Jacobians of reducible curves
up vote
6
down vote
favorite
Let $C$ be a connected, reduced but reducible projective curve over an algebraically closed field. Denote by $pi:tilde{C}to C$ its normalisation. There are two standard references for computing $J(C)=rm{Pic}^0(C)$, namely Harris-Morrison's 'Moduli of Curves' section 5.B and Bosch-Lütkebohmert-Raynaud's 'Néron Models' section 9.2.
I am happy with the computations of the Jacobian for the node, the cusp and for compact type curves, but am experiencing some confusion applying the occasionally unclear text of the above references to more complicated cases.
Take for example the case of three distinct smooth rational curves meeting at one point - i.e. a star-shaped cubic in $mathbb{P}^2$. The arithmetic genus is one (as it's a plane cubic) whereas the normalisation consists of three disjoint $mathbb{P}^1$'s. For any $C$ as in the first paragraph we have an induced exact sequence $$0tomathcal{O}^*_Ctopi_*mathcal{O}^*_{tilde{C}}to Fto0$$ where $F$ is a torsion sheaf supported on the singular locus of $C$. Harris-Morrison in exercise 5.11(2) give a formula for computing how many $mathbb{G}_m$ and $mathbb{G}_a$ factors the sheaf $F$ comprises of. In their notation, if $C$ is the star-shaped cubic with singularity $p$, one finds $b=2$ as the number of branches is 3, whereas $dim(pi_*mathcal{O}^*_{tilde{C}}/mathcal{O}^*_C)_p$ should be ''the contribution at $p$ to the difference in arithmetic genus between $C$ and $tilde{C}$''. This would imply $a+b=1$ meaning $a=-1$ which is already confusing and my first point in need of clarification.
Moving on under the assumption $a=0$ though, the cohomology of the above sequence gives $$0tomathbb{G}_mtomathbb{G}_m^3torm{H}^0(F)to J(C)to J(tilde{C})to 0.$$ $J(tilde{C})=0$ as it consists of disjoint rational curves whereas the above computation gives $rm{H}^0(F)=mathbb{G}_m^2$. This leads to $J(C)=0$ by a dimension count from the sequence. This is at odds with my expectation that the Jacobian of $C$ should be one dimensional.
Some further comments:
A further way to see that $rm{H}^0(F)=mathbb{G}_m^2$ in the above example is the following. The stalk at $p$ of the first short exact sequence above is as follows: the middle group is generated by three functions $f_1,f_2,f_3$, each defined on a neighbourhood of a preimage of $p$ on each of the $mathbb{P}^1$'s of the normalisation, and it contains $mathcal{O}^*_{C,p}$ those functions so that $f_1(p)=f_2(p)=f_3(p)$. The quotient $F_p$ should then be given by $(frac{f_1(p)}{f_2(p)}, frac{f_2(p)}{f_3(p)})$ ie $mathbb{G}_m^2$ - although there may be a mistake here.
Unraveling Propositions 9 and 10 (and the definition of the intermediate curve $X'$) in the 'Néron Models' section, one expects copies of $mathbb{G}_a$ to appear in $F$ whenever the singularity admits a resolution which is a homeomorphism (eg a cusp) in the sense that there's only one preimage. Whenever this is not the case, we expect a copy of $mathbb{G}_m$ in $F$ for the number of branches at the singularity, minus 1 (the long sequence in the middle of p249).
algebraic-geometry
|
show 3 more comments
up vote
6
down vote
favorite
Let $C$ be a connected, reduced but reducible projective curve over an algebraically closed field. Denote by $pi:tilde{C}to C$ its normalisation. There are two standard references for computing $J(C)=rm{Pic}^0(C)$, namely Harris-Morrison's 'Moduli of Curves' section 5.B and Bosch-Lütkebohmert-Raynaud's 'Néron Models' section 9.2.
I am happy with the computations of the Jacobian for the node, the cusp and for compact type curves, but am experiencing some confusion applying the occasionally unclear text of the above references to more complicated cases.
Take for example the case of three distinct smooth rational curves meeting at one point - i.e. a star-shaped cubic in $mathbb{P}^2$. The arithmetic genus is one (as it's a plane cubic) whereas the normalisation consists of three disjoint $mathbb{P}^1$'s. For any $C$ as in the first paragraph we have an induced exact sequence $$0tomathcal{O}^*_Ctopi_*mathcal{O}^*_{tilde{C}}to Fto0$$ where $F$ is a torsion sheaf supported on the singular locus of $C$. Harris-Morrison in exercise 5.11(2) give a formula for computing how many $mathbb{G}_m$ and $mathbb{G}_a$ factors the sheaf $F$ comprises of. In their notation, if $C$ is the star-shaped cubic with singularity $p$, one finds $b=2$ as the number of branches is 3, whereas $dim(pi_*mathcal{O}^*_{tilde{C}}/mathcal{O}^*_C)_p$ should be ''the contribution at $p$ to the difference in arithmetic genus between $C$ and $tilde{C}$''. This would imply $a+b=1$ meaning $a=-1$ which is already confusing and my first point in need of clarification.
