On the number of cosets of a sublattice in a lattice Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Index of a sublattice in a lattice and a homomorphism between them$mathrmcard(mathbbZ^n/MmathbbZ^n) = |det(M)|$?What is a coset in a lattice and how many they are?Proof: $detpmatrixlangle v_i , v_j rangleneq0$ $iff v_1,dots,v_n~textl.i.$Trace of tensor productsmallest integer contained in sublattice $Rightarrow$ $L'=[q,rtau+s]$Prove that there is a basis of a lattice $Lambda$ s.t. a reflection is of a certain formGeometrically describe these Cosets and form a bijection with the Orbit-Stabilizer relation.Consider $f: mathbb Z_8 rightarrow mathbb Z_4$. List the cosets of the kernal of $f$Finding the matrix representation of a certain functionTrue/false: There are $4$ linearly independent vectors $v_1,v_2,v_3,v_4 in mathbbR^3$proof of orthogonality of the wavefunctions on the lattice-torusHow would I answer the following question about the determinant of a matrix?
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On the number of cosets of a sublattice in a lattice
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Index of a sublattice in a lattice and a homomorphism between them$mathrmcard(mathbbZ^n/MmathbbZ^n) = |det(M)|$?What is a coset in a lattice and how many they are?Proof: $detpmatrixlangle v_i , v_j rangleneq0$ $iff v_1,dots,v_n~textl.i.$Trace of tensor productsmallest integer contained in sublattice $Rightarrow$ $L'=[q,rtau+s]$Prove that there is a basis of a lattice $Lambda$ s.t. a reflection is of a certain formGeometrically describe these Cosets and form a bijection with the Orbit-Stabilizer relation.Consider $f: mathbb Z_8 rightarrow mathbb Z_4$. List the cosets of the kernal of $f$Finding the matrix representation of a certain functionTrue/false: There are $4$ linearly independent vectors $v_1,v_2,v_3,v_4 in mathbbR^3$proof of orthogonality of the wavefunctions on the lattice-torusHow would I answer the following question about the determinant of a matrix?
$begingroup$
In several questions, e.g. 1, 2, it has been asked why the index of a sublattice $MmathbbZ^n$ in $mathbbZ^n$ is equal to $det(M)$, that is $|mathbbZ^n/MmathbbZ^n|=det(M)$. The answer to this question seems to be quite theoretical, and I was wondering if there was a more intuitive approach similar to the explanation in 1 (specifically I have trouble understanding the statements "The cosets of L are precisely (a,b)+L, where (a,b) is an integer point that belongs to B(X). The number of those points is the area of B(X), which is det(B).")
Moreover in my current studies I often encounter the situation of a "general" lattice $AmathbbZ^nsubsetmathbbZ^n$ being quotiented by a sublattice $MmathbbZ^n$ as $AmathbbZ^n/MmathbbZ^n$ and I have reason to believe that $|AmathbbZ^n/MmathbbZ^n|=fracdet(M)det(A)$. I have not been able to find a reference for it, and it might not be true. However as an example we have for
beginalign*
A=(v_1 v_2 v_3 v_4)=beginpmatrix 3 & 1 & 1 & 1\ -1 & 3 & -1 & 1\ -1 & 1 & 3 & -1\ -1 & -1 & 1 & 3 endpmatrix text and M=beginpmatrix 3 & 3 & 3 & 3\ 3 & 3 & -3 & -3\ -3 & 3 & 3 & -3\ -3 & 3 & -3 & 3 endpmatrix
endalign*
that $fracdet(M)det(A)=9$ and one can show that the nontrivial cosets of $M$ are given by $Mpm v_i$, $i=1,2,3,4$.
abstract-algebra matrices group-theory integer-lattices
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
$endgroup$
add a comment |
$begingroup$
In several questions, e.g. 1, 2, it has been asked why the index of a sublattice $MmathbbZ^n$ in $mathbbZ^n$ is equal to $det(M)$, that is $|mathbbZ^n/MmathbbZ^n|=det(M)$. The answer to this question seems to be quite theoretical, and I was wondering if there was a more intuitive approach similar to the explanation in 1 (specifically I have trouble understanding the statements "The cosets of L are precisely (a,b)+L, where (a,b) is an integer point that belongs to B(X). The number of those points is the area of B(X), which is det(B).")
Moreover in my current studies I often encounter the situation of a "general" lattice $AmathbbZ^nsubsetmathbbZ^n$ being quotiented by a sublattice $MmathbbZ^n$ as $AmathbbZ^n/MmathbbZ^n$ and I have reason to believe that $|AmathbbZ^n/MmathbbZ^n|=fracdet(M)det(A)$. I have not been able to find a reference for it, and it might not be true. However as an example we have for
beginalign*
A=(v_1 v_2 v_3 v_4)=beginpmatrix 3 & 1 & 1 & 1\ -1 & 3 & -1 & 1\ -1 & 1 & 3 & -1\ -1 & -1 & 1 & 3 endpmatrix text and M=beginpmatrix 3 & 3 & 3 & 3\ 3 & 3 & -3 & -3\ -3 & 3 & 3 & -3\ -3 & 3 & -3 & 3 endpmatrix
endalign*
that $fracdet(M)det(A)=9$ and one can show that the nontrivial cosets of $M$ are given by $Mpm v_i$, $i=1,2,3,4$.
abstract-algebra matrices group-theory integer-lattices
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
$endgroup$
add a comment |
$begingroup$
In several questions, e.g. 1, 2, it has been asked why the index of a sublattice $MmathbbZ^n$ in $mathbbZ^n$ is equal to $det(M)$, that is $|mathbbZ^n/MmathbbZ^n|=det(M)$. The answer to this question seems to be quite theoretical, and I was wondering if there was a more intuitive approach similar to the explanation in 1 (specifically I have trouble understanding the statements "The cosets of L are precisely (a,b)+L, where (a,b) is an integer point that belongs to B(X). The number of those points is the area of B(X), which is det(B).")
