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Hodge conjecture

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Unsolved problem in geometry
Millennium Prize Problems
Topological features of a spaceX{\displaystyle X}, such as a hole (labelled byA{\displaystyle A}) are usually detected usingsingular (co)homology, where the presence of a non-zero class[α]Hsingk(X){\displaystyle [\alpha ]\in H_{\text{sing}}^{k}(X)} indicates the spaceX{\displaystyle X} has a (dimensionk{\displaystyle k}) hole. Such a class is represented by a(co)chain ofsimplices, depicted by the red polygon built out of 1-simplices (line segments) on the left. This class detects the holeA{\displaystyle A} by looping around it. In this case, there is in fact apolynomial equation whose zero set, depicted in green on the right, also detects the hole by looping around it. TheHodge conjecture generalises this statement to higher dimensions.

Inmathematics, theHodge conjecture is a major unsolved problem inalgebraic geometry andcomplex geometry that relates thealgebraic topology of anon-singularcomplexalgebraic variety to its subvarieties.

In simple terms, the Hodge conjecture asserts that the basic topological information like the number of holes in certaingeometric spaces,complex algebraic varieties, can be understood by studying the possible nice shapes sitting inside those spaces, which look likezero sets ofpolynomial equations. The latter objects can be studied usingalgebra and thecalculus ofanalytic functions, and this allows one to indirectly understand the broad shape and structure of often higher-dimensional spaces which cannot be otherwise easily visualized.

More specifically, the conjecture states that certainde Rham cohomology classes are algebraic; that is, they are sums ofPoincaré duals of thehomology classes of subvarieties. It was formulated by the Scottish mathematicianWilliam Vallance Douglas Hodge as a result of a work in between 1930 and 1940 to enrich the description of de Rham cohomology to include extra structure that is present in the case of complex algebraic varieties. It received little attention before Hodge presented it in an address during the 1950International Congress of Mathematicians, held inCambridge, Massachusetts. The Hodge conjecture is one of theClay Mathematics Institute'sMillennium Prize Problems, with a prize of $1,000,000 US for whoever can prove or disprove the Hodge conjecture.

Motivation

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Main article:Hodge theory § Hodge theory for complex projective varieties

LetX be acompactcomplex manifold of complex dimensionn. ThenX is anorientablesmooth manifold of real dimension2n{\displaystyle 2n}, so itscohomology groups lie in degrees zero through2n{\displaystyle 2n}. AssumeX is aKähler manifold, so that there is a decomposition on its cohomology with complexcoefficients

Hn(X,C)=p+q=nHp,q(X),{\displaystyle H^{n}(X,\mathbb {C} )=\bigoplus _{p+q=n}H^{p,q}(X),}

whereHp,q(X){\displaystyle H^{p,q}(X)} is the subgroup of cohomology classes which are represented byharmonic forms of type(p,q){\displaystyle (p,q)}. That is, these are the cohomology classes represented bydifferential forms which, in some choice of local coordinatesz1,,zn{\displaystyle z_{1},\ldots ,z_{n}}, can be written as aharmonic function times

dzi1dzipdz¯j1dz¯jq.{\displaystyle dz_{i_{1}}\wedge \cdots \wedge dz_{i_{p}}\wedge d{\bar {z}}_{j_{1}}\wedge \cdots \wedge d{\bar {z}}_{j_{q}}.}

SinceX is a compact oriented manifold,X has afundamental class, and soX can be integrated over.

LetZ be a complex submanifold ofX of dimensionk, and leti:ZX{\displaystyle i\colon Z\to X} be theinclusion map. Choose a differential formα{\displaystyle \alpha } of type(p,q){\displaystyle (p,q)}. We can integrateα{\displaystyle \alpha } overZ using thepullback functioni{\displaystyle i^{*}},

Ziα.{\displaystyle \int _{Z}i^{*}\alpha .}

To evaluate this integral, choose a point ofZ and call itz=(z1,,zk){\displaystyle z=(z_{1},\ldots ,z_{k})}. The inclusion ofZ inX means that we can choose a localbasis onX and havezk+1==zn=0{\displaystyle z_{k+1}=\cdots =z_{n}=0} (rank-nullity theorem). Ifp>k{\displaystyle p>k}, thenα{\displaystyle \alpha } must contain somedzi{\displaystyle dz_{i}} wherezi{\displaystyle z_{i}} pulls back to zero onZ. The same is true fordz¯j{\displaystyle d{\bar {z}}_{j}} ifq>k{\displaystyle q>k}. Consequently, this integral is zero if(p,q)(k,k){\displaystyle (p,q)\neq (k,k)}.

