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Motivation:

Consider an arbitrary conic section in $\mathbb{R}^2$ given by

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

Now consider the map $$\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^2, \phi_{\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} }(x,y) = \left( \frac{ax + by + c}{dx+ey+f}, \frac{gx+hy+i}{dx+ey+f} \right) $$

This map sends a conic section to a conic section. This is easy enough to verify with algebra by observing that if for some (x,y):

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

Then there exists an $x'$ and $y'$ s.t. (if $\phi$ is invertible) $$ A \phi(x')^2 + B\phi(x')\phi(y') + C\phi(y')^2 + D\phi(x') + E\phi(y') + F = 0$$

And therefore

$$ A \left( \frac{ax' + by' + c}{dx'+ey'+f} \right)^2 + B \left( \frac{ax' + by' + c}{dx'+ey'+f} \right) \left( \frac{gx'+hy'+i}{dx'+ey'+f} \right) + C \left( \frac{gx'+hy'+i}{dx'+ey'+f} \right)^2 + D \left( \frac{ax' + by' + c}{dx'+ey'+f} \right) + E \left( \frac{gx'+hy'+i}{dx'+ey'+f} \right) + F = 0 $$

But by multiplying out all the denominators and grouping like powers of $(x')^n(y')^m$ we can clearly see that $(x',y')$ itself lies on some conic section.

So then (it feels clear to me that) we can conclude $\phi$ sends a conic section to a section.

The question:

Now is this set of $\phi$ the ONLY continuous functions $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that send conics to conics?

Or do there exist some more exotic functions with this property?

It does seem this family of functions is well known as the set of homographies which I found out about via this question.

Sidharth Ghoshal
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  • The complex analysis tag is here only because of the apparent relationship of these functions to Mobius transforms. – Sidharth Ghoshal May 31 '24 at 17:52
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    A similar question for Möbius transformations: https://math.stackexchange.com/q/4243057/42969 . – Martin R May 31 '24 at 18:25
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    I think you could write the question a little bit better. For example, your map $\phi$ is not well-defined on the line $dx +ey+f=0$, so it does not map conics to conics, but more like conics minus some points to conics minus some points. I would try to write it up as a question in $P\mathbb R^2$ – Aitor Iribar Lopez Jun 02 '24 at 09:07
  • @AitorIribarLopez by $P\mathbb{R}^2$ do you mean $\mathbb{RP}^2$ which has a model as the unit sphere with opposite points identified and which every point in $\mathbb{R}^2$ can be mapped to via $(x,y) \rightarrow (x,y,1) \rightarrow \text{intersection with surface of sphere by line passing through}$? – Sidharth Ghoshal Jun 02 '24 at 20:32
  • @SidharthGhoshal yes – Aitor Iribar Lopez Jun 03 '24 at 11:01

1 Answers1

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This is a partial result towards an answer.

We can try a brute force approach to enumerate all functions which preserve conics by definition. We begin by recalling an ODE that all conics satisfy.

$$ C[y] = 9(y')^2 (y''''') -45(y'')(y''')(y'''') +40(y''')^3 = 0 $$

So given a parametric curve as $x(t)=t, y(t)$ as long as the $y(t)$ obeys the above equation then locally this curve is tracing out a conic.

Let us now consider some arbitrary function $\mathbb{R}^2 \rightarrow \mathbb{R}^2 : \phi = (\phi_x(x,y), \phi_y(x,y))$

This function has two components and accepts an $(x,y)$ pair and produces as $\phi_x, \phi_y$ output. If $\phi$ sends conics to conics then that means that if $(x(t), y(t))$ form a conic then $\phi_x(x(t), y(t)), \phi_y(x(t), y(t))$ must also form a conic. If we define a curve $(t, y(t))$ to form a conic IFF it obeys the equation above then the conclusion we have is that

$$ C[\phi_y] = C[y]*F(x,y,\phi_x ,\phi_y... ) $$

It's clear then that if $C[y] = 0$ then $C[\phi_y] = 0$, for any choice of $F$. So if $\phi_y$ obeys ANY such equation then it necessarily sends conics to conics.

Putting this together we have that

$$ 9(\phi_y(t, y(t))')^2 (\phi_y(t, y(t))''''') -45(\phi_y(t, y(t))'')(\phi_y(t, y(t))''')(\phi_y(t, y(t))'''') +40(\phi_y(t, y(t))''')^3 = \left( 9(y')^2 (y''''') -45(y'')(y''')(y'''') +40(y''')^3 \right) F(x,y,\phi_x, \phi_y, ... ) $$

Let's allow $F=1$ for now. We will revisit it later:

$$ 9(\phi_y(t, y(t))')^2 (\phi_y(t, y(t))''''') -45(\phi_y(t, y(t))'')(\phi_y(t, y(t))''')(\phi_y(t, y(t))'''') +40(\phi_y(t, y(t))''')^3 = \left( 9(y')^2 (y''''') -45(y'')(y''')(y'''') +40(y''')^3 \right) $$

Recall that these derivatives can be expanded via partial derivatives such as: $\phi_y(t, y(t))' = \partial_x[\phi_y] + \partial_y[\phi_y] y'(t) $

Thus the ENTIRE LHS can be expanded in terms of partial derivatives to look something as follows

$$ F_1 + F_2 y + F_3 y' + F_4 (y')^2 + ... $$

I.E. a polynomial in the derivatives of $y$ with coefficient $F_i$ being a function of partial derivatives of $\phi_y$.

This LHS has to equal $ \left( 9(y')^2 (y''''') -45(y'')(y''')(y'''') +40(y''')^3 \right) $ and the only way that can happen is if the $F_i$ corresponding to the $(y')^2 (y''''')$ term is equal to $9$, the $(y'')(y''')(y'''')$ term is equal to $-45$, the $(y''')^3$ is equal to $40$ and the rest are zero. So we end up with a massive system of partial differential equations almost ALL set to 0 except 3 lines:

$$ F_1 (\phi_y, \partial_x[\phi_y], \partial_y[\phi_y], ... ) = 0 \\ F_2 (\phi_y, \partial_x[\phi_y], \partial_y[\phi_y], ... ) = 0 \\ \vdots \\ F_{k_1} (\phi_y, \partial_x[\phi_y], \partial_y[\phi_y], ... ) = 9 \\ \vdots \\ F_{k_2} (\phi_y, \partial_x[\phi_y], \partial_y[\phi_y], ... ) = -45 \\ \vdots \\ F_{k_3} (\phi_y, \partial_x[\phi_y], \partial_y[\phi_y], ... ) = 45 \\ \vdots $$

The solutions to THIS system are examples of functions which send conics to conics. Now some obervations:

  1. As a system with WAY MORE equations than unknowns it is surprising ANY solutions exist at all.

  2. This assumes $\phi_x(x,y) = x$ we know a larger solution space exists than this but we need a more generic $C$ operator to represent them

  3. This is letting $F=1$. There are infinitely many other choices of $F$ that can be tried but they form a finite dimensional space in derivatives of $y$. If you pick a bad $F$ then this becomes unsolvable for obvious reasons.

Sidharth Ghoshal
  • 18,359