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Normal bundle to a section of $mathcalProj$ of a rank 2 vector bundle
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I am stuck at solving 20.2.I from Vakil's notes.
The set up is as follows: suppose $X$ is a scheme, $mathcalV$ is a rank 2 locally free sheaf.
Then given a short exact sequence $$0 to mathcalS to mathcalV to mathcalQ to 0 $$ with $mathcalS, mathcalQ$ being invertible sheaves, we obtain a section $$sigma: X cong mathbbP(mathcalQ) to mathbbP(mathcalV)$$ by projectivizing the sequence.
Remark. Vakil''s convention for projectivization is $$mathbbP(mathcalV) = mathcalProj (Sym^bullet mathcalV)$$ without taking the dual (17.2.3).
The task is to show that the normal bundle $mathcalN_sigma(X)/mathbbP(mathcalV)$ is isomorphic to $mathcalQ otimes mathcalS^vee$.
Following the hint, we can assume $mathcalS cong mathcalO_X$.
To get a trivialization for $mathcalN_sigma(X)/mathbbP(mathcalV)$, take set $U subset X$ where $mathcalQ$ is trivial. Then the exact sequence splits $$0 to mathcalO_U to mathcalO_U^oplus 2 to mathcalO_U to 0,$$where the last map is just the projection onto the second factor and the section thus identifies with $$U to U times mathbbP^1$$ $$p mapsto (p,[0,1]).$$
We see that therefore the image is the locus $V(x_0)$ with $x_0 in Gamma(U, mathcalO_U times mathbbP^1(1))$. Moreover the section factors through $U times U_1$ ($U_1$ being the standard chart for projective line), and we can identify the conormal ideal as $$fracx_0/1mathcalO_U[x_0/1]x_0/1^2mathcalO_U[x_0/1] cong mathcalO_U.$$
The last step in the hint is to take trivializing cover $U_i$ for $mathcalQ$ with cocycle $g_ij$ and show that it also serves as a cocylce for the normal bundle.
beginarray
00 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0\
& & uparrowid && uparrow && uparrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalV|_U_ij & stackrellongrightarrow & mathcalQ|_U_ij & stackrellongrightarrow 0\
& & downarrowid && downarrow && downarrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0.
endarray
The outer right square is
beginarray
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij \
uparrow_beginbmatrix 1 & a_ij \ 0 & g_ij endbmatrix && uparrowg_ij \
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij
endarray
So it seems like when I take the corresponding projectivization
beginarray
UU_ij times mathbbP^1 & stackrellongleftarrow & X \
downarrow && downarrowid \
U_ij times mathbbP^1 & stackrellongleftarrow & X
endarray
section $x_0 in Gamma(U_ij times mathbbP^1, mathcalO_U_ij times mathbbP^1(1))$ downstairs pulls back to $x_0$ upstairs again, thus not producing interesting cocycle in the trivialization for the normal bundle. (And as far as I can tell, in order to get the correct cocylcle, we need $x_0$ to pullback to $g_ij^-1 x_0$.)
Can someone point to a mistake in the argument above?
algebraic-geometry vector-bundles
asked Jul 19 at 14:23
Bananeen
596212
add a comment |Â
up vote
1
down vote
favorite
I am stuck at solving 20.2.I from Vakil's notes.
The set up is as follows: suppose $X$ is a scheme, $mathcalV$ is a rank 2 locally free sheaf.
Then given a short exact sequence $$0 to mathcalS to mathcalV to mathcalQ to 0 $$ with $mathcalS, mathcalQ$ being invertible sheaves, we obtain a section $$sigma: X cong mathbbP(mathcalQ) to mathbbP(mathcalV)$$ by projectivizing the sequence.
Remark. Vakil''s convention for projectivization is $$mathbbP(mathcalV) = mathcalProj (Sym^bullet mathcalV)$$ without taking the dual (17.2.3).
The task is to show that the normal bundle $mathcalN_sigma(X)/mathbbP(mathcalV)$ is isomorphic to $mathcalQ otimes mathcalS^vee$.
Following the hint, we can assume $mathcalS cong mathcalO_X$.
