Any representable moduli problem of elliptic curves is rigid

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EDIT: As I now tried to provide an answer to this problem, I would appreciate any proof verification of my reasoning. Thank you very much!

In Katz and Mazur's book (available here, page 109, page 60 in the pdf), it is stated as an obvious statement that any representable moduli problem is rigid. Actually, being new to these notions, I fail to see how one can show this. Let me sum up the situation.

We consider the category $(textbfEll)$ of elliptic curves over a base (maps are cartesian squares of elliptic curves). Let $mathcalP$ be a moduli problem for elliptic curves (that is a contravariant functor from $(textbfEll)$ to $(textbfSet)$). Assume that $mathcalP$ is representable: there exists an elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a functorial isomorphism $$mathcalP(E/S)simeqoperatornameHom_(textbfEll)(E/S,mathbbE/mathcalM(mathcalP))$$
for any object $E/S$ of $(textbfEll)$.

The statement we need to prove is the following.

Let $E/S$ be an object of $(textbfEll)$ given with an element $ain mathcalP(E/S)$. Let $epsilon:E/Srightarrow E/S$ be an automorphism. By functoriality, it induces a map $mathcalP(E/S) rightarrow mathcalP(E/S)$. Assume that $a$ is a fixed point of this map. Then, $epsilon$ is the identity.

This statement is said to be explicitly visible from the description of representable problems. Thus I feel like the proof should be quite straightforward, but I fail to see the point. Actually, I may not understand to what extent "to be representable" is a powerful statement in this context. Could somebody please clarify this point to me? I am thankful in advance for your help.

proof-verification algebraic-geometry elliptic-curves moduli-space

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edited Jul 23 at 23:21

asked Jul 23 at 7:17

Suzet

2,213427

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up vote
0
down vote

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EDIT: As I now tried to provide an answer to this problem, I would appreciate any proof verification of my reasoning. Thank you very much!

In Katz and Mazur's book (available here, page 109, page 60 in the pdf), it is stated as an obvious statement that any representable moduli problem is rigid. Actually, being new to these notions, I fail to see how one can show this. Let me sum up the situation.

We consider the category $(textbfEll)$ of elliptic curves over a base (maps are cartesian squares of elliptic curves). Let $mathcalP$ be a moduli problem for elliptic curves (that is a contravariant functor from $(textbfEll)$ to $(textbfSet)$). Assume that $mathcalP$ is representable: there exists an elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a functorial isomorphism $$mathcalP(E/S)simeqoperatornameHom_(textbfEll)(E/S,mathbbE/mathcalM(mathcalP))$$
for any object $E/S$ of $(textbfEll)$.

The statement we need to prove is the following.

Let $E/S$ be an object of $(textbfEll)$ given with an element $ain mathcalP(E/S)$. Let $epsilon:E/Srightarrow E/S$ be an automorphism. By functoriality, it induces a map $mathcalP(E/S) rightarrow mathcalP(E/S)$. Assume that $a$ is a fixed point of this map. Then, $epsilon$ is the identity.

This statement is said to be explicitly visible from the description of representable problems. Thus I feel like the proof should be quite straightforward, but I fail to see the point. Actually, I may not understand to what extent "to be representable" is a powerful statement in this context. Could somebody please clarify this point to me? I am thankful in advance for your help.

proof-verification algebraic-geometry elliptic-curves moduli-space

share|cite|improve this question

edited Jul 23 at 23:21

asked Jul 23 at 7:17

Suzet

2,213427

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

EDIT: As I now tried to provide an answer to this problem, I would appreciate any proof verification of my reasoning. Thank you very much!

In Katz and Mazur's book (available here, page 109, page 60 in the pdf), it is stated as an obvious statement that any representable moduli problem is rigid. Actually, being new to these notions, I fail to see how one can show this. Let me sum up the situation.

We consider the category $(textbfEll)$ of elliptic curves over a base (maps are cartesian squares of elliptic curves). Let $mathcalP$ be a moduli problem for elliptic curves (that is a contravariant functor from $(textbfEll)$ to $(textbfSet)$). Assume that $mathcalP$ is representable: there exists an elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a functorial isomorphism $$mathcalP(E/S)simeqoperatornameHom_(textbfEll)(E/S,mathbbE/mathcalM(mathcalP))$$
for any object $E/S$ of $(textbfEll)$.

The statement we need to prove is the following.

Let $E/S$ be an object of $(textbfEll)$ given with an element $ain mathcalP(E/S)$. Let $epsilon:E/Srightarrow E/S$ be an automorphism. By functoriality, it induces a map $mathcalP(E/S) rightarrow mathcalP(E/S)$. Assume that $a$ is a fixed point of this map. Then, $epsilon$ is the identity.

