When is a vector bundle morphism a vector bundle?
$begingroup$
When does a morphism $f:E'to E$ of vector bundles over the same base $B$ make $E'$ a vector bundle over $E$? Definitely $f$ has to be surjective, and this seems like it is also sufficient. Because any point of $B$ has a neighborhood $U$ over which $E'$ and $E$ are both trivial. Then it seems to me that $E'$ is trivialized over the open set $p^{-1}(U)subseteq E$, where $p$ is the projection $Eto B$.
(Maybe it's a simple question, but just wanted to double check).
algebraic-geometry vector-bundles
asked Dec 13 '18 at 17:37
nopnop
413
$endgroup$
|
show 1 more comment
$begingroup$
When does a morphism $f:E'to E$ of vector bundles over the same base $B$ make $E'$ a vector bundle over $E$? Definitely $f$ has to be surjective, and this seems like it is also sufficient. Because any point of $B$ has a neighborhood $U$ over which $E'$ and $E$ are both trivial. Then it seems to me that $E'$ is trivialized over the open set $p^{-1}(U)subseteq E$, where $p$ is the projection $Eto B$.
(Maybe it's a simple question, but just wanted to double check).
algebraic-geometry vector-bundles
asked Dec 13 '18 at 17:37
nopnop
413
$endgroup$
$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48
|
show 1 more comment
$begingroup$
When does a morphism $f:E'to E$ of vector bundles over the same base $B$ make $E'$ a vector bundle over $E$? Definitely $f$ has to be surjective, and this seems like it is also sufficient. Because any point of $B$ has a neighborhood $U$ over which $E'$ and $E$ are both trivial. Then it seems to me that $E'$ is trivialized over the open set $p^{-1}(U)subseteq E$, where $p$ is the projection $Eto B$.
(Maybe it's a simple question, but just wanted to double check).
algebraic-geometry vector-bundles
asked Dec 13 '18 at 17:37
nopnop
413
$endgroup$
When does a morphism $f:E'to E$ of vector bundles over the same base $B$ make $E'$ a vector bundle over $E$? Definitely $f$ has to be surjective, and this seems like it is also sufficient. Because any point of $B$ has a neighborhood $U$ over which $E'$ and $E$ are both trivial. Then it seems to me that $E'$ is trivialized over the open set $p^{-1}(U)subseteq E$, where $p$ is the projection $Eto B$.
(Maybe it's a simple question, but just wanted to double check).
algebraic-geometry vector-bundles
algebraic-geometry vector-bundles
asked Dec 13 '18 at 17:37
nopnop
413
asked Dec 13 '18 at 17:37
nopnop
413
edited Dec 14 '18 at 20:45
nop
asked Dec 13 '18 at 17:37
nopnop
413
asked Dec 13 '18 at 17:37
nopnop
413
asked Dec 13 '18 at 17:37
nopnop
413
413
$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48
|
show 1 more comment
$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48
$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48
|
show 1 more comment
1 Answer
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$begingroup$
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'cong Eoplus ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $ker(f)to B$ along $Eto B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.
answered Dec 13 '18 at 20:44
BenBen
4,23421334
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add a comment |
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$begingroup$
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'cong Eoplus ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $ker(f)to B$ along $Eto B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.
answered Dec 13 '18 at 20:44
BenBen
4,23421334
$endgroup$
add a comment |
$begingroup$
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'cong Eoplus ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $ker(f)to B$ along $Eto B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.
answered Dec 13 '18 at 20:44
BenBen
4,23421334
$endgroup$
add a comment |
$begingroup$
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'cong Eoplus ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $ker(f)to B$ along $Eto B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.
answered Dec 13 '18 at 20:44
BenBen
4,23421334
$endgroup$
Yes, that's true. [No, not quite, see below.] It's a bit easier to see if $B$ is paracompact, say, since then any surjective vector bundle surjection on $B$ splits, i.e., $E'cong Eoplus ker(f)$ such that $f$ gets identified with the projection and this is the pull-back of the vector bundle $ker(f)to B$ along $Eto B$.
Edit: Roland is right. If $f$ is not split, then it's wrong. On the other hand, every splitting gives rise to a vector bundle structure, unique up to isomorphism.
answered Dec 13 '18 at 20:44
BenBen
4,23421334
edited Dec 14 '18 at 13:00
answered Dec 13 '18 at 20:44
BenBen
4,23421334
answered Dec 13 '18 at 20:44
BenBen
4,23421334
answered Dec 13 '18 at 20:44
BenBen
4,23421334
4,23421334
add a comment |
add a comment |
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$begingroup$
If $B={*}$ is just a point, then $E,E'$ are just vector spaces. Now if $f:E'to E$ is a linear map, and $ein E$, what is the vector space structure on $f^{-1}(e)$ ? What would be the zero element ? You need to put an additional structure which won't be canonical, and in certain context may not even exists. Indeed, if $f:E'to E$ is to be a vector bundle, you need in particular (that's enough actually) the zero section : $s:Eto E'$. So you need $f$ to split. If you are talking about algebraic or holomorphic bundle, such a splitting does not necessarily exist.
$endgroup$
– Roland
Dec 14 '18 at 8:37
$begingroup$
Dear @Roland, you are right, it needs a splitting, but are you sure that the structure really depends on the splitting? It seems to me that different splittings give rise to isomorphic structures.
$endgroup$
– Ben
Dec 14 '18 at 9:52
$begingroup$
@Ben Up to isomorphism, yes different splitting give isomorphic structures and even canonically isomorphic structure. Yet these structures are different. So $E'to E$ is not a vector bundle in a unique way (and sometimes not a vector bundle at all). In fact $E'$ is a bundle of affine spaces parallels to $ker f$. The choice of the origin completely determines the vector space structure. Given two such choices represented by two sections $s,s'$, then the translation by $s-s'$ is indeed an isomorphism of the vector space structure given by $s$ to the one given by $s'$.
$endgroup$
– Roland
Dec 14 '18 at 11:38
$begingroup$
@Roland Thanks for your comment. I understand what you mean. Do you know of any cases in which we can be ensured a splitting? By the way, I mean algebraic bundles.
$endgroup$
– nop
Dec 14 '18 at 16:44
$begingroup$
Also, I thought any variety is compact (because Noetherian), so Ben's comment would suggest $f$ splits?
$endgroup$
– nop
Dec 14 '18 at 16:48