The tree of $SL_2$ over a local field
$begingroup$
I’m studying Chapter II: $mathbf{SL}_2$ of Serre’s book “Trees”. In paragraph 1, Serre defines the tree of $SL_2$ over a local field $K$. In particular, he considers the set of $mathcal O$-lattices of $K^2$ (where $mathcal O={xin Kvert nu(x)geq 0}$ is the ring of integers of $(K,nu)$) modulo the action of $K^{times}:=Ksetminus{0}$ by left multiplication. The set of lattice classes such is the set of vertices of a combinatorial graph $X$.
In $mathbf{Theorem 1}$, he proves that $X$ is a tree: connectedness is clear, while the fact that $X$ simply connected is not well explained. At some point (pag. 70) he says:
To prove that $X$ is a tree, it now suffices to show that, if $Lambda_0,dots,Lambda _n$, $ngeq 1$ is the sequence of vertices in a path without backtracking m in $X$, then $Lambda_0neq Lambda _n$. In fact we shall show (by induction on n) that $d(Lambda_0,Lambda _n)=n$.
From what we have said above, we can find representatives $L_iin Lambda _i$ such that $L_{i+1}subset L_i$ and $l(L_i/L_{i+1})=1$. We have $l(L_0/L_n)=n$ and we want to show $L_nnotsubset L_0pi$. By induction hypothesis we have $L_{n-1}notsubset L_0pi$.
The lattices $L_n$ and $L_{n-2}pi$ are the inverse images of two lines in the $k=mathcal O/pimathcal O$-plane $L_{n-1}/L_{n-1}pi$.
I don’t understand what does it mean by $mathit{lines}$. I thought that, since $L_nsubset L_{n-1}$ and $L_{n-1}/L_{n-1}pisimeq k^2$, $L_n $ is canonically embedded in it; but what about $L_{n-2}pi$?
Note. I’m assuming $ K$ to be commutative.
graph-theory trees p-adic-number-theory bruhat-tits-theory
asked Jan 16 at 17:30
BlumerBlumer
428210
$endgroup$
add a comment |
$begingroup$
I’m studying Chapter II: $mathbf{SL}_2$ of Serre’s book “Trees”. In paragraph 1, Serre defines the tree of $SL_2$ over a local field $K$. In particular, he considers the set of $mathcal O$-lattices of $K^2$ (where $mathcal O={xin Kvert nu(x)geq 0}$ is the ring of integers of $(K,nu)$) modulo the action of $K^{times}:=Ksetminus{0}$ by left multiplication. The set of lattice classes such is the set of vertices of a combinatorial graph $X$.
In $mathbf{Theorem 1}$, he proves that $X$ is a tree: connectedness is clear, while the fact that $X$ simply connected is not well explained. At some point (pag. 70) he says:
To prove that $X$ is a tree, it now suffices to show that, if $Lambda_0,dots,Lambda _n$, $ngeq 1$ is the sequence of vertices in a path without backtracking m in $X$, then $Lambda_0neq Lambda _n$. In fact we shall show (by induction on n) that $d(Lambda_0,Lambda _n)=n$.
From what we have said above, we can find representatives $L_iin Lambda _i$ such that $L_{i+1}subset L_i$ and $l(L_i/L_{i+1})=1$. We have $l(L_0/L_n)=n$ and we want to show $L_nnotsubset L_0pi$. By induction hypothesis we have $L_{n-1}notsubset L_0pi$.
The lattices $L_n$ and $L_{n-2}pi$ are the inverse images of two lines in the $k=mathcal O/pimathcal O$-plane $L_{n-1}/L_{n-1}pi$.
I don’t understand what does it mean by $mathit{lines}$. I thought that, since $L_nsubset L_{n-1}$ and $L_{n-1}/L_{n-1}pisimeq k^2$, $L_n $ is canonically embedded in it; but what about $L_{n-2}pi$?
Note. I’m assuming $ K$ to be commutative.
graph-theory trees p-adic-number-theory bruhat-tits-theory
asked Jan 16 at 17:30
BlumerBlumer
428210
$endgroup$
$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
1
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
1
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10
add a comment |
$begingroup$
I’m studying Chapter II: $mathbf{SL}_2$ of Serre’s book “Trees”. In paragraph 1, Serre defines the tree of $SL_2$ over a local field $K$. In particular, he considers the set of $mathcal O$-lattices of $K^2$ (where $mathcal O={xin Kvert nu(x)geq 0}$ is the ring of integers of $(K,nu)$) modulo the action of $K^{times}:=Ksetminus{0}$ by left multiplication. The set of lattice classes such is the set of vertices of a combinatorial graph $X$.
In $mathbf{Theorem 1}$, he proves that $X$ is a tree: connectedness is clear, while the fact that $X$ simply connected is not well explained. At some point (pag. 70) he says:
To prove that $X$ is a tree, it now suffices to show that, if $Lambda_0,dots,Lambda _n$, $ngeq 1$ is the sequence of vertices in a path without backtracking m in $X$, then $Lambda_0neq Lambda _n$. In fact we shall show (by induction on n) that $d(Lambda_0,Lambda _n)=n$.
From what we have said above, we can find representatives $L_iin Lambda _i$ such that $L_{i+1}subset L_i$ and $l(L_i/L_{i+1})=1$. We have $l(L_0/L_n)=n$ and we want to show $L_nnotsubset L_0pi$. By induction hypothesis we have $L_{n-1}notsubset L_0pi$.
