"Motive"의 두 판 사이의 차이

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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)
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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}
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H^1_{dR}(\mathbb{G}_m)=\mathbb{C}\frac{dz}{z}
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De Rham isomorphism
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H^1(\mathbb{G}_m,Z) \times H^1_{dR}(\mathbb{G}_m) \to \mathbb{C} is a perfect pairing
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(\gamma,\omega) \to \int_{\gamma}\omega
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i.e. H^1_{dR}(\mathbb{G}_m) = ^1(\mathbb{G}_m,Z)\otimes_{\mathbb{Z}}\mathbb{C}
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question. under this isomorphism, \frac{dz}{z} = c\times <math>\gamma_0^{\vee}</math>  what is c?
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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}