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

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 same.

X smooth, alternating sum of dimension is independent of l.