"Motive"의 두 판 사이의 차이
130번째 줄: | 130번째 줄: | ||
H^0(E,\Omega^1_E) = \mathbb{C}\cdot \frac{dx}{2y} | H^0(E,\Omega^1_E) = \mathbb{C}\cdot \frac{dx}{2y} | ||
− | \omega_{\alpha}=\int_{\alpha}\frac{dx}{2y}, | + | \omega_{\alpha}=\int_{\alpha}\frac{dx}{2y}, \omega_{\beta}=\int_{\beta}\frac{dx}{2y} \in \mathbb{C} |
− | \omega_{\ | + | These are linearly independent over real numbers so we get a lattice \Lambda=\mathbb{Z}\omega_{\alpha}+\mathbb{Z}\omega_{\beta}\subset \mathbb{C} |
+ | |||
+ | |||
+ | |||
+ | \int E(C)\to \mathbb{C}/\Lambda is an isomorphism | ||
+ | |||
+ | inverse map : Weierstrass \wp-function | ||
+ | |||
+ | |||
+ | |||
+ | abelian varieties | ||
+ | |||
+ | \mathbb{Z}-linear category |
2010년 12월 3일 (금) 09:02 판
geometry roughly= cohomology
example
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,\mathbb{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}\frac{dz}{z} = 2\pi i
Etale cohomology
exponential map : \mathbb{C}\to \mathbb{C}^{*}
H^1(\mathbb{G}_m,\mathbb{Z}) = Hom(\pi_1(\mathbb{C}^{*}),\mathb{Z})
Let l be a prime.
H^1_{et}(\mathbb{G}_m,\mathbb{Q}_{l}) is a 1-dimensional \mathbb{Q}_{l} vector space on which Gal(\bar{\mathbb{Q}}/\mathbb{Q}) acts.
We get a character called the cyclotomic character.
general picture
k field (Q,F_q,C,\cdots)
from (separable finit type k-schemes) to category of motives
- Betti cohomology Vec over Q (Hodeg structure)
- de Rham cohomology Vec over k if char k = 0 (graded vector space)
- if l\neq char(k) etal cohomology vec over \mathbb{Q}_l (Galois representation)
- crystalline cohomology Vec over \mathbb{Q}_p
(category of motives) can do linear algebra
\mathbb{Q}-linear \otimes-category
bigger picture obtained when we compare cohomologies
Betti <-> de Rham , Hodge theory
crystalline(de Rham) <-> etale, p-adic Hodge theory
What we like in linear algebra :
1 dimension
2 f : V\to V, characteristic polynomial
something we don't know :
X over (k = \bar{k}), char(k)\neq 0
for all l prime to characteristic, dim_{\mathbb{Q}_l H^i_{et}(X,\mathbb{Q}_l)
We don't know how to show that these numbers are independent of l.
we know that if X over k is smooth and proper,
k=\bar{\mathbb{F}_q}, then we know that these numbers are independent of l (Deligne-Weil II, trace formula for etale cohomology)
X smooth, alternating sum of dimension, \sum(-1)^i dim_{\mathbb{Q}_l H^i_{et}(X,\mathbb{Q}_l) is independent of l. (intersection theory of cycles)
ex : elliptic curve
E : y^2=x^3-Ax-B, \Delta\neq 0 , A,B in \mathbb{Q}
over complex numbers, let \alpha, \beta generators H_1
H^0(E,\Omega^1_E) = \mathbb{C}\cdot \frac{dx}{2y}
\omega_{\alpha}=\int_{\alpha}\frac{dx}{2y}, \omega_{\beta}=\int_{\beta}\frac{dx}{2y} \in \mathbb{C}
These are linearly independent over real numbers so we get a lattice \Lambda=\mathbb{Z}\omega_{\alpha}+\mathbb{Z}\omega_{\beta}\subset \mathbb{C}
\int E(C)\to \mathbb{C}/\Lambda is an isomorphism
inverse map : Weierstrass \wp-function
abelian varieties
\mathbb{Z}-linear category