en.wikipedia.org/wiki/Hodge_conjecture
2 corrections found
i ∗ : X → Z {\displaystyle i^{\*}\colon X\to Z}
The pullback is written in the wrong direction. For an inclusion i: Z → X, the induced pullback goes from forms or cohomology on X to forms or cohomology on Z, not from X to Z as a map of spaces.
Full reasoning
For a map or inclusion (i: Z \to X), the pullback is the induced map on forms, sheaves, or cohomology in the opposite direction. In particular, standard references describe pullback on differential forms as a map from forms on the target to forms on the source, e.g. (i^\Omega_X^p \to \Omega_Z^p), or on cohomology as (i^:H^(X)\to H^(Z)).
So the article's notation i^*: X → Z is incorrect: that looks like a map of spaces, but the space map is (i:Z\to X). The pullback is not a map (X\to Z); it is a contravariant map on forms/cohomology induced by (i).
2 sources
- Section 50.2 (07HX): The de Rham complex — The Stacks project
In particular, if f : Y -> X is a morphism of schemes over a base scheme S, then there is a map of complexes Ω^•_{X/S} -> f_*Ω^•_{Y/S}. Linearizing, for every p we obtain a canonical map f^*Ω^p_{X/S} -> Ω^p_{X'/S'}.
- Pullback Map — Wolfram MathWorld
A pullback map is a map induced in the opposite direction from a function or morphism... For a smooth map and a differential r-form on [the target], the pullback is a differential r-form on [the source].
this integral is zero if ( p , q ) ≠ ( n , n )
The indices are wrong here: for a complex submanifold Z of dimension k, the integral can only be nonzero for forms of type (k,k), not (n,n), where n is the dimension of the ambient manifold X.
Full reasoning
Here (Z) has complex dimension (k), so it has real dimension (2k). Standard integration theory says that on an (m)-dimensional manifold, only top-degree (m)-forms are integrated. Thus on (Z), only forms of total degree (2k) can contribute to the integral.
Also, for a complex submanifold one can choose local coordinates so that (Z) is given by (z_{k+1}=\cdots=z_n=0). In those coordinates, a form of type ((p,q)) restricts to zero unless (p\le k) and (q\le k). Combining this with the top-degree requirement (p+q=2k), the only possible nonzero case is ((p,q)=(k,k)).
So the article's statement this integral is zero if (p,q) ≠ (n,n) uses the wrong indices: the correct distinguished bidegree is ((k,k)), determined by the dimension of (Z), not ((n,n)), determined by the ambient space (X).
3 sources
- Form Integration — Wolfram MathWorld
A differential k-form can be integrated on an n-dimensional manifold... Since [the integrand] is a top-dimensional form... the integral of the n-form is well-defined.
- Normal form — Encyclopedia of Mathematics
This result is equivalent to the Implicit function theorem. In particular, it shows that the image of an immersion locally looks like a coordinate subspace.
- Manifold — Encyclopedia of Mathematics
The existence of a local parametrization is provided by the implicit-function theorem from the condition of maximal rank of the Jacobi matrix of the system.