This statement is said to be explicitly visible from the description of representable problems. Thus I feel like the proof should be quite straightforward, but I fail to see the point. Actually, I may not understand to what extent "to be representable" is a powerful statement in this context. Could somebody please clarify this point to me? I am thankful in advance for your help.

proof-verification algebraic-geometry elliptic-curves moduli-space

share|cite|improve this question

edited Jul 23 at 23:21

asked Jul 23 at 7:17

Suzet

2,213427

EDIT: As I now tried to provide an answer to this problem, I would appreciate any proof verification of my reasoning. Thank you very much!

In Katz and Mazur's book (available here, page 109, page 60 in the pdf), it is stated as an obvious statement that any representable moduli problem is rigid. Actually, being new to these notions, I fail to see how one can show this. Let me sum up the situation.

We consider the category $(textbfEll)$ of elliptic curves over a base (maps are cartesian squares of elliptic curves). Let $mathcalP$ be a moduli problem for elliptic curves (that is a contravariant functor from $(textbfEll)$ to $(textbfSet)$). Assume that $mathcalP$ is representable: there exists an elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a functorial isomorphism $$mathcalP(E/S)simeqoperatornameHom_(textbfEll)(E/S,mathbbE/mathcalM(mathcalP))$$
for any object $E/S$ of $(textbfEll)$.

The statement we need to prove is the following.

Let $E/S$ be an object of $(textbfEll)$ given with an element $ain mathcalP(E/S)$. Let $epsilon:E/Srightarrow E/S$ be an automorphism. By functoriality, it induces a map $mathcalP(E/S) rightarrow mathcalP(E/S)$. Assume that $a$ is a fixed point of this map. Then, $epsilon$ is the identity.

This statement is said to be explicitly visible from the description of representable problems. Thus I feel like the proof should be quite straightforward, but I fail to see the point. Actually, I may not understand to what extent "to be representable" is a powerful statement in this context. Could somebody please clarify this point to me? I am thankful in advance for your help.

proof-verification algebraic-geometry elliptic-curves moduli-space

share|cite|improve this question

edited Jul 23 at 23:21

asked Jul 23 at 7:17

Suzet

2,213427

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 23:21

edited Jul 23 at 23:21

edited Jul 23 at 23:21

asked Jul 23 at 7:17

Suzet

2,213427

asked Jul 23 at 7:17

Suzet

2,213427

asked Jul 23 at 7:17

Suzet

2,213427

2,213427

add a comment | 

add a comment | 

1 Answer
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0
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accepted

I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $mathcalP$ is represented by a universal elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a universal element $a_0in mathcalP(mathbbE/mathcalM(mathcalP))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $ain mathcalP(E/S)$ is the same as to give a morphism $E/Srightarrow mathbbE/mathcalM(mathcalP)$ in $(textbfEll)$ such that the image of $a_0$ by the induced map $mathcalP(mathbbE/mathcalM(mathcalP))rightarrow mathcalP(E/S)$ is precisely $a$. That is the data of a morphism $phi: Srightarrow mathcalM(mathcalP)$ and an isomorphism $iota: Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $operatornameAut(E/S)$ acts on $mathcalP(E/S)$ by keeping $phi$ as it is and by composing $iota$ on the right. To say that an automorphism $epsilon$ has a fixed point means that $iotacircepsilon=iota$ as isomorphisms of elliptic curves $Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ over $S$.

Thus, $iota(epsilon-1)=0$. Because $iota$ is an isomorphism, $operatornamedeg(iota)not = 0$, so necessarily $operatornamedeg(epsilon -1)=0$, that is $epsilon - 1=0$.

share|cite|improve this answer

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
  – Distracted Kerl
  22 hours ago
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
1

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oldest

votes

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oldest

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votes

up vote
0
down vote



accepted

I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $mathcalP$ is represented by a universal elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a universal element $a_0in mathcalP(mathbbE/mathcalM(mathcalP))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $ain mathcalP(E/S)$ is the same as to give a morphism $E/Srightarrow mathbbE/mathcalM(mathcalP)$ in $(textbfEll)$ such that the image of $a_0$ by the induced map $mathcalP(mathbbE/mathcalM(mathcalP))rightarrow mathcalP(E/S)$ is precisely $a$. That is the data of a morphism $phi: Srightarrow mathcalM(mathcalP)$ and an isomorphism $iota: Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $operatornameAut(E/S)$ acts on $mathcalP(E/S)$ by keeping $phi$ as it is and by composing $iota$ on the right. To say that an automorphism $epsilon$ has a fixed point means that $iotacircepsilon=iota$ as isomorphisms of elliptic curves $Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ over $S$.

