Motive

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

Let k be a 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) etale 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\neq0 , A,B in\mathbb{Q}

over complex numbers, let\alpha,\betagenerators 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}

\intE(C)\to\mathbb{C}/\Lambdais an isomorphism

inverse map : Weierstrass\wp-function

abelian varieties form a\mathbb{Z}-linear category. So take a tensor with\mathbb{Q}

(abelian varieties)\otimes\mathbb{Q}  = (category of ab. varieties up to isogeny) . these are\mathbb{Q}-linear

this is inside the category of motives.

 

 

http://en.wikipedia.org/wiki/Motive_(algebraic_geometry)

 

 

Feynman motive