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Пишет chervov ([info]chervov) в [info]ljr_math
@ 2006-08-21 18:47:00
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Vopros pro gomotipii
Tut u moego druga fizika voznik vopros:

> Dano gladkoe lokal,no trivial'noe rassloenie nad sferoi

> p: M \to S^n.

> Sloi odnosvyaznii.
> Dani dva secheniya etogo rassloeniya

> s_0, s_1: S^n \to M, p \cdot s_i = id. ( i = 0,1 )

> Izvesto, chto secheniya homotopni
> KAK OTOBRAZHENIYA, t.e. suschestvuet homotopiya

> h: [0,1] \times S^n \to M, h(0,x)=s_0(x), h(1,x)= s_1(x), gde x \in S^n.

> Vopros: homotopni li eti secheniya KAK SECHENIYA?
> T.e., ih mozhno soedinit' v prostranstve sechenii.
> V zapisi eto oznachaet:
> suschestvuet li homotopiya

> g: [0,1] \times S^n \to M, g(0,x)=s_0(x), g(1,x)= s_1(x), dlya kotoroi
> p \cdot g(t,x) = x ?

Mne kazetsya chto da, Gosha Sharygin schitaet chto net :)
Chto dumaet narod ?

(Добавить комментарий)


[info]shribavavsenahu
2006-08-22 16:29 (ссылка)
Da, pohozhe, eto verno. Rassuzhdenie takoe. Pust' $n>1$. Predstavim nashu sferu kak ob"edinenie dvuh sharov: $S^n=B_1\cup B_2$. Pust' u rassloeniya est' secheniya $s$ i $s'$. Trivializuem rassloenie nad $B_1$: $M|B_1=B_1\times F$, gde $F$ -- sloi. Zamenyaya $s$ i $s'$ na gomotopnye im v klasse sechenii, schitaem, chto $s=s'$ na $B_2$, $s=s'=*$ na granice $S^{n-1}$ sharov $B_1$ i $B_2$ ($*$ -- otmechennaya tochka $F$).

Nad $B_1$ $s$, sootv., $s'$ imeet vid $id\times f$, sootv., $id\times f'$, gde $f,f'$ -- otobrazheniya $B_1\to F$, ravnye $*$ na $S^{n-1}$. Pust' $g,g'$ -- elementy $\pi_n(F)$, sootvetstvuyuschie $f,f'$. Esli $s\sim s'$ kak otobrazhenie, to ih obrazy v $\pi_n(M)$ korrektno opredeleny i ravny (tut my pol'zuemsya tem, chto $F$ i $S^n$ odnosvyazny, a znachit, $M$ tozhe). No eti obrazy otlichayutsya drug ot druga na obraz $g-g'$ v $\pi_n(M)$, otkuda obraz $g-g'$ v $\pi_n(M)$ nulevoi. Znachit $g=g'$ (tak kak u nashego rassloeniya est' secheniya, tochnaya gomotopicheskaya posledovatel'nost' rasscheplyaetsya), poetomu $f$ gomotopno $f'$ v klasse otobrazhenii $B_1\to F$, ravnyh $*$ na granice, a znachit, $s$ gomotopno $s'$ kak sechenie.

V sluchae $n=1$ prohodit to zhe rassuzhdenie, no na etot raz $g=g'$ sleduet iz odnosvyaznosti $F$.

Chto-to vrode togo.

(Ответить) (Ветвь дискуссии)

spasibo za otvet
(Анонимно)
2006-08-24 23:05 (ссылка)
1. Spasibo za otvet.
2. Hochu utochnit':
Vi pishete:
> > schitaem, chto $s=s'$ na $B_2$, $s=s'=*$ na granice $S^{n-1}$
>> sharov $B_1$ i $B_2$ ($*$ -- otmechennaya tochka $F$).
Vopros: razve takoe vozmozhno POSLE TRIVIALIZACII OTDEL'NO na kazhdom share ?

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: spasibo za otvet
[info]shribavavsenahu
2006-08-25 16:04 (ссылка)
Da, Vy pravy, navernoe, eto nuzhno kak-to poyasnit'.

My vsegda mozhem schitat', chto nashi secheniya $s$ i $s'$ sovpadayut na $B_2$.

