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If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
Update: To make the question more concrete, consider one of the easiest (comparing to the other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $p\ne m\ne 0$ the equation $P(x,y)=p$ P(x,y)=m$ has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this.
Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have
$|(x/y-a)|\cdot |(x/y-b)|\cdot |(x/y-c)|=O(1)/|y^3|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $|x/y-a|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $|x/y-a|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction.
The question is then: is there an asymptotic geometry proof of the Thue theorem.
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edited Aug 13 2011 at 11:34 |
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
Update: To make the question more concrete, consider one of the easiest (comparing to other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $p\ne 0$ the equation $P(x,y)=p$ has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this.
Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have
$|(x/y-a)|\cdot |(x/y-b)|\cdot |(x/y-c)|=O(1)/|y^3|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $|x/y-a|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $|x/y-a|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction.
The question is then: is there an asymptotic geometry proof of the Thue theorem.
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edited Jan 7 2011 at 10:34 |
Update: To make the question more concrete, consider one of the easiest (comparing to other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and is not a product of two polynomials with integer coefficients, then for every integer $p\ne 0$ the equation $P(x,y)=p$ has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this. Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $|(x/y-a)|\cdot |(x/y-b)|\cdot |(x/y-c)|=O(1)/|y^3|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $|x/y-a|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $|x/y-a|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. The question is then: is there an asymptotic geometry proof of the Thue theorem.
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edited Jan 6 2011 at 15:04 |
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edited Jan 6 2011 at 2:14 |
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the smallest largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
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edited Jan 5 2011 at 23:59 |
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$(it : given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the "displacement" number)smallest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
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edited Jan 5 2011 at 23:27 |
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$ (it is the "displacement" number).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to the free a hyperbolic group is finite (the latter fact follows because the asymptotic cone of the free a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
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asked Jan 5 2011 at 23:19 |
Infinitely many solutions of a diophantine equation
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.
Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$ (it is the "displacement" number).
Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.
The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to the free group is finite (the latter fact follows because the asymptotic cone of the free group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).
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