Thus, $iota(epsilon-1)=0$. Because $iota$ is an isomorphism, $operatornamedeg(iota)not = 0$, so necessarily $operatornamedeg(epsilon -1)=0$, that is $epsilon - 1=0$.

share|cite|improve this answer

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
  – Distracted Kerl
  22 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $mathcalP$ is represented by a universal elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a universal element $a_0in mathcalP(mathbbE/mathcalM(mathcalP))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $ain mathcalP(E/S)$ is the same as to give a morphism $E/Srightarrow mathbbE/mathcalM(mathcalP)$ in $(textbfEll)$ such that the image of $a_0$ by the induced map $mathcalP(mathbbE/mathcalM(mathcalP))rightarrow mathcalP(E/S)$ is precisely $a$. That is the data of a morphism $phi: Srightarrow mathcalM(mathcalP)$ and an isomorphism $iota: Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $operatornameAut(E/S)$ acts on $mathcalP(E/S)$ by keeping $phi$ as it is and by composing $iota$ on the right. To say that an automorphism $epsilon$ has a fixed point means that $iotacircepsilon=iota$ as isomorphisms of elliptic curves $Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ over $S$.

Thus, $iota(epsilon-1)=0$. Because $iota$ is an isomorphism, $operatornamedeg(iota)not = 0$, so necessarily $operatornamedeg(epsilon -1)=0$, that is $epsilon - 1=0$.

share|cite|improve this answer

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
  – Distracted Kerl
  22 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $mathcalP$ is represented by a universal elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a universal element $a_0in mathcalP(mathbbE/mathcalM(mathcalP))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $ain mathcalP(E/S)$ is the same as to give a morphism $E/Srightarrow mathbbE/mathcalM(mathcalP)$ in $(textbfEll)$ such that the image of $a_0$ by the induced map $mathcalP(mathbbE/mathcalM(mathcalP))rightarrow mathcalP(E/S)$ is precisely $a$. That is the data of a morphism $phi: Srightarrow mathcalM(mathcalP)$ and an isomorphism $iota: Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $operatornameAut(E/S)$ acts on $mathcalP(E/S)$ by keeping $phi$ as it is and by composing $iota$ on the right. To say that an automorphism $epsilon$ has a fixed point means that $iotacircepsilon=iota$ as isomorphisms of elliptic curves $Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ over $S$.

Thus, $iota(epsilon-1)=0$. Because $iota$ is an isomorphism, $operatornamedeg(iota)not = 0$, so necessarily $operatornamedeg(epsilon -1)=0$, that is $epsilon - 1=0$.

share|cite|improve this answer

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $mathcalP$ is represented by a universal elliptic curve $mathbbE/mathcalM(mathcalP)$ together with a universal element $a_0in mathcalP(mathbbE/mathcalM(mathcalP))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $ain mathcalP(E/S)$ is the same as to give a morphism $E/Srightarrow mathbbE/mathcalM(mathcalP)$ in $(textbfEll)$ such that the image of $a_0$ by the induced map $mathcalP(mathbbE/mathcalM(mathcalP))rightarrow mathcalP(E/S)$ is precisely $a$. That is the data of a morphism $phi: Srightarrow mathcalM(mathcalP)$ and an isomorphism $iota: Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $operatornameAut(E/S)$ acts on $mathcalP(E/S)$ by keeping $phi$ as it is and by composing $iota$ on the right. To say that an automorphism $epsilon$ has a fixed point means that $iotacircepsilon=iota$ as isomorphisms of elliptic curves $Exrightarrowsim mathbbEtimes_mathcalM(mathcalP)S$ over $S$.

Thus, $iota(epsilon-1)=0$. Because $iota$ is an isomorphism, $operatornamedeg(iota)not = 0$, so necessarily $operatornamedeg(epsilon -1)=0$, that is $epsilon - 1=0$.

share|cite|improve this answer

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 22:40

edited Jul 23 at 22:40

edited Jul 23 at 22:40

answered Jul 23 at 11:51

Suzet

2,213427

answered Jul 23 at 11:51

Suzet

2,213427

answered Jul 23 at 11:51

Suzet

2,213427

2,213427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
  – Distracted Kerl
  22 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
  – Distracted Kerl
  22 hours ago
  

  
  
  
  

Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
– Distracted Kerl
22 hours ago

Dear Suzet, I think you can just simplify the $i$ in your equation $icircepsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $operatornameAut(E/S)$ on the set of homs and then applying the description. (In essence the same thing).
– Distracted Kerl
22 hours ago

add a comment | 

 

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