Psevdogaliley Fazo
Quyidagi formulalar orqali R fazosida psevdogaliley metrikasi d p ( x, y ) aniqlanadi:
n
( x  y ) 2 ,
agar
1
 1
d p2 ( x, y )  
 p
n
2
2
agar
  i  2 ( xi  yi )   j  p 1 ( x j  y j ) ,
Bu yerda x  ( x1 , x2 ,...., xn ) , y  ( y1 , y2 ,...., yn )  R n .
x1  y1 ;
x1  y1 ;
Juftlik ( R n , d p ) psevdogaliley fazo deb ataladi va p n deb belgilanadi.
Quyidagi belgilarni kiritamiz:
U n   e1 :   R,
Vn   i 2  i ei :  i  R
n
Endi GL(n, R ) dagi quyidagi pastki guruhni ko’rib chiqamiz:
Gn   g  GL ( n, R ) : gU n  U n , gVn  Vn 
Agar g  ( g ij )  Gn , unda g (e1 )  ( y1,0,...,0) U n , shuning uchun gi1  0 har qanday
i=2,3,…,n uchun.
Endi quyidagi to’plamni ko’rib chiqamiz :
( n, p )   g  Gn : g11  1, g V g |V . Agar g  (n, p ) va x, y  R n bo’lib, x1  y1  0,
n
n
u holda gx, gy vektorlarning birinchi koordinatalari nol bo’ladi. Vn uchun e2 ,..., en
bazisni tanlab , uni R n1 bilan identifikatsiya qilamiz. Bu shuni anglatadiki (n, p ) -bu
psevdogaliley fazosining o’zgarishlar guruhi.
Mahsusu psedogaliley guruh:
S (n, p )   g  (n, p ) : det( g )  1
Yo’l x (t )  x j (t ) j n berilgan bo’lib, quyidagini kiritamiz :
M n 1 ( x (t ))  ( x j ( j 1) (t )) i , j  2 n yo’l x(t )   regulyar deyiladi, agar:
t  (0,1)
det M n1 ( x(t ))  0
Teorema 4. Agar x(t ) va y (t ) (n, p ) - ekvivalent (mos ravishda S (n, p ) ekvivalent ) bo’lsa, u holda va faqat shartlar quyidagi tengliklar bajarilganda bajariladi:
(4)
y1 (t )   x1 (t );
M n11 ( x(t ))  M n1(1) ( x(t ))  M n11 ( y (t ))  M n1(1) ( y (t ));
(5)
(6)
M n1 ( x(t ))  I  M n1 ( x (t ))  M n1 ( y (t ))  I  M n1 ( y (t )).
Barcha t  (0,1) uchun (mos ravishda, tengliklar (5), (6) va quyidagi tengliklar bajariladi):
yn (t )  xn (t );
(7)
det M n 1 ( x(t ))  det M n 1 ( y (t ));
(8)
2 va 4- teoremalar yordamida ( n, p ) uchun ekvivalentlik (mos ravishda, S  ( n, p )
uchun ekvivalentlik) mezoni -  - reguliyar yo’llari uchun quyidagicha aniqlaymiz.
Teoreme 5.  - regulyar yo’llar x (t ) va y (t ) ( n, p ) ga ekvivalent (yani S  ( n, p ) ga
ekvivalent) bo’ladi faqat va faqat quyidagi tenglik (4) va tenglik
2
p
2
p
2
p
p
 ( x (t ))   ( x (t ))   ( y (t ))   ( y (t ))
(k )
i
(k )
i
i2
(k )
i
i  p 1
2
(k )
i
i2
i  p 1
(9)
bajarilganda, barcha t  (0,1) va k  0,1, 2,..., n  2 uchun (yani (7), (8) va (9) tengliklar
bajarilganda) bo’ladi.
Teorema 6. Faraz qilaylik x (t ) va y (t ) - R da shunday yo’llarki, x (t ) va y (t )   regulyar yo’llardir. Unda:
n
(i)
Yo’llar x (t ) va y (t ) R (n, p ) ga ekvivalent bo‘ladi, agar va faqat agar
quyidagi tengliklar bajarilganda:
(10)
y1(1) (t )   x1(1) (t );
(1)
n
p
2
2
p
p
2
p
 ( xi( k ) (t ))   ( xi( k ) (t ))   ( yi( k ) (t ))   ( yi( k ) (t ))
i2
i  p 1
i2
(1)
2
i  p 1
(11)
barcha t  (0,1) , k  0,1, 2,..., n  1 lar uchun
n
(ii) Yo’llar x (t ) va y (t ) R (n, p ) ga ekvivalent bo’ladi, agar va faqat agar
tengliklar (10), (11) k  0,1, 2,..., n  2 da va quyidagilar bajarilganda:
y1(1) (t )  x1(1) (t );
det M n(1)1 ( x(t ))  det M n(1)1 ( y(t ))
Har bir t  (0,1) uchun.
Yo’llarning bazi noyevkilid geometriyadagi ekvivalentligi
n
Endi R da quyidag tenglik bilan aniqlanadigan metrikani ko’rib chiqamiz:
( x  y ) 2 ,
 1 1

