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Need to show that if $f$ is a glide reflection then there is only one line $L$ such that $f(L) = L$

What I know is that a glide reflection is an isometry

$$f(z)=a\bar{z}+b,$$ such that $|a|=1$ and $a\bar{b}+b\neq0$.

Now assume that two lines $L_1$ and $L_2$ such that are axes for this glide reflection. Take $x_1 \in L_1$ and $x_2 \in L_2$. Since glide reflection maps given point to some new "location", then

$$f(x_1)=a\bar{x_1}+b=a\bar{x_2}+b=f(x_2)$$

But then $\bar{x_1}=\bar{x_2}$ and consequently $x_1=x_2$ and two lines are the same.

Could the suffice for a proof? Thanks!

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So your conclusion is that if $(x_1,x_2)\in L_1\times L_2$, then $x_1=x_2$. That sounds problematic already :) Rather than showing that, you should focus on showing that $f(L_1)=f(L_2)$. –  rschwieb Apr 27 '13 at 13:27
    
I'm rather stuck. could provide a hint how I should start doing that? –  Sarunas Apr 27 '13 at 14:22
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2 Answers 2

up vote 1 down vote accepted

What the proof idea in your question really shows is that $f$ is injective.

One problem here is the fact that it isn't easy to describe a “line” in the complex number plane. At least not as easy as say the real Euclidean plane. Most formalisms don't provide a nice way to compare lines for equality.

One possible approach would be the following:

  1. Take an arbitrary point $z\in\mathbb C$.
  2. Apply the transformation once, which gives you $f(z)$
  3. Apply the transformation a second time, resulting in $f(f(z))$
  4. Compute the differences $f(z)-z$ and $f(f(z))-z$
  5. If all three points lie on a single line, then the second difference will be in the same direction as the first
  6. You can express this by requiring that the fraction between the two differences is a real number (and will in fact be $\frac12$ in this case)

Solving the resulting equation for z should result in a condition which restricts $z$ to lie on a single line. The step 5 above is only a necessary condition for $f(L)=L$, not a sufficient one, so on the whole this will proove that there can be no more than a single fixed line. Showing that the axis of reflection expressed in this way actually is a fixed line should be rather easy by comparison.

Edit:
I attempted to provide more details on the computation. The computation is quite lengthy, and turns to using real variables fairly quickly, but I hope it will be possible to understand my idea from this.

$$ \mathbb R \ni r = \frac{f(z)-z}{f(f(z))-z} = \frac{a\bar z + b - z}{a(\bar a z + \bar b) + b - z} = \frac{a\bar z + b - z}{a\bar b + b} \\ r(a\bar b + b) = a\bar z + b - z $$

Now let's move to real-valued variables: $a = a_x+a_yi, b = b_x+b_yi$ and $z = z_x + z_yi$. Splitting the resulting equation into real and imaginary part results in the following two real-valued equations (transformed with the help of a computer algebra system):

$$ \begin{pmatrix} (-a_xb_x - a_yb_y - b_x) & (a_x - 1) & a_y \\ (a_xb_y - a_yb_x - b_y) & a_y & (-a_x-1) \end{pmatrix} \cdot\begin{pmatrix}r \\ z_x \\ z_y\end{pmatrix} = \begin{pmatrix}b_x \\ b_y\end{pmatrix} $$

This system is underdefined, and will in general have a 1-dimensional solution space, corresponding to your 1-dimensional fixed line $L$. When I used sage to solve the above, I got the following solution, depending on an arbitrarily chosen parameter $r$:

\begin{align*} z_x &= \frac{(2r - 1)a_yb_y + (a_x^2r + a_y^2r + (2r - 1)a_x + r - 1)b_x}{a_x^2 + a_y^2 - 1} \\ z_y &= \frac{(2r - 1)a_yb_x + (a_x^2r + a_y^2r - (2r - 1)a_x + r - 1)b_y}{a_x^2 + a_y^2 - 1} \end{align*}

If you look closely, you'll notice however that the denominator is zero since $\lvert a\rvert=1$. For this reason, you cannot simply choose $r$ arbitrarily.

But there might still be solutions, for which the numerator of these must be zero as well. Solving for that numerator, with the added knowledge of $a_x^2+a_y^2=1$, gives the condition $r=\frac12$. So you know that if there is any chance for a solution, you'll have to try $r=\frac12$. That gives the system of equations

$$\begin{pmatrix} a_x - 1 & a_y \\ a_y & -a_x-1 \end{pmatrix} \cdot\begin{pmatrix} z_x \\ z_y \end{pmatrix} =\frac12\begin{pmatrix} - a_xb_x - a_yb_y + b_x \\ a_xb_y - a_yb_x + b_y \end{pmatrix}$$

There is a unique solution to this:

\begin{align*} z_x &= \tfrac12 b_x & z_y &= \tfrac12 b_y \end{align*}

But a single fixed solution isn't appropriate here, so let's compute the determinant of the matrix to see when there will be more than a single solution.

