Euclidean Isometry

Question

This is part of my homework problems, it is not for assignment nothing to hand in first of all, i just dont get how to go about this proof. If i can see it it would be good.

Question is

Show that a euclidean isometry has exactly

one line of fixed points, or a single fixed point, or no fixed points, and a parallel family of invariant lines, or no fixed points and a single invariant line

if you could prove any of the four i would be really thank full then i would try doing the remaining three.

Thanks in advance

Answer 1

Make a case distinction. Show that for each of the four kinds of isometries, one of the described properties hold. For example, a rotation will always have the center of rotation as a fixed point. Similar arguments hold for the other three kinds, assigning one of the described properties to each kind of isometry.

For bonus points, you might discuss the identity transformation as a special fifth case, and argue whether or not any of the mentioned properties apply to that one.

If that kind of proof is too informal for your context, then you'll have to be more specific about what kind of formalisms you're using in this area, or find your own way to put this informal matching into semi-formal proofs.

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Edit: As your comment states that you are supposed to formulate this using complex numbers, you can start by enumerating all the arithmetic operations you can perform on the complex plane without affecting relative distances, i.e. all the isometries. You will find the following result:

1. Addition of a complex number (i.e. translation)
2. Multiplication by a complex number of absolute value one (i.e. rotation around the origin)
3. Conjugation of the complex plane (i.e. reflection along the real axis)

The rather hard part is showing that this list is enough, i.e. that any other isomentry can be composed from these. I'd consider that proof easier for matrices, but it is doable for complex numbers as well. I'd try something like this:

1. If an isometry reverses orientation, then you can apply a conjugation and end up with an orientation-preserving operation.
2. If the isometry maps the origin to some other point, you can subtract that difference to obtain an operation which fixes the origin.
3. If the isometry preserves orientation and fixes the origin, it has to be a rotation, so you can apply the reverse rotation to end up with the identity.
4. The above three steps combined form the reverse of the original isometry, so if you apply their inverse in reverse order, you obtain the original isometry.

For this reason, any isometry in $\mathbb C$ will be of the following form:

$$ f: a + bi \mapsto e^{i\varphi} (a \pm bi) + (x + yi) \qquad a,b,x,y,\varphi\in\mathbb R $$

You may notice that my proof would have put the conjugation as the last step, whereas my $(a-bi)$ would be the first step. This is all right, because I can move an outer conjugation to that inner position if I reverse the sign of both $\varphi$ and $y$ in the process.

So now you can solve $f(a+bi) = a+bi$, performing case distinctions where necessary, in order to obtain the fixed points. Expressing fixed lines using complex numbers will be more difficult, but by then you might have gotten the hang of this.

Answer 2

For example, take the translation isometry, which is defined by $T(v)=v+p$ where $p$ is a fixed vector in $\mathbb{R^2}$. You're looking for fixed points, so we set up $T(v)=v$, or $v+p=v$ which has no solutions (makes sense, moving points in one direction doesn't keep anything fixed).

What about reflections? Hint: reflections by definition occur over a line. What do you think happens to the points on the line over which you are reflecting?

There's also the identity, which does nothing. The fixed points of this should be self explanatory.

Rotations: Rotations occur by rotating every point about a point $c$ in the plane by a fixed angle. There's two cases to consider here. The first is if $\theta=2\pi k$, where $k$ is a positive integer. Otherwise...

Glide reflections are a combination of reflections and translations. This can be treated similarly to the above.

Answer 3

There is no point in proving any of the four, as there is only a single statement that can be used to classify an arbitrary isometry (of the euclidean plane) as one of four types of isometries:

1. There are reflections; they have their exis as fixed points (and a family of invariant lines)
2. There are rotations; they have the rotation center as fixed point (and possibly invariant lines)
3. There are translations; they have no fixed points, but every line parallel to the directoin of translation is invariant
4. There are glide reflections; they have no fixed points, but the glide axis is invariant
5. There is identity, for which every point is a fixed point

It seems that the problem statement failed to rule out the last possibility.

How can one show that any isometry $f$ belongs to one of these four (or rather five) types?

Assume $f$ has at least two fixed points $A,B$. Then any point $P$ oon $AB$ is uniquely determined by its distances to $A$ and $B$. Since $f$ leaves this distant unchanged, necessarily $f(P)=P$, i.e. $AB$ is a line of fixed points. If we assum ethat $C$ is another fixed point not on $AB$, then any point $P$ is uniquely determined by its distances to $A,B,C$, hence $f(P)=P$ and $f$ is the identity. Thus so far we have that $f$ with more than one fixed point is either of the first or of the fifth type.

Apart from that there are the isometries with exactly one fixed point, i.e. type 2, and then there remain the ones with no fixed point at all. For the latter we need to show that there is either one invariant line or a parallel family.

Can there be no invariant line at all? Consider a point $A$ and $f(A)$ and $f(f(A))$. If these three are collinear, than the line uniquely determined by the two (distinct!) points $A$ and $f(A)$ is the same as the one detemined by $f(A)$ and $f(f(A))$. We conclude that this line is invariant. Therefore, assume that $A,f(A), f(f(A))$ are not collinear. Then the center $B$ of the circumcircle of the triangle $Af(A)f(f(A))$ is either invariant (which we excluded) or is reflected at $f(A)f(f(A))$, but then the center of the parallelogram $f(f(A))Bf(A)f(B)$ turns out to be a fixed point. Therefore, there is at least one invariant line. Either there is only this invariantline and we have type 5.

Or there is at least one more invariant line. If these two invariant lines intersect, then their intersection is a fixed point, which we have ruled out. Therefore any two invariant lines are parallel. Let $A, B$ be on different invariant lines. Then $ABf(B)f(A)$ has two parallel sides and the other two sides have equal length. Eihter the sides of equal length intersect (and that intersection must be a fixed point) or they dont't and we have a parallelogram. Then if $C$ is any point, we see from the abundant lot of congruent triangles that $Cf(C)$ is parallel to $Af(A)$ (and $Bf(B)$, hence every parallel line is invariant.