The Hodge conjecture then (loosely) asks:

Which cohomology classes inHk,k(X){\displaystyle H^{k,k}(X)} come from complex subvarietiesZ?

Statement of the Hodge conjecture

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Let

Hdgk(X)=H2k(X,Q)Hk,k(X).{\displaystyle \operatorname {Hdg} ^{k}(X)=H^{2k}(X,\mathbb {Q} )\cap H^{k,k}(X).}

We call this the group ofHodge classes of degree 2k onX.

The modern statement of the Hodge conjecture is

Hodge conjecture. LetX be a non-singular complex projective manifold. Then every Hodge class onX is a linear combination with rational coefficients of the cohomology classes of complex subvarieties ofX.

A projective complex manifold is a complex manifold which can be embedded incomplex projective space. Because projective space carries a Kähler metric, theFubini–Study metric, such a manifold is always a Kähler manifold. ByChow's theorem, a projective complex manifold is also a smooth projective algebraic variety, that is, it is the zero set of a collection of homogeneous polynomials.

Reformulation in terms of algebraic cycles

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Another way of phrasing the Hodge conjecture involves the idea of an algebraic cycle. Analgebraic cycle onX is a formal combination of subvarieties ofX; that is, it is something of the form

iciZi.{\displaystyle \sum _{i}c_{i}Z_{i}.}

The coefficients are usually taken to be integral or rational. We define the cohomology class of an algebraic cycle to be the sum of the cohomology classes of its components. This is an example of the cycle class map of de Rham cohomology, seeWeil cohomology. For example, the cohomology class of the above cycle would be

ici[Zi].{\displaystyle \sum _{i}c_{i}[Z_{i}].}

Such a cohomology class is calledalgebraic. With this notation, the Hodge conjecture becomes

LetX be a projective complex manifold. Then every Hodge class onX is algebraic.

The assumption in the Hodge conjecture thatX be algebraic (projective complex manifold) cannot be weakened. In 1977,Steven Zucker showed that it is possible to construct a counterexample to the Hodge conjecture as complex tori with analytic rational cohomology of type(p,p){\displaystyle (p,p)}, which is not projective algebraic. (see appendix B ofZucker (1977))

Known cases of the Hodge conjecture

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See Theorem 1 of Bouali.[1]

Low dimension and codimension

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The first result on the Hodge conjecture is due toLefschetz (1924). In fact, it predates the conjecture and provided some of Hodge's motivation.

Theorem (Lefschetz theorem on (1,1)-classes) Any element ofH2(X,Z)H1,1(X){\displaystyle H^{2}(X,\mathbb {Z} )\cap H^{1,1}(X)} is the cohomology class of adivisor onX{\displaystyle X}. In particular, the Hodge conjecture is true forH2{\displaystyle H^{2}}.

A very quick proof can be given usingsheaf cohomology and theexponential exact sequence. (The cohomology class of a divisor turns out to equal to its firstChern class.) Lefschetz's original proof proceeded bynormal functions, which were introduced byHenri Poincaré. However, theGriffiths transversality theorem shows that this approach cannot prove the Hodge conjecture for higher codimensional subvarieties.

By theHard Lefschetz theorem, one can prove:[2]

Theorem. If for somep<n2{\displaystyle p<{\frac {n}{2}}} the Hodge conjecture holds for Hodge classes of degreep{\displaystyle p}, then the Hodge conjecture holds for Hodge classes of degree2np{\displaystyle 2n-p}.

Combining the above two theorems implies that Hodge conjecture is true for Hodge classes of degree2n2{\displaystyle 2n-2}. This proves the Hodge conjecture whenX{\displaystyle X} has dimension at most three.

