second theorem of Minkowski proof
I am wondering if anyone have a proof of second theorem of Minkowski.
He says that if $vol(K) = 2^{n} det(L)$ and $K$ is compact and symmetric and convex then $K$ contains a non zero lattice point. I would appreciate your help.
Thanks!
integer-lattices vector-lattices
asked Dec 10 '18 at 20:55
user99812
62
add a comment |
I am wondering if anyone have a proof of second theorem of Minkowski.
He says that if $vol(K) = 2^{n} det(L)$ and $K$ is compact and symmetric and convex then $K$ contains a non zero lattice point. I would appreciate your help.
Thanks!
integer-lattices vector-lattices
asked Dec 10 '18 at 20:55
user99812
62
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21
add a comment |
I am wondering if anyone have a proof of second theorem of Minkowski.
He says that if $vol(K) = 2^{n} det(L)$ and $K$ is compact and symmetric and convex then $K$ contains a non zero lattice point. I would appreciate your help.
Thanks!
integer-lattices vector-lattices
asked Dec 10 '18 at 20:55
user99812
62
I am wondering if anyone have a proof of second theorem of Minkowski.
He says that if $vol(K) = 2^{n} det(L)$ and $K$ is compact and symmetric and convex then $K$ contains a non zero lattice point. I would appreciate your help.
Thanks!
integer-lattices vector-lattices
integer-lattices vector-lattices
asked Dec 10 '18 at 20:55
user99812
62
asked Dec 10 '18 at 20:55
user99812
62
edited Dec 10 '18 at 21:08
asked Dec 10 '18 at 20:55
user99812
62
asked Dec 10 '18 at 20:55
user99812
62
asked Dec 10 '18 at 20:55
user99812
62
62
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21
add a comment |
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21
add a comment |
1 Answer
1
active
oldest
votes
Let me take the slightly stronger condition $mathrm{vol}(K) > 2^n det(L)$, rather than $mathrm{vol}(K) = 2^n det(L)$. (You can then use the fact that $K$ is compact to argue that equality actually suffices.) By applying a linear transformation, let's also assume that $L$ is the integer lattice $mathbb{Z}^n$, so that we simply have $mathrm{vol}(K) > 2^n$. (This step isn't necessary at all, but I think it makes the proof easier to understand.)
For vectors $x,x' in mathbb{R}^n$, let's write $x equiv x' mathbb{Z}^n$ if $x - x' in mathbb{Z}^n$. (I.e. $x$ and $x'$ have coordinates with the same fractional part.) Suppose that there exists distinct $x, x' in K/2$ such that $x equiv x' bmod mathbb{Z}^n$. Then, $x - x' in mathbb{Z}^n$ by definition, and $x - x' neq 0$ because $x$ and $x'$ are distinct. However, by symmetry, $-x' in K/2$, and by convexity, $(x-x')/2 in K/2$ so that $x - x'$ is a non-zero lattice vector in $K$.
It therefore remains to find such an $x, x' in K/2$. To that end, we simply notice that for every $x in mathbb{R}^n$, there is a unique representative $f(x) equiv x bmod mathbb{Z}^n$ with $f(x) in [0,1)^n$. Notice that the map $x$ is a piecewise combination of volume-preserving maps. So, if $f$ is an injective mapping when restricted to $K/2$, then we must have
$$mathrm{vol}(f(K/2)) = mathrm{vol}(K/2) = 2^{-n}mathrm{vol}(K) > 1
; .
$$
But, this can't be true, since the entire image of $f$ is $[0,1)^n$, which has volume $1$. Therefore, $f$ is not injective over $K/2$. I.e., there exist distinct $x,x' in K/2$ such that $f(x) = f(x')$, as needed.
Edit: For what it's worth, this is typically referred to as Minkowski's first theorem. His second theorem finds a set of linearly independent vectors satisfying a certain constraint.
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034453%2fsecond-theorem-of-minkowski-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let me take the slightly stronger condition $mathrm{vol}(K) > 2^n det(L)$, rather than $mathrm{vol}(K) = 2^n det(L)$. (You can then use the fact that $K$ is compact to argue that equality actually suffices.) By applying a linear transformation, let's also assume that $L$ is the integer lattice $mathbb{Z}^n$, so that we simply have $mathrm{vol}(K) > 2^n$. (This step isn't necessary at all, but I think it makes the proof easier to understand.)
