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

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

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

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

MHS for L

1

H

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

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

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

The End

Thanks for your attention!

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