Why are central isometries all linear transformations?

Question

http://maths.mq.edu.au/~chris/geometry/CHAP06%20Isometries.pdf

I'm struggling with theorem 2: central isometries are linear transformations.

I don't see how 'parallel lines stay parallel' is true a priori. Could anybody explain this proof?

Answer 1

If I understood you correctly, you accept that isometries map straight lines to straight lines. To be more precise, they map straight lines onto straight lines.

Now, let $T$ be your isometry and let $r$ and $s$ be two (distinct) parallel straight lines. Then there is a plane $p$ containig them both. And, since $T$ is an isometry, $T(p)$ is also a plane. If $T(r)$ and $T(s)$ were not parallel, then, since both of them are in the plane $T(p)$, thet would intersect. But that's impossible, because $r\cap s=\emptyset$ and $T$ is one-to-one, since it is an isometry.

Answer 2

I think you need to show that an isometry maps planes to planes. For this, take three points in a plane $\pi_1$ forming a right angle triangle at $A$. Say this maps under $T$ to $A'$, $B'$, $C'$. Because of Pythagoras, the triangle $A'B'C'$ has a right angle at $A'$. Consider $P$ a point in the plane $\pi_1$, and $Q$, $R$ the projections of $P$ on the lines $AB$, respectively $AC$. We have the equality between distances: $$PQ^2 + PR^2 = PA^2$$ It follows that for the point $P'= T(P)$, $Q'=T(Q)$, $R'=T(R)$ we have $$P'Q'^2 + P'R'^2 = P'A'^2$$ Now, the point $P'$ is on the line $A'B'$, and the point $Q'$ is on the line $A'C'$. Because of the equality above, $P'$ has to lie in the plane $A'B'C'$ (and $Q'$, $R'$ have to be the projections of $P'$ onto the lines $A'B'$, respectively $A'C'$) otherwise the LHS would be bigger.

Answer 3

To show that isometries map line segments to line segments, it helps to know that the line segment between two distinct points $a$ and $b$ has the form $$ Line(a,b) = \{x | d(a,x) + d(x,b) = d(a,b)\}. $$ So $$ \begin{split} f(Line(a,b)) & = \{f(x) | d(a,x) + d(x,b) = d(a,b)\} \\ &= \{f(x) | d(f(a),f(x)) + d(f(x),f(b)) = d(f(a),f(b))\} \\ &= \{z | d(f(a),z) + d(z,f(b)) = d(f(a),f(b))\} \\ &= Line(f(a), f(b)). \end{split} $$ I did use the fact that isometries from $R^n$ to $R^n$ are surjective.