Views
5 years ago

# SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

## 92 integer values;

92 integer values; hence, a sufficient condition that (3.3) fails is that for some odd order element o S n , Z~(o) - ~ Z'i¢(o) is an odd integer. The above condition replaces sigGU(F0)sp) by Zy¢ which only requires a determination of J(F0)sp. However, the next lemma shows that this only involves a computation in the local algebra Q(F0)sp (recall this is the quotient Cx,k/I(F 0) where I(F0) is the ideal generated by the coordinate functions of F0). Lemma 3.5 : /f F0(x,~) : ~n+l,0 ~ R n ,0 is weighted homogeneous then .9(F0)Sp "~ Q0~0)Sp and B(F0)sb ~, Q(F0)sb as G-representations. Via this lemma we can define three mod 2 invariants which give quick if crude guarantees for the existence of branches. For semi-weighted homogeneous F define j(F) = dimlR(Q(F0)sp) mod 2 d(F) --- dimN(Q(f0)sd) rood 2 and if F is semi-weighted homogeneous for bifurcation equivalence, we also define b(F) = dirn~(Q(FO)sb) mod 2. The invariant d(F) turns out to be the mod 2 degree of f; however, the invariants are all independent, as can be seen from their evaluation for specific examples. These invariants are related to mod 2 versions of various numbers already considered. We decompose the set of half-branches B' = B+ u B_ by the sign of ~. on the half branches and B' = B e u B o by sign(det(dxF)) (on B e sign(det(dxF)) > 0). Corollary 7 : Let F(x,g) : R n+l,0 ~ ~n ,0 be a semi-weighted homogeneous germ which is also semi-weighted homogeneous /'or bifurcation equivalence (here there is no mention of a group). i) the number of nontrivial branches z j(F) mod2 //) card(B+) - 1 m j(F) + b(F) mod 2 iii) card(Be) - 1 z j(F) + d(F) rood 2 Together these invariants yield a sufficient condition for the existence of nontrivial branches. Corollary 8 : Let F(x,g) : ~n+l,0 ----* [~n ,0 be as in corollary 7. If then there is a nonUivial branch for F. O(F), b(F), d(F)-l) ~ 0 ~ (7/2Z) 3 Hence, if F is G-equivariant and semi-weighted homogeneous (including for bifurcation equivalence) and G' is an isotropy subgroup with dim~fFix(G')) > 1 then F' =

93 F I Fix(G') x ~ : Fix(G') x \$t ---o Fix(G') has the same properties. Then a sufficient condition that there is a nontrivial branch in Fix(G') is that OFF'), b(F'), d(F3-1) # 0 (here one must be careful to use F' in determining s, Sp, etc.). In the next section we shall examine a number of applications of these results. § 4 Examples Example 4.1: Let W(D4) denote the Weyl group which is generated by the group S 4 of permutations on {Xl ..... x 4} together with the transformations x i I--o ± x i of determinant = 1. This group has a natural action on [I 4. The generic equivariant bifucation germ F : ~15,0 \$14,0 for this action of W(D 4) is given by (4.2) F(x,X) = (Xl3+0tx2x3x4+~.x 1 ..... x43+O~XlX2X3+~.x4) Field and Richardson [FR] and IF] have shown that the maximal isotropy subgroup conjecture fails for this group (and all W(Dm) m ~ 4). We shall also see that the presence of orbits with submaxirnal isotropy can be detected via the reduced methods of §3. We compare the character gj with the sum of the characters of the permutation representations for the orbits with maximal isotropy via the homomorphism 8 G. Weights are assigned a i -- wt(xi) = 1 and wt(g) -- 2, so that F is weighted homogeneous with all d i = 3. Hence, s -- 4.3-4 - 8 and so Sp -- 8 - 2 = 6. Using the coordinate functions for F, we see that we can replace terms in Q(F) 6 ( = 5(F)6) involving xi3 by terms involving lower powers of x i. Hence we can give a basis for QfF)6 using monomials which naturally decompose into bases for representations of G --- W(D4). Table 4.3 ~3 K2xi2 Kxix j ~,XlX2X3X4 Kxi2xjxk Kxi2xj2 xi2xj2xk 2 xi2xj2xkx£ dim 1 4 6 1 12 6 4 6 char 1 2+ q) 2 + 2 9 1 4 + 4 9 2 + 2 9 2 + 9 2 + 2 9 Here for each monomial m = ~,~3xCL the subspace spanned by the monomials obtained by applying permutations to m yield a representation of dimension given by the second row of the table (also it is to be understood that for terms such as xix j, i ~ j); this representation has a modular character given by the third row. Also, ~(W(D4)) is generated by the two modular characters given by

