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A linear form can not be too small on rational points
Clash Royale CLAN TAG#URR8PPP
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Notations.
Let $xi=(xi_1,xi_2,xi_3,xi_4)inmathbb( Rsetminusmathbb Q)^4$ such that $xi_1xi_4-xi_2xi_3ne 0$ and the $xi_i$ are linearly independent over $mathbb Q$.
I have the following linear form:
$$beginmatrixLcolon & mathbb R^6 & to & mathbb R \ &(eta_1,ldots,eta_6) & mapsto & eta_1(xi_1xi_4-xi_2xi_3)-eta_2xi_4+eta_3xi_3-eta_4xi_2+eta_5xi_1-eta_6.endmatrix$$
We consider the norm $VertcdotVert$ to be the euclidean norm on $mathbb R^6$.
The problem.
I am interesting in finding a constant $gamma>0$, such that if we choose $xi$ properly, then the resulting linear form $L_xi$ will verify:
$$forall etainmathbb Z^6setminus0,quad L_xi(eta)geqslant frac cVertetaVert^gammagcd(eta_1,ldots,eta_6),$$
where $c=c_xi$ is a constant which depends only on $xi$.
The conjecture.
There are hopes for this to be true, since if we choose $xi$ properly (for instance badly approximated by rationals), then for $etainmathbb Z^6setminus0$, $L_xi(eta)$ will have troubles being too small.
I believe that the constant $gamma=2$ would work for a fine choice of $xi$.
Additional remarks.
This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated.
I do believe that $gamma=2$ would work (and it would be the best), but any proof that would work for a $gamma<4$ would be great.
real-analysis irrational-numbers rational-numbers diophantine-approximation linear-form
asked 3 hours ago
E. Joseph
11.4k82755
 |Â
show 2 more comments
up vote
2
down vote
favorite
Notations.
Let $xi=(xi_1,xi_2,xi_3,xi_4)inmathbb( Rsetminusmathbb Q)^4$ such that $xi_1xi_4-xi_2xi_3ne 0$ and the $xi_i$ are linearly independent over $mathbb Q$.
I have the following linear form:
$$beginmatrixLcolon & mathbb R^6 & to & mathbb R \ &(eta_1,ldots,eta_6) & mapsto & eta_1(xi_1xi_4-xi_2xi_3)-eta_2xi_4+eta_3xi_3-eta_4xi_2+eta_5xi_1-eta_6.endmatrix$$
We consider the norm $VertcdotVert$ to be the euclidean norm on $mathbb R^6$.
The problem.
I am interesting in finding a constant $gamma>0$, such that if we choose $xi$ properly, then the resulting linear form $L_xi$ will verify:
$$forall etainmathbb Z^6setminus0,quad L_xi(eta)geqslant frac cVertetaVert^gammagcd(eta_1,ldots,eta_6),$$
where $c=c_xi$ is a constant which depends only on $xi$.
The conjecture.
There are hopes for this to be true, since if we choose $xi$ properly (for instance badly approximated by rationals), then for $etainmathbb Z^6setminus0$, $L_xi(eta)$ will have troubles being too small.
I believe that the constant $gamma=2$ would work for a fine choice of $xi$.
Additional remarks.
This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated.
I do believe that $gamma=2$ would work (and it would be the best), but any proof that would work for a $gamma<4$ would be great.
real-analysis irrational-numbers rational-numbers diophantine-approximation linear-form
asked 3 hours ago
E. Joseph
11.4k82755
Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
2
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Notations.
Let $xi=(xi_1,xi_2,xi_3,xi_4)inmathbb( Rsetminusmathbb Q)^4$ such that $xi_1xi_4-xi_2xi_3ne 0$ and the $xi_i$ are linearly independent over $mathbb Q$.
I have the following linear form:
$$beginmatrixLcolon & mathbb R^6 & to & mathbb R \ &(eta_1,ldots,eta_6) & mapsto & eta_1(xi_1xi_4-xi_2xi_3)-eta_2xi_4+eta_3xi_3-eta_4xi_2+eta_5xi_1-eta_6.endmatrix$$
We consider the norm $VertcdotVert$ to be the euclidean norm on $mathbb R^6$.
The problem.
