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# An Introduction to Lawson Homology, II

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## An Introduction to Lawson Homology, II

Mark E. Walker University of Nebraska – Lincoln

December 2010

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Review of Lawson homology

ForX quasi-projective over C:

Zr(X) =topological abelian group ofr-cycles on X LrHm(X) :=πm−2rZr(X)

There are maps:

HmM(X,Z(r))→LrHm(X)→HmBM(X(C),Z(r)) The left-hand map is an isomorphism withZ/n-coefficients.

The right-hand map is the topic of Suslin’s Conjecture (see below).

Additional comment: These are maps of (non-finitely

generated) MHS’s, whereHmM(X,Z(r))has the trivial MHS. The MHS ofLrHm(X) is induced by MHS onHsing(Cr,e(X)).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Review of Lawson homology

ForX quasi-projective over C:

Zr(X) =topological abelian group ofr-cycles on X LrHm(X) :=πm−2rZr(X)

There are maps:

HmM(X,Z(r))→LrHm(X)→HmBM(X(C),Z(r)) The left-hand map is an isomorphism withZ/n-coefficients.

The right-hand map is the topic of Suslin’s Conjecture (see below).

Additional comment: These are maps of (non-finitely

generated) MHS’s, whereHmM(X,Z(r))has the trivial MHS.

The MHS ofLrHm(X)is induced by MHS on Hsing(Cr,e(X)).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Related theories

I mentioned (but did not define)morphic cohomology, LH(X), yesterday.

There is also a version ofK-theory that uses algebraic equivalence, calledsemi-topological K-theory: Let Grass =`

nGrass(Cn). ForX projective, Kqsemi(X) :=πq

Maps(X,Grass)h+

whereh+denotes “homotopy theoretic group completion” of the homotopy-commutativeH-spaceMaps(X,Grass).

For example,K0semi(X) =K0(X)/(alg. equiv.).

“Every formal property one might expect involvingLH, LH andKsemidoes indeed hold.”

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Some of the properties of K

semi

There are natural maps

Kn(X)→Knsemi(X)→ku−n(X(C)).

Kn(X,Z/m)−→K= nsemi(X,Z/m) form >0.

There is a Chern character isomorphism

ch:Knsemi(X)Q−→ ⊕= LqH2q−n(X,Q).

For X smooth, there is an Atiyah-Hirzebruch spectral sequence

E2p,q=L−qHp−q⇒K−p−qsemi (X), which degenerates upon tensoring with Q.

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Review of Lawson homology and related theories Suslin’s Conjecture

Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Suslin’s Conjecture for Lawson/morphic (co)homology

Conjecture (Suslin’s Conjecture — Lawson form) For a smooth, quasi-projective varietyX, the map

LrHm(X)→Hmsing(X(C),Z(r))

is an isomorphism form≥d+r and a monomorphism for m=d+r−1.

“Suslin’s Conjecture = Bloch-Kato (really,

Beilinson-Lichtenbaum) withZ-coefficients (over C)”:

LtHn(X) ∼=? HnZar(X, tr≤tZ), whereπ: (V ar/C)analytic→(V ar/C)Zar.

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Review of Lawson homology and related theories Suslin’s Conjecture

Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## (Thin) Evidence: Cases where Suslin’s Conjecture is known

For codimension one cycles. Proof is by explicit calculation ofLdim(X)−1H(X).

For L0H (by Dold-Thom Theorem) and Ldim(X)H

(trivially). In particular, it’s known for all surfaces.

For special varieties, such as toric varieties, cellular varieties, linear varieties (that are smooth).

Certain hyper-surfaces of dim. 3[Voineagu]

With finite coefficients — i.e., Bloch-Kato is known [Voevodsky].

The cohomological form holds for π0 [Bloch-Ogus]

Ld−tH2(d−t)(X) =LtH2t(X)∼=H2tZar(X, tr≤tZ)

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Review of Lawson homology and related theories Suslin’s Conjecture

Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Finite generation of Lawson homology

Suslin’s Conjecture predicts, in particular, that

LrHm(X) is finitely generated form≥dim(X) +r−1.

The converse is also known: letCrH(X)be the “cone” of the map fromLrH toHBM, so that

· · · →CrHm(X)LrHm(X)HmBM(X)CrHm−1(X)→ · · ·. Voevodsky’s Bloch-Kato⇒ CrHm(X,Z/n) = 0 for

m≥dim(X) +r−1, and henceCrHm(X,Z)is a divisible group in this range.

