second theorem of Minkowski proof












1














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







share|cite|improve this question
























  • 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
















1














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







share|cite|improve this question
























  • 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














1












1








1







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 21:08

























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


















  • 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










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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.






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    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 19:41









        Noah Stephens-Davidowitz

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