Mixing
Understanding notation in Fulton's Intersection Theory
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In Fulton Intersection Theory Second Edition there is the definition of "degree" of a zero cycle. I am referring to Definition 1.4 where he says
If $X$ is a complete scheme and $alpha=sum_P n_P[P]$ is a zero cycle on $X$ the degree $text{deg}(alpha)$ is defined to be $sum_P n_P[R(P):K]$ where the second term is the degree of the extension of the respective residue fields.
He adds that degree is also denoted as $int_X(alpha)$.
I understand the definition but I don't understand the integral notation. I would like to understand the motivation behind this. Since this theory of intersection on schemes is inspired by the theory of intersection on complex manifolds(on which we can integrate classes of cocycles on classes of homology) I am led to believe that the notation reflects a historical usage of the same notion of degree.
I think of cycles and more precisely the classes of cycles(that is, the elements of $A_k(X)$ as in Fulton) to be analogous to the classes of cycles in the homology(as in the singular homology $H_{2k}(X')$ of a complex manifold $X'$). In differential geometry we learnt to integrate only Cohomology classes of the top form on a manifold. Therefore I wonder if the notation suggests some sort of duality.
But I am not sure if my intuition is flawed. In any case the notation is not clear to me.
algebraic-geometry intersection-theory
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
$endgroup$
add a comment |
$begingroup$
In Fulton Intersection Theory Second Edition there is the definition of "degree" of a zero cycle. I am referring to Definition 1.4 where he says
If $X$ is a complete scheme and $alpha=sum_P n_P[P]$ is a zero cycle on $X$ the degree $text{deg}(alpha)$ is defined to be $sum_P n_P[R(P):K]$ where the second term is the degree of the extension of the respective residue fields.
He adds that degree is also denoted as $int_X(alpha)$.
I understand the definition but I don't understand the integral notation. I would like to understand the motivation behind this. Since this theory of intersection on schemes is inspired by the theory of intersection on complex manifolds(on which we can integrate classes of cocycles on classes of homology) I am led to believe that the notation reflects a historical usage of the same notion of degree.
I think of cycles and more precisely the classes of cycles(that is, the elements of $A_k(X)$ as in Fulton) to be analogous to the classes of cycles in the homology(as in the singular homology $H_{2k}(X')$ of a complex manifold $X'$). In differential geometry we learnt to integrate only Cohomology classes of the top form on a manifold. Therefore I wonder if the notation suggests some sort of duality.
But I am not sure if my intuition is flawed. In any case the notation is not clear to me.
algebraic-geometry intersection-theory
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
$endgroup$
add a comment |
$begingroup$
In Fulton Intersection Theory Second Edition there is the definition of "degree" of a zero cycle. I am referring to Definition 1.4 where he says
If $X$ is a complete scheme and $alpha=sum_P n_P[P]$ is a zero cycle on $X$ the degree $text{deg}(alpha)$ is defined to be $sum_P n_P[R(P):K]$ where the second term is the degree of the extension of the respective residue fields.
He adds that degree is also denoted as $int_X(alpha)$.
I understand the definition but I don't understand the integral notation. I would like to understand the motivation behind this. Since this theory of intersection on schemes is inspired by the theory of intersection on complex manifolds(on which we can integrate classes of cocycles on classes of homology) I am led to believe that the notation reflects a historical usage of the same notion of degree.
I think of cycles and more precisely the classes of cycles(that is, the elements of $A_k(X)$ as in Fulton) to be analogous to the classes of cycles in the homology(as in the singular homology $H_{2k}(X')$ of a complex manifold $X'$). In differential geometry we learnt to integrate only Cohomology classes of the top form on a manifold. Therefore I wonder if the notation suggests some sort of duality.
But I am not sure if my intuition is flawed. In any case the notation is not clear to me.
algebraic-geometry intersection-theory
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
$endgroup$
In Fulton Intersection Theory Second Edition there is the definition of "degree" of a zero cycle. I am referring to Definition 1.4 where he says
If $X$ is a complete scheme and $alpha=sum_P n_P[P]$ is a zero cycle on $X$ the degree $text{deg}(alpha)$ is defined to be $sum_P n_P[R(P):K]$ where the second term is the degree of the extension of the respective residue fields.
He adds that degree is also denoted as $int_X(alpha)$.
I understand the definition but I don't understand the integral notation. I would like to understand the motivation behind this. Since this theory of intersection on schemes is inspired by the theory of intersection on complex manifolds(on which we can integrate classes of cocycles on classes of homology) I am led to believe that the notation reflects a historical usage of the same notion of degree.