Thus, ifLrHm(X) is finitely generated for

m≥dim(X) +r−1, thenCrHm(X) = 0 in this range, and hence Suslin’s Conjecture holds.

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Behavior of correspondences on Lawson homology

A mapY → Cr(X) (for example, an inclusion) determines a cycle

Γ  //

p

rel. dim. rIIIIIIIIYII\$\$×X

Y

and hence a map on cycles spaces

Γ :Z0(Y)→ Zr(X) determined by

y7→Γy =p−1(y)∈ Zr(X).

Applyingπm−2r gives

Γ :Hm−2rsing (Y(C)) =L0Hm−2r(Y)→LrHm(X).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Lifting elements to singular cohomology

Given an element

α∈LrHm(X) =πm−2rZr(X) it lifts toα˜ ∈Hm−2rsing (Cr,e(X)(C)) along the maps

Hm−2rsing (Cr,e(X)(C))→Hm−2rsing (Zr(X))πm−2rZr(X).

Using singular Lefschetz,α˜ lifts to

˜˜

α∈Hm−2rsing (Y(C))→LrHm(X) for someY ⊂ Cr,e(X) with dim(Y)≤m−2r.

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Lifting along correspondences

Thus we have a surjection M

dim(Y)≤m−2r

Hm−2rsing (Y)LrHm(X)

where each mapHm−2rsing (Y)→LrHm(X) is the map associated to an equi-dimensional correspondence Γ : Y _ _ _//X of rel. dim. r:

Hm−2rsing (Y)∼=L0Hm−2r(Y)−→LΓ rHm(X).

Remark: This surjection can be used to understand MHS on LrHm(X).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Lifting along correspondences

Thus we have a surjection M

dim(Y)≤m−2r

Hm−2rsing (Y)LrHm(X)

where each mapHm−2rsing (Y)→LrHm(X) is the map associated to an equi-dimensional correspondence Γ : Y _ _ _//X of rel. dim. r:

Hm−2rsing (Y)∼=L0Hm−2r(Y)−→LΓ rHm(X).

Remark: This surjection can be used to understand MHS on LrHm(X).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Lifting to smooth varieties, using Hodge theory

For such a correspondenceΓ, the composition Hm−2rsing (Y)−→LΓ rHm(X)→Hmsing(X)

coincides with the map on singular cohomology induced byΓ:

Γ :Hm−2rsing (Y)→Hmsing(X)

WhenX is smooth, lettingY˜ →Y be a resolution of singularities, Hodge theory gives:

im

Hm−2rsing (Y)→Hmsing(X)

=

im

Hm−2rsing ( ˜Y)→Hm−2rsing (Y)→Hmsing(X)

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## Characterizing image in singular homology

Proposition (Friedlander-Mazur)

ForX smooth and projective, the topological filtration is contained in the “correspondence” filtration: Every element of

FrtopHmsing(X) := im LrHm(X)→Hmsing(X(C)) is contained in the image of

Γ :Hm−2rsing (W(C))→Hmsing(X(C)) whereW is smooth withdim(W)≤m−2r andΓ is a correspondence.

(SinceHmsing(X(C))is f.g., a single pair W,Γsuffices.)

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Weak form of Suslin’s Conjecture

Recall Suslin’s conjecture predicts

LrHm(X)−→H= msing(X(C)), form≥dim(X) +r.

Conjecture (Weak form of Suslin’s Conjecture (or Friedlander-Mazur Conjecture))

For a smooth, projective varietyX, the map LrHm(X)→Hmsing(X(C)) is onto form≥dim(X) +r.

In particular, the weak Suslin conjecture predicts:

Lm−dim(X)Hm(X)Hmsing(X)

is onto, for allm. (Ifm <dim(X), let Lm−dim(X):=L0.)

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## Consequence of weak Suslin conjecture

Using the Proposition concerning the lifting of elements of im(LrHm→Hmsing)along correspondences:

Proposition

Assume the weak form of Suslin’s Conjecture holds forX. Let d= dim(X).

Then for each integerm, there is a smooth, projective variety Y of dim 2d−m and a correspondenceΓ : Y _ _ _//X of rel.

dim. m−dsuch that

Γ :H2d−msing (Y(C))Hmsing(X(C)) is onto.

The existence of such aY,Γ turns out to be a very strong condition onX. In fact....

