colon [u:v:w]\mapsto\frac{-v}{u+w}+i\frac{\sqrt{w^2-u^2-v^2}}{u+w}$$

The disk $\D^2_\mathsf{Coefs}$ here is fully contained in the affine patch $w=1$ and so we may take the domain to be the unit disk centered at $\vec{0}$ in $\R^2={(u,v)}$ giving the expression below, which is the usual transformation from the Klein disk ${(u,v)\mid u^2+v^2<1}$ to the upper half plane ${z\mid \mathsf{Im}(z)>0}$ up to possibly a reflection of the domain/codomain, depending on your source.

$$\Roots\colon (u,v)\mapsto \frac{-v}{1+u}+i\frac{\sqrt{1-u^2-v^2}}{1+u}$$

After a linear change of coordinates, the quadratic formula provides the usual identification of the Klein model with the upper half plane model of hyperbolic space.

Appendix

Here’s the easy lemma that was used in the proof: an equivariant map between two spaces each equipped with a transitive $G$ action is determined by its value at a point.

Denote $f(x)=\phi(x)=y$. Let $z\in X$, and choose $g\in G$ such that $z=g.x$.
Then evaluating $f, \phi$ on $z$ gives $f(z)=f(g.x)=g.f(x)=g.y$ and similarly $\phi(z)=\phi(g.x)=g.\phi(x)=g.y$. Thus $f(z)=\phi(z)$ so $f=\phi$.