Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Congruence transformations (isometries) and similarity transformations (isometries + dilations) should be constructable. What about other affine transformations? Other conformal mappings?

edit: by constructable, I mean given the defining information for the transformation in a geometric way (e.g. a dilation requires a center and a ratio, so the given could be a point and two segments), can you construct the image of a point under the transformation from its preimage?

share|improve this question

4 Answers 4

up vote 3 down vote accepted

edit (2010-07-26): The question is much more involved than I'd originally thought. As implied in the question, I knew that congruence and similarity transformations are constructible. Immediately below this section is my original answer, which only demonstrates congruence transformations and was intended more to give an idea of what an answer might look like (since, at the time, there was another answer that was not particularly helpful). In the last section of this answer is my justification that all affine transformations of the plane are constructible. In re-reading that now, I realize that I'd assumed the ability to construct a point, say $P'$, on a line, say $\overleftrightarrow{RP}$, such that $\frac{RP'}{RP}$ is equal to some known ratio. This is equivalent to being able to construct the dilation of $P$ by the known ratio about center $R$. I've added the construction of such a dilation below the congruence transformation section.

edit (2012-01-28): A conversation with some colleagues reminded me about this problem and in starting to ask them about it, I realized I'd completely missed that all Möbius transformations are constructible. Since any Möbius transformation can be expressed as a composition of translation, reflection, inversion, dilation, rotation, and translation (I think there's a typical decomposition that's roughly in that order, hence my listing translation twice). The only one of these that I have not yet shown is constructible is inversion, so I have appended that construction.


As a partial answer, here are constructions of the basic congruence transformations, assuming basic construction techniques like constructing a line parallel or perpendicular to a given line through a given point and angle-copying:

  1. reflection Given a point $P$ and a line $\ell$, construct the line perpendicular to $\ell$ through $P$, and construct the circle centered at the intersection of this new line and $\ell$ and passing through $P$. The image of $P$ under a reflection over the line $\ell$ is the point of intersection of the circle and the new line (the one not at $P$).

  2. translation Given a point $P$ and a vector $\overrightarrow{AB}$ (from $A$ to $B$), construct the line through $A$ and $P$, the line through $B$ parallel to line $\overleftrightarrow{PA}$, and the line through $P$ parallel to $\overrightarrow{AB}$. The image of $P$ under translation by vector $\overrightarrow{AB}$ is the intersection of the two constructed parallels.

  3. rotation Given a point $P$, a center of rotation $R$, and an $\angle ABC$ (from $A$ to $C$), construct the line through $P$ and $R$, copy $\angle ABC$ such that the copy of $B$ coincides with $R$ and the copy of $A$ is on ray $\overrightarrow{RP}$ and let the copy of $C$ be called $D$, construct the circle with center at $R$ and passing through $P$. The image of $P$ under rotation by $\angle ABC$ about point $R$ is the intersection of the circle with ray $\overrightarrow{RD}$.


dilation Given a point $P$, a center of dilation $R$, and a ratio $\frac{AC}{AB}$ (where point $C$ lies on ray $\overrightarrow{AB}$), translate $B$ to $B'$ and $C$ to $C'$ by the translation that takes $A$ to $R$, construct line $\overleftrightarrow{B'P}$, construct the line through $C'$ parallel to $\overleftrightarrow{B'P}$, and construct ray $\overrightarrow{RP}$. The image of $P$ under a dilation about $R$ by a factor of $\frac{AC}{AB}$ is the intersection of ray $\overrightarrow{RP}$ and the line through $C'$ parallel to $\overleftrightarrow{B'P}$.


All affine transformations are constructible. Per MathWorld and Wikipedia, an affine transformation of the plane is a transformation of the plane that preserves collinearity and preserves ratios of distances on any given line.

First, to show that affine transformations preserve parallelism, suppose that two lines $\overleftrightarrow{MN}$ and $\overleftrightarrow{PQ}$ are parallel, and that their images, lines $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$, intersect at $X'$, the image of $X$. Since affine transformations preserve collinearity, $X$ must be on $\overleftrightarrow{MN}$ and on $\overleftrightarrow{PQ}$, which is a contradiction, so $\overleftrightarrow{M'N'}$ and $\overleftrightarrow{P'Q'}$ cannot intersect. Thus, affine transformations preserve parallelism.

An affine transformation is determined by a $\triangle ABC$ and its image, $\triangle A'B'C'$ (per MathWorld; Wikipedia talks about defining an affine transformation by a parallelogram and its image, which is equivalent since affine transformations preserve parallelism). Given point $P$ and triangles $\triangle ABC$ and $\triangle A'B'C'$, construct line $\ell_1$ through $P$ parallel to $\overline{AB}$ and line $\ell_2$ through $P$ and parallel to $\overline{AC}$, call the intersection of $\ell_1$ with $\overline{AC}$ $I_1$ and call the intersection of $\ell_2$ with $\overline{AB}$ $I_2$, extend $\overline{A'B'}$ past $B'$ to a point $I'_2$ such that $\frac{AB}{AI_2}=\frac{A'B'}{A'I'_2}$, extend $\overline{A'C'}$ past $C'$ to a point $I'_1$ such that $\frac{AC}{AI_1}=\frac{A'C'}{A'I'_1}$, construct line $\ell'_1$ through $I'_1$ parallel to $\overline{A'B'}$ and line $\ell'_2$ through $I'_2$ parallel to $\overline{A'C'}$. The image of $P$ under the affine transformation mapping $\triangle ABC$ onto $\triangle A'B'C'$ is the intersection of lines $\ell'_1$ and $\ell'_2$.


