How to prove that two $p$-adic lattices are isomorphic?
$begingroup$
Let $mathbb{Z}_{p}$ be the p-adic number.
a pair $(L,<>)$ is called lattice if $L$ be a free $mathbb{Z}_{p}$ module of finite rank and $<>:L×L to mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist isomorphism of $mathbb{Z}_{p}$ module $L_{1} to L_{2}$ preserving $<>$.
let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$begin{pmatrix}0&2^k&\2^k&0&end{pmatrix},begin{pmatrix}2^{k+1}&2^k&\2^k&2^{k+1}&end{pmatrix}$.
How to prove $X_{1}oplus X_{1} cong X_{2}oplus X_{2}$ and write this isomorphism explicitly ?
abstract-algebra number-theory quadratic-forms p-adic-number-theory integer-lattices
asked Oct 8 '18 at 5:37
neronero
34917
$endgroup$
|
show 3 more comments
$begingroup$
Let $mathbb{Z}_{p}$ be the p-adic number.
a pair $(L,<>)$ is called lattice if $L$ be a free $mathbb{Z}_{p}$ module of finite rank and $<>:L×L to mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist isomorphism of $mathbb{Z}_{p}$ module $L_{1} to L_{2}$ preserving $<>$.
let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$begin{pmatrix}0&2^k&\2^k&0&end{pmatrix},begin{pmatrix}2^{k+1}&2^k&\2^k&2^{k+1}&end{pmatrix}$.
How to prove $X_{1}oplus X_{1} cong X_{2}oplus X_{2}$ and write this isomorphism explicitly ?
abstract-algebra number-theory quadratic-forms p-adic-number-theory integer-lattices
asked Oct 8 '18 at 5:37
neronero
34917
$endgroup$
$begingroup$
Do you not just want to find an isomorphism $X_1 simeq X_2$?
$endgroup$
– Torsten Schoeneberg
Oct 10 '18 at 22:56
2
$begingroup$
@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
$endgroup$
– Jyrki Lahtonen
Oct 11 '18 at 9:03
1
$begingroup$
@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
$endgroup$
– Torsten Schoeneberg
Oct 16 '18 at 5:46
$begingroup$
@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
$endgroup$
– Jyrki Lahtonen
Dec 12 '18 at 8:58
1
$begingroup$
@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:47
|
show 3 more comments
$begingroup$
Let $mathbb{Z}_{p}$ be the p-adic number.
a pair $(L,<>)$ is called lattice if $L$ be a free $mathbb{Z}_{p}$ module of finite rank and $<>:L×L to mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist isomorphism of $mathbb{Z}_{p}$ module $L_{1} to L_{2}$ preserving $<>$.
let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$begin{pmatrix}0&2^k&\2^k&0&end{pmatrix},begin{pmatrix}2^{k+1}&2^k&\2^k&2^{k+1}&end{pmatrix}$.
How to prove $X_{1}oplus X_{1} cong X_{2}oplus X_{2}$ and write this isomorphism explicitly ?
abstract-algebra number-theory quadratic-forms p-adic-number-theory integer-lattices
asked Oct 8 '18 at 5:37
neronero
34917
$endgroup$
Let $mathbb{Z}_{p}$ be the p-adic number.
a pair $(L,<>)$ is called lattice if $L$ be a free $mathbb{Z}_{p}$ module of finite rank and $<>:L×L to mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist isomorphism of $mathbb{Z}_{p}$ module $L_{1} to L_{2}$ preserving $<>$.
let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$begin{pmatrix}0&2^k&\2^k&0&end{pmatrix},begin{pmatrix}2^{k+1}&2^k&\2^k&2^{k+1}&end{pmatrix}$.
How to prove $X_{1}oplus X_{1} cong X_{2}oplus X_{2}$ and write this isomorphism explicitly ?
abstract-algebra number-theory quadratic-forms p-adic-number-theory integer-lattices
abstract-algebra number-theory quadratic-forms p-adic-number-theory integer-lattices
asked Oct 8 '18 at 5:37
neronero
34917
asked Oct 8 '18 at 5:37
neronero
34917
asked Oct 8 '18 at 5:37
neronero
34917
asked Oct 8 '18 at 5:37
neronero
34917
asked Oct 8 '18 at 5:37
neronero
34917
34917
$begingroup$
Do you not just want to find an isomorphism $X_1 simeq X_2$?
