Is there a morphism from a locally free sheaf to the dual twisted by determinant bundle?
$begingroup$
Let $X$ be a smooth projective variety over the field $mathbb{C}$. Let $E$ be a locally free sheaf of rank $r$ and $F=E^{vee}$, the dual locally free sheaf. If rank $E$=2, then we have the isomorphism $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?
algebraic-geometry
asked Jan 8 at 10:08
user349424user349424
34317
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add a comment |
$begingroup$
Let $X$ be a smooth projective variety over the field $mathbb{C}$. Let $E$ be a locally free sheaf of rank $r$ and $F=E^{vee}$, the dual locally free sheaf. If rank $E$=2, then we have the isomorphism $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?
algebraic-geometry
asked Jan 8 at 10:08
user349424user349424
34317
$endgroup$
add a comment |
$begingroup$
Let $X$ be a smooth projective variety over the field $mathbb{C}$. Let $E$ be a locally free sheaf of rank $r$ and $F=E^{vee}$, the dual locally free sheaf. If rank $E$=2, then we have the isomorphism $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?
algebraic-geometry
asked Jan 8 at 10:08
user349424user349424
34317
$endgroup$
Let $X$ be a smooth projective variety over the field $mathbb{C}$. Let $E$ be a locally free sheaf of rank $r$ and $F=E^{vee}$, the dual locally free sheaf. If rank $E$=2, then we have the isomorphism $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?
algebraic-geometry
algebraic-geometry
asked Jan 8 at 10:08
user349424user349424
34317
asked Jan 8 at 10:08
user349424user349424
34317
asked Jan 8 at 10:08
user349424user349424
34317
asked Jan 8 at 10:08
user349424user349424
34317
asked Jan 8 at 10:08
user349424user349424
34317
34317
add a comment |
add a comment |
1 Answer
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To give a morphism $E to E^vee otimes det(E)$ is the same as to give a bilinear form on $E$ with values in $det(E)$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that
$$
det(E^vee otimes det(E)) cong det(E^vee) otimes det(E)^r cong det(E)^{r-1}
$$
is isomorphic to $det(E)$. So, this can be true only if $det(E) = O_X$, or if $det(E)$ is a point of order $r-2$ on $Pic^0(X)$.
Note also that there may be no nontrivial morphisms. For instance, take $X = mathbb{P}^1$ and $E = O(-1) oplus O(-1) oplus O(-1)$. Then $E^vee otimes det(E) cong O(-2) oplus O(-2) oplus O(-2)$ and $Hom(E,E^vee otimes det(E)) = 0$.
answered Jan 8 at 11:16
SashaSasha
5,208139
$endgroup$
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
To give a morphism $E to E^vee otimes det(E)$ is the same as to give a bilinear form on $E$ with values in $det(E)$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that
$$
det(E^vee otimes det(E)) cong det(E^vee) otimes det(E)^r cong det(E)^{r-1}
$$
is isomorphic to $det(E)$. So, this can be true only if $det(E) = O_X$, or if $det(E)$ is a point of order $r-2$ on $Pic^0(X)$.
Note also that there may be no nontrivial morphisms. For instance, take $X = mathbb{P}^1$ and $E = O(-1) oplus O(-1) oplus O(-1)$. Then $E^vee otimes det(E) cong O(-2) oplus O(-2) oplus O(-2)$ and $Hom(E,E^vee otimes det(E)) = 0$.
answered Jan 8 at 11:16
SashaSasha
5,208139
$endgroup$
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
add a comment |
$begingroup$
To give a morphism $E to E^vee otimes det(E)$ is the same as to give a bilinear form on $E$ with values in $det(E)$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that
$$
det(E^vee otimes det(E)) cong det(E^vee) otimes det(E)^r cong det(E)^{r-1}
$$
is isomorphic to $det(E)$. So, this can be true only if $det(E) = O_X$, or if $det(E)$ is a point of order $r-2$ on $Pic^0(X)$.
Note also that there may be no nontrivial morphisms. For instance, take $X = mathbb{P}^1$ and $E = O(-1) oplus O(-1) oplus O(-1)$. Then $E^vee otimes det(E) cong O(-2) oplus O(-2) oplus O(-2)$ and $Hom(E,E^vee otimes det(E)) = 0$.
answered Jan 8 at 11:16
SashaSasha
5,208139
$endgroup$
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
add a comment |
$begingroup$
To give a morphism $E to E^vee otimes det(E)$ is the same as to give a bilinear form on $E$ with values in $det(E)$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that
$$
det(E^vee otimes det(E)) cong det(E^vee) otimes det(E)^r cong det(E)^{r-1}
$$
is isomorphic to $det(E)$. So, this can be true only if $det(E) = O_X$, or if $det(E)$ is a point of order $r-2$ on $Pic^0(X)$.
Note also that there may be no nontrivial morphisms. For instance, take $X = mathbb{P}^1$ and $E = O(-1) oplus O(-1) oplus O(-1)$. Then $E^vee otimes det(E) cong O(-2) oplus O(-2) oplus O(-2)$ and $Hom(E,E^vee otimes det(E)) = 0$.
answered Jan 8 at 11:16
SashaSasha
5,208139
$endgroup$
To give a morphism $E to E^vee otimes det(E)$ is the same as to give a bilinear form on $E$ with values in $det(E)$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that
$$
det(E^vee otimes det(E)) cong det(E^vee) otimes det(E)^r cong det(E)^{r-1}
$$
is isomorphic to $det(E)$. So, this can be true only if $det(E) = O_X$, or if $det(E)$ is a point of order $r-2$ on $Pic^0(X)$.
Note also that there may be no nontrivial morphisms. For instance, take $X = mathbb{P}^1$ and $E = O(-1) oplus O(-1) oplus O(-1)$. Then $E^vee otimes det(E) cong O(-2) oplus O(-2) oplus O(-2)$ and $Hom(E,E^vee otimes det(E)) = 0$.
answered Jan 8 at 11:16
SashaSasha
5,208139
edited Jan 8 at 13:17
answered Jan 8 at 11:16
SashaSasha
5,208139
answered Jan 8 at 11:16
SashaSasha
5,208139
answered Jan 8 at 11:16
SashaSasha
5,208139
5,208139
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
add a comment |
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
Is the above condition for injectivity or to give a morphism?
$endgroup$
– user349424
Jan 8 at 12:57
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
In other words, is there a morphism which is not injective?
$endgroup$
– user349424
Jan 8 at 12:59
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
$begingroup$
There may be no nonzero morphisms, I edited the answer to include an example.
$endgroup$
– Sasha
Jan 8 at 13:18
add a comment |
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