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From the system $H^{-T}CH^{-1} = kC$ where $C = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}$, I can obtain 6 constraints on the 10 unknown elements, 9 from $H$ and one $k$. Then I can deduce that this system has (10 - 6) four degrees of freedom.

It is clear to me that rotations about the origin and reflections about a line through the origin will preserve the unit circle and hence account for two of the parameters.

According to this answer there exist two other types of transformation, hyperbolic rotations about the x and y axes in the 3D vector space, which also preserve the unit circle at the origin. This gives a total of 4 independent parameters.

My questions then are

  1. Can $H$ be either a three-parameter or a four-parameter family of transformations depending on whether we let $k$ vary or not?
  2. Could someone please explain how it was deduced that hyperbolic rotations on the 3D vector space preserve the unit circle at the origin in the 2D projective space? The answer did not make much sense to me.

Any help would be greatly appreciated!

ImmenselyHappy
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  • Equality of the matrices introduced unnecessary constraints. The correct condition is $H^{-T}CH^{-1}=kC$ for some unknown $k$. – amd Dec 19 '19 at 17:34
  • Would that matter though as $H$ and $H/\sqrt{k}$ represent the same transformation. By taking $k = 1$ am I going to miss out on finding some transformations? – ImmenselyHappy Dec 19 '19 at 17:52
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    Yes, you are going to miss some, as you’ve discovered. I’ve run into cases in which with strict equality in the matrix equations, the resulting system of equations became inconsistent. It’s the same as if you were looking for a homography $H$ that preserves a particular line: that line is an eigenvector of $H^{-T}$, but you shouldn’t assume that the corresponding eigenvalue is $1$. – amd Dec 19 '19 at 18:12
  • @amd I've updated the question to add the scaling factor. Thanks. – ImmenselyHappy Dec 19 '19 at 18:42
  • Incidentally, that answer you linked is suspect. If you actually apply the listed transformations to $\operatorname{diag}(-1,1,1)$, you don’t get a multiple of that matrix back. There is a boost that preserves the metric (maps the unit circle to itself), but it’s not either of the two listed there. – amd Dec 19 '19 at 19:05
  • If the rotation is around the z axes and the hyperbolic rotations are around the x and y axes then it preserves diag(1, 1, -1). I'm not sure why diag(-1, 1, 1) is used. The 3 transformations given in this answer work. – ImmenselyHappy Dec 19 '19 at 19:18

1 Answers1

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  1. As $H$ represents an equivalence class, it is no specialization to set $k$ as 1. Hence the system has only 3 (9 - 6) degrees of freedom.

  2. The crux of the other answer is that an equivalent way to solve for this H would be to solve the following system of equation in 3D. $$\begin{pmatrix} H & 0 \\\ 0 & 1 \end{pmatrix} \begin{pmatrix} C & 0 \\\ 0 & 1 \end{pmatrix} \begin{pmatrix} H^T & 0 \\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} C & 0 \\\ 0 & 1 \end{pmatrix}$$ or, $$\begin{pmatrix} HCH^{T} & 0 \\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} C & 0 \\\ 0 & 1 \end{pmatrix}$$

    This means that the set of projective transformations in 2D that preserve a unit circle $C$ at the origin is the same as the set of affine transformations that preserves the quadric represented by $\begin{pmatrix} C & 0 \\\ 0 & 1 \end{pmatrix}$ in 3D.

I have written about this in more detail in case anyone needs it.

ImmenselyHappy
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