$endgroup$
– Torsten Schoeneberg
Oct 10 '18 at 22:56
2
$begingroup$
@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
$endgroup$
– Jyrki Lahtonen
Oct 11 '18 at 9:03
1
$begingroup$
@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
$endgroup$
– Torsten Schoeneberg
Oct 16 '18 at 5:46
$begingroup$
@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
$endgroup$
– Jyrki Lahtonen
Dec 12 '18 at 8:58
1
$begingroup$
@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:47
|
show 3 more comments
$begingroup$
Do you not just want to find an isomorphism $X_1 simeq X_2$?
$endgroup$
– Torsten Schoeneberg
Oct 10 '18 at 22:56
2
$begingroup$
@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
$endgroup$
– Jyrki Lahtonen
Oct 11 '18 at 9:03
1
$begingroup$
@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
$endgroup$
– Torsten Schoeneberg
Oct 16 '18 at 5:46
$begingroup$
@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
$endgroup$
– Jyrki Lahtonen
Dec 12 '18 at 8:58
1
$begingroup$
@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:47
$begingroup$
Do you not just want to find an isomorphism $X_1 simeq X_2$?
$endgroup$
– Torsten Schoeneberg
Oct 10 '18 at 22:56
$begingroup$
Do you not just want to find an isomorphism $X_1 simeq X_2$?
$endgroup$
– Torsten Schoeneberg
Oct 10 '18 at 22:56
2
2
$begingroup$
@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
$endgroup$
– Jyrki Lahtonen
Oct 11 '18 at 9:03
$begingroup$
@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
$endgroup$
– Jyrki Lahtonen
Oct 11 '18 at 9:03
1
1
$begingroup$
@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
$endgroup$
– Torsten Schoeneberg
Oct 16 '18 at 5:46
$begingroup$
@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
$endgroup$
– Torsten Schoeneberg
Oct 16 '18 at 5:46
$begingroup$
@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
$endgroup$
– Jyrki Lahtonen
Dec 12 '18 at 8:58
$begingroup$
@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
$endgroup$
– Jyrki Lahtonen
Dec 12 '18 at 8:58
1
1
$begingroup$
@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:47
$begingroup$
@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 13:47
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I fiddled a little with your $mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 oplus X_1$ with $X_2 oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = mathbb{Z}[zeta_3]$, which is a quadratic ring extension of $mathbb{Z}_2$---defined by$$zeta_3^2 + zeta_3 + 1 = 0$$---the inner product being given by$$langle a, brangle = aoverline{b} + boverline{a},$$where the overline is the automorphism of $A$ sending $zeta_3$ to its inverse and each element of $mathbb{Z}_2$ to itself. Hence $X_2 oplus X_2$ can similarly be identified with $A oplus A$. Now it is easy to see that $A$ has an element $u$ with $uoverline{u} = -1$. Then the subgroup$$H = {(x, u.x) mid x in A}$$of $A$ is totally isotropic, and so is$$I = {(x, u.x.zeta_3) mid x in A},$$while $A, oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A oplus A$ is as a lattice isomorphic to $X_1 oplus X_1$.
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
$endgroup$
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
add a comment |
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$begingroup$
I fiddled a little with your $mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 oplus X_1$ with $X_2 oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = mathbb{Z}[zeta_3]$, which is a quadratic ring extension of $mathbb{Z}_2$---defined by$$zeta_3^2 + zeta_3 + 1 = 0$$---the inner product being given by$$langle a, brangle = aoverline{b} + boverline{a},$$where the overline is the automorphism of $A$ sending $zeta_3$ to its inverse and each element of $mathbb{Z}_2$ to itself. Hence $X_2 oplus X_2$ can similarly be identified with $A oplus A$. Now it is easy to see that $A$ has an element $u$ with $uoverline{u} = -1$. Then the subgroup$$H = {(x, u.x) mid x in A}$$of $A$ is totally isotropic, and so is$$I = {(x, u.x.zeta_3) mid x in A},$$while $A, oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A oplus A$ is as a lattice isomorphic to $X_1 oplus X_1$.
