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I recently asked for natural topologies on the set of lines in $\mathbb R^2$. Now I'm aiming for a similar question on the set $S_p$ of conic sections in $\mathbb R^2$ sharing the same focus $p$ (but not necessary having the same major axis). The situation is an idealization of the ecliptic plane and all stuff in the solar system. Are there natural topologies on this set?

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  • $\begingroup$ Plane conics can be parametrized by $6$ parameters $[a: ... : f]$. So it is naturally homeomorphic to $\mathbb P^5$, the set of lines passing through the origin of $\mathbb R^6$. $\endgroup$ – user171326 Jul 5 '16 at 10:51
  • $\begingroup$ Sorry I didn't read your message correctly, you didn't want all the conics. But conics having a given focus is a subset of $\mathbb P^5$ so you can give the induced topology to these conics. $\endgroup$ – user171326 Jul 5 '16 at 10:53
  • $\begingroup$ @N.H. I think I got the idea. $\endgroup$ – Lehs Jul 5 '16 at 11:36
  • $\begingroup$ @N.H. Though an anomali compared with the case of lines, since that space was homeomorphic to a punctured projective plane. $\endgroup$ – Lehs Jul 5 '16 at 11:50
  • $\begingroup$ Yes. The space of conics is also not exactly $\mathbb P^5$ since there are double lines, and even the equation $(ax + by + c)^2$ was considered as conics, even if its set of points is simply a line. So you have to remove an hypersurface for only keep the smooth conics (i.e without double lines or pairs of lines). $\endgroup$ – user171326 Jul 5 '16 at 11:59
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It depends a lot on what exactly you want to call a conic. My first impulse was along the same lines as what N.H. wrote in a comment (and later in an answer): six numbers to describe a conic, but scalar multiples describe the same conic, so this looks like $\mathbb P^5$. But then N.H. went on to exclude degenerate conics, while I'd take the opposite approach of actually including degenerate conics. So in a first approximation, I'd say a conic is defined as a set of points $\{p\in\mathbb P^2\;|\;p^TAp=0\}$, described by a symmetric matrix

$$A=\begin{pmatrix}a&c&d\\c&b&e\\d&e&f\end{pmatrix}$$

(Some of my coefficients differ from the N.H.'s formula by a factor of two, but that makes no difference for the topology.) The matrix must not be the zero matrix, but apart from that I'd include everything else: the case of $\det(A)=0$ where the conic degenerates to a pair of lines, and $\operatorname{rank}(A)=1$ where these lines coincide to form a double line, and also cases like the identity matrix where the only solutions have complex coordinates. But that's my personal view, and the choice here will have a huge impact on the resulting topology.

Actually, the scenario just described is just half the truth. Via projective duality, it makes equal sense to describe a conic not in terms of incident points, but in terms of tangent lines. You'd say a line $l$ is a tangent to a conic if it satisfies $l^TBl=0$. In the non-degenerate case, $B$ would simply be any multiple of the inverse of $A$. But in the degenerate case, that definition breaks down. Instead one requires $(A,B)$ to form a primal-dual pair, i.e. $A\cdot B=\lambda\mathbb 1$ for some $\lambda$ which may even be zero. If $A$ has rank two, then $B$ can still be computed as the adjoint of $A$. But if $A$ has rank one (a double line), then $B$ can no longer be derived from it, since any pair of (real, complex, distinct or coinciding) points on that double line can serve as the components which make up $B$. I honestly don't know how to express this space of all conics in terms of topology.

I do know however how to express foci. A point $p$ is a focus if both the lines connecting it to the ideal circle points $(1,\pm i,0)$ are tangents to the conic. Since for the sake of topology the actual location of the origin is irrelevant, I'd consider $p=(0,0,1)$. Joining that to the circle points you get

$$t_{1,2}=\begin{pmatrix}0\\0\\1\end{pmatrix}\times\begin{pmatrix}1\\\pm i\\0\end{pmatrix}=\begin{pmatrix}\mp i\\1\\0\end{pmatrix}$$

Now it is easier to look at the dual matrix. In order to avoid confusion, I'll use different letters for its entries.

$$B=\begin{pmatrix} g & k & l \\ k & h & m \\ l & m & n \end{pmatrix} \qquad t_{1,2}^TBt_{1,2} = (\mp i,1,0)\cdot \begin{pmatrix} g & k & l \\ k & h & m \\ l & m & n \end{pmatrix}\cdot \begin{pmatrix}\mp i\\1\\0\end{pmatrix} =(h-g)\mp i(2k) $$

So if the point $p$ should be a focus, then the two equations above have to be satisfied. If all your coefficients are real, that means you need $(h-g)=0$ and $k=0$ for the dual matrix. So in this view, you are essentially imposing a restriction to the intersection of two hyperplanes in this space of all conics.

It is possible to translate the conditions back into equivalent conditions for the primal matrix, at least for the non-degenerate case. If $B=\operatorname{adj}(A)$ then you have

$$ 0=h-g= \begin{vmatrix}a&d\\d&f\end{vmatrix}- \begin{vmatrix}b&e\\e&f\end{vmatrix}= (a-b)f-d^2+e^2 \qquad 0=k= -\begin{vmatrix}c&e\\d&f\end{vmatrix}= de-cf $$

This looks more complicated, but since the space of dual conics should have the same topology as the space of primal conics (unless you include degenerate conics but consider degenerate conics with the same primal matrix to be the same no matter the dual), this should still essentially be a restriction to two subspaces as in the dual view, at least topologically speaking.

If you want to exclude degenerate conics, you could express that using either of the following inequalities:

$$0\neq\det(A)=abf+2cde-ae^2-bd^2-fc^2\\ 0\neq\det(B)=ghn+2klm-gm^2-hl^2-nk^2$$

Last but not least, if you want to exclude conics with no real points on them, you'd look at the eigenvalues of the matrix. If two of the eigenvalues have one sign and a third eigenvalue has the opposite sign, then you have a real and non-degenerate conic. If all three signs are the same, you have a non-degenerate conic with no real solutions. If one or two eigenvalues are zero, then you have a degenerate conic. I don't know how to translate these eigenvalue considerations into topology.

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Consider as "conics" the zero set of equation $ax^2 + by^2 + cxy + dx + ey + f = 0$.

Since $(a,b,c,d,e,f)$ and $(\lambda a, \lambda b, \lambda c, \lambda d, \lambda e, \lambda f)$ define the same equation for $\lambda \neq 0$, a conic define naturally a point in $\mathbb P^5$, the real projective space of dimension $5$.

Not every point in $\mathbb P ^5$ gives you a conic, so you have to remove the points $(a,b,c,d,e,f)$ where $abc = 0$, and where $b^2 - 4ac = 0$. The last case gives you degenerate conics (union of two lines, or even a single line). Since you are in real coordinate I guess you have to remove some other points (for example the "conic" $x^2 + y^2 + 1 = 0$ has no real point).

Finally, $\mathbb P ^5$ has a natural topology (the quotient topology), so for a given point $p$, the conics with focus $p$ form a variety in the space of conics and has therefore a natural topology.

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  • $\begingroup$ I have my doubts about your conditions for excluding degenerate conics. See my answer for the condition I'd use. The term $b^2-4ac$ looks like the minor one would use to classify a conic as a parabola, except that you have the roles of $b$ and $c$ exchanged in your equation (and I followed that example to make comparing answers easier). $\endgroup$ – MvG Jul 5 '16 at 18:46

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