Moving on under the assumption $a=0$ though, the cohomology of the above sequence gives $$0tomathbb{G}_mtomathbb{G}_m^3torm{H}^0(F)to J(C)to J(tilde{C})to 0.$$ $J(tilde{C})=0$ as it consists of disjoint rational curves whereas the above computation gives $rm{H}^0(F)=mathbb{G}_m^2$. This leads to $J(C)=0$ by a dimension count from the sequence. This is at odds with my expectation that the Jacobian of $C$ should be one dimensional.
Some further comments:
A further way to see that $rm{H}^0(F)=mathbb{G}_m^2$ in the above example is the following. The stalk at $p$ of the first short exact sequence above is as follows: the middle group is generated by three functions $f_1,f_2,f_3$, each defined on a neighbourhood of a preimage of $p$ on each of the $mathbb{P}^1$'s of the normalisation, and it contains $mathcal{O}^*_{C,p}$ those functions so that $f_1(p)=f_2(p)=f_3(p)$. The quotient $F_p$ should then be given by $(frac{f_1(p)}{f_2(p)}, frac{f_2(p)}{f_3(p)})$ ie $mathbb{G}_m^2$ - although there may be a mistake here.
Unraveling Propositions 9 and 10 (and the definition of the intermediate curve $X'$) in the 'Néron Models' section, one expects copies of $mathbb{G}_a$ to appear in $F$ whenever the singularity admits a resolution which is a homeomorphism (eg a cusp) in the sense that there's only one preimage. Whenever this is not the case, we expect a copy of $mathbb{G}_m$ in $F$ for the number of branches at the singularity, minus 1 (the long sequence in the middle of p249).
algebraic-geometry
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
1
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54
|
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $C$ be a connected, reduced but reducible projective curve over an algebraically closed field. Denote by $pi:tilde{C}to C$ its normalisation. There are two standard references for computing $J(C)=rm{Pic}^0(C)$, namely Harris-Morrison's 'Moduli of Curves' section 5.B and Bosch-Lütkebohmert-Raynaud's 'Néron Models' section 9.2.
I am happy with the computations of the Jacobian for the node, the cusp and for compact type curves, but am experiencing some confusion applying the occasionally unclear text of the above references to more complicated cases.
Take for example the case of three distinct smooth rational curves meeting at one point - i.e. a star-shaped cubic in $mathbb{P}^2$. The arithmetic genus is one (as it's a plane cubic) whereas the normalisation consists of three disjoint $mathbb{P}^1$'s. For any $C$ as in the first paragraph we have an induced exact sequence $$0tomathcal{O}^*_Ctopi_*mathcal{O}^*_{tilde{C}}to Fto0$$ where $F$ is a torsion sheaf supported on the singular locus of $C$. Harris-Morrison in exercise 5.11(2) give a formula for computing how many $mathbb{G}_m$ and $mathbb{G}_a$ factors the sheaf $F$ comprises of. In their notation, if $C$ is the star-shaped cubic with singularity $p$, one finds $b=2$ as the number of branches is 3, whereas $dim(pi_*mathcal{O}^*_{tilde{C}}/mathcal{O}^*_C)_p$ should be ''the contribution at $p$ to the difference in arithmetic genus between $C$ and $tilde{C}$''. This would imply $a+b=1$ meaning $a=-1$ which is already confusing and my first point in need of clarification.
Moving on under the assumption $a=0$ though, the cohomology of the above sequence gives $$0tomathbb{G}_mtomathbb{G}_m^3torm{H}^0(F)to J(C)to J(tilde{C})to 0.$$ $J(tilde{C})=0$ as it consists of disjoint rational curves whereas the above computation gives $rm{H}^0(F)=mathbb{G}_m^2$. This leads to $J(C)=0$ by a dimension count from the sequence. This is at odds with my expectation that the Jacobian of $C$ should be one dimensional.
Some further comments:
A further way to see that $rm{H}^0(F)=mathbb{G}_m^2$ in the above example is the following. The stalk at $p$ of the first short exact sequence above is as follows: the middle group is generated by three functions $f_1,f_2,f_3$, each defined on a neighbourhood of a preimage of $p$ on each of the $mathbb{P}^1$'s of the normalisation, and it contains $mathcal{O}^*_{C,p}$ those functions so that $f_1(p)=f_2(p)=f_3(p)$. The quotient $F_p$ should then be given by $(frac{f_1(p)}{f_2(p)}, frac{f_2(p)}{f_3(p)})$ ie $mathbb{G}_m^2$ - although there may be a mistake here.
Unraveling Propositions 9 and 10 (and the definition of the intermediate curve $X'$) in the 'Néron Models' section, one expects copies of $mathbb{G}_a$ to appear in $F$ whenever the singularity admits a resolution which is a homeomorphism (eg a cusp) in the sense that there's only one preimage. Whenever this is not the case, we expect a copy of $mathbb{G}_m$ in $F$ for the number of branches at the singularity, minus 1 (the long sequence in the middle of p249).
algebraic-geometry
Let $C$ be a connected, reduced but reducible projective curve over an algebraically closed field. Denote by $pi:tilde{C}to C$ its normalisation. There are two standard references for computing $J(C)=rm{Pic}^0(C)$, namely Harris-Morrison's 'Moduli of Curves' section 5.B and Bosch-Lütkebohmert-Raynaud's 'Néron Models' section 9.2.