I think of cycles and more precisely the classes of cycles(that is, the elements of $A_k(X)$ as in Fulton) to be analogous to the classes of cycles in the homology(as in the singular homology $H_{2k}(X')$ of a complex manifold $X'$). In differential geometry we learnt to integrate only Cohomology classes of the top form on a manifold. Therefore I wonder if the notation suggests some sort of duality.
But I am not sure if my intuition is flawed. In any case the notation is not clear to me.
algebraic-geometry intersection-theory
algebraic-geometry intersection-theory
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
asked Jan 30 at 16:32
GrobberGrobber
1,44911622
1,44911622
add a comment |
add a comment |
1 Answer
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The Chow group is indeed a kind of homology theory (more precisely a Borel-Moore homology but on a complete space, standard homology and Borel-Moore homology coincide). It has indeed the same kind of functoriality as a Borel-Moore homology (covariant with respect to proper mapping and contravariant with respect to open inclusion), and the same kind of properties (localization exact sequence...).
On the other hand, the degree map is really a map $int:H_0(X)tomathbb{Z}$ (from a standard, that is not Borel-Moore, homology theory). Indeed, in a standard homology, a 0-cycle is indeed a formal finite sum of points with multiplicities and the degree is the sum of all the multiplicities. Alternatively, this is the map $H_0(X)to H_0(pt)=mathbb{Z}$. (Note that we do require covariance with $f:Xto pt$ so this is only possible with standard homology, or with Borel-Moore homology if $X$ is proper).
On an oriented manifold $X$ of dimension $d$, there is a fundamental class $[X]in H^{BM}_d(X)$ such that the cap-product with $[X]$ induces for any $n$ isomorphisms $H^n_c(X)oversetsimto H_{d-n}(X)$. If $X$ is compact, then $[X]in H_d^{BM}=H_d(X)$, and $H_c^n(X)=H^n(X)$, these maps are then the usual Poincaré isomorphisms $H^n(X)oversetsimto H_{d-n}(X)$.
If use only the group of top degree and by composition with the degree map, we get the map : $H^d_c(X)to H_0(X)overset{int}tomathbb{Z}$.
Using $mathbb{R}$ as coefficients and using the identification with de Rham cohomology, this map is indeed the integration of compactly supported top differential forms.
answered Jan 30 at 18:07
RolandRoland
7,44911015
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$begingroup$
The Chow group is indeed a kind of homology theory (more precisely a Borel-Moore homology but on a complete space, standard homology and Borel-Moore homology coincide). It has indeed the same kind of functoriality as a Borel-Moore homology (covariant with respect to proper mapping and contravariant with respect to open inclusion), and the same kind of properties (localization exact sequence...).
On the other hand, the degree map is really a map $int:H_0(X)tomathbb{Z}$ (from a standard, that is not Borel-Moore, homology theory). Indeed, in a standard homology, a 0-cycle is indeed a formal finite sum of points with multiplicities and the degree is the sum of all the multiplicities. Alternatively, this is the map $H_0(X)to H_0(pt)=mathbb{Z}$. (Note that we do require covariance with $f:Xto pt$ so this is only possible with standard homology, or with Borel-Moore homology if $X$ is proper).
On an oriented manifold $X$ of dimension $d$, there is a fundamental class $[X]in H^{BM}_d(X)$ such that the cap-product with $[X]$ induces for any $n$ isomorphisms $H^n_c(X)oversetsimto H_{d-n}(X)$. If $X$ is compact, then $[X]in H_d^{BM}=H_d(X)$, and $H_c^n(X)=H^n(X)$, these maps are then the usual Poincaré isomorphisms $H^n(X)oversetsimto H_{d-n}(X)$.
If use only the group of top degree and by composition with the degree map, we get the map : $H^d_c(X)to H_0(X)overset{int}tomathbb{Z}$.
Using $mathbb{R}$ as coefficients and using the identification with de Rham cohomology, this map is indeed the integration of compactly supported top differential forms.
answered Jan 30 at 18:07
RolandRoland
7,44911015
$endgroup$
add a comment |
$begingroup$
The Chow group is indeed a kind of homology theory (more precisely a Borel-Moore homology but on a complete space, standard homology and Borel-Moore homology coincide). It has indeed the same kind of functoriality as a Borel-Moore homology (covariant with respect to proper mapping and contravariant with respect to open inclusion), and the same kind of properties (localization exact sequence...).
On the other hand, the degree map is really a map $int:H_0(X)tomathbb{Z}$ (from a standard, that is not Borel-Moore, homology theory). Indeed, in a standard homology, a 0-cycle is indeed a formal finite sum of points with multiplicities and the degree is the sum of all the multiplicities. Alternatively, this is the map $H_0(X)to H_0(pt)=mathbb{Z}$. (Note that we do require covariance with $f:Xto pt$ so this is only possible with standard homology, or with Borel-Moore homology if $X$ is proper).
