Mixing
pushforward and pullback sheaf is isomorphism?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I am reading Hartshorne's book chapter 5 (on surface) and I have a question:
on page 371, proposition 2.3, it says:
Let $X$ be surface, $C$ curve, $pi:Xto C$ ruled surface. $f$ be a fiber, $sigma$ be a section of $pi$, $C_0=sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:
$L(D')cong pi^*pi_* L(D')$
I don't know how this comes from? Can someone helps me? Thanks in advance.
algebraic-geometry
asked Jul 16 at 21:30
User X
848
add a comment |Â
up vote
4
down vote
favorite
I am reading Hartshorne's book chapter 5 (on surface) and I have a question:
on page 371, proposition 2.3, it says:
Let $X$ be surface, $C$ curve, $pi:Xto C$ ruled surface. $f$ be a fiber, $sigma$ be a section of $pi$, $C_0=sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:
$L(D')cong pi^*pi_* L(D')$
I don't know how this comes from? Can someone helps me? Thanks in advance.
algebraic-geometry
asked Jul 16 at 21:30
User X
848
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Oh I see. Thank you!
– User X
Jul 17 at 8:02
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am reading Hartshorne's book chapter 5 (on surface) and I have a question:
on page 371, proposition 2.3, it says:
Let $X$ be surface, $C$ curve, $pi:Xto C$ ruled surface. $f$ be a fiber, $sigma$ be a section of $pi$, $C_0=sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:
$L(D')cong pi^*pi_* L(D')$
I don't know how this comes from? Can someone helps me? Thanks in advance.
algebraic-geometry
asked Jul 16 at 21:30
User X
848
I am reading Hartshorne's book chapter 5 (on surface) and I have a question:
on page 371, proposition 2.3, it says:
Let $X$ be surface, $C$ curve, $pi:Xto C$ ruled surface. $f$ be a fiber, $sigma$ be a section of $pi$, $C_0=sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:
$L(D')cong pi^*pi_* L(D')$
I don't know how this comes from? Can someone helps me? Thanks in advance.
algebraic-geometry
asked Jul 16 at 21:30
User X
848
asked Jul 16 at 21:30
User X
848
asked Jul 16 at 21:30
User X
848
asked Jul 16 at 21:30
User X
848
848
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Oh I see. Thank you!
– User X
Jul 17 at 8:02
add a comment |Â
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Oh I see. Thank you!
– User X
Jul 17 at 8:02
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Oh I see. Thank you!
– User X
Jul 17 at 8:02
Oh I see. Thank you!
– User X
Jul 17 at 8:02
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
By Lemma 2.1 of the same section, $pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*leftrightarrows f_*$, we have a map $pi^*pi_*L(D') to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $Hom(pi^*pi*L(D'),L(D'))cong Hom(pi_*L(D'),pi_*L(D'))$. Since $pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $mathcalO_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.
answered Jul 16 at 22:12
KReiser
7,60511230
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
By Lemma 2.1 of the same section, $pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*leftrightarrows f_*$, we have a map $pi^*pi_*L(D') to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $Hom(pi^*pi*L(D'),L(D'))cong Hom(pi_*L(D'),pi_*L(D'))$. Since $pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $mathcalO_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.
answered Jul 16 at 22:12
KReiser
7,60511230
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
add a comment |Â
up vote
3
down vote
accepted
By Lemma 2.1 of the same section, $pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*leftrightarrows f_*$, we have a map $pi^*pi_*L(D') to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $Hom(pi^*pi*L(D'),L(D'))cong Hom(pi_*L(D'),pi_*L(D'))$. Since $pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $mathcalO_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.
answered Jul 16 at 22:12
KReiser
7,60511230
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
By Lemma 2.1 of the same section, $pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*leftrightarrows f_*$, we have a map $pi^*pi_*L(D') to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $Hom(pi^*pi*L(D'),L(D'))cong Hom(pi_*L(D'),pi_*L(D'))$. Since $pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $mathcalO_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.
answered Jul 16 at 22:12
KReiser
7,60511230
By Lemma 2.1 of the same section, $pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*leftrightarrows f_*$, we have a map $pi^*pi_*L(D') to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $Hom(pi^*pi*L(D'),L(D'))cong Hom(pi_*L(D'),pi_*L(D'))$. Since $pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $mathcalO_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.
answered Jul 16 at 22:12
KReiser
7,60511230
edited Jul 17 at 7:17
answered Jul 16 at 22:12
KReiser
7,60511230
answered Jul 16 at 22:12
KReiser
7,60511230
answered Jul 16 at 22:12
KReiser
7,60511230
7,60511230
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
add a comment |Â
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
Isn't $L(D')$ a sheaf on $X$?
– Andrew
Jul 17 at 5:34
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
@Andrew fixed! Thank you for your attention to detail.
– KReiser
Jul 17 at 7:19
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853879%2fpushforward-and-pullback-sheaf-is-isomorphism%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Please see my new answer - I had mixed up which way the upper and lower stars had been placed and gave an incorrect solution.
– KReiser
Jul 17 at 7:18
Oh I see. Thank you!
– User X
Jul 17 at 8:02