To get a trivialization for $mathcalN_sigma(X)/mathbbP(mathcalV)$, take set $U subset X$ where $mathcalQ$ is trivial. Then the exact sequence splits $$0 to mathcalO_U to mathcalO_U^oplus 2 to mathcalO_U to 0,$$where the last map is just the projection onto the second factor and the section thus identifies with $$U to U times mathbbP^1$$ $$p mapsto (p,[0,1]).$$
We see that therefore the image is the locus $V(x_0)$ with $x_0 in Gamma(U, mathcalO_U times mathbbP^1(1))$. Moreover the section factors through $U times U_1$ ($U_1$ being the standard chart for projective line), and we can identify the conormal ideal as $$fracx_0/1mathcalO_U[x_0/1]x_0/1^2mathcalO_U[x_0/1] cong mathcalO_U.$$
The last step in the hint is to take trivializing cover $U_i$ for $mathcalQ$ with cocycle $g_ij$ and show that it also serves as a cocylce for the normal bundle.
beginarray
00 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0\
& & uparrowid && uparrow && uparrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalV|_U_ij & stackrellongrightarrow & mathcalQ|_U_ij & stackrellongrightarrow 0\
& & downarrowid && downarrow && downarrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0.
endarray
The outer right square is
beginarray
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij \
uparrow_beginbmatrix 1 & a_ij \ 0 & g_ij endbmatrix && uparrowg_ij \
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij
endarray
So it seems like when I take the corresponding projectivization
beginarray
UU_ij times mathbbP^1 & stackrellongleftarrow & X \
downarrow && downarrowid \
U_ij times mathbbP^1 & stackrellongleftarrow & X
endarray
section $x_0 in Gamma(U_ij times mathbbP^1, mathcalO_U_ij times mathbbP^1(1))$ downstairs pulls back to $x_0$ upstairs again, thus not producing interesting cocycle in the trivialization for the normal bundle. (And as far as I can tell, in order to get the correct cocylcle, we need $x_0$ to pullback to $g_ij^-1 x_0$.)
Can someone point to a mistake in the argument above?
algebraic-geometry vector-bundles
asked Jul 19 at 14:23
Bananeen
596212
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am stuck at solving 20.2.I from Vakil's notes.
The set up is as follows: suppose $X$ is a scheme, $mathcalV$ is a rank 2 locally free sheaf.
Then given a short exact sequence $$0 to mathcalS to mathcalV to mathcalQ to 0 $$ with $mathcalS, mathcalQ$ being invertible sheaves, we obtain a section $$sigma: X cong mathbbP(mathcalQ) to mathbbP(mathcalV)$$ by projectivizing the sequence.
Remark. Vakil''s convention for projectivization is $$mathbbP(mathcalV) = mathcalProj (Sym^bullet mathcalV)$$ without taking the dual (17.2.3).
The task is to show that the normal bundle $mathcalN_sigma(X)/mathbbP(mathcalV)$ is isomorphic to $mathcalQ otimes mathcalS^vee$.
Following the hint, we can assume $mathcalS cong mathcalO_X$.
To get a trivialization for $mathcalN_sigma(X)/mathbbP(mathcalV)$, take set $U subset X$ where $mathcalQ$ is trivial. Then the exact sequence splits $$0 to mathcalO_U to mathcalO_U^oplus 2 to mathcalO_U to 0,$$where the last map is just the projection onto the second factor and the section thus identifies with $$U to U times mathbbP^1$$ $$p mapsto (p,[0,1]).$$
We see that therefore the image is the locus $V(x_0)$ with $x_0 in Gamma(U, mathcalO_U times mathbbP^1(1))$. Moreover the section factors through $U times U_1$ ($U_1$ being the standard chart for projective line), and we can identify the conormal ideal as $$fracx_0/1mathcalO_U[x_0/1]x_0/1^2mathcalO_U[x_0/1] cong mathcalO_U.$$
The last step in the hint is to take trivializing cover $U_i$ for $mathcalQ$ with cocycle $g_ij$ and show that it also serves as a cocylce for the normal bundle.
beginarray
00 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0\
& & uparrowid && uparrow && uparrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalV|_U_ij & stackrellongrightarrow & mathcalQ|_U_ij & stackrellongrightarrow 0\
& & downarrowid && downarrow && downarrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0.