On an oriented manifold $X$ of dimension $d$, there is a fundamental class $[X]in H^{BM}_d(X)$ such that the cap-product with $[X]$ induces for any $n$ isomorphisms $H^n_c(X)oversetsimto H_{d-n}(X)$. If $X$ is compact, then $[X]in H_d^{BM}=H_d(X)$, and $H_c^n(X)=H^n(X)$, these maps are then the usual Poincaré isomorphisms $H^n(X)oversetsimto H_{d-n}(X)$.
If use only the group of top degree and by composition with the degree map, we get the map : $H^d_c(X)to H_0(X)overset{int}tomathbb{Z}$.
Using $mathbb{R}$ as coefficients and using the identification with de Rham cohomology, this map is indeed the integration of compactly supported top differential forms.
answered Jan 30 at 18:07
RolandRoland
7,44911015
$endgroup$
add a comment |
$begingroup$
The Chow group is indeed a kind of homology theory (more precisely a Borel-Moore homology but on a complete space, standard homology and Borel-Moore homology coincide). It has indeed the same kind of functoriality as a Borel-Moore homology (covariant with respect to proper mapping and contravariant with respect to open inclusion), and the same kind of properties (localization exact sequence...).
On the other hand, the degree map is really a map $int:H_0(X)tomathbb{Z}$ (from a standard, that is not Borel-Moore, homology theory). Indeed, in a standard homology, a 0-cycle is indeed a formal finite sum of points with multiplicities and the degree is the sum of all the multiplicities. Alternatively, this is the map $H_0(X)to H_0(pt)=mathbb{Z}$. (Note that we do require covariance with $f:Xto pt$ so this is only possible with standard homology, or with Borel-Moore homology if $X$ is proper).
On an oriented manifold $X$ of dimension $d$, there is a fundamental class $[X]in H^{BM}_d(X)$ such that the cap-product with $[X]$ induces for any $n$ isomorphisms $H^n_c(X)oversetsimto H_{d-n}(X)$. If $X$ is compact, then $[X]in H_d^{BM}=H_d(X)$, and $H_c^n(X)=H^n(X)$, these maps are then the usual Poincaré isomorphisms $H^n(X)oversetsimto H_{d-n}(X)$.
If use only the group of top degree and by composition with the degree map, we get the map : $H^d_c(X)to H_0(X)overset{int}tomathbb{Z}$.
Using $mathbb{R}$ as coefficients and using the identification with de Rham cohomology, this map is indeed the integration of compactly supported top differential forms.
answered Jan 30 at 18:07
RolandRoland
7,44911015
$endgroup$
The Chow group is indeed a kind of homology theory (more precisely a Borel-Moore homology but on a complete space, standard homology and Borel-Moore homology coincide). It has indeed the same kind of functoriality as a Borel-Moore homology (covariant with respect to proper mapping and contravariant with respect to open inclusion), and the same kind of properties (localization exact sequence...).
On the other hand, the degree map is really a map $int:H_0(X)tomathbb{Z}$ (from a standard, that is not Borel-Moore, homology theory). Indeed, in a standard homology, a 0-cycle is indeed a formal finite sum of points with multiplicities and the degree is the sum of all the multiplicities. Alternatively, this is the map $H_0(X)to H_0(pt)=mathbb{Z}$. (Note that we do require covariance with $f:Xto pt$ so this is only possible with standard homology, or with Borel-Moore homology if $X$ is proper).
On an oriented manifold $X$ of dimension $d$, there is a fundamental class $[X]in H^{BM}_d(X)$ such that the cap-product with $[X]$ induces for any $n$ isomorphisms $H^n_c(X)oversetsimto H_{d-n}(X)$. If $X$ is compact, then $[X]in H_d^{BM}=H_d(X)$, and $H_c^n(X)=H^n(X)$, these maps are then the usual Poincaré isomorphisms $H^n(X)oversetsimto H_{d-n}(X)$.
If use only the group of top degree and by composition with the degree map, we get the map : $H^d_c(X)to H_0(X)overset{int}tomathbb{Z}$.
Using $mathbb{R}$ as coefficients and using the identification with de Rham cohomology, this map is indeed the integration of compactly supported top differential forms.
answered Jan 30 at 18:07
RolandRoland
7,44911015
answered Jan 30 at 18:07
RolandRoland
7,44911015
answered Jan 30 at 18:07
RolandRoland
7,44911015
answered Jan 30 at 18:07
RolandRoland
7,44911015
7,44911015
add a comment |
add a comment |
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