I am interesting in finding a constant $gamma>0$, such that if we choose $xi$ properly, then the resulting linear form $L_xi$ will verify:
$$forall etainmathbb Z^6setminus0,quad L_xi(eta)geqslant frac cVertetaVert^gammagcd(eta_1,ldots,eta_6),$$
where $c=c_xi$ is a constant which depends only on $xi$.
The conjecture.
There are hopes for this to be true, since if we choose $xi$ properly (for instance badly approximated by rationals), then for $etainmathbb Z^6setminus0$, $L_xi(eta)$ will have troubles being too small.
I believe that the constant $gamma=2$ would work for a fine choice of $xi$.
Additional remarks.
This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated.
I do believe that $gamma=2$ would work (and it would be the best), but any proof that would work for a $gamma<4$ would be great.
real-analysis irrational-numbers rational-numbers diophantine-approximation linear-form
asked 3 hours ago
E. Joseph
11.4k82755
Notations.
Let $xi=(xi_1,xi_2,xi_3,xi_4)inmathbb( Rsetminusmathbb Q)^4$ such that $xi_1xi_4-xi_2xi_3ne 0$ and the $xi_i$ are linearly independent over $mathbb Q$.
I have the following linear form:
$$beginmatrixLcolon & mathbb R^6 & to & mathbb R \ &(eta_1,ldots,eta_6) & mapsto & eta_1(xi_1xi_4-xi_2xi_3)-eta_2xi_4+eta_3xi_3-eta_4xi_2+eta_5xi_1-eta_6.endmatrix$$
We consider the norm $VertcdotVert$ to be the euclidean norm on $mathbb R^6$.
The problem.
I am interesting in finding a constant $gamma>0$, such that if we choose $xi$ properly, then the resulting linear form $L_xi$ will verify:
$$forall etainmathbb Z^6setminus0,quad L_xi(eta)geqslant frac cVertetaVert^gammagcd(eta_1,ldots,eta_6),$$
where $c=c_xi$ is a constant which depends only on $xi$.
The conjecture.
There are hopes for this to be true, since if we choose $xi$ properly (for instance badly approximated by rationals), then for $etainmathbb Z^6setminus0$, $L_xi(eta)$ will have troubles being too small.
I believe that the constant $gamma=2$ would work for a fine choice of $xi$.
Additional remarks.
This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated.
I do believe that $gamma=2$ would work (and it would be the best), but any proof that would work for a $gamma<4$ would be great.
real-analysis irrational-numbers rational-numbers diophantine-approximation linear-form
real-analysis irrational-numbers rational-numbers diophantine-approximation linear-form
asked 3 hours ago
E. Joseph
11.4k82755
asked 3 hours ago
E. Joseph
11.4k82755
edited 2 hours ago
asked 3 hours ago
E. Joseph
11.4k82755
asked 3 hours ago
E. Joseph
11.4k82755
asked 3 hours ago
E. Joseph
11.4k82755
11.4k82755
Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
2
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago
 |Â
show 2 more comments
Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
2
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago
Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
2
2
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago
 |Â
show 2 more comments
2 Answers
2
active
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votes
up vote
3
down vote
I'm afraid the converse is true. For any $zeta_1, ldots, zeta_k in mathbbR$ and any integer $n > 0$, there exist integers $n_0, ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1zeta_1 + ldots + n_kzeta_k| leq n^-k$ (this is plainly the pigeonhole principle applied to the set of fractional parts $m_1zeta_1 + ldots + m_kzeta_k$ for all positive integers $m_1, ldots, m_k$ with values at most $n$). In your case, $k = 5$.
answered 1 hour ago
metamorphy
1,50911
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
add a comment |Â
up vote
3
down vote
I think this is a direct application of a Theorem of Kleinbock and Margulis
https://arxiv.org/pdf/math/9810036.pdf
Let me give a little more detail. Let $f$ be the following map from $mathbbR^4 to mathbbR^5$
$$f(xi_1,xi_2,xi_3,xi_4)=(xi_1xi_4-xi_2xi_3,-xi_4,xi_3,-xi_2,xi_1).$$
It is not hard to check that partial derivatives $(partial f/partial xi_i)_1leq ileq 4$ together with $partial^2 f/partial xi_1partialxi_4$ span $mathbbR^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $xi=(xi_1,..,xi_4)$, $f(xi)$ is not very well approximable, meaning that for all $epsilon>0$, there exist only finitely many integer vectors $q in mathbbZ^5$ such that there exist a $pin mathbbZ$ such that
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) leq 1.$$
Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_epsilon>0$ such that for all $q in mathbbZ^5$ and $pin mathbbZ$,
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) geq c_epsilon.$$
Since $langle q, f(xi) rangle + p=L_xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.