endarray
The outer right square is
beginarray
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij \
uparrow_beginbmatrix 1 & a_ij \ 0 & g_ij endbmatrix && uparrowg_ij \
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij
endarray
So it seems like when I take the corresponding projectivization
beginarray
UU_ij times mathbbP^1 & stackrellongleftarrow & X \
downarrow && downarrowid \
U_ij times mathbbP^1 & stackrellongleftarrow & X
endarray
section $x_0 in Gamma(U_ij times mathbbP^1, mathcalO_U_ij times mathbbP^1(1))$ downstairs pulls back to $x_0$ upstairs again, thus not producing interesting cocycle in the trivialization for the normal bundle. (And as far as I can tell, in order to get the correct cocylcle, we need $x_0$ to pullback to $g_ij^-1 x_0$.)
Can someone point to a mistake in the argument above?
algebraic-geometry vector-bundles
asked Jul 19 at 14:23
Bananeen
596212
I am stuck at solving 20.2.I from Vakil's notes.
The set up is as follows: suppose $X$ is a scheme, $mathcalV$ is a rank 2 locally free sheaf.
Then given a short exact sequence $$0 to mathcalS to mathcalV to mathcalQ to 0 $$ with $mathcalS, mathcalQ$ being invertible sheaves, we obtain a section $$sigma: X cong mathbbP(mathcalQ) to mathbbP(mathcalV)$$ by projectivizing the sequence.
Remark. Vakil''s convention for projectivization is $$mathbbP(mathcalV) = mathcalProj (Sym^bullet mathcalV)$$ without taking the dual (17.2.3).
The task is to show that the normal bundle $mathcalN_sigma(X)/mathbbP(mathcalV)$ is isomorphic to $mathcalQ otimes mathcalS^vee$.
Following the hint, we can assume $mathcalS cong mathcalO_X$.
To get a trivialization for $mathcalN_sigma(X)/mathbbP(mathcalV)$, take set $U subset X$ where $mathcalQ$ is trivial. Then the exact sequence splits $$0 to mathcalO_U to mathcalO_U^oplus 2 to mathcalO_U to 0,$$where the last map is just the projection onto the second factor and the section thus identifies with $$U to U times mathbbP^1$$ $$p mapsto (p,[0,1]).$$
We see that therefore the image is the locus $V(x_0)$ with $x_0 in Gamma(U, mathcalO_U times mathbbP^1(1))$. Moreover the section factors through $U times U_1$ ($U_1$ being the standard chart for projective line), and we can identify the conormal ideal as $$fracx_0/1mathcalO_U[x_0/1]x_0/1^2mathcalO_U[x_0/1] cong mathcalO_U.$$
The last step in the hint is to take trivializing cover $U_i$ for $mathcalQ$ with cocycle $g_ij$ and show that it also serves as a cocylce for the normal bundle.
beginarray
00 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0\
& & uparrowid && uparrow && uparrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalV|_U_ij & stackrellongrightarrow & mathcalQ|_U_ij & stackrellongrightarrow 0\
& & downarrowid && downarrow && downarrow\
0 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow & mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij & stackrellongrightarrow 0.
endarray
The outer right square is
beginarray
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij \
uparrow_beginbmatrix 1 & a_ij \ 0 & g_ij endbmatrix && uparrowg_ij \
mathcalO_U_ij^oplus 2 & stackrellongrightarrow & mathcalO_U_ij
endarray
So it seems like when I take the corresponding projectivization
beginarray
UU_ij times mathbbP^1 & stackrellongleftarrow & X \
downarrow && downarrowid \
U_ij times mathbbP^1 & stackrellongleftarrow & X
endarray
section $x_0 in Gamma(U_ij times mathbbP^1, mathcalO_U_ij times mathbbP^1(1))$ downstairs pulls back to $x_0$ upstairs again, thus not producing interesting cocycle in the trivialization for the normal bundle. (And as far as I can tell, in order to get the correct cocylcle, we need $x_0$ to pullback to $g_ij^-1 x_0$.)
