Is the geometrical meaning of cup product still valid for subvarieties?
$begingroup$
It is known that cup product is Poincaré dual to the intersection. I'm referring to the following fact: if $X$ is a closed, oriented smooth manifold and $A, B$ are transverse-intersecting oriented submanifolds of codimension $i, j$ respectively, then
$$[A cap B]^* = [A]^* smile [B]^* in H^{i+j}(X)space ,$$
where the asterisk denotes Poincaré dual.
My question: is the same true if we take $A, B$ to be transverse-intersecting algebraic varieties? (And does that even make sense? I think that an algebraic subvariety defines an homology class given by the pushforward of the inclusion of the top class, and therefore it makes sense; but correct me if I'm wrong).
For context: I'm studying Schubert calculus, and I want to use this fact when $A, B$ are Schubert varieties, but I think Schubert varieties aren't smooth manifolds in general, since they contain singular points.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
$endgroup$
add a comment |
$begingroup$
It is known that cup product is Poincaré dual to the intersection. I'm referring to the following fact: if $X$ is a closed, oriented smooth manifold and $A, B$ are transverse-intersecting oriented submanifolds of codimension $i, j$ respectively, then
$$[A cap B]^* = [A]^* smile [B]^* in H^{i+j}(X)space ,$$
where the asterisk denotes Poincaré dual.
My question: is the same true if we take $A, B$ to be transverse-intersecting algebraic varieties? (And does that even make sense? I think that an algebraic subvariety defines an homology class given by the pushforward of the inclusion of the top class, and therefore it makes sense; but correct me if I'm wrong).
For context: I'm studying Schubert calculus, and I want to use this fact when $A, B$ are Schubert varieties, but I think Schubert varieties aren't smooth manifolds in general, since they contain singular points.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
$endgroup$
add a comment |
$begingroup$
It is known that cup product is Poincaré dual to the intersection. I'm referring to the following fact: if $X$ is a closed, oriented smooth manifold and $A, B$ are transverse-intersecting oriented submanifolds of codimension $i, j$ respectively, then
$$[A cap B]^* = [A]^* smile [B]^* in H^{i+j}(X)space ,$$
where the asterisk denotes Poincaré dual.
My question: is the same true if we take $A, B$ to be transverse-intersecting algebraic varieties? (And does that even make sense? I think that an algebraic subvariety defines an homology class given by the pushforward of the inclusion of the top class, and therefore it makes sense; but correct me if I'm wrong).
For context: I'm studying Schubert calculus, and I want to use this fact when $A, B$ are Schubert varieties, but I think Schubert varieties aren't smooth manifolds in general, since they contain singular points.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
$endgroup$
It is known that cup product is Poincaré dual to the intersection. I'm referring to the following fact: if $X$ is a closed, oriented smooth manifold and $A, B$ are transverse-intersecting oriented submanifolds of codimension $i, j$ respectively, then
$$[A cap B]^* = [A]^* smile [B]^* in H^{i+j}(X)space ,$$
where the asterisk denotes Poincaré dual.
My question: is the same true if we take $A, B$ to be transverse-intersecting algebraic varieties? (And does that even make sense? I think that an algebraic subvariety defines an homology class given by the pushforward of the inclusion of the top class, and therefore it makes sense; but correct me if I'm wrong).
For context: I'm studying Schubert calculus, and I want to use this fact when $A, B$ are Schubert varieties, but I think Schubert varieties aren't smooth manifolds in general, since they contain singular points.
algebraic-geometry algebraic-topology schubert-calculus
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
edited Jan 2 at 0:50
Matt Samuel
38.7k63769
38.7k63769
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
asked Jul 10 '17 at 13:52
un umile appassionatoun umile appassionato
367316
367316
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
This is basically the approach via stratifolds developed by Kreck; see
Kreck, Matthias.
Differential algebraic topology.
From stratifolds to exotic spheres. Graduate Studies in Mathematics, 110. American Mathematical Society, Providence, RI, 2010
and
Bunke, Ulrich; Kreck, Matthias; Schick, Thomas. A geometric description of differential cohomology. Ann. Math. Blaise Pascal 17 (2010), no. 1, 1–16.
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is basically the approach via stratifolds developed by Kreck; see
Kreck, Matthias.
Differential algebraic topology.
From stratifolds to exotic spheres. Graduate Studies in Mathematics, 110. American Mathematical Society, Providence, RI, 2010
and
Bunke, Ulrich; Kreck, Matthias; Schick, Thomas. A geometric description of differential cohomology. Ann. Math. Blaise Pascal 17 (2010), no. 1, 1–16.
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
$endgroup$
add a comment |
$begingroup$
This is basically the approach via stratifolds developed by Kreck; see
Kreck, Matthias.
Differential algebraic topology.
From stratifolds to exotic spheres. Graduate Studies in Mathematics, 110. American Mathematical Society, Providence, RI, 2010
and
Bunke, Ulrich; Kreck, Matthias; Schick, Thomas. A geometric description of differential cohomology. Ann. Math. Blaise Pascal 17 (2010), no. 1, 1–16.
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
$endgroup$
add a comment |
$begingroup$
This is basically the approach via stratifolds developed by Kreck; see
Kreck, Matthias.
Differential algebraic topology.
From stratifolds to exotic spheres. Graduate Studies in Mathematics, 110. American Mathematical Society, Providence, RI, 2010
and
Bunke, Ulrich; Kreck, Matthias; Schick, Thomas. A geometric description of differential cohomology. Ann. Math. Blaise Pascal 17 (2010), no. 1, 1–16.
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
$endgroup$
This is basically the approach via stratifolds developed by Kreck; see
Kreck, Matthias.
Differential algebraic topology.
From stratifolds to exotic spheres. Graduate Studies in Mathematics, 110. American Mathematical Society, Providence, RI, 2010
and
Bunke, Ulrich; Kreck, Matthias; Schick, Thomas. A geometric description of differential cohomology. Ann. Math. Blaise Pascal 17 (2010), no. 1, 1–16.
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
edited Jul 10 '17 at 14:08
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
answered Jul 10 '17 at 14:01
Mikhail KatzMikhail Katz
30.7k14398
30.7k14398
add a comment |
add a comment |
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