The Lefschetz theorem on (1,1)-classes also implies that if all Hodge classes are generated by the Hodge classes of divisors, then the Hodge conjecture is true:

Corollary. If the algebraHdg(X)=kHdgk(X){\displaystyle \operatorname {Hdg} ^{*}(X)=\bigoplus \nolimits _{k}\operatorname {Hdg} ^{k}(X)} is generated byHdg1(X){\displaystyle \operatorname {Hdg} ^{1}(X)}, then the Hodge conjecture holds forX{\displaystyle X}.

Hypersurfaces

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By the strong and weakLefschetz theorem, the only non-trivial part of the Hodge conjecture forhypersurfaces is the degreem part (i.e., the middle cohomology) of a 2m-dimensional hypersurfaceXP2m+1{\displaystyle X\subset \mathbf {P} ^{2m+1}}. If the degreed is 2, i.e.,X is aquadric, the Hodge conjecture holds for allm. Form=2{\displaystyle m=2}, i.e.,fourfolds, the Hodge conjecture is known ford5{\displaystyle d\leq 5}.[3]

Abelian varieties

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For mostabelian varieties, the algebra Hdg*(X) is generated in degree one, so the Hodge conjecture holds. In particular, the Hodge conjecture holds for sufficiently general abelian varieties, for products of elliptic curves, and for simple abelian varieties of prime dimension.[4][5][6] However,Mumford (1969) constructed an example of an abelian variety where Hdg2(X) is not generated by products of divisor classes.Weil (1977) generalized this example by showing that whenever the variety hascomplex multiplication by animaginary quadratic field, then Hdg2(X) is not generated by products of divisor classes.Moonen & Zarhin (1999) proved that in dimension less than 5, either Hdg*(X) is generated in degree one, or the variety has complex multiplication by an imaginary quadratic field. In the latter case, the Hodge conjecture is only known in special cases.

Generalizations

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The integral Hodge conjecture

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Hodge's original conjecture was:

Integral Hodge conjecture. LetX be a projective complex manifold. Then every cohomology class inH2k(X,Z)Hk,k(X){\displaystyle H^{2k}(X,\mathbb {Z} )\cap H^{k,k}(X)} is the cohomology class of an algebraic cycle with integral coefficients onX.

This is now known to be false. The first counterexample was constructed byAtiyah & Hirzebruch (1961). UsingK-theory, they constructed an example of a torsion cohomology class—that is, a cohomology classα such that = 0 for some positive integern—which is not the class of an algebraic cycle. Such a class is necessarily a Hodge class.Totaro (1997) reinterpreted their result in the framework ofcobordism and found many examples of such classes.

The simplest adjustment of the integral Hodge conjecture is:

Integral Hodge conjecture modulo torsion. LetX be a projective complex manifold. Then every cohomology class inH2k(X,Z)Hk,k(X){\displaystyle H^{2k}(X,\mathbb {Z} )\cap H^{k,k}(X)} is the sum of a torsion class and the cohomology class of an algebraic cycle with integral coefficients onX.

Equivalently, after dividingH2k(X,Z)Hk,k(X){\displaystyle H^{2k}(X,\mathbb {Z} )\cap H^{k,k}(X)} by torsion classes, every class is the image of the cohomology class of an integral algebraic cycle. This is also false.Kollár (1992) found an example of a Hodge classα which is not algebraic, but which has an integral multiple which is algebraic.

Rosenschon & Srinivas (2016) have shown that in order to obtain a correct integral Hodge conjecture, one needs to replace Chow groups, which can also be expressed asmotivic cohomology groups, by a variant known asétale (orLichtenbaum)motivic cohomology. They show that the rational Hodge conjecture is equivalent to an integral Hodge conjecture for this modified motivic cohomology.

The Hodge conjecture for Kähler varieties

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A natural generalization of the Hodge conjecture would ask:

Hodge conjecture for Kähler varieties, naive version. LetX be a complex Kähler manifold. Then every Hodge class onX is a linear combination with rational coefficients of the cohomology classes of complex subvarieties ofX.