Can someone point to a mistake in the argument above?
algebraic-geometry vector-bundles
asked Jul 19 at 14:23
Bananeen
596212
edited Jul 19 at 14:33
asked Jul 19 at 14:23
Bananeen
596212
asked Jul 19 at 14:23
Bananeen
596212
asked Jul 19 at 14:23
Bananeen
596212
596212
add a comment |Â
add a comment |Â
1 Answer
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Typically, this is done much more directly as follows. If we have a non-zero section $mathcalO_Xto V$, this gives a non-zero section of $mathcalO_P(V)(1)$, using the surjection $pi^*VtomathcalO_P(V)(1)$, where $pi:P(V)to X$ is the natural projection. If $Ysubset P(V)$ is the divisor where this section vanishes, it is clear that $N_Y/P(V)=mathcalO_P(V)(1)|Y$. In your case, chasing diagrams, it is immediate that $mathcalO_P(V)(1)|Y=pi^*Q|Y$, where you have started with $0tomathcalO_Xto Vto Qto 0$.
answered Jul 19 at 16:02
Mohan
11k1816
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Typically, this is done much more directly as follows. If we have a non-zero section $mathcalO_Xto V$, this gives a non-zero section of $mathcalO_P(V)(1)$, using the surjection $pi^*VtomathcalO_P(V)(1)$, where $pi:P(V)to X$ is the natural projection. If $Ysubset P(V)$ is the divisor where this section vanishes, it is clear that $N_Y/P(V)=mathcalO_P(V)(1)|Y$. In your case, chasing diagrams, it is immediate that $mathcalO_P(V)(1)|Y=pi^*Q|Y$, where you have started with $0tomathcalO_Xto Vto Qto 0$.
answered Jul 19 at 16:02
Mohan
11k1816
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
add a comment |Â
up vote
2
down vote
Typically, this is done much more directly as follows. If we have a non-zero section $mathcalO_Xto V$, this gives a non-zero section of $mathcalO_P(V)(1)$, using the surjection $pi^*VtomathcalO_P(V)(1)$, where $pi:P(V)to X$ is the natural projection. If $Ysubset P(V)$ is the divisor where this section vanishes, it is clear that $N_Y/P(V)=mathcalO_P(V)(1)|Y$. In your case, chasing diagrams, it is immediate that $mathcalO_P(V)(1)|Y=pi^*Q|Y$, where you have started with $0tomathcalO_Xto Vto Qto 0$.
answered Jul 19 at 16:02
Mohan
11k1816
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Typically, this is done much more directly as follows. If we have a non-zero section $mathcalO_Xto V$, this gives a non-zero section of $mathcalO_P(V)(1)$, using the surjection $pi^*VtomathcalO_P(V)(1)$, where $pi:P(V)to X$ is the natural projection. If $Ysubset P(V)$ is the divisor where this section vanishes, it is clear that $N_Y/P(V)=mathcalO_P(V)(1)|Y$. In your case, chasing diagrams, it is immediate that $mathcalO_P(V)(1)|Y=pi^*Q|Y$, where you have started with $0tomathcalO_Xto Vto Qto 0$.
answered Jul 19 at 16:02
Mohan
11k1816
Typically, this is done much more directly as follows. If we have a non-zero section $mathcalO_Xto V$, this gives a non-zero section of $mathcalO_P(V)(1)$, using the surjection $pi^*VtomathcalO_P(V)(1)$, where $pi:P(V)to X$ is the natural projection. If $Ysubset P(V)$ is the divisor where this section vanishes, it is clear that $N_Y/P(V)=mathcalO_P(V)(1)|Y$. In your case, chasing diagrams, it is immediate that $mathcalO_P(V)(1)|Y=pi^*Q|Y$, where you have started with $0tomathcalO_Xto Vto Qto 0$.
answered Jul 19 at 16:02
Mohan
11k1816
answered Jul 19 at 16:02
Mohan
11k1816
answered Jul 19 at 16:02
Mohan
11k1816
answered Jul 19 at 16:02
Mohan
11k1816
11k1816
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
add a comment |Â
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
Thank you, Mohan, for your reply. I think I've seen you suggesting a similar way in another thread and it indeed gives a slick solution. However, I thought that would still be beneficial to understand the nitty-gritty details, hidden in the machinery, following Vakil's hint.
– Bananeen
Jul 19 at 16:53
add a comment |Â
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