$$\begin{vmatrix} a_x - 1 & a_y \\ a_y & -a_x-1 \end{vmatrix} = 1-a_x^2-a_y^2 = 0$$

So for $\lvert a\rvert=1$ the two equations are linearily dependent. There are two possible cases. Eithere there is no solution at all, or you have one real degree of freedom in solving one, and then the other one will be solved automatically. We can show that the second is the case here, by combining the two rows of the right hand side:

$$ a_y(- a_xb_x - a_yb_y + b_x) - (a_x-1)(a_xb_y - a_yb_x + b_y) = 0 $$

Strictly speaking we should have handled the case $a=1$ separately, as in that case both factors in the above comparison would be equal to $0$. But for $a=1$ the system becomes a lot simpler, so showing that a commn solution is still possible is left as an excercise.

So by now we basically have a single equation describing our line:

$$ L = \left\{ z_x + z_yi \in\mathbb C \;\middle\vert\; z_x,z_y\in\mathbb R, (a_x - 1)z_x + a_yz_y = \frac12(- a_xb_x - a_yb_y + b_x) \right\}$$

So far we have shown that any point on this line will be collinear with its first and second image under $f$. What remains to be shown is that these images do in fact lie on the same line. You can achieve this by plugging $z''=f(f(z))$ into the equation:

\begin{gather*} (a_x - 1)z''_x + a_yz''_y = \\ = (a_x - 1)(a_x^2z_x + a_y^2z_x + a_xb_x + a_yb_y + b_x) + a_y(a_x^2z_y + ay^2*z_y - a_xb_y + a_yb_x + b_y) = \\ = (a_x - 1)z_x + a_yz_y = \frac12(- a_xb_x - a_yb_y + b_x) \end{gather*}

So now you know that if $z$ lies on $L$, then $f(f(z))$ will lie on $L$ as well, and $f(z)$ will lie on the same line as $z$ and $f(f(z))$, so it will lie on $L$ as well. Therefore $L$ is a fixed line. It is the only such fixed line, since the above computations did not provide any alternatives.

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What do you mean by "requiring that the fraction between the two differences is a real"? That is $\frac{f(z)-z}{f(f(z))-z}=\frac{1}{2}$ Right? But how would I require that? Multiplying the fraction by the conjugate of the denominator or by taking the real part of the fraction? Thanks! –  Sarunas Apr 28 '13 at 7:37
    
As I tinkered with this on paper, I was amazed at how I groped to express lines, and now I'm relieved to see in your answer that this isn't out of the ordinary :P I only got so far as deducing that the axis is parallel to the position vector for $b+a\overline{b}$. –  rschwieb Apr 28 '13 at 12:04
    
@Sarunas: I've added a lot of computation details to my answer. Hope it is more useful now. Pretty would be something different, I agree, but at least it appears to get the job done. –  MvG Apr 29 '13 at 7:19
    
thank you! but... despite even if I get the idea, the details are way out of my skill level to think them on my own :) –  Sarunas Apr 30 '13 at 8:53
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Here is what I arrived at.

We produce the axis with some vector algebra.

By computing $f(f(z))=z+b+a\overline{b}$, we can see the expected result that the translation that occurs along the axis is by the nonzero amount $t=(b+a\overline{b})/2$. The translation helps us by showing us the direction of the axis, but we would still like to find a point on the axis.

We will know when we've found such a point $x$ if $f(x)-x=t$, because that will indicate that the point was translated along the axis of reflection and experienced no reflection across the axis.

Working with a little vector algebra you can compute that $x=(b-a\overline{b})/4$ yields $f(x)=(a\overline{b}+3b)/4$, and $f(x)-x=t$. I found these two points by considering $0$ and $f(0)$ and drawing a parallelogram with sides the vector $t$, and I surmised the axis of reflection went through this parallelogram (parallel to $t$).

So, basic vector algebra says that this axis is $L=\{\lambda t+x\mid \lambda\in \Bbb R\}$. From this you can compute that $f(\lambda t+x)=(\lambda+\frac{1}{2})t+x\in L$.


Finally, I wanted to offer an argument for uniqueness that doesn't require computation. I haven't tried to translate it into a computational argument, but in principal the idea should translate.

Suppose there are two axes $L_1$ and $L_2$ such that $f(L_i)=L_i$. By the hypotheses, $b\neq 0$, so $f$ has no fixed points. Note that if the two axes intersect at a point $p$, then in order for $p$ to stay on both lines, it must be a fixed point of $f$. Thus we have a contradiction unless the two axes are parallel.

Now if they are parallel but distinct, $L_1$ must be reflected to the "other side" of $L_2$. But since both lines are fixed under $f$, this is an absurdity. So in fact, $L_1$ and $L_2$ have to be the same line.

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