Theorem AS1
The affine symmetry group of E0 is E0(2) = {t : t(x) = Ax, with A orthogonal}.
Proof
We know that the elements of E0(2) are symmetries of E0.
It remains only to show that there are no affine symmetries.
Suppose that t is an affine symmetry.
We know that affine transformations map centres to centres, so t(O) = (O).
Now let P be any point on E0. Since t maps E0 to
E0, t(P) = Q on E0 and so
t maps OP to OQ, a segment of the same length, i.e. |t(OP)| =|OP| = 1.
For any segment AB, there is a radius OP parallel to AB. Then, since t is affine,
it preserves the ratio of the parallel segments AB, OP. Thus,
|t(AB)| = |AB|.
Since AB is any segment, we see that t is an isometry, so tεE(2).
Since t
maps E0 to itself, it belongs to E0(2).
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