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In a complex (projective) plane CP2 using homogeneous coordinates $(x, y, z)$ what is the group of (projective) transformations that leave the complex circle $$x^2 + y^2 = R^2 z^2 $$ invariant as a whole?

Michael Hardy
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Gerard
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1 Answers1

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A circle (or in fact any non-degenerate conic) is isomorphic to a projective line. So a circle in $\mathbb{CP}^2$ is isomeorphic to $\mathbb{CP}^1$, and the projective transformations which leave it fixed are isomorphic to the projective transformations of $\mathbb{CP}^1$, i.e. the Möbius transformations.

A projective transformation of the line is uniquely determined by three points and their images. So if you fix three distinct points $A,B,C$ on your circle, you can describe any element of your group by specifying three distinct points $A',B',C'$ on that circle which are the images of $A,B,C$.

To find the resulting transformation in $\mathbb{CP}^2$, remember that any such transformation is uniquely determined by four points and their images, while no three points in either quadruple may be collinear. You already have three suitable pairs of points, so all you need is a fourth point. I suggest you take the one point on the circle in harmonic position. To find that, construct the tangents to the circle in $A$ and $B$, connect their point of intersection with $C$ and label the other intersection of that line with the circle as $D$. Do the same for the image points.

For details on computing a transformation give pairs of preimage and image points, see this post of mine. For details of computing $D$ resp. $D'$, feel free to post that as a separate question and notify me in a comment.

Community
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MvG
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  • I missed the Möbius transformation, so that is a good starting point. On the other hand I am looking for the group of plane transfromations. E.g. I have found that all harmonic perspectivities (centre P, axis p) leave the circle invariant as long as (P, p) are polar w.r.t. the circle. – Gerard Mar 03 '15 at 16:14
  • @Gerard: Each line transformation fully determines a corresponding plane transformation, as detailed in my (currently) third paragraph. Since a harmonic perspectivity with polarity is already fully determined by the point, and a point in the plane has only two degrees of freedom while a Möbius transformation has three, this only covers some of the relevant transformations. More precisely, threse perspectivities are already determined by their two fixed points, whereas a Möbius transformation still has a one-parameter family for every pair of fixed points. (All degrees of freedom are complex.) – MvG Mar 03 '15 at 16:47
  • I will try to figure this out. My guess now is that a Möbius transf. corresponds in $\infty^2$ cases to a harmonic perspectivity and that I am still missing another plane transfromation, perhaps the one where the 2 invariant points on the circle coincide. – Gerard Mar 03 '15 at 19:29