Más contenido relacionado La actualidad más candente (20) Similar a One way to see higher dimensional surface (20) One way to see higher dimensional surface2. •
• GLn (R), GLn (C)
• GLn (R), O(n)
{( ) }
a, b
• SL2 (R) = |ab − cd = 1
c, d
• R4 ab − cd = 1
3. • Mn (R) : n
• Mn (R) : n
• GLn (R) :
= {g ∈ Mn (R) | detg ̸= 0}
• SLn (R) = {g ∈ Mn (R) | detg = 1}
• SLn (C) = {g ∈ Mn (C) | detg = 1}
• O(n) : = {g ∈ Mn (R) | gt g = In }
• SO(n) : = {g ∈ Mn (R) | gt g = In , detg = 1}
• U(n) : = {g ∈ Mn (C) | g∗ g = In }
• SU(n) : = {g ∈ Mn (C) | g∗ g = In , detg = 1}
5. { }
∑
n+1
n
• S = 1 n+1
(x , . . . , x )∈R n+1
| i 2
(x ) = 1
i=1
6. 1
{( )}
cos θ sin θ
• SO(2) = = S1
− sin θ cos θ
{ }
∑
n+1
• O(n + 1)/O(n) = S = n 1 n+1
(x , . . . x )| i 2
(x ) = 1
i=1
0
.
• O(n) ∋ A →
A .
. ∈ O(n + 1) O(n)
0
0 ··· 0 1
O(n + 1)
• O(3, R) R3 /O(2, R) {(0, 0, z)} = S1 × R
•
• SL2 (R) = S1 × R2
7. 2(1/2)
• H = {a + bi + cj + dk|a, b, c, d ∈ R}
• i = j = k2 = −1, ij = k, jk = i, ki = j, ji = −k,
2 2
kj = −i, ik = −j
• (e.g. ij ̸= ji)
• z = a + bi + cj + dk ∈ H
• ¯ = a − bi − cj − dk
z
• |z|2 = z¯
z
• S3 = {(a, b, c, d) ∈ R4 | z = a + bi + cj + dk |z| = 1}
8. 2(2/2)
• ( )
a + bi c + di
• z = a + bi + cj + dk ↔
−c + di −a + bi
• → S3 H SU(2)
Figure:
9. • S3 4 →
Figure: Dimensions
10. 1
• f : R2 → R f(x, y) = x2 − y3
• f −1 (0) : x = 0
• {(x, y) ∈ R2 |x2/3 + y2/3 = 1}
• http://www-history.mcs.st-and.ac.uk/Curves/Curves.html
11. f : Rr → R a ∈ Rr
∂f
(a)(1 ≤ i ≤ r) 0
∂xi
f −1 (f(a))
12. 1
• f : R2 → R f(x, y) = x2 − y2
• ∂x f = 2x, ∂y f = −2y
• ∂x f = ∂y f = 0 (x, y) = (0, 0)
• f −1 (0) : 2
• f −1 (c) (c ̸= 0) :
13. 2
• det : Mn (R) → R ; g → detg
• SLn (R) = det−1 (1) SLn (R)
14. 2
•
• (Rn or Cn ) R R
• S1 = {(x, y) ∈ R2 | x2 + y2 = 1},
• D = {(x, y) ∈ R2 | y ≥ 0}
15. ∑
n
• f : Mn (R) → R ; g → Tr(t gg) f(g) = 2
gij
i,j=1
• g ∈ O(n) −1
f(g) = n √ O(n) ⊂ f (n)
O(n) Mn (R) n
16. • O(p, q) = {g ∈ Mn (R) | t gIp,q g = Ip,q }(p + q = n) q>0
1
..
.
1
• Ip,q =
−1
..
.
{( ) −1 }
cosh θ sinh θ
• O(1, 1) = ∈ R |0 ≤ θ < 2π
4
sinh θ cosh θ
• O(3, 1)
18. 1
• On (R) −1 1
(= SO(n))
• π0 (On (R)) = Z2
• dimZ π0 (X) = #{X/ ∼}
• x∼y⇔x y
19. (2/2)
• g g′ On (R)
g g′ c c
• c : [−1, 1] → On (R) ; c(−1) = g, c(1) = g′
• det ◦ c
• (det ◦ c)(−1) = det(g) = −1
• (det ◦ c)(1) = det(g′ ) = 1
• (det ◦ c)(a) = 0 a ∈ (−1, 1)
• c(a) ̸∈ On (R) c
20. 2
Figure:
• π0 ( ) = Z2
22. • π1 (X, x0 ) = {f : [0, 1] → X|f(0) = f(1) = x0 }/ ∼
• f∼g⇔f g
•
• http://www.slideshare.net/KentaOono/pf-iseminar
• π1 (X, x0 ) = 0 ⇒ 1
23. • X
1 1 I→X
p = I(0) q = I(1)
···
• www.math.sci.hokudai.ac.jp/
˜
ishikawa/isoukika/isoukika5.pdf
24. • πn (X, x0 ) = {f : [0, 1]n → X|f(a) = x0 ∀a ∈ ∂([0, 1]n )}/ ∼
• f∼g↔f g
25. •
• :π1 (SOn (R)) = Z/2Z = {0, 1} (if n ≥ 3)
26. (1/2)
• su(n) =
= {X ∈ Mn (C)|X + ∗ X = 0}
• su(2) 3 R3
• g ∈ SU(2), X ∈ su(2) Adg (X) = gXg−1
gXg−1 ∈ su(2)
• Adg su(2)
• Adg ∈ SO(3)
• SU(2) → SO(3) ; g → Adg
• 2 1 X = Adg g
X 2
28. • (dimR Mn (R) = n2 )
•
• (Sn )
•
X → πn (X), Hn (X)
•
• C∞ (M) = {f : M → R|f }
• f:M→R
•
• f : [−1, 1] → M =
• f : S1 → M, =