inversion Given a point $P$ and a circle centered at $O$, construct ray $\overrightarrow{OP}$, let $X$ be the point of intersection of $\overrightarrow{OP}$ with the circle. The image of $P$ under an inversion through the circle is the image of $X$ under a dilation by $\frac{OX}{OP}$ centered at $O$.

share|improve this answer
    
Why is this downvoted? Is this incorrect? –  Casebash Jul 24 '10 at 6:49
    
If there's something wrong with this answer, please let me know. –  Isaac Jul 24 '10 at 7:05
4  
@Harry: You know that if you downvote someone for no reason they usually get sympathy upvotes and actually gain reputation –  Casebash Jul 24 '10 at 8:51
    
Points here are meaningless to me. –  user126 Jul 24 '10 at 8:58

Couldn't all transformation which send each point (x,y) to another point (x',y') which can be computed from the first one by performing only the four operations and extraction of square root?

share|improve this answer
    
I'm pretty sure that's a subset of all conformal mappings, but without a bit clearer definition of what can be used with +, -, *, and /, and sqrt(), I'm not sure about constructability. Could you give some examples of what you have in mind (and perhaps some examples of what doesn't fit your description)? –  Isaac Jul 26 '10 at 6:06

All Möbius transformations (i.e., translations, rotations, dilations, and transversions) can certainly be constructed since they are just compositions of inversions (which are of course generalized reflections) by the Cartan-Dieudonne theorem. Indeed, these are just the transformations that relate all the possible geometric primitives (in this case generalized circles) to each other.

However, the above claim that all affine transformations can be constructed appears misleading to me. All triangles are equivalent modulo an affine transformation, but it is well known that only a dense subset of all real-valued angles is constructible with a compass and straightedge. For instance, no triangle with a 20° angle can be constructed, so no existing triangle can be transformed into one with a 20° angle. If one is presented with two arbitrary triangles as above, then they can be used to define a particular affine transformation, but neither of the triangles may be themselves constructible and so a generic affine transformation is not constructible from generalized circles with a compass and straightedge.

share|improve this answer
    
The bit about affine transformations may be misleading, but I don't think it's incorrect—to construct the affine transformation taking one particular triangle onto a second particular triangle that has a 20° angle requires being given that second triangle, so not constructing the 20° angle from a simpler set of given objects. –  Isaac Jan 28 '13 at 21:59
    
Right, I didn't mean to suggest that it's wrong. I just read the question as "What transformations can be constructed from scratch with a compass and straightedge?", which is more restrictive than your actual question. I suppose it would also be possible to combine a generic affine transformation with an inversion to create a generic projective transformation as well? –  Andrew Shevchuk Jan 28 '13 at 22:15

Projective Transformations

Projective transformations of the (projective) plane $\mathbb{RP}^2$ can be constructed using only straightedge. Defining input are four preimage points and their corresponding images.

Four points $A$ through $D$ in general position, together with their image points $A'$ through $D'$ again in general position, uniquely define a projective transformation. For every point $E$, the line $AE$ intersects $BC$ in a given point, and $AD$ intersects $BC$ in another point. Together with the points $B$ and $C$ themselves, this gives four points on the line $AB$. By a similar construction, four points on $AD$ can be constructed, using the connections $BE$ and $BC$. The cross ratio of four points remains the same under projective transformations. On the image side, the lines $B'C'$ and $A'D'$ already have both their endpoints and their intersection already defined. So the fourth point which has the same cross ratio is uniquely defined. Using at most two pserspectivities, one can transfer the cross ratio from the preimage to the image line. Doing so for both lines gives two points, which connected to $A$ resp. $B$ will intersect in the image point $E'$.

Computations

You can choose a projective basis of four points, corresponding to the following homogenous coordinates:

\begin{align*} A&=\begin{pmatrix}1\\0\\0\end{pmatrix} & B&=\begin{pmatrix}0\\1\\0\end{pmatrix} & C&=\begin{pmatrix}0\\0\\1\end{pmatrix} & D&=\begin{pmatrix}1\\1\\1\end{pmatrix} \end{align*}

In common embeddings of the projective plane, two of these would be “at infinity”, in which case the computations outlined below would involve things like parallel lines, for which you'd probably need some form of compass. But you can also choose a projectively transformed situation (or a different embedding, which is a different view on the same thing) so that all four points are finite, and you can do most of the stuff I'll outline below using straightedge alone.

You can obtain lines serving as coordinate axes from the points given above. You can also project any point in the plane onto these axes. There are constructions given by von Staudt which perform additions and multiplications on a projective line. For addition you need points for “zero” and “infinity”, which along the $x$ axis would be $C$ and $A$. For multiplication you also need a point at position “one”, which would be the intersection of $AC$ with $BD$.

So given a certain projective reference frame (the four points $A$ through $D$), you can do the following:

  • construct points on a line corresponding to original coordinates
  • construct points on that same line corresponding to arbitrary rational numbers
  • constructively perform elementary arithmetic operations like addition, subtraction, multiplication and division
  • use these operations to compute/construct new coordinates
  • turn these coordinates back into a point in the plane

Unless you hit the special case of parallel lines resp. points at infinity, all of the above can be done using just a straightedge. So many things you could compute from the coordinates can in fact be constructed. I guess this is what @mau meant in his answer. Since you asked about compass and straightedge, you won't have to worry about the special case of parallel lines either, and you can add taking squeare roots to your sets of primitive operations.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.