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Beilinson’s Theorem

Theorem (Beilinson)

The validity of all Grothendieck’s standard conjectures overC is equivalent to the following property: For each smooth, projectiveX, there is a Y andΓ as above such that

Γ :H2d−msing (Y(C))Hmsing(X(C)) is onto.

Corollary (Beilinson)

The weak form of Suslin’s Conjecture is equivalent to the validity of all of Grothendieck’s standard conjectures overC.

Note: The⇐direction of the Corollary was originally shown by Friedlander-Mazur.

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## Should we believe Suslin’s Conjecture?

Beilinson’s result makes it clear Suslin’s Conjecture is a very strong conjecture. Its weak form is equivalent to

Grothendieck’s standard conjectures.

Perhaps the strong form of Suslin’s conjecture is simply false.

The first unknown case occurs for1-cycles on a smooth projective3-dimensional variety X:

Question

For a smooth, projective3-dimensional variety X, is πm−2Z1(X) =:L1Hm(X)→Hmsing(X(C)) one-to-one form≥3?

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## Mixed Hodge structures for Lawson homology

Assume (for simplicity)X is projective. Then we have a surjection

M

Y,dim(Y)≤m−2r

Hm−2rsing (Y(C))LrHm(X)

of (non f.g.) MHS’s (and where the maps are given by correspondences).

Thus,LrHm(X)has same Hodge type as Hm−2rsing of a union of (highly singular) varieties of dimensionm−2r.

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1

3

## (X ))

For example,

M

Y,dim(Y)=1

H1sing(Y(C))L1H3(X)

and soL1H3(X) has Hodge type: (0,0),(−1,0),(0,−1).

If we assumedim(X) = 3, then Suslin’s conjecture predicts L1H3(X)H3sing(X(C),Z(1))∼=Hsing3 (X,Z(2)) and the target has Hodge type(−1,0),(0,−1).

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## A Conjecture concerning Hodge type

Conjecture

ForX smooth, projective of dimension 3, the Lawson group L1H3(X)has Hodge type (−1,0),(0,−1).

A proof (or counter-example) of just this conjecture would represent highly significant progress.

Note that the validity of this conjecture implies: Conjecture

ForX smooth, projective of dimension 3, the map H3M(X,Z(1))→L1H3(X) is a torsion map.

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## A Conjecture concerning Hodge type

Conjecture

ForX smooth, projective of dimension 3, the Lawson group L1H3(X)has Hodge type (−1,0),(0,−1).

A proof (or counter-example) of just this conjecture would represent highly significant progress.

Note that the validity of this conjecture implies:

Conjecture

ForX smooth, projective of dimension 3, the map H3M(X,Z(1))→L1H3(X) is a torsion map.

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Review of Lawson homology and related theories Suslin’s Conjecture Correspondences Beilinson’s Theorem More on Suslin’s (strong) conjeture

## dim(X ) = 3, Y ⊂ X , dim(Y ) = 2, U = X \ Y

L1H4(Y)

=

dim(Y) = 2 //H4sing(Y)

L1H4(X)

GSConjonto

SuslinConj⇒∼= //H4sing(X)

L1H4(U)

SuslinConj⇒∼= //Hsing4 (U)

L1H3(Y)

=

dim(Y) = 2 //H3sing(Y)

L1H3(X)

SuslinConj1-1 //H3sing(X)

L1H3(U) //Hsing3 (U) 20 / 23

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## Passing to function fields

LetC(X) := lim−→U⊂XU. Proposition

LrHm(C(X)) = 0 ifm < d+r.

For example, L1H3(SpecC) = 0 ifdim(X) = 3.

Assuming Grothendieck Standard Conjectures:

L1H4(C(X))→H4sing(C(X))SuslinC−→0L1H3(X)→H3sing(X).

Challenge

Find a good method of describing/constructing elements of

Hmsing(C(X)) =Hsing2d−m(C(X)).

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## Toy examples

With finite coefficients:

Hmsing(C(X),Z/n) =Lm−dim(X)Hm(C(X),Z/n)

=HM2d−m(C(X),Z/n(2d−m))

=K2d−mMilnor(C(X))/n For Hsing1 :

lim−→

Y

Hsing,Y1 (X)→Hsing1 (X)→Hsing1 (C(X))

→lim−→

Y

Hsing,Y2 (X)→Hsing2 (X)

Since Hsing,Y1 (X) = 0 andHsing,Y2 (X)∼=H2 dim(Ysing )(Y)= free abelian group on integral components ofY:

0→Hsing1 (X)→Hsing1 (C(X))→Z1(X)hom0→0

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## The End

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