Why not add cuspidal curves in the moduli space of stable curves?picard group of moduli of elliptic r-prym curvesWhat is the difference between the moduli space of curves and the moduli space of orbi-curves?Evaluation maps for moduli of stable mapsUniversal curve of stacks of stable curvereference request for homotopy exact sequence of moduli stacks of curvesModuli problem of stable nodal curves over the integersModuli of stable and semistable $G$-Higgs bundles on curvesWhat automorphic forms are expected to occur in the zeta function of moduli space of curves?
Why not add cuspidal curves in the moduli space of stable curves?
picard group of moduli of elliptic r-prym curvesWhat is the difference between the moduli space of curves and the moduli space of orbi-curves?Evaluation maps for moduli of stable mapsUniversal curve of stacks of stable curvereference request for homotopy exact sequence of moduli stacks of curvesModuli problem of stable nodal curves over the integersModuli of stable and semistable $G$-Higgs bundles on curvesWhat automorphic forms are expected to occur in the zeta function of moduli space of curves?
$begingroup$
Let $mathcalM_g,n$ be the moduli space (stack) of stable smooth curves of genus $g$ with $n$ marked points over $mathbbC. $ It's known that by adding stable nodal curves to $mathcalM_g,n$, the resulting space $overlinemathcalM_g,n$ is compact. But why is it so? For example, consider the following family of elliptic curves in $mathcalM_1,1$, $$y^2=x^3+t,$$ where $t in mathbbC^*$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $overlinemathcalM_1,1$ compact?
ag.algebraic-geometry algebraic-curves moduli-spaces
asked 9 hours ago
Yuhang ChenYuhang Chen
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$endgroup$
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show 4 more comments
$begingroup$
Let $mathcalM_g,n$ be the moduli space (stack) of stable smooth curves of genus $g$ with $n$ marked points over $mathbbC. $ It's known that by adding stable nodal curves to $mathcalM_g,n$, the resulting space $overlinemathcalM_g,n$ is compact. But why is it so? For example, consider the following family of elliptic curves in $mathcalM_1,1$, $$y^2=x^3+t,$$ where $t in mathbbC^*$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $overlinemathcalM_1,1$ compact?
ag.algebraic-geometry algebraic-curves moduli-spaces
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
$endgroup$
– Asvin
8 hours ago
1
$begingroup$
Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
$endgroup$
– Asvin
1 hour ago
|
show 4 more comments
$begingroup$
Let $mathcalM_g,n$ be the moduli space (stack) of stable smooth curves of genus $g$ with $n$ marked points over $mathbbC. $ It's known that by adding stable nodal curves to $mathcalM_g,n$, the resulting space $overlinemathcalM_g,n$ is compact. But why is it so? For example, consider the following family of elliptic curves in $mathcalM_1,1$, $$y^2=x^3+t,$$ where $t in mathbbC^*$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $overlinemathcalM_1,1$ compact?
ag.algebraic-geometry algebraic-curves moduli-spaces
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $mathcalM_g,n$ be the moduli space (stack) of stable smooth curves of genus $g$ with $n$ marked points over $mathbbC. $ It's known that by adding stable nodal curves to $mathcalM_g,n$, the resulting space $overlinemathcalM_g,n$ is compact. But why is it so? For example, consider the following family of elliptic curves in $mathcalM_1,1$, $$y^2=x^3+t,$$ where $t in mathbbC^*$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $overlinemathcalM_1,1$ compact?
ag.algebraic-geometry algebraic-curves moduli-spaces
ag.algebraic-geometry algebraic-curves moduli-spaces
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
asked 9 hours ago
Yuhang ChenYuhang Chen
712 bronze badges
712 bronze badges
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yuhang Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
$endgroup$
– Asvin
8 hours ago
1
$begingroup$
Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
$endgroup$
– Asvin
1 hour ago
|
show 4 more comments
3
$begingroup$
I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
$endgroup$
– Asvin
8 hours ago
1
$begingroup$
Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
$endgroup$
– Asvin
1 hour ago
3
3
$begingroup$
I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
$endgroup$
– Asvin
8 hours ago
$begingroup$
I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
$endgroup$
– Asvin
8 hours ago
1
1
$begingroup$
Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
2 hours ago
$begingroup$
@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
$endgroup$
– Asvin
1 hour ago
$begingroup$
@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
$endgroup$
– Asvin
1 hour ago
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If you add cuspidal curves, then $overlinemathcalM_1,1$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 mboxand y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $mathbbC^ast$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $mathbbC^ast to overlinemathcalM_1,1$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $mathrmSpec mathbbC[t^pm 1]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
$endgroup$
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
$endgroup$
– Yuhang Chen
28 mins ago
add a comment
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1 Answer
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$begingroup$
If you add cuspidal curves, then $overlinemathcalM_1,1$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 mboxand y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $mathbbC^ast$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $mathbbC^ast to overlinemathcalM_1,1$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $mathrmSpec mathbbC[t^pm 1]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
$endgroup$
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
$endgroup$
– Yuhang Chen
28 mins ago
add a comment
|
$begingroup$
If you add cuspidal curves, then $overlinemathcalM_1,1$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 mboxand y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $mathbbC^ast$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $mathbbC^ast to overlinemathcalM_1,1$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $mathrmSpec mathbbC[t^pm 1]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
$endgroup$
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
$endgroup$
– Yuhang Chen
28 mins ago
add a comment
|
$begingroup$
If you add cuspidal curves, then $overlinemathcalM_1,1$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 mboxand y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $mathbbC^ast$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $mathbbC^ast to overlinemathcalM_1,1$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $mathrmSpec mathbbC[t^pm 1]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
$endgroup$
If you add cuspidal curves, then $overlinemathcalM_1,1$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 mboxand y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $mathbbC^ast$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $mathbbC^ast to overlinemathcalM_1,1$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $mathrmSpec mathbbC[t^pm 1]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
answered 7 hours ago
David E SpeyerDavid E Speyer
111k10 gold badges293 silver badges565 bronze badges
111k10 gold badges293 silver badges565 bronze badges
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
$endgroup$
– Yuhang Chen
28 mins ago
add a comment
|
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
$endgroup$
– Yuhang Chen
28 mins ago
1
1
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
$endgroup$
– Qfwfq
5 hours ago
$begingroup$
Could you explain better "The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover"? I would've said that what is meant to accomodate this phenomenon is exactly the use of the étale topology on the base category (whether we use algebraic spaces as c.m.s.'s or we go full stacky); and that the definition of c.m.s. is meant to avoid stackiness (automorphisms) while still retaining the same geometric points and some geometry. Is my intuition incorrect about this?
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– Qfwfq
5 hours ago
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Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
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– Yuhang Chen
28 mins ago
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Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $mathbbC^* to overlinemathcalM_1,1$ you mentioned, on the level of coarse moduli spaces.
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– Yuhang Chen
28 mins ago
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I believe this has to do with the fact that $mathscr M_g,n$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $mathbb C[[t]]$, the family will become either a node or good reduction.
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– Asvin
8 hours ago
1
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Also, a cusp would have a $mathbbC^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford.
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– Qfwfq
5 hours ago
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@Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $mathbbC^*$, even with a marked point at the cusp?
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– Yuhang Chen
2 hours ago
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@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
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– Yuhang Chen
2 hours ago
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@yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve.
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– Asvin
1 hour ago