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?













8












$begingroup$


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?










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  • 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
    7 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
    4 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
    1 hour ago










  • $begingroup$
    @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
    $endgroup$
    – Yuhang Chen
    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
    40 mins ago















8












$begingroup$


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?










share|cite|improve this question







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
    7 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
    4 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
    1 hour ago










  • $begingroup$
    @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
    $endgroup$
    – Yuhang Chen
    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
    40 mins ago













8












8








8


2



$begingroup$


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?










share|cite|improve this question







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






share|cite|improve this question







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.










share|cite|improve this question







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.








share|cite|improve this question




share|cite|improve this question






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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 8 hours ago









Yuhang ChenYuhang Chen

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642 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
    7 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
    4 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
    1 hour ago










  • $begingroup$
    @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
    $endgroup$
    – Yuhang Chen
    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
    40 mins ago












  • 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
    7 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
    4 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
    1 hour ago










  • $begingroup$
    @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
    $endgroup$
    – Yuhang Chen
    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
    40 mins 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
7 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
7 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
4 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
4 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
1 hour 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
1 hour ago












$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
1 hour ago




$begingroup$
@Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"?
$endgroup$
– Yuhang Chen
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
40 mins 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
40 mins ago










1 Answer
1






active

oldest

votes


















18














$begingroup$

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".






share|cite|improve this answer









$endgroup$










  • 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
    4 hours ago













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









18














$begingroup$

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".






share|cite|improve this answer









$endgroup$










  • 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
    4 hours ago
















18














$begingroup$

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".






share|cite|improve this answer









$endgroup$










  • 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
    4 hours ago














18














18










18







$begingroup$

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".






share|cite|improve this answer









$endgroup$



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".







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 6 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
    4 hours ago













  • 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
    4 hours 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
4 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?
$endgroup$
– Qfwfq
4 hours ago












Yuhang Chen is a new contributor. Be nice, and check out our Code of Conduct.









draft saved

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Yuhang Chen is a new contributor. Be nice, and check out our Code of Conduct.












Yuhang Chen is a new contributor. Be nice, and check out our Code of Conduct.











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