I am happy with the computations of the Jacobian for the node, the cusp and for compact type curves, but am experiencing some confusion applying the occasionally unclear text of the above references to more complicated cases.
Take for example the case of three distinct smooth rational curves meeting at one point - i.e. a star-shaped cubic in $mathbb{P}^2$. The arithmetic genus is one (as it's a plane cubic) whereas the normalisation consists of three disjoint $mathbb{P}^1$'s. For any $C$ as in the first paragraph we have an induced exact sequence $$0tomathcal{O}^*_Ctopi_*mathcal{O}^*_{tilde{C}}to Fto0$$ where $F$ is a torsion sheaf supported on the singular locus of $C$. Harris-Morrison in exercise 5.11(2) give a formula for computing how many $mathbb{G}_m$ and $mathbb{G}_a$ factors the sheaf $F$ comprises of. In their notation, if $C$ is the star-shaped cubic with singularity $p$, one finds $b=2$ as the number of branches is 3, whereas $dim(pi_*mathcal{O}^*_{tilde{C}}/mathcal{O}^*_C)_p$ should be ''the contribution at $p$ to the difference in arithmetic genus between $C$ and $tilde{C}$''. This would imply $a+b=1$ meaning $a=-1$ which is already confusing and my first point in need of clarification.
Moving on under the assumption $a=0$ though, the cohomology of the above sequence gives $$0tomathbb{G}_mtomathbb{G}_m^3torm{H}^0(F)to J(C)to J(tilde{C})to 0.$$ $J(tilde{C})=0$ as it consists of disjoint rational curves whereas the above computation gives $rm{H}^0(F)=mathbb{G}_m^2$. This leads to $J(C)=0$ by a dimension count from the sequence. This is at odds with my expectation that the Jacobian of $C$ should be one dimensional.
Some further comments:
A further way to see that $rm{H}^0(F)=mathbb{G}_m^2$ in the above example is the following. The stalk at $p$ of the first short exact sequence above is as follows: the middle group is generated by three functions $f_1,f_2,f_3$, each defined on a neighbourhood of a preimage of $p$ on each of the $mathbb{P}^1$'s of the normalisation, and it contains $mathcal{O}^*_{C,p}$ those functions so that $f_1(p)=f_2(p)=f_3(p)$. The quotient $F_p$ should then be given by $(frac{f_1(p)}{f_2(p)}, frac{f_2(p)}{f_3(p)})$ ie $mathbb{G}_m^2$ - although there may be a mistake here.
Unraveling Propositions 9 and 10 (and the definition of the intermediate curve $X'$) in the 'Néron Models' section, one expects copies of $mathbb{G}_a$ to appear in $F$ whenever the singularity admits a resolution which is a homeomorphism (eg a cusp) in the sense that there's only one preimage. Whenever this is not the case, we expect a copy of $mathbb{G}_m$ in $F$ for the number of branches at the singularity, minus 1 (the long sequence in the middle of p249).
algebraic-geometry
algebraic-geometry
edited Nov 22 at 14:14
asked Nov 22 at 9:51
bey
312
312
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
1
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54
|
show 3 more comments
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
1
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
1
1
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54
|
show 3 more comments
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008926%2fjacobians-of-reducible-curves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008926%2fjacobians-of-reducible-curves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
A way to reduce your question to the case of nodal curves that you are familiar with, could be to perform a stable reduction to a degeneration of $C$. This is also explained in the book of Harris and Morrison, Ch. 3 if I remember correctly.
– James
Nov 22 at 14:36
@James I don't see how that helps. In particular, C is already a (non-semistable) degeneration of smooth cubics. Taking a semistable model of the complement of this fibre gives me an unrelated family where the central fibre is nodal
– bey
Nov 22 at 15:48
It helps because the central fibre of the stable reduction has the same Jacobian as the curve you started with. Moreover, the central fibre of the stable reduction admits only nodes as singularities which makes the computation of the Jacobian simpler.
– James
Nov 22 at 18:23
1
The arithmetic genus of a curve is usually defined as $1-chi(mathcal{O})$. So, the disjoint union of several smooth genus zero curves has arithmetic genus depending on the number of components and not necessarily zero.
– Mohan
Nov 22 at 18:50
@Mohan wonderful, this solves my problem, thanks. So now $p_a(tilde{C})=chi(mathcal{O}_{tilde{C}}=-2$ which means the difference in arithmetic genus is 3. Two dimensions are accounted for by the $mathbb{G}_m^2$ coming from the three branches, and there's an extra $mathbb{G}_a$, making $J(tilde{C})=mathbb{G}_a$.
– bey
Nov 22 at 20:54