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Type Equivalence

Two different homomorphisms from E to {1,-1} can give rise to patterns with exactly the same symmetries and antisymmetries. For example, with E = p2 = {t1, t2, R}, where t1 and t2 are independent translations and R is a half-turn, consider the two homomorphisms

P1(t1) = -1, P1(t2) = 1, P1(R) = 1
and
P2(t1) = -1, P2(t2) = 1, P2(R)= -1.

These are different homomorphisms, but every function whose symmetries arise from either one can be seen to have an alternating grid of positive and negating two-centers; the patterns seem to have the same symmetry. This is because the composition t1 R is a half-turn, which must be positive if both t1 and R are negating.

The thing that matters is that half of the two-centers are negating and the others are positive. Finding an isomorphism of E that carries one homomorphism to the other solves the apparent problem.

The isomorphism

i(tj) = tj, i(R) = t1 R

satisfies P1(i(k)) = P2(k) for all k in E.

We need to be one step more precise. Many of these infinite groups can be isomorphic to subgroups of themselves. The isomorphism playing the role of i in the formula above will not literally go from a single group of isometries to itself.

Thus we say that if E and E' are isomorphic to the same wallpaper group, and hence to each other via the isomorphism i, then the homomorphisms P1 (from E' to {1,-1}) and P2 (from E to {1,-1}) give rise to the same wallpaper type if, and only if, P1 o i = P2.

A wallpaper type is an equivalence class of homomorphisms under this sense of equivalence. A function has a given wallpaper type if there is a homomorphism in that equivalence class for which the function satisfies the equation above.

Our next task is to count these equivalence classes.


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Communications in Visual Mathematics, vol 1, no 1, August 1998.
Copyright © 1998, The Mathematical Association of America. All rights reserved.
Created: 08 Jul 1998 --- Last modified: Sep 30, 2003 9:26:25 AM
Comments to: CVM@maa.org