2
d p ( x , y )   ( xn  y n ) 2 ,
 p
n2
2
2
  i  2 ( xi  yi )   i  p 1 ( x j  y j ) ,
agar
agar
x1  y1 ;
agar
x1  y1 , xn  yn ;
x1  y1 , xn  yn ;
Tanlaymiz:
U n   e1 :   R


 n 1

   i e i :  i  R, i  2,..., n  2 
i2
 i2

n
Wn   e :   R
Vn  Lin e i 
n 1
Va GL (n, R ) GL (n, R ) guruhida quyidagi kichik guruhni ko’rib chiqamiz:
Gn'   g  GL ( n, R ) : gU n  U n , gVn  Vn , gWn  Wn  .
Agar g  ( g ij )in, j 1  Gn' bo’lsa, u holda g (e1 )   y1 , 0,..., 0  U n , g (en )  0, 0,..., yn   Wn va shuning
uchun gi1  ( g (e1 ), ei )  0 agar i  2,..., n va gin  ( g (en ), ei )  0 agar i  1, 2,..., n  1 .
Matritsa g  Gn quyidagi ko’rinishga ega:
'
 g11 g12
g1n 1 0 


g12
g 2 n 1 0 
0
 , g  0,
g 
g nn  0
11


g n 1n 1 0 
 0 g n 12


gn2
g nn 1 g nn 
0
.
(12)
Aftidan, har qanday g  GL (n, R ) , (12) ko’rinishga ega bo’lsa, quyidagi tenglik o’rinli
bo’ladi:
gU n  U n , gVn  Vn , gWn  Wn .
Quyidagicha to’plamni ko’rib chiqmiz:
p
RnO(n, p)   g  Gn' : g11  1, g nn  1va g Vn  bu Vn da soxta ortogonal o ' zgarish .
Bazis e2 ,..., en 1 Vn bo’lsin va Vn  R n2 indentifikatsiya qilinsin. Shunda, har qanday
g p RnO(n, p) uchun o’zgarish:
h  (hij )in,j1 2  g V
n
Gruppaning elementi bo’ladi: O  n  2, p  1 bunda:
hij  (he j , ei )  ( ge j , ei ) barcha i, j  2, ... , n  2 uchun.
Demak,
p
RnO(n, p )   g  ( g ij )in, j 1  GL(n, R ) : g11  1, g i1  0, i  2,...., n,
g nn  1, g in  0, i  1,..., n  1, gVn  Vn , ( g ij ) in,j11  O ( n  2, p  1)  .
Taqdimot 2. O’zgartirish g  ( g ij )in, j 1  G n quyidagi to’plamga p RnO(n, R) tegishli bo’ladi
faqat va faqat
d p2 ( gx, gy )  d p2 ( x, y )
barcha
x, y  R n
Uchun bajarilganda.
Isbot. Agar g  ( g ij )in, j 1 p RnO (n, p ) ,
x  ( x1 ,..., xn )T y  ( y1 ,..., yn )T  R n , va x1  y1 barcha i  2,..., n  1 uchun
xi  1, g11  0 faqat i  2,..., n da, va g nn  1, gin  0 bo’lsa, unda:
d p2 ( x, y )  ( x1  y1 ) 2  d p2 ( x, y )  ( xn  yn ) 2 .
Demak,
 n g x

 j 1 1 j j

 n 1 g x

  j 1 2 j j
   x   n 1 g x , ...,  n 1 g
x ,
j 2 2 j j
j 1 ( n 1) j j 

  1
gx  


n 1
 xn   j 1 g nj x j
 n 1
 

  j 1 g ( n  1) j x j 
 n

  j 1 g nj x j

 n g y

 j 1 1 j j

 n 1 g y

  j 1 2 j j
   y   n 1 g y , ...,  n 1 g
y ,
j 2 2 j j
j 1 ( n 1) j j 

  1
gy  

 

n 1
 xn   j 1 g nj y j
 n 1
 

  j 1 g ( n  1) j y j 
 n

  j 1 g nj y j

Va shu sababli:
d p2 ( gx, gy )  ( x1  y1 ) 2  d p2 ( x, y ). d p2 ( gx, gy )  ( xn  yn ) 2  d p2 ( x, y ).