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Let $y=x^2$ and $x^2+y^2=2$ be a parabola and circle in the standard embedding plane $\{[x:y:1]\mid x,y\in\mathbf{R} \}\subset \mathbf{RP}^2$. Give a projective transformation $t_A$ which

  • maps the parabola onto the circle

  • leaves the $Y$-axis invariant

  • leaves the intersection points of the parabola and circle invariant.

How many such transformations exist?

Let $A=\begin{bmatrix}a&b&c\\d&e&f \\ g&h&i \end{bmatrix}\in GL_3( \mathbf{R})$ be the matrix representation. Condition (b) gives $b=h=0$. The points of intersection of the parabola and circle in the embedding plane are $(1,1,1)$ and $(-1,1,1)$. Condition (c) gives three systems of equations from which we get $a=1,c=0,d=0$ and $e+f=1$, $g+i=1$. Now matrix $A$ becomes $\begin{bmatrix}1&0&0 \\ 0&e&1-e \\ g & 0 & 1-g \end{bmatrix}$. For condition (a), we have

$$A\begin{bmatrix}x\\x^2\\1 \end{bmatrix}=\begin{bmatrix}x \\ ex^2+1-e \\ gx+1-g\end{bmatrix}$$ so we must have $g=0$ and $x^2+(ex^2+(1-e))^2=2$ for all $x$ is not even a valid equation, even though it gives $e=0$ but then $A$ is singular. Maybe my mistake involves not taking into account the fact that the coordinates are homogeneous.

Could someone provide some help?

rae306
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  • Your suspicion is correct. By insisting on equality you’re effectively solving for an affinity instead of a projectivity. – amd Oct 23 '18 at 19:09
  • As far as uniqueness goes, you can answer that by counting degrees of freedom. – amd Oct 23 '18 at 19:12
  • @amd I have written the whole thing out while taking into account the homogeneous coordinates, but this becomes a big mess. Could you help? – rae306 Oct 23 '18 at 19:55
  • $b=h=0$ maps the point $(0:1:0)$ onto itself, but that isn’t the correct conclusion to draw from condition (b), which says that the line $(1:0:0)$ gets mapped onto itself. This affects the first column of $A$, not the second. – amd Oct 23 '18 at 20:19

1 Answers1

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Consideration without any computation

See this post where I explain how in general you have 3 real degrees of freedom when mapping one conic to another. From your third condition you know that you have two fixed points, accounting for two of these degrees of freedom. From the second point you know that the two intersection between the $y$ axis and the parabola (one of them at infinity) must map to the two intersections with the circle. If one maps to one then the other will map to the other thanks to the symmetry of the setup. But which maps to which is open, so you have 2 choices.

Computation with homogeneous coordinates done right

If you want to stick to your approach, but take homogeneous coordinates into account the right way, you get something like this:

$$ A(1:1:1)=(a+b+c):(d+e+f):(g+h+i)=1:1:1 \\ \Rightarrow\quad a+b+c=d+e+f=g+h+i\\ A(-1:1:1)=(-a+b+c):(-d+e+f):(-g+h+i)=-1:1:1 \\ \Rightarrow\quad -a+b+c=d-e-f=g-h-i $$

for the two fixed points and

$$ A(0:0:1)=c:f:i=0:\pm\sqrt2:1 \\ \Rightarrow\quad c=0, f=\pm\sqrt2i \\ A(0:1:0)=b:e:h=0:\mp\sqrt2:1 \\ \Rightarrow\quad b=0,e=\mp\sqrt2h $$

In total you count $2+2+2+2=8$ linear equations, enough to determine all variables up to a common factor which is without geometric relevance. You see the two cases in the $\pm\sqrt2$ there. Strictly speaking this is just expressing your third condition and the pointwise version of the second.

Verifying the map between conics

That the resulting transformation $A$ indeed maps the parabola to the conic is probably best verified in a final step:

$$A^T\cdot\begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix}\cdot A =\lambda\begin{bmatrix}2&0&0\\0&0&-1\\0&-1&0\end{bmatrix}$$

In other words, if you put a (preimage) point $p$ into the quadratic form of the parabola on the right hand side of the equation, it is zero if and only if the transformed point $Ap$ makes the equation of the circle zero, i.e. the central matrix on the left.

A concrete solution

For readers who need a concrete solution to help understand the above, here is one. But don't cheat your learning experience by just copying that!

$$A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\begin{bmatrix}4\mp2\sqrt2&0&0\\0&4\mp3\sqrt2&\pm\sqrt2\\0&3\mp2\sqrt2&1\end{bmatrix}$$

For completeness: Why considering the whole line is no good

You could also express the second condition literally. The line $1:0:0$ (i.e. $1x+0y+0=0$) is to remain fixed. Lines are mapped using the inverse matrix, or equivalently the adjugate which is eqivalent up to an irrelevant factor. You have

$$\operatorname{adj}(A)=\begin{bmatrix} ei-fh & ch-bi & bf-ce \\ fg-di & ai-cg & cd-af \\ dh-eg & bg-ah & ae-bd \end{bmatrix}\\ \operatorname{adj}(A)(1:0:0)=ei-fh:fg-di:dh-eg=1:0:0 \\ \Rightarrow\quad fg-di=dh-eg=0 $$

So this gives only two conditions, not four as what we got above from mapping explicit points on that axis. What is worse is that the condition is no longer linear. The condition for mapping the parabola to the circle written above is also quadratic in $A$, and if you use the general line equation you need that condition in full. So on the whole I'd rather avoid this approach and instead stick to the points of intersection.

MvG
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  • Thanks very much for your answer. Why do you consider the images of $(0:0:1)$ and $(0:1:0)$ and where does the $\sqrt{2}$ come from? And where do you take the parabola which is turned into a circle? Could you elaborate? – rae306 Oct 23 '18 at 23:04
  • $(0:0:1)$ and $(0:1:0)$ are the intersections of the parabola $x^2-yz=0$ with the line $x=0$: you get all the points where $x$ is zero and either $y=0$ or $z=0$. Geometrically you get the origin and the point at infinity in the vertical direction. And intersecting the circle $x^2+y^2-2z^2$ with $x=0$ you get $y=\pm\sqrt2z$. That $\sqrt2$ is the radius of the circle, since a typical equation for the circle around the origin would be $x^2+y^2=r^2$. – MvG Oct 24 '18 at 05:32
  • @MvG You say "you could also express the second condition literally. The line $1:0:0$ (i.e. $1x+0y+0=0$) is to remain fixed". But isn't the Y-axis represented by the line (projective point) $0:1:0$? Feel like my understanding of the Y-axis is incorrect. – Nanoputian Nov 27 '20 at 22:17
  • @Nanoputian: You need to distinguish between the $y$ axis as a line and points on it. You can interpret each vector as either a point or a line. $(0:1:0)$ as a point would be the point at infinity in the vertical direction, where all vertical lines meet. See also https://math.stackexchange.com/q/2930038/35416 which explains the equation of a vertical line in a bit more detail. – MvG Dec 14 '20 at 09:45