For vectors $x,x' in mathbb{R}^n$, let's write $x equiv x' mathbb{Z}^n$ if $x - x' in mathbb{Z}^n$. (I.e. $x$ and $x'$ have coordinates with the same fractional part.) Suppose that there exists distinct $x, x' in K/2$ such that $x equiv x' bmod mathbb{Z}^n$. Then, $x - x' in mathbb{Z}^n$ by definition, and $x - x' neq 0$ because $x$ and $x'$ are distinct. However, by symmetry, $-x' in K/2$, and by convexity, $(x-x')/2 in K/2$ so that $x - x'$ is a non-zero lattice vector in $K$.
It therefore remains to find such an $x, x' in K/2$. To that end, we simply notice that for every $x in mathbb{R}^n$, there is a unique representative $f(x) equiv x bmod mathbb{Z}^n$ with $f(x) in [0,1)^n$. Notice that the map $x$ is a piecewise combination of volume-preserving maps. So, if $f$ is an injective mapping when restricted to $K/2$, then we must have
$$mathrm{vol}(f(K/2)) = mathrm{vol}(K/2) = 2^{-n}mathrm{vol}(K) > 1
; .
$$
But, this can't be true, since the entire image of $f$ is $[0,1)^n$, which has volume $1$. Therefore, $f$ is not injective over $K/2$. I.e., there exist distinct $x,x' in K/2$ such that $f(x) = f(x')$, as needed.
Edit: For what it's worth, this is typically referred to as Minkowski's first theorem. His second theorem finds a set of linearly independent vectors satisfying a certain constraint.
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
add a comment |
Let me take the slightly stronger condition $mathrm{vol}(K) > 2^n det(L)$, rather than $mathrm{vol}(K) = 2^n det(L)$. (You can then use the fact that $K$ is compact to argue that equality actually suffices.) By applying a linear transformation, let's also assume that $L$ is the integer lattice $mathbb{Z}^n$, so that we simply have $mathrm{vol}(K) > 2^n$. (This step isn't necessary at all, but I think it makes the proof easier to understand.)
For vectors $x,x' in mathbb{R}^n$, let's write $x equiv x' mathbb{Z}^n$ if $x - x' in mathbb{Z}^n$. (I.e. $x$ and $x'$ have coordinates with the same fractional part.) Suppose that there exists distinct $x, x' in K/2$ such that $x equiv x' bmod mathbb{Z}^n$. Then, $x - x' in mathbb{Z}^n$ by definition, and $x - x' neq 0$ because $x$ and $x'$ are distinct. However, by symmetry, $-x' in K/2$, and by convexity, $(x-x')/2 in K/2$ so that $x - x'$ is a non-zero lattice vector in $K$.
It therefore remains to find such an $x, x' in K/2$. To that end, we simply notice that for every $x in mathbb{R}^n$, there is a unique representative $f(x) equiv x bmod mathbb{Z}^n$ with $f(x) in [0,1)^n$. Notice that the map $x$ is a piecewise combination of volume-preserving maps. So, if $f$ is an injective mapping when restricted to $K/2$, then we must have
$$mathrm{vol}(f(K/2)) = mathrm{vol}(K/2) = 2^{-n}mathrm{vol}(K) > 1
; .
$$
But, this can't be true, since the entire image of $f$ is $[0,1)^n$, which has volume $1$. Therefore, $f$ is not injective over $K/2$. I.e., there exist distinct $x,x' in K/2$ such that $f(x) = f(x')$, as needed.
Edit: For what it's worth, this is typically referred to as Minkowski's first theorem. His second theorem finds a set of linearly independent vectors satisfying a certain constraint.
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
add a comment |
Let me take the slightly stronger condition $mathrm{vol}(K) > 2^n det(L)$, rather than $mathrm{vol}(K) = 2^n det(L)$. (You can then use the fact that $K$ is compact to argue that equality actually suffices.) By applying a linear transformation, let's also assume that $L$ is the integer lattice $mathbb{Z}^n$, so that we simply have $mathrm{vol}(K) > 2^n$. (This step isn't necessary at all, but I think it makes the proof easier to understand.)