answered 1 hour ago
user120527
1,142211
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I'm afraid the converse is true. For any $zeta_1, ldots, zeta_k in mathbbR$ and any integer $n > 0$, there exist integers $n_0, ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1zeta_1 + ldots + n_kzeta_k| leq n^-k$ (this is plainly the pigeonhole principle applied to the set of fractional parts $m_1zeta_1 + ldots + m_kzeta_k$ for all positive integers $m_1, ldots, m_k$ with values at most $n$). In your case, $k = 5$.
answered 1 hour ago
metamorphy
1,50911
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
add a comment |Â
up vote
3
down vote
I'm afraid the converse is true. For any $zeta_1, ldots, zeta_k in mathbbR$ and any integer $n > 0$, there exist integers $n_0, ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1zeta_1 + ldots + n_kzeta_k| leq n^-k$ (this is plainly the pigeonhole principle applied to the set of fractional parts $m_1zeta_1 + ldots + m_kzeta_k$ for all positive integers $m_1, ldots, m_k$ with values at most $n$). In your case, $k = 5$.
answered 1 hour ago
metamorphy
1,50911
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I'm afraid the converse is true. For any $zeta_1, ldots, zeta_k in mathbbR$ and any integer $n > 0$, there exist integers $n_0, ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1zeta_1 + ldots + n_kzeta_k| leq n^-k$ (this is plainly the pigeonhole principle applied to the set of fractional parts $m_1zeta_1 + ldots + m_kzeta_k$ for all positive integers $m_1, ldots, m_k$ with values at most $n$). In your case, $k = 5$.
answered 1 hour ago
metamorphy
1,50911
I'm afraid the converse is true. For any $zeta_1, ldots, zeta_k in mathbbR$ and any integer $n > 0$, there exist integers $n_0, ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1zeta_1 + ldots + n_kzeta_k| leq n^-k$ (this is plainly the pigeonhole principle applied to the set of fractional parts $m_1zeta_1 + ldots + m_kzeta_k$ for all positive integers $m_1, ldots, m_k$ with values at most $n$). In your case, $k = 5$.
answered 1 hour ago
metamorphy
1,50911
answered 1 hour ago
metamorphy
1,50911
answered 1 hour ago
metamorphy
1,50911
answered 1 hour ago
metamorphy
1,50911
1,50911
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
add a comment |Â
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
If this happens to be true, it would be unfortunate indeed, since it would prove the converse. Though I am not so convince what you state holds...
– E. Joseph
1 hour ago
add a comment |Â
up vote
3
down vote
I think this is a direct application of a Theorem of Kleinbock and Margulis
https://arxiv.org/pdf/math/9810036.pdf
Let me give a little more detail. Let $f$ be the following map from $mathbbR^4 to mathbbR^5$
$$f(xi_1,xi_2,xi_3,xi_4)=(xi_1xi_4-xi_2xi_3,-xi_4,xi_3,-xi_2,xi_1).$$
It is not hard to check that partial derivatives $(partial f/partial xi_i)_1leq ileq 4$ together with $partial^2 f/partial xi_1partialxi_4$ span $mathbbR^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $xi=(xi_1,..,xi_4)$, $f(xi)$ is not very well approximable, meaning that for all $epsilon>0$, there exist only finitely many integer vectors $q in mathbbZ^5$ such that there exist a $pin mathbbZ$ such that
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) leq 1.$$
Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_epsilon>0$ such that for all $q in mathbbZ^5$ and $pin mathbbZ$,
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) geq c_epsilon.$$
Since $langle q, f(xi) rangle + p=L_xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.