Teper' pust' $s_0$ -- lyuboe iz sechenii $s,s'$. My hotim zamenit' $s_0$ na gomotopnoe emu v klasse sechenii, tak chtoby $s_0|S^{n-1}=\partial B_1=\partial B_2$ bylo ravno $*$ v vybrannoi vyshe trivializacii $M$ nad $B_1$. Eto mozhno sdelat', potomu chto

1. Otobrazhenie $S^{n-1}\to F$, kotoroe poluchaetsya, kogda my vybiraem trivializaciyu nad $B_1$, styagivaemo (tak kak prodolzhaetsya na shar). Znachit, ogranichenie $s_0|S^{n-1}$ mozhno progomotopirovat' v klasse sechenii $M$ nad $S^{n-1}$, tak chtoby ono v trivializacii $B_1$ bylo postoyannym i ravnym $*$.

2. Para $(S^{n-1},B^n)$ -- korassloenie (= para Borsuka). Otsyuda sleduet, chto esli u nas est' sechenie $t$ trivial'nogo rassloeniya $B^n\times F$ i gomotopiya ogranicheniya $t_1=t|S^{n-1}$ v kakoe-to drugoe sechenie $t_1':S^{n-1}\to S^{n-1}\times F$, to etu gomotopiyu mozhno prodolzhit' do gomotopii (v klasse sechenii) mezhdu $t$ i $t'$, gde $t'$ sechenie $B^n\times F$, prodolzhayuschee $t_1'$.

Primerno tak.

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: spasibo za otvet
(Анонимно)
2006-08-27 21:15 (ссылка)
1. Esche raz spasibo za podderzhku!
Pozvol'te, ya snachala utochnyu, verno li ya Vas ponyal.

2. Na 1-m etape Vi homotopiruete sechenie na S^{n-1}
tak, chtobi v trivializacii nad B_1 ono stalo *.
Zatem etu homotopiyu vi rasprostranyaete (v trivializacii!) na vse B_1.
(Kstati, mozhno pri etom sdelat' tak, chto odno iz sechenii
progomotopiruetsya na B_1 v *.)
Zatem Vi vozvraschaete homotopiyu na S^{n-1}
v homotopiyu na S^{n-1} \subset B_2 kak
nad podmnogoobraziem i rasprostranyaete ee do
homotopii na vsem B_2.
Tak?

3. Poidem dal'she. A dal'she Vi rassuzhdaete tak, kak bud-to
DANNAYA homotopiya secheniya s v s' kak otobrazheniya
prohodit zavedomo dvazhdi INVARIANTNO:
a) ona postoyanna na B_2,
b) ona s B_1 "ne zalazit" B_2
Vot esli eto tak, to mi imeem pravo
sosredotochit'sya na homotopii nad B_1 i vse poluchaetsya.
No, pozhaluista, ob'yasnite, POCHEMU dopuscheniya
a) i b) verni.

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: spasibo za otvet
[info]shribavavsenahu
2006-08-28 00:02 (ссылка)
>1. Esche raz spasibo za podderzhku!

Vsegda pozhaluista!

>Pozvol'te, ya snachala utochnyu, verno li ya Vas ponyal.
>
>2. Na 1-m etape Vi homotopiruete sechenie na S^{n-1}
>tak, chtobi v trivializacii nad B_1 ono stalo *.
>Zatem etu homotopiyu vi rasprostranyaete (v trivializacii!) na vse B_1.
>(Kstati, mozhno pri etom sdelat' tak, chto odno iz sechenii
>progomotopiruetsya na B_1 v *.)

Da. No zametim, chto ne oba odnovremenno.

>Zatem Vi vozvraschaete homotopiyu na S^{n-1}
>v homotopiyu na S^{n-1} \subset B_2 kak
>nad podmnogoobraziem i rasprostranyaete ee do
>homotopii na vsem B_2.

A vot etogo uzhe ya ne ponimayu. S $B_2$ my postupaem tochno tak zhe, kak i s $B_1$: my trivializuem rassloenie nad $B_2$, u nas est' gomotopiya sechenii nad $S^{n-1}$ i my prodolzhaem ee do gomotopii sechenii nad $B_2$. Tak kak ishodnye secheniya sovpadali na $B_2$, to, chto poluchitsya v konce gomotopii tozhe budet tam sovpadat'.

