second theorem of Minkowski proof
$begingroup$
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
user99812user99812
62
$endgroup$
add a comment |
$begingroup$
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
user99812user99812
62
$endgroup$
$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
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And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21
add a comment |
$begingroup$
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
user99812user99812
62
$endgroup$
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
user99812user99812
62
asked Dec 10 '18 at 20:55
user99812user99812
62
edited Dec 10 '18 at 21:08
user99812
asked Dec 10 '18 at 20:55
user99812user99812
62
asked Dec 10 '18 at 20:55
user99812user99812
62
asked Dec 10 '18 at 20:55
user99812user99812
62
62
$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
$begingroup$
And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21
add a comment |
$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
$begingroup$
And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21
$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
$begingroup$
And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21
add a comment |
1 Answer
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$begingroup$
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-DavidowitzNoah Stephens-Davidowitz
1636
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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-DavidowitzNoah Stephens-Davidowitz
1636
$endgroup$
add a comment |
$begingroup$
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-DavidowitzNoah Stephens-Davidowitz
1636
$endgroup$
add a comment |
$begingroup$
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-DavidowitzNoah Stephens-Davidowitz
1636
$endgroup$
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-DavidowitzNoah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-DavidowitzNoah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-DavidowitzNoah Stephens-Davidowitz
1636
answered Dec 14 '18 at 19:41
Noah Stephens-DavidowitzNoah Stephens-Davidowitz
1636
1636
add a comment |
add a comment |
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$begingroup$
I think you need some more hypotheses on $K$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 21:05
$begingroup$
yeah edited thanks, do you have an idea?@LordSharktheUnknown
$endgroup$
– user99812
Dec 10 '18 at 21:09
$begingroup$
And, what is $L$?
$endgroup$
– Berci
Dec 10 '18 at 21:19
$begingroup$
$L$ is a lattice @Berci
$endgroup$
– user99812
Dec 10 '18 at 21:21