"Motive"의 두 판 사이의 차이
21번째 줄: | 21번째 줄: | ||
H^0(\mathbb{G}_m,Z)=Z | H^0(\mathbb{G}_m,Z)=Z | ||
− | H^1(\mathbb{G}_m,Z)=Z , this is dual to H_1(\mathbb{G}_m,Z) | + | H^1(\mathbb{G}_m,Z)=Z , this is dual to H_1(\mathbb{G}_m,Z) we can call the generator as <math>\gamma_0^{\vee}</math> where <math>\gamma_0</math> is the homology generator. |
29번째 줄: | 29번째 줄: | ||
H^0_{dR}(\mathbb{G}_m)=\mathbb{C} | H^0_{dR}(\mathbb{G}_m)=\mathbb{C} | ||
− | H^1_{dR}(\mathbb{G}_m)=\mathbb{C} | + | H^1_{dR}(\mathbb{G}_m)=\mathbb{C}\frac{dz}{z} |
+ | |||
+ | |||
+ | |||
+ | De Rham isomorphism | ||
+ | |||
+ | H^1(\mathbb{G}_m,Z) \times H^1_{dR}(\mathbb{G}_m) \to \mathbb{C} is a perfect pairing | ||
+ | |||
+ | (\gamma,\omega) \to \int_{\gamma}\omega | ||
+ | |||
+ | i.e. H^1_{dR}(\mathbb{G}_m) = ^1(\mathbb{G}_m,Z)\otimes_{\mathbb{Z}}\mathbb{C} | ||
+ | |||
+ | |||
+ | |||
+ | question. under this isomorphism, \frac{dz}{z} = c\times <math>\gamma_0^{\vee}</math> what is c? | ||
+ | |||
+ | c = \int_{\gamma_0} |
2010년 12월 3일 (금) 08:22 판
geometry roughly= cohomology
examples
circle S^1
Betti cohomolgy (singular cohomology)
H^0(S^1,Z)=Z
H^1(S^1,Z)=Z
\mathbb{G}_m = \mathbb{C}^{x} = \mathbb{C}/{0} same homotopy class as S^1
Betti cohomology is same
H^0(\mathbb{G}_m,Z)=Z
H^1(\mathbb{G}_m,Z)=Z , this is dual to H_1(\mathbb{G}_m,Z) we can call the generator as \(\gamma_0^{\vee}\) where \(\gamma_0\) is the homology generator.
de Rham cohomology
H^0_{dR}(\mathbb{G}_m)=\mathbb{C}
H^1_{dR}(\mathbb{G}_m)=\mathbb{C}\frac{dz}{z}
De Rham isomorphism
H^1(\mathbb{G}_m,Z) \times H^1_{dR}(\mathbb{G}_m) \to \mathbb{C} is a perfect pairing
(\gamma,\omega) \to \int_{\gamma}\omega
i.e. H^1_{dR}(\mathbb{G}_m) = ^1(\mathbb{G}_m,Z)\otimes_{\mathbb{Z}}\mathbb{C}
question. under this isomorphism, \frac{dz}{z} = c\times \(\gamma_0^{\vee}\) what is c?
c = \int_{\gamma_0}