Skip to main content
MathOverflow

Return to Question

minor typos
Source Link
Martin Sleziak
Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$)

I am studystudying a little about distinguished triangle and iI want to know that what iI write below is correct or not!

weWe have naruralnatural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

here iHere I used the fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is iI think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$.Now Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$)

I am study a little about distinguished triangle and i want to know that what i write below is correct or not!

we have narural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

here i used fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is i think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$.Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$)

I am studying a little about distinguished triangle and I want to know that what I write below is correct or not!

We have natural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

Here I used the fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is I think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$. Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

added 1053 characters in body
Source Link
Tom
Tom
  • 71
  • 4

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but)

I don'tam study a little about distinguished triangle and i want to know whythat what i write below is correct or not!

we have narural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) thanks we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

here i used fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is i think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$.Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but I don't know why) thanks.

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$)

I am study a little about distinguished triangle and i want to know that what i write below is correct or not!

we have narural morphism $f^*\Omega_Y \to\Omega_C(p_1+...+p_n)$ if we consider these two term in two complex (put these two term in $0$-place and other elements of complex are $0$) we will have distinguished triangle

$[...\to 0 \to f^*\Omega_Y \to 0 \to ...] \to [... \to 0 \to (0,f^*\Omega_Y,0),0) \to (f^*\Omega_Y,0,\Omega_C(p_1+...+p_n) \to0 \to...] \to [...\to 0\to(f^*\Omega_Y,0)\to (0,\Omega_C(p_1+...+p_n))\to 0 \to...]$

here i used fact that if we have $f:K^* \to L^*$(map between complexes) then we have distinguished triangle $K^* \to Cyl(f) \to C(f) \to K[1]^*$

Now the first term of distinguished triangle is clear.the third term is same as $ f^*\Omega_Y \to \Omega_C(p_1+...+p_n)$ (via definition of map in $C(f)$ which i didnt imply) and the first is i think will be isomorph to complex [$\Omega_C(p_1+...+p_n)]$.Now if we apply $Ext(-,O_C) $ we will get the long exact sequence in my question.

I want to know is it correct or not?

minor typos
Source Link
Martin Sleziak
Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40

Generalizing of normal sheaf via short exacexact sequence

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but I dontdon't know why) thanks.

Generalizing of normal sheaf via short exac sequence

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but I dont know why) thanks.

Generalizing of normal sheaf via short exact sequence

I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but I don't know why) thanks.

Source Link
Tom
Tom
  • 71
  • 4