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Consider the transformation:

$T \begin{bmatrix} x \\ y \\ \end{bmatrix} $ $\mapsto$$\frac{\sqrt{2}}{2} \begin{bmatrix} x+y-1 \\ -x+y+1 \\ \end{bmatrix} $

Show that $T$ is an isometry, then write the matrix of the projectivity $\tau:\mathbb{P^2} \to \mathbb{P^2} $ where $\mathbb{P^2} = P(\mathbb{R^3})$. Write also $\tau^{-1}$.

To answer to the first question I think I have just to show that it has been obtained as composition of projectivities of the form: $z \mapsto az$ and $z \mapsto z+b$

For the second one, should I find a projective frame for the input and one other for the output? Such as:

$P_1 \begin{bmatrix} 1 \\ 0 \\0\\ \end{bmatrix} $, $P_2 \begin{bmatrix} 0 \\ 1 \\0\\ \end{bmatrix} $, $P_3 \begin{bmatrix} 0 \\ 0 \\1\\ \end{bmatrix} $, $P_4 \begin{bmatrix} 1 \\ 1 \\1\\ \end{bmatrix} $ and $Q_1 \begin{bmatrix} \frac{\sqrt{2}}{2} \\ 0 \\\frac{\sqrt{2}}{2}\\ \end{bmatrix} $, $Q_2 \begin{bmatrix} \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\0\\ \end{bmatrix} $, $Q_3 \begin{bmatrix} -\frac{\sqrt{2}}{2} \\ 0 \\\frac{\sqrt{2}}{2}\\ \end{bmatrix} $, $Q_4 \begin{bmatrix} \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\\sqrt{2}\\ \end{bmatrix} $

Where

$Q_1 \in x+y-1$;

$Q_2 \in x+y-1 \cap -x+y+1$

$Q_3 \in -x+y+1$

$Q_4 = Q_1+Q_2+Q_3$

So, the matrix which represents $\tau$ would be: \begin{bmatrix} \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2} \\ 0&\frac{\sqrt{2}}{2}&0 \\ \frac{\sqrt{2}}{2}&0&\sqrt{2} \end{bmatrix}

Am I right? What about $\tau^{-1}$? Thank you everyone

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  • $\begingroup$ Asking you to write the matrix of the transformation is a bit misleading. The matrix is homogeneous, so any nonzero scalar multiple of it also represents $T$. $\endgroup$ – amd May 24 '18 at 0:58
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To show that the transformation is an isometry, you might be best served demonstrating that it preserves length. So if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ and compute the distance between them, and then do the same for the transformed points, the distances should match. You can actually avoid a few square roots, and all the hassle that goes with them, by comparing squared distances instead. So show that the following equality always holds:

\begin{align*} (x_1-x_2)^2+(y_1-y_2)^2&= \left(\frac{\sqrt2}2(x_1+y_1-1)-\frac{\sqrt2}2(x_2+y_2-1)\right)^2 \\&+ \left(\frac{\sqrt2}2(-x_1+y_1+1)-\frac{\sqrt2}2(-x_2+y_2+1)\right)^2 \end{align*}

You can show that a transformation is an isometry by decomposing it into more elementary isometries like translation, rotation or reflection. But I'd guess the naïve approach I just suggested is easier in most cases to most readers. And the $z\mapsto az$ you mentioned sounds like it might have a scaling component you don't want in an isometry, so that only would make sense if you identify vectors with complex numbers, require $\lvert a\rvert=1$ and are OK with not covering reflections.

To turn the operation into a transformation matrix, I must say that while I welcome the general approach of mapping four points, I consider it overkill in this scenario. Just start with the general form of the matrix, using the fact that any affine transformation can be written with $0,0,1$ in the last row:

$$\begin{pmatrix}m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ 0 & 0 & 1\end{pmatrix}\cdot \begin{pmatrix}x\\y\\1\end{pmatrix} = \begin{pmatrix} \tfrac{\sqrt2}2(x+y-1) \\ \tfrac{\sqrt2}2(-x+y+1) \\ 1 \end{pmatrix}$$

Now the right hand side is fairly simple: it has one component that is linear in $x$, so put the coefficient of that into the first column. Then there is a component that is linear in $y$, so put the coefficient of that in the second column. And lastly there is a constant term, which goes into the third column. Done. If you want you can pick a convenient scalar multiple of the matrix in the end, e.g. to avoid fractions.

For the inverse operation, just computing the matrix inverse would work. Often the adjoint is even easier to compute since it avoids divisions, and it differs from the inverse only in a common factor which is irrelevant for projective geometry.

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  • $\begingroup$ Thank you, really clear and exhaustive! $\endgroup$ – Arcticmonkey May 26 '18 at 12:53

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