Download e-book for kindle: A lecture on 5-fold symmetry and tilings of the plane by Penrose R.

By Penrose R.

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However, we no longer have a form on [Be 77], g GF(q). V is an irreducible g(ux, vx) = g(u, v) v 2 . 1_ipear form on (1. e. llorphislD of order replace V V over "g on V over inti- K. We construct our K[C]. For a reference see - 53 - "- Proposition. 2) field of is the anihilator of K. If "- then K = H0"1<[e] (V,V) V K is -a-finite extension -------- in K[e] A ~ Ho~(V,V), and then K[e]/I=A. -~_t_rJvial, and a g(ua,v) = g(u,va ) u,v E V for all a E A. and a ProoE: a : e ... e be the antiautomorphism of o Let a x E e.

4). Finally, part (iv) follows from part (v) and the nonsingularity of G. This proves the Lemma. 7) Theorem. Assume that an irreducible K[G]-module. - so that K K V is a finite field; whi~~js is 8 : V x V ~ K by fi~ed G. Set is a nonsingular "- K HomK[G] (V,V) is a finite extension field of occurs: (l) g There is a nonsingular classical form on the and V g where (ii) K -+ K 't": The form g ~ g 1:g is the trace mapping. 6). 6) by setting "- is nontrivial on et an~ is l-I. K a is h. B and that is symme- A tric or symplectic.

Fixes the form ~ commutes x T= T. Let Choose c E F where Note that -1 = cq+l Let Then (-v,u). so that 2( q+l) then i\s above let q+1. In add it ion, of order q+l EeE divides GF(q) [~] F 0 (u,v)~= defined by -1. (v,vc» (a+b~)(a-b~) dim g and T = Iu S. 011 E f}. 1/2 dim ~2 = -1, S. vb~) - 49 - g (u,v) + a 2g (u,v) + b 2g (U,Il) + abg(u,Il~)+abg(u~2,v~) 0 0 0 go(u,v)(l+a Zt-b 2) + ab(g(u,v~) -g(u,v~» o . W Therefore, S. is a maximal isotropic subs pace of We now prove two propositions which will complete the proof of the theorem.

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