This is too optimistic, because there are not enough subvarieties to make this work. A possible substitute is to ask instead one of the two following questions:

Hodge conjecture for Kähler varieties, vector bundle version. LetX be a complex Kähler manifold. Then every Hodge class onX is a linear combination with rational coefficients of Chern classes of vector bundles onX.
Hodge conjecture for Kähler varieties, coherent sheaf version. LetX be a complex Kähler manifold. Then every Hodge class onX is a linear combination with rational coefficients of Chern classes of coherent sheaves onX.

Voisin (2002) proved that the Chern classes of coherent sheaves give strictly more Hodge classes than the Chern classes of vector bundles and that the Chern classes of coherent sheaves are insufficient to generate all the Hodge classes. Consequently, the only known formulations of the Hodge conjecture for Kähler varieties are false.

The generalized Hodge conjecture

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Hodge made an additional, stronger conjecture than the integral Hodge conjecture. Say that a cohomology class onX is ofco-level c (coniveau c) if it is the pushforward of a cohomology class on ac-codimensional subvariety ofX. The cohomology classes of co-level at leastc filter the cohomology ofX, and it is easy to see that thecth step of the filtrationNcHk(X,Z) satisfies

NcHk(X,Z)Hk(X,Z)(Hkc,c(X)Hc,kc(X)).{\displaystyle N^{c}H^{k}(X,\mathbf {Z} )\subseteq H^{k}(X,\mathbf {Z} )\cap (H^{k-c,c}(X)\oplus \cdots \oplus H^{c,k-c}(X)).}

Hodge's original statement was:

Generalized Hodge conjecture, Hodge's version.NcHk(X,Z)=Hk(X,Z)(Hkc,c(X)Hc,kc(X)).{\displaystyle N^{c}H^{k}(X,\mathbf {Z} )=H^{k}(X,\mathbf {Z} )\cap (H^{k-c,c}(X)\oplus \cdots \oplus H^{c,k-c}(X)).}

Grothendieck (1969) observed that this cannot be true, even with rational coefficients, because the right-hand side is not always a Hodge structure. His corrected form of the Hodge conjecture is:

Generalized Hodge conjecture.NcHk(X,Q) is the largest sub-Hodge structure ofHk(X,Z) contained inHkc,c(X)Hc,kc(X).{\displaystyle H^{k-c,c}(X)\oplus \cdots \oplus H^{c,k-c}(X).}

This version is open.

Algebraicity of Hodge loci

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The strongest evidence in favor of the Hodge conjecture is the algebraicity result ofCattani, Deligne & Kaplan (1995). Suppose that we vary the complex structure ofX over a simply connected base. Then the topological cohomology ofX does not change, but the Hodge decomposition does change. It is known that if the Hodge conjecture is true, then the locus of all points on the base where the cohomology of a fiber is a Hodge class is in fact an algebraic subset, that is, it is cut out by polynomial equations. Cattani, Deligne & Kaplan (1995) proved that this is always true, without assuming the Hodge conjecture.

See also

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References

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  1. ^Bouali, Johann (2024-10-07). "Degeneration of families of projective hypersurfaces and Hodge conjecture".arXiv:2401.03465v13 [math.AG].
  2. ^Shioda, Tetsuji (July 13–24, 1981)."What is known about the Hodge Conjecture?". In S. Iitaka (ed.).Advanced Studies in Pure Mathematics. Algebraic Varieties and Analytic Varieties. Vol. 1. Tokyo, Japan: Mathematical Society of Japan. p. 58.doi:10.2969/aspm/00110000.ISBN 9784864970594.
  3. ^James Lewis:A Survey of the Hodge Conjecture, 1991, Example 7.21
  4. ^Mattuck, Arthur (1958). "Cycles on abelian varieties".Proceedings of the American Mathematical Society.9 (1):88–98.doi:10.1090/S0002-9939-1958-0098752-1.JSTOR 2033404.
  5. ^"Algebraic cycles and poles of zeta functions".ResearchGate. Retrieved2015-10-23.
  6. ^Tankeev, Sergei G (1988-01-01). "Cycles on simple abelian varieties of prime dimension over number fields".Mathematics of the USSR-Izvestiya.31 (3):527–540.Bibcode:1988IzMat..31..527T.doi:10.1070/im1988v031n03abeh001088.

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