The lattices $L_n$ and $L_{n-2}pi$ are the inverse images of two lines in the $k=mathcal O/pimathcal O$-plane $L_{n-1}/L_{n-1}pi$.
I don’t understand what does it mean by $mathit{lines}$. I thought that, since $L_nsubset L_{n-1}$ and $L_{n-1}/L_{n-1}pisimeq k^2$, $L_n $ is canonically embedded in it; but what about $L_{n-2}pi$?
Note. I’m assuming $ K$ to be commutative.
graph-theory trees p-adic-number-theory bruhat-tits-theory
asked Jan 16 at 17:30
BlumerBlumer
428210
$endgroup$
I’m studying Chapter II: $mathbf{SL}_2$ of Serre’s book “Trees”. In paragraph 1, Serre defines the tree of $SL_2$ over a local field $K$. In particular, he considers the set of $mathcal O$-lattices of $K^2$ (where $mathcal O={xin Kvert nu(x)geq 0}$ is the ring of integers of $(K,nu)$) modulo the action of $K^{times}:=Ksetminus{0}$ by left multiplication. The set of lattice classes such is the set of vertices of a combinatorial graph $X$.
In $mathbf{Theorem 1}$, he proves that $X$ is a tree: connectedness is clear, while the fact that $X$ simply connected is not well explained. At some point (pag. 70) he says:
To prove that $X$ is a tree, it now suffices to show that, if $Lambda_0,dots,Lambda _n$, $ngeq 1$ is the sequence of vertices in a path without backtracking m in $X$, then $Lambda_0neq Lambda _n$. In fact we shall show (by induction on n) that $d(Lambda_0,Lambda _n)=n$.
From what we have said above, we can find representatives $L_iin Lambda _i$ such that $L_{i+1}subset L_i$ and $l(L_i/L_{i+1})=1$. We have $l(L_0/L_n)=n$ and we want to show $L_nnotsubset L_0pi$. By induction hypothesis we have $L_{n-1}notsubset L_0pi$.
The lattices $L_n$ and $L_{n-2}pi$ are the inverse images of two lines in the $k=mathcal O/pimathcal O$-plane $L_{n-1}/L_{n-1}pi$.
I don’t understand what does it mean by $mathit{lines}$. I thought that, since $L_nsubset L_{n-1}$ and $L_{n-1}/L_{n-1}pisimeq k^2$, $L_n $ is canonically embedded in it; but what about $L_{n-2}pi$?
Note. I’m assuming $ K$ to be commutative.
graph-theory trees p-adic-number-theory bruhat-tits-theory
graph-theory trees p-adic-number-theory bruhat-tits-theory
asked Jan 16 at 17:30
BlumerBlumer
428210
asked Jan 16 at 17:30
BlumerBlumer
428210
edited Jan 16 at 21:31
Blumer
asked Jan 16 at 17:30
BlumerBlumer
428210
asked Jan 16 at 17:30
BlumerBlumer
428210
asked Jan 16 at 17:30
BlumerBlumer
428210
428210
$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
1
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
1
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10
add a comment |
$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
1
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
1
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10
$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
1
1
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
1
1
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10
add a comment |
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$begingroup$
Thank you, I added the missing part. Now it should be clear (for you).
$endgroup$
– Blumer
Jan 16 at 17:56
1
$begingroup$
What is $L_n$ ? Any $mathcal{O}$-module between $L_{n-1}$ and $pi L_{n-1}$ ? If $k = mathbf{F}_p,pi = p$ then $L_{n-1}/pi L_{n-1} cong mathbf{F}_ptimes mathbf{F}_p$ so any non-trivial subgroup is $cong (a,b)mathbf{F}_p$ (a line). For other $k$ the same works because the $mathcal{O}$-module condition implies $L_{n}/pi L_{n-1} $ is a $k$-vector space and $L_{n-1}/pi L_{n-1} $ is $2$-dimensional $k$-vector space, so $L_{n-1} supsetneq L_n supsetneq pi L_{n-1} $ implies $L_n/ pi L_{n-1}$ is one-dimensional $k$-vector space (a line)
$endgroup$
– reuns
Jan 16 at 18:00
$begingroup$
@reuns I don’t understand why is $L_n/pi L_{n-1}$ even defined. Why is $pi L_{n-1}$ an $mathcal O$-submodule of $L_n$?
$endgroup$
– Blumer
Jan 16 at 21:36
1
$begingroup$
You didn't specify what hypothesis you have on $L_{n-1},L_n$. What I assumed is some groups $L_{n-1} supsetneq L_n supsetneq pi L_{n-1}$ and all of them are free rank $2$ sub $mathcal{O}$-modules of $K^2$. So $L_n/pi L_{n-1}$ is a quotient of groups. Also $pi L_{n-1}$ is a sub-$mathcal{O}$-module of $L_n$ and $L_n/pi L_{n-1}$ is a quotient of (free) $mathcal{O}$-modules. So $L_n/pi L_{n-1}$ is an $mathcal{O}$-module and since $pi L_n = 0$ in the quotient it is a $mathcal{O}/(pi)$-module
$endgroup$
– reuns
Jan 17 at 11:10