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
$endgroup$
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
add a comment |
$begingroup$
I fiddled a little with your $mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 oplus X_1$ with $X_2 oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = mathbb{Z}[zeta_3]$, which is a quadratic ring extension of $mathbb{Z}_2$---defined by$$zeta_3^2 + zeta_3 + 1 = 0$$---the inner product being given by$$langle a, brangle = aoverline{b} + boverline{a},$$where the overline is the automorphism of $A$ sending $zeta_3$ to its inverse and each element of $mathbb{Z}_2$ to itself. Hence $X_2 oplus X_2$ can similarly be identified with $A oplus A$. Now it is easy to see that $A$ has an element $u$ with $uoverline{u} = -1$. Then the subgroup$$H = {(x, u.x) mid x in A}$$of $A$ is totally isotropic, and so is$$I = {(x, u.x.zeta_3) mid x in A},$$while $A, oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A oplus A$ is as a lattice isomorphic to $X_1 oplus X_1$.
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
$endgroup$
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
add a comment |
$begingroup$
I fiddled a little with your $mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 oplus X_1$ with $X_2 oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = mathbb{Z}[zeta_3]$, which is a quadratic ring extension of $mathbb{Z}_2$---defined by$$zeta_3^2 + zeta_3 + 1 = 0$$---the inner product being given by$$langle a, brangle = aoverline{b} + boverline{a},$$where the overline is the automorphism of $A$ sending $zeta_3$ to its inverse and each element of $mathbb{Z}_2$ to itself. Hence $X_2 oplus X_2$ can similarly be identified with $A oplus A$. Now it is easy to see that $A$ has an element $u$ with $uoverline{u} = -1$. Then the subgroup$$H = {(x, u.x) mid x in A}$$of $A$ is totally isotropic, and so is$$I = {(x, u.x.zeta_3) mid x in A},$$while $A, oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A oplus A$ is as a lattice isomorphic to $X_1 oplus X_1$.
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
$endgroup$
I fiddled a little with your $mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 oplus X_1$ with $X_2 oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = mathbb{Z}[zeta_3]$, which is a quadratic ring extension of $mathbb{Z}_2$---defined by$$zeta_3^2 + zeta_3 + 1 = 0$$---the inner product being given by$$langle a, brangle = aoverline{b} + boverline{a},$$where the overline is the automorphism of $A$ sending $zeta_3$ to its inverse and each element of $mathbb{Z}_2$ to itself. Hence $X_2 oplus X_2$ can similarly be identified with $A oplus A$. Now it is easy to see that $A$ has an element $u$ with $uoverline{u} = -1$. Then the subgroup$$H = {(x, u.x) mid x in A}$$of $A$ is totally isotropic, and so is$$I = {(x, u.x.zeta_3) mid x in A},$$while $A, oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A oplus A$ is as a lattice isomorphic to $X_1 oplus X_1$.
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
answered Dec 14 '18 at 18:27
Get Off The InternetGet Off The Internet
1,457316
answered Dec 14 '18 at 18:27
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Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
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– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
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Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
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– Jyrki Lahtonen
Dec 14 '18 at 20:03
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Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
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– Jyrki Lahtonen
Dec 15 '18 at 9:42
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Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
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– Jyrki Lahtonen
Dec 15 '18 at 9:45
add a comment |
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
$begingroup$
Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:02
1
1
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure!
$endgroup$
– Jyrki Lahtonen
Dec 14 '18 at 20:03
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Question: Are you sure that $Aoplus A$ is the direct sum $Hoplus I$? The matrix $$pmatrix{1&ucr 1&uzeta_3cr}$$ has determinant $u(zeta_3-1)$. A problem with this is that $u=(1pmsqrt{-7})/2$ is not a 2-adic unit.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:42
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
$begingroup$
Never mind! One of them actually is a unit - for an appropriate choice of $sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $Bbb{Q}_2$ :-)
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 9:45
add a comment |
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Do you not just want to find an isomorphism $X_1 simeq X_2$?
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– Torsten Schoeneberg
Oct 10 '18 at 22:56
2
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@Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $sqrt{-7}inBbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it.
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– Jyrki Lahtonen
Oct 11 '18 at 9:03
1
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@JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly.
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– Torsten Schoeneberg
Oct 16 '18 at 5:46
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@TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it.
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– Jyrki Lahtonen
Dec 12 '18 at 8:58
1
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@san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(sqrt{-7},0))in X_2oplus X_2$ is isotropic. Recall that $sqrt ninBbb{Z}_2$, $n$ a square-free integer, if and only if $nequiv1pmod 8$.
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– Jyrki Lahtonen
Dec 14 '18 at 13:47