Views
5 years ago

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

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

186 coordinates, but not

186 coordinates, but not of the continuous ones. There are 3 cases: n = 3, 4, >_ 5. For each of them we give a list of the normal forms of top-cod < "1 and their universal unfolding, a flowchart, where the solution of a recognition problem can be found following the arrows to a particular normal form, and two tables of complicated coefficients and algebraic datas. We have tried to avoid as many redundancies as possible by referring back to previous tables. Theorem 7 (Classification) The normal forms up to topological codimension I can be found in List I, 2 or 3, for n = 3, 4 or ~_ 5 respectively. Unless specified in Table ~, the coe~wients el and ~ are given by the sign of their respective derivative which is an invariant of the changes of coordinates. In particular they must be nonzero. Theorem 8 (Recognition problem) The flowcharts, Figure 5,6 or 7 for n = 3, 4 or > 5 respectively, summarize the recognition problem. For a given germ it is solved folloudng the arrows lcadCng to a normal form. The common condition to all the bifurcation problems isp0 = 0. Proofs. The proofs of these two theorems follow the classical framework. Explicit changes of coordinates are performed for the lower order terms and then Theorem 4 is used (cf. Table 3) to cut off the higher order terms. Then Theorem 6 is used to evaluate the Cc°- codimension and an explicit inspection shows if the topological codimension of the normal form can be reduced, n The topological properties of the normal forms and their universal unfolding can be studied via Damon's theory [5]. In particular, in addition to the distinguished values given by the recognition problem (cf. Figure 5,6 and 7), we only find a few other special cases. For n -- 3, in the case VIII, f -- 0 is a transition value for the unfolding, even if it is not important for the recognition problem. For n -- 4 the coefficient v for the cases IV,V can vanish without any topological effect. case normal form top( C°°)-cod univ.unf . I (~a, e, O) 0 id. B" (6a 2 + exA, e, O) 0 id. ///" (~a 2 + ~I,X 2, ~, O) 1 +(a, O, O) v (~a 3 + e~ ;~, ~, o) 1 +(,~a, o, o) VI (~a, elA, e) 0 id. VaT (~a, eta 2, ~) 1 +(0, a, 0) vM (f,, + ~a2 + ,1:~, ~a, ,) 1(2) +(,,a, o, o) IX (~a, e2u + elA, ~4v + ~3A) I +(0, O, a) List 1. Normal forms and universal unfoldings for n = 3.

case I II 1V V VI VII IX case I U 117 N n=5 n=6 n>7 V YII V//ir I"X 187 normal form top(C°°)-cod univ.unf. cf.n = 3 (h u + 5a 2 + elA, 6, O) 0(1) id. (hu+~a2+eiA2, e,O) 1(2) +(a, 0, 0) (62u + vu 2 + ~a 2 + 61A,6,0) 1 +(ont, 0,0) (hu + 5a 3 + vau + 6,A, 6, O) 1(2) +(eta, O, O) ef.n = 3 cf.n = 3 (,3,, + .IA, 62a, 6) 1 +(oa, O, O) cf.n=3 List 2. Normal forms and universal unfoldings for n = 4. normal form top(C°°)-cod univ.unf. el.n= 3 (e2n + ~a 2 + etA, e, O) 0 id. (e2u + ~a 2 + 61.X 2, 6, O) 1 +(or, O, O) (e2u 2 + ga 2 "4" elA, e,O) 1 +(ccu, O,O) (j,,2 + 6a2 + 6, ~,, 6, o) 1(2) +(,~,,, o, o) (62u 2 + ku 3 + 6a 2 + 61;~, 6, o) 1(2) +(o,u, o, o) (e2n + g an +ifa 3 + elA, e, O) 1(2) +(era, O, O) ef.n =3 ef.n = 3 ef.n = 4 ef.n = 3 List 3. Normal forms and universal unfoldings for n > 5.

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 ...
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 ...
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 ...
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 KLÖ'ZOJ T ll ll ll ll klözo) 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 ...
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
Reading grade 6 2.A.5.b - mdk12
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 ...
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
5 b 5 6 10 c 12 a 5 8 A a 13 b Classroom Exercises - flip@mrflip.com
Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...
EAS 540 - STABLE ISOTOPE GEOCHEMISTRY Thrs., BUS B-5; 6 ...
Panel B 4:45-5:30 Round 2 Panel A 5:30-6:15 | Panel B 6:15-7 ...
TEST ANSWERS Version A 1. B 2. B 3. E 4. E 5. D 6. B 7. C 8. B 9. A ...
Solutions to Quiz 5 Sample B - Loyola University Chicago
(0 12 3¡) 4(5 6 3 798 9 %@4a7(6 b cedgfhfhiqps rti u)vxwtwty
1. Nyelvismereti feladatsor 1. C 2. B 3. D 4. D 5. A 6. B 7. B 8. C ... - Itk
1 A 2 B 3 A 4 B 5 6 7 8 A 9 B 10B½ 11 A 12 13 14 B 15 A 16 B 17 A ...
¡' &) (10 , 2 354 6&7 98@ ) AC B$ E DG FH I 6 &Q ... - Université Lille 1
V 5 1 5 B 6 L 4 X P T S F
Friday Warm-up A 5:00-5:30 PM, Warm-up B 5:30-6:00 PM Starts at ...
Midterm 1 KEY MC Answers 1. B 2. B 3. A 4. D 5. A 6. D 7. C 8. A 9 ...
1. Nyelvismereti feladatsor 1. A 2. B 3. A 4. A 5. D 6. D 7. B 8. C 9 ... - Itk
9 6030 EC 10 6030 B 11 6030 R 12 6030 N 5 6031 EC 6 6031 B 7 ...
10 20 30 40 50 3 e d c 5 1 2 4 8 7 g f 6 B A a b