>Tak?
>
>3. Poidem dal'she. A dal'she Vi rassuzhdaete tak, kak bud-to
>DANNAYA homotopiya secheniya s v s' kak otobrazheniya
>prohodit zavedomo dvazhdi INVARIANTNO:
>a) ona postoyanna na B_2,
>b) ona s B_1 "ne zalazit" B_2

Net, etogo ya ne utverzhdayu. Razberem sluchai $n>1$. V etom sluchae $M$ odnosvyazno, i lyubomu otobrazheniyu $f:S^n\to M$ (v chastnosti, nashim secheniyam) sootvetstvuet korrektno opredelennyi element $[f]$ gruppy $\pi_n(M)$. Tak kak $s\sim s'$ kak otobrazhenie, imeem $[s]=[s']$. S drugoi storony, po opredeleniyu slozheniya v $\pi_n(M)$ i pol'zuyas' tem, chto $s=s'$ na $B_2$, $s=s'=*$ na $S^{n-1}$, poluchaem $[s]=[\tilde s]+i_*([g]), [s']=[\tilde s]+i_*([g'])$, gde $i_*:\pi_n(F)\to\pi_n(M)$ inducirovano vlozheniem $i:F\subset M$, a $\tilde s$ -- sechenie, ravnoe $s$ i $s'$ na $B_2$ i tozhdestvenno ravnoe $*$ na $B_1$ (v vybrannoi s samogo nachala trivializacii $M$ nad $B_1$.

V sluchae $n=1$ vse analogichno.

Gde-to tak

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: spasibo za otvet
(Анонимно)
2006-09-01 22:42 (ссылка)
еще раз спасибо за помощь

автор вопроса послал Вам письмо на shribavavsenahu@hotmail.com

(Ответить) (Уровень выше)

Vozrazenia Goshi
[info]chervov
2006-08-25 13:50 (ссылка)
Mozhet byt', tut vse i pravil'no. No est' nekotorye somnitel'nye detali.
Naprimer, nejasno, kak predlagaetsja deformirovat' sechenija na odnoj
polusfere. da ewe tak, chtoby poluchit' zadannoe otobrazhenie na granice
-- ved' pri jetom nado budet chto-to deformirovat' na granice vtoroj
polusfery, sledovatel'no, nado budet vse prodolzhat' do gomotopii na
vsej sfere. Vozmozhno, pravda, jeta trudnost' preodolima.

Voobwe zhe gomotopicheskie klassy sechenij rassloenija
klassificirujutsja n-mernymi gomologijami bazy s koefficientami v
n-1-mernyh (ili n+1-mernyh?) gomotopijah sloja (1-1 sootvetstvie v
nekotorom smysle). Pri jetom mozhet okazat'sja, chto total'noe
prostranstvo stjagivaemo, no ni baza, ni sloj ne stjagivaemy. Togda vse
otobrazhenija v total'noe prostranstvo gomotopny, no sechenij oche'
mnogo raznyh. U sfery, pravda, prostye gomologii, tak chto ochen'
verojatno, chto vse sechenija gomotopny. Kstati, rassuzhdenie, kotoroe
tebe prislali, po-suwestvu, povtorjaet postroenie vysheupomjanutogo
1-1-sootvetstvija.

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: Vozrazenia Goshi
[info]shribavavsenahu
2006-08-25 20:05 (ссылка)
Secheniya rassloenii ne klassificiruyutsya gomologiyami bazy s koefficientami v gomotopiyah sloya. V sluchae $S^n, n>2$ v $H^n(S^n,\pi_{n-1}(F))$ lezhit prepyatstvie k suschestvovaniyu secheniya. V obschem sluchae vse slozhnee, tak kak prepyatstvii celaya seriya.

Voobsche, dlya togo chtoby znat' gomotopicheskuyu klassifikaciyu sechenii, nedostatochno znat' gomotopicheskie tipy bazy i sloya. Naprimer, u trivial'nogo rassloeniya $S^4\times S^3\to S^4$ est' secheniya, a u kvaternionnogo rassloeniya Hopfa $S^3\subset S^7\to S^4$ -- net (eto glavnoe SU(2)-rassloenie, i esli by u nego byli secheniya, ono bylo by trivial'nym).

Chto-to v takom duhe.

(Ответить) (Уровень выше) (Ветвь дискуссии)

Re: Vozrazenia Goshi
[info]chervov
2006-08-28 15:25 (ссылка)
From Gosha

Vse verno, v strogom smysle slova prepjatstvij k suwestvovaniju sechenij bol'she (celaja serija). Chto kasaetsja ih klassofikacii, to tam tozhe voznikaet celaja serija klassov, pozvoljajuwih razlichat' sechenija (ved' poslojnuju gomotopiju mezhdu sechenijami mozhno rassmatrivat' kak sechenie inducirovannogo rassloenija na S^n\times[0;1]).

Rassuzhdenie, pozvoljauwee sdvinut' sechenija drug v druga ja podrobno ne proverjal, no, kazhetsja, ono pravil'noe.

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