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Chern class of a vector bundle and the associated projective space bundle
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Canonical Vector Bundle associated to a complete intersectionSeemingly contradictional facts on whether Chern classes determine a line bundle or not.Projective bundle is projective?Computing degrees of projective varieties via Chern classesChern classes of the associated vector bundle of a branched coveringTangent bundle of product of projective spacesChern class computation of push forward.How can I compute the chern roots of a vector bundle using the characteristic classes?Universal property of the tautological line bundleEvery cycle class a Chern class?
$begingroup$
I have a very basic question regarding Chern classes. Let $X$ be a smooth projective variety and $mathcalE$ a vector bundle on it. Let $pi:mathbbP(mathcalE)to X$ denote the projective space bundle over $X$ associated to $mathcalE$.
How are the Chern classes $c_i$ of the vector bundle $mathcalE$ on $X$ related to the Chern classes $c_i'$ of the projective variety $mathbbP(mathcalE)$, i.e. the Chern classes of the tangent bundle of $mathbbP(mathcalE)$? My naive hope would be that we simply have $c_i'=pi^* c_i$. Is that true? Is there a good reference?
algebraic-geometry vector-bundles
asked Apr 13 at 14:31
HansHans
1,682616
$endgroup$
add a comment |
$begingroup$
I have a very basic question regarding Chern classes. Let $X$ be a smooth projective variety and $mathcalE$ a vector bundle on it. Let $pi:mathbbP(mathcalE)to X$ denote the projective space bundle over $X$ associated to $mathcalE$.
How are the Chern classes $c_i$ of the vector bundle $mathcalE$ on $X$ related to the Chern classes $c_i'$ of the projective variety $mathbbP(mathcalE)$, i.e. the Chern classes of the tangent bundle of $mathbbP(mathcalE)$? My naive hope would be that we simply have $c_i'=pi^* c_i$. Is that true? Is there a good reference?
algebraic-geometry vector-bundles
asked Apr 13 at 14:31
HansHans
1,682616
$endgroup$
1
$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago
add a comment |
$begingroup$
I have a very basic question regarding Chern classes. Let $X$ be a smooth projective variety and $mathcalE$ a vector bundle on it. Let $pi:mathbbP(mathcalE)to X$ denote the projective space bundle over $X$ associated to $mathcalE$.
How are the Chern classes $c_i$ of the vector bundle $mathcalE$ on $X$ related to the Chern classes $c_i'$ of the projective variety $mathbbP(mathcalE)$, i.e. the Chern classes of the tangent bundle of $mathbbP(mathcalE)$? My naive hope would be that we simply have $c_i'=pi^* c_i$. Is that true? Is there a good reference?
algebraic-geometry vector-bundles
asked Apr 13 at 14:31
HansHans
1,682616
$endgroup$
I have a very basic question regarding Chern classes. Let $X$ be a smooth projective variety and $mathcalE$ a vector bundle on it. Let $pi:mathbbP(mathcalE)to X$ denote the projective space bundle over $X$ associated to $mathcalE$.
How are the Chern classes $c_i$ of the vector bundle $mathcalE$ on $X$ related to the Chern classes $c_i'$ of the projective variety $mathbbP(mathcalE)$, i.e. the Chern classes of the tangent bundle of $mathbbP(mathcalE)$? My naive hope would be that we simply have $c_i'=pi^* c_i$. Is that true? Is there a good reference?
algebraic-geometry vector-bundles
algebraic-geometry vector-bundles
asked Apr 13 at 14:31
HansHans
1,682616
asked Apr 13 at 14:31
HansHans
1,682616
asked Apr 13 at 14:31
HansHans
1,682616
asked Apr 13 at 14:31
HansHans
1,682616
asked Apr 13 at 14:31
HansHans
1,682616
1,682616
1
$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago
add a comment |
1
$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago
1
1
$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= mathcalO oplus mathcalO$ over $mathbbCP^1$. Then $$mathbbP(E) cong mathbbCP^1 times mathbbCP^1.$$
Now, note that $c_2(E)=0$ since $E$ is a trivial bundle, but $int_mathbbP(E) c_2(TmathbbP(E)) = 4$, since this is equal to the topological Euler characteristic of $mathbbP(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $mathbbP(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $mathbbP(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
answered Apr 13 at 14:44
Nick LNick L
1,412210
$endgroup$
add a comment |
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$begingroup$
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= mathcalO oplus mathcalO$ over $mathbbCP^1$. Then $$mathbbP(E) cong mathbbCP^1 times mathbbCP^1.$$
Now, note that $c_2(E)=0$ since $E$ is a trivial bundle, but $int_mathbbP(E) c_2(TmathbbP(E)) = 4$, since this is equal to the topological Euler characteristic of $mathbbP(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $mathbbP(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $mathbbP(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
answered Apr 13 at 14:44
Nick LNick L
1,412210
$endgroup$
add a comment |
$begingroup$
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= mathcalO oplus mathcalO$ over $mathbbCP^1$. Then $$mathbbP(E) cong mathbbCP^1 times mathbbCP^1.$$
Now, note that $c_2(E)=0$ since $E$ is a trivial bundle, but $int_mathbbP(E) c_2(TmathbbP(E)) = 4$, since this is equal to the topological Euler characteristic of $mathbbP(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $mathbbP(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $mathbbP(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
answered Apr 13 at 14:44
Nick LNick L
1,412210
$endgroup$
add a comment |
$begingroup$
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= mathcalO oplus mathcalO$ over $mathbbCP^1$. Then $$mathbbP(E) cong mathbbCP^1 times mathbbCP^1.$$
Now, note that $c_2(E)=0$ since $E$ is a trivial bundle, but $int_mathbbP(E) c_2(TmathbbP(E)) = 4$, since this is equal to the topological Euler characteristic of $mathbbP(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $mathbbP(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $mathbbP(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
answered Apr 13 at 14:44
Nick LNick L
1,412210
$endgroup$
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= mathcalO oplus mathcalO$ over $mathbbCP^1$. Then $$mathbbP(E) cong mathbbCP^1 times mathbbCP^1.$$
Now, note that $c_2(E)=0$ since $E$ is a trivial bundle, but $int_mathbbP(E) c_2(TmathbbP(E)) = 4$, since this is equal to the topological Euler characteristic of $mathbbP(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $mathbbP(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $mathbbP(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
answered Apr 13 at 14:44
Nick LNick L
1,412210
edited Apr 13 at 14:57
answered Apr 13 at 14:44
Nick LNick L
1,412210
answered Apr 13 at 14:44
Nick LNick L
1,412210
answered Apr 13 at 14:44
Nick LNick L
1,412210
1,412210
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$begingroup$
The answer is correct, but one can say more. There is a formula for the tangent bundle of a projective bundle that relates it to the pullback of the tangent bundle from below AND the relative tangent bundle. So there is a formula, but the formula makes it clear that what you are asking will almost never hold.
$endgroup$
– aginensky
Apr 13 at 20:10
$begingroup$
Hi @aginensky ! That sounds interesting, can you maybe state this formula in a new answer or point me to a reference?
$endgroup$
– Hans
2 days ago