Moreover in my current studies I often encounter the situation of a "general" lattice $AmathbbZ^nsubsetmathbbZ^n$ being quotiented by a sublattice $MmathbbZ^n$ as $AmathbbZ^n/MmathbbZ^n$ and I have reason to believe that $|AmathbbZ^n/MmathbbZ^n|=fracdet(M)det(A)$. I have not been able to find a reference for it, and it might not be true. However as an example we have for
beginalign*
A=(v_1 v_2 v_3 v_4)=beginpmatrix 3 & 1 & 1 & 1\ -1 & 3 & -1 & 1\ -1 & 1 & 3 & -1\ -1 & -1 & 1 & 3 endpmatrix text and M=beginpmatrix 3 & 3 & 3 & 3\ 3 & 3 & -3 & -3\ -3 & 3 & 3 & -3\ -3 & 3 & -3 & 3 endpmatrix
endalign*
that $fracdet(M)det(A)=9$ and one can show that the nontrivial cosets of $M$ are given by $Mpm v_i$, $i=1,2,3,4$.
abstract-algebra matrices group-theory integer-lattices
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
$endgroup$
In several questions, e.g. 1, 2, it has been asked why the index of a sublattice $MmathbbZ^n$ in $mathbbZ^n$ is equal to $det(M)$, that is $|mathbbZ^n/MmathbbZ^n|=det(M)$. The answer to this question seems to be quite theoretical, and I was wondering if there was a more intuitive approach similar to the explanation in 1 (specifically I have trouble understanding the statements "The cosets of L are precisely (a,b)+L, where (a,b) is an integer point that belongs to B(X). The number of those points is the area of B(X), which is det(B).")
Moreover in my current studies I often encounter the situation of a "general" lattice $AmathbbZ^nsubsetmathbbZ^n$ being quotiented by a sublattice $MmathbbZ^n$ as $AmathbbZ^n/MmathbbZ^n$ and I have reason to believe that $|AmathbbZ^n/MmathbbZ^n|=fracdet(M)det(A)$. I have not been able to find a reference for it, and it might not be true. However as an example we have for
beginalign*
A=(v_1 v_2 v_3 v_4)=beginpmatrix 3 & 1 & 1 & 1\ -1 & 3 & -1 & 1\ -1 & 1 & 3 & -1\ -1 & -1 & 1 & 3 endpmatrix text and M=beginpmatrix 3 & 3 & 3 & 3\ 3 & 3 & -3 & -3\ -3 & 3 & 3 & -3\ -3 & 3 & -3 & 3 endpmatrix
endalign*
that $fracdet(M)det(A)=9$ and one can show that the nontrivial cosets of $M$ are given by $Mpm v_i$, $i=1,2,3,4$.
abstract-algebra matrices group-theory integer-lattices
abstract-algebra matrices group-theory integer-lattices
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
asked Apr 2 at 9:27
plebmaticianplebmatician
458212
458212
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
You can apply the second isomorphism theorem for groups to see that you are correct:
$$fracBbb Z / MBbb ZABbb Z / MBbb Z cong fracBbb ZABbb Z.$$
Therefore
$$fracdet M = det A.$$
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
$endgroup$
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can apply the second isomorphism theorem for groups to see that you are correct:
$$fracBbb Z / MBbb ZABbb Z / MBbb Z cong fracBbb ZABbb Z.$$
Therefore
$$fracdet M = det A.$$
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
$endgroup$
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
add a comment |
$begingroup$
You can apply the second isomorphism theorem for groups to see that you are correct:
$$fracBbb Z / MBbb ZABbb Z / MBbb Z cong fracBbb ZABbb Z.$$
Therefore
$$fracdet M = det A.$$
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
$endgroup$
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
add a comment |
$begingroup$
You can apply the second isomorphism theorem for groups to see that you are correct:
$$fracBbb Z / MBbb ZABbb Z / MBbb Z cong fracBbb ZABbb Z.$$
Therefore
$$fracdet M = det A.$$
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
$endgroup$
You can apply the second isomorphism theorem for groups to see that you are correct:
$$fracBbb Z / MBbb ZABbb Z / MBbb Z cong fracBbb ZABbb Z.$$
Therefore
$$fracdet M = det A.$$
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
answered Apr 2 at 9:40
Lukas KoflerLukas Kofler
1,4312520
1,4312520
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
add a comment |
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic?
$endgroup$
– plebmatician
Apr 2 at 12:03
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
$begingroup$
What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach.
$endgroup$
– Lukas Kofler
Apr 2 at 12:32
add a comment |
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