answered 1 hour ago
user120527
1,142211
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
add a comment |Â
up vote
3
down vote
I think this is a direct application of a Theorem of Kleinbock and Margulis
https://arxiv.org/pdf/math/9810036.pdf
Let me give a little more detail. Let $f$ be the following map from $mathbbR^4 to mathbbR^5$
$$f(xi_1,xi_2,xi_3,xi_4)=(xi_1xi_4-xi_2xi_3,-xi_4,xi_3,-xi_2,xi_1).$$
It is not hard to check that partial derivatives $(partial f/partial xi_i)_1leq ileq 4$ together with $partial^2 f/partial xi_1partialxi_4$ span $mathbbR^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $xi=(xi_1,..,xi_4)$, $f(xi)$ is not very well approximable, meaning that for all $epsilon>0$, there exist only finitely many integer vectors $q in mathbbZ^5$ such that there exist a $pin mathbbZ$ such that
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) leq 1.$$
Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_epsilon>0$ such that for all $q in mathbbZ^5$ and $pin mathbbZ$,
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) geq c_epsilon.$$
Since $langle q, f(xi) rangle + p=L_xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.
answered 1 hour ago
user120527
1,142211
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I think this is a direct application of a Theorem of Kleinbock and Margulis
https://arxiv.org/pdf/math/9810036.pdf
Let me give a little more detail. Let $f$ be the following map from $mathbbR^4 to mathbbR^5$
$$f(xi_1,xi_2,xi_3,xi_4)=(xi_1xi_4-xi_2xi_3,-xi_4,xi_3,-xi_2,xi_1).$$
It is not hard to check that partial derivatives $(partial f/partial xi_i)_1leq ileq 4$ together with $partial^2 f/partial xi_1partialxi_4$ span $mathbbR^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $xi=(xi_1,..,xi_4)$, $f(xi)$ is not very well approximable, meaning that for all $epsilon>0$, there exist only finitely many integer vectors $q in mathbbZ^5$ such that there exist a $pin mathbbZ$ such that
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) leq 1.$$
Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_epsilon>0$ such that for all $q in mathbbZ^5$ and $pin mathbbZ$,
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) geq c_epsilon.$$
Since $langle q, f(xi) rangle + p=L_xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.
answered 1 hour ago
user120527
1,142211
I think this is a direct application of a Theorem of Kleinbock and Margulis
https://arxiv.org/pdf/math/9810036.pdf
Let me give a little more detail. Let $f$ be the following map from $mathbbR^4 to mathbbR^5$
$$f(xi_1,xi_2,xi_3,xi_4)=(xi_1xi_4-xi_2xi_3,-xi_4,xi_3,-xi_2,xi_1).$$
It is not hard to check that partial derivatives $(partial f/partial xi_i)_1leq ileq 4$ together with $partial^2 f/partial xi_1partialxi_4$ span $mathbbR^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $xi=(xi_1,..,xi_4)$, $f(xi)$ is not very well approximable, meaning that for all $epsilon>0$, there exist only finitely many integer vectors $q in mathbbZ^5$ such that there exist a $pin mathbbZ$ such that
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) leq 1.$$
Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_epsilon>0$ such that for all $q in mathbbZ^5$ and $pin mathbbZ$,
$$|langle q, f(xi) rangle + p|. |q|^5(1+epsilon) geq c_epsilon.$$
Since $langle q, f(xi) rangle + p=L_xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.
answered 1 hour ago
user120527
1,142211
answered 1 hour ago
user120527
1,142211
answered 1 hour ago
user120527
1,142211
answered 1 hour ago
user120527
1,142211
1,142211
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
add a comment |Â
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things).
– E. Joseph
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
As metamorphy pointed out, 5 is not enough, but $5+epsilon$ will do
– user120527
1 hour ago
add a comment |Â
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Wouldn’t your assumptions imply that the property holds therefore for all $eta in mathbbR^6backslash 0$? In this case, I guess the statement is false — taking a $mathbbR^6$ vector orthogonal to the one defined by $(xi_1xi_4-xi_2 xi_3, -xi_4,xi_3,-xi_2,xi_1,-1)$ would do the job
– João Ramos
3 hours ago
@JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $etainmathbb R^6setminus0$. Is this the case juste because $L_xi$ and $VertcdotVert$ are continuous?
– E. Joseph
2 hours ago
I would say so... as both the norm and the functional are continuous - and the constants bounding $L_xi$ from below are only dependent on $xi$ -, one can take limits to a general $eta$.
– João Ramos
2 hours ago
2
I think that, instead of the norm (in $c/||eta||^gamma$), there must be something involving denominators of $eta$. Otherwise, replacing $eta$ with $eta/N$ (with large natural $N$) leads to absurd.
– metamorphy
2 hours ago
Like metamorphy states, there’s got to be something that measures rationality of $eta$, otherwise a that much general statement has to be false
– João Ramos
2 hours ago