For vectors $x,x' in mathbb{R}^n$, let's write $x equiv x' mathbb{Z}^n$ if $x - x' in mathbb{Z}^n$. (I.e. $x$ and $x'$ have coordinates with the same fractional part.) Suppose that there exists distinct $x, x' in K/2$ such that $x equiv x' bmod mathbb{Z}^n$. Then, $x - x' in mathbb{Z}^n$ by definition, and $x - x' neq 0$ because $x$ and $x'$ are distinct. However, by symmetry, $-x' in K/2$, and by convexity, $(x-x')/2 in K/2$ so that $x - x'$ is a non-zero lattice vector in $K$.
It therefore remains to find such an $x, x' in K/2$. To that end, we simply notice that for every $x in mathbb{R}^n$, there is a unique representative $f(x) equiv x bmod mathbb{Z}^n$ with $f(x) in [0,1)^n$. Notice that the map $x$ is a piecewise combination of volume-preserving maps. So, if $f$ is an injective mapping when restricted to $K/2$, then we must have
$$mathrm{vol}(f(K/2)) = mathrm{vol}(K/2) = 2^{-n}mathrm{vol}(K) > 1
; .
$$
But, this can't be true, since the entire image of $f$ is $[0,1)^n$, which has volume $1$. Therefore, $f$ is not injective over $K/2$. I.e., there exist distinct $x,x' in K/2$ such that $f(x) = f(x')$, as needed.
Edit: For what it's worth, this is typically referred to as Minkowski's first theorem. His second theorem finds a set of linearly independent vectors satisfying a certain constraint.
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
Let me take the slightly stronger condition $mathrm{vol}(K) > 2^n det(L)$, rather than $mathrm{vol}(K) = 2^n det(L)$. (You can then use the fact that $K$ is compact to argue that equality actually suffices.) By applying a linear transformation, let's also assume that $L$ is the integer lattice $mathbb{Z}^n$, so that we simply have $mathrm{vol}(K) > 2^n$. (This step isn't necessary at all, but I think it makes the proof easier to understand.)
For vectors $x,x' in mathbb{R}^n$, let's write $x equiv x' mathbb{Z}^n$ if $x - x' in mathbb{Z}^n$. (I.e. $x$ and $x'$ have coordinates with the same fractional part.) Suppose that there exists distinct $x, x' in K/2$ such that $x equiv x' bmod mathbb{Z}^n$. Then, $x - x' in mathbb{Z}^n$ by definition, and $x - x' neq 0$ because $x$ and $x'$ are distinct. However, by symmetry, $-x' in K/2$, and by convexity, $(x-x')/2 in K/2$ so that $x - x'$ is a non-zero lattice vector in $K$.
It therefore remains to find such an $x, x' in K/2$. To that end, we simply notice that for every $x in mathbb{R}^n$, there is a unique representative $f(x) equiv x bmod mathbb{Z}^n$ with $f(x) in [0,1)^n$. Notice that the map $x$ is a piecewise combination of volume-preserving maps. So, if $f$ is an injective mapping when restricted to $K/2$, then we must have
$$mathrm{vol}(f(K/2)) = mathrm{vol}(K/2) = 2^{-n}mathrm{vol}(K) > 1
; .
$$
But, this can't be true, since the entire image of $f$ is $[0,1)^n$, which has volume $1$. Therefore, $f$ is not injective over $K/2$. I.e., there exist distinct $x,x' in K/2$ such that $f(x) = f(x')$, as needed.
Edit: For what it's worth, this is typically referred to as Minkowski's first theorem. His second theorem finds a set of linearly independent vectors satisfying a certain constraint.
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-Davidowitz
1636
1636
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034453%2fsecond-theorem-of-minkowski-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I think you need some more hypotheses on $K$.
– Lord Shark the Unknown
Dec 10 '18 at 21:05
yeah edited thanks, do you have an idea?@LordSharktheUnknown
– user99812
Dec 10 '18 at 21:09
And, what is $L$?
– Berci
Dec 10 '18 at 21:19
$L$ is a lattice @Berci
– user99812
Dec 10 '18 at 21:21