Page 1 Page 2 D B 5 6 D MOTOR Mo~, lmotowon T R T ...
B E R L I N C H A U S S E E S T R A S S E 5 7 - 6 1 B ... - Glen Leddy
Page 1 7 6 5 | 4 L 3 | 2 | 1 CK T ll ll ll )l KLÃ'ZOJ T ll ll ll ll klÃ¶zo) A B ...
1 5 . T T O - F O R U M 1 2 . T T O - F O R U M O c t o b e r 1 6 ...
1O 9 I B I 7 I 6 5 4 I 3 2 1
D e se m b e r 0 5, V olum e 0 6, Issue 11 w w w .ta n go n o tica s.co m
P P P NP NP 12 3 4 5 6 7 8 W 9 10 11 12 13 14 W 15 16 0 17 A C B ...
DVAA: Flat Galaxies (by R.A.) A B C D E F G H I J K L 1 2 3 4 5 6 7 8 ...
H O T E L - R E S TA U R A N T- B A R S e e s t r a s s e 2 2 5 6 3 ...
a 1 2 a b b a 3 4 a b b a 5 6 a b b a 7 8 a b b a 9 10 ... - CodeMath.com
6-5 Practice B cc.pdf - MrWalkerHomework
Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...
8 * 5 * 4 * 3 * 2 * 1 A 8 A 7 A 6 A 5 A 4 A 3 A 2 B 8 B 7 B 6 B 5 B 4 B 3 ...
5 b 5 6 10 c 12 a 5 8 A a 13 b Classroom Exercises - flip@mrflip.com
EAS 540 - STABLE ISOTOPE GEOCHEMISTRY Thrs., BUS B-5; 6 ...
Solutions to Quiz 5 Sample B - Loyola University Chicago
Panel B 4:45-5:30 Round 2 Panel A 5:30-6:15 | Panel B 6:15-7 ...
Â¡' &) (10 , 2 354 6&7 98@ ) AC B\$ E DG FH I 6 &Q ... - UniversitÃ© Lille 1
TEST ANSWERS Version A 1. B 2. B 3. E 4. E 5. D 6. B 7. C 8. B 9. A ...
V 5 1 5 B 6 L 4 X P T S F
(0 12 3Â¡) 4(5 6 3 798 9 %@4a7(6 b cedgfhfhiqps rti u)vxwtwty
Friday Warm-up A 5:00-5:30 PM, Warm-up B 5:30-6:00 PM Starts at ...
1. NC 2. L 3. NP 4. D 5. F 6. A 7. M a) b) b) 8. T 9 ... - Bangladesh Bank
1. Nyelvismereti feladatsor 1. C 2. B 3. D 4. D 5. A 6. B 7. B 8. C ... - Itk
1 2 (a) 2 (b) 3 4 5 6 7 8 9 10 11 - Travel Insurance
1. 2. 3. 4. 5. 6. 7. 8. Meeting A. M B. Pl Approval Awards a Public He ...
1 2 (a) 2 (b) 3 4 5 6 7 8 9 10 11 - Travel Insurance
A. LISTENING. 1-5 B. VOCABULARY 6-15 C. GRAMMAR. 16-25
1 2 3 4 5 6 A B C D Last Name FirstName Major Advisor ... - Alumni