Are there Hermitian Unitary matrices U and V generating $mathbb{Z}/2 ast mathbb{Z}/2$?
Are there involutory unitary matrices U and V such that the group generated by U and V is isomorphic to $mathbb{Z}/2 ast mathbb{Z}/2$? If so, how many such pairs of matrices are there? Is there a known way to classify or at least generate examples of such pairs? Alternatively, can you prove no such pairs exist?
I am mostly interested in the case where U and V are $N times N$ matrices with $N = 2^n$ for positive integers $n$, but would also be interested in any special cases.
Thoughts so far: If U and V are involutions and unitary, then each must be Hermitian also. Thus, each of U and V must be a matrix with all eigenvalues equal to $pm 1$.
Also, since $langle U, V rangle cong mathbb{Z}/2 ast mathbb{Z}/2$ and since U and V are involutions, $UVUV = UVU^{-1}V^{-1} = [U,V] neq I$, so U and V cannot commute, and in fact they also cannot anticommute since that would mean $[U,V]^2 = I$, which would again fail our requirements.
Another line of thinking comes from an answer to this question (Matrices which are both unitary and Hermitian) which states that a matrix U is Hermitian and Unitary if and only if $U = 2P - I$, for some orthogonal projection $P$. I'm not sure if this helps.
Any ideas?
linear-algebra matrices group-theory
asked Nov 29 '18 at 3:21
user334137user334137
515210
add a comment |
Are there involutory unitary matrices U and V such that the group generated by U and V is isomorphic to $mathbb{Z}/2 ast mathbb{Z}/2$? If so, how many such pairs of matrices are there? Is there a known way to classify or at least generate examples of such pairs? Alternatively, can you prove no such pairs exist?
I am mostly interested in the case where U and V are $N times N$ matrices with $N = 2^n$ for positive integers $n$, but would also be interested in any special cases.
Thoughts so far: If U and V are involutions and unitary, then each must be Hermitian also. Thus, each of U and V must be a matrix with all eigenvalues equal to $pm 1$.
Also, since $langle U, V rangle cong mathbb{Z}/2 ast mathbb{Z}/2$ and since U and V are involutions, $UVUV = UVU^{-1}V^{-1} = [U,V] neq I$, so U and V cannot commute, and in fact they also cannot anticommute since that would mean $[U,V]^2 = I$, which would again fail our requirements.
Another line of thinking comes from an answer to this question (Matrices which are both unitary and Hermitian) which states that a matrix U is Hermitian and Unitary if and only if $U = 2P - I$, for some orthogonal projection $P$. I'm not sure if this helps.
Any ideas?
linear-algebra matrices group-theory
asked Nov 29 '18 at 3:21
user334137user334137
515210
5
You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22
add a comment |
Are there involutory unitary matrices U and V such that the group generated by U and V is isomorphic to $mathbb{Z}/2 ast mathbb{Z}/2$? If so, how many such pairs of matrices are there? Is there a known way to classify or at least generate examples of such pairs? Alternatively, can you prove no such pairs exist?
I am mostly interested in the case where U and V are $N times N$ matrices with $N = 2^n$ for positive integers $n$, but would also be interested in any special cases.
Thoughts so far: If U and V are involutions and unitary, then each must be Hermitian also. Thus, each of U and V must be a matrix with all eigenvalues equal to $pm 1$.
Also, since $langle U, V rangle cong mathbb{Z}/2 ast mathbb{Z}/2$ and since U and V are involutions, $UVUV = UVU^{-1}V^{-1} = [U,V] neq I$, so U and V cannot commute, and in fact they also cannot anticommute since that would mean $[U,V]^2 = I$, which would again fail our requirements.
Another line of thinking comes from an answer to this question (Matrices which are both unitary and Hermitian) which states that a matrix U is Hermitian and Unitary if and only if $U = 2P - I$, for some orthogonal projection $P$. I'm not sure if this helps.
Any ideas?
linear-algebra matrices group-theory
asked Nov 29 '18 at 3:21
user334137user334137
515210
Are there involutory unitary matrices U and V such that the group generated by U and V is isomorphic to $mathbb{Z}/2 ast mathbb{Z}/2$? If so, how many such pairs of matrices are there? Is there a known way to classify or at least generate examples of such pairs? Alternatively, can you prove no such pairs exist?
I am mostly interested in the case where U and V are $N times N$ matrices with $N = 2^n$ for positive integers $n$, but would also be interested in any special cases.
Thoughts so far: If U and V are involutions and unitary, then each must be Hermitian also. Thus, each of U and V must be a matrix with all eigenvalues equal to $pm 1$.
Also, since $langle U, V rangle cong mathbb{Z}/2 ast mathbb{Z}/2$ and since U and V are involutions, $UVUV = UVU^{-1}V^{-1} = [U,V] neq I$, so U and V cannot commute, and in fact they also cannot anticommute since that would mean $[U,V]^2 = I$, which would again fail our requirements.
Another line of thinking comes from an answer to this question (Matrices which are both unitary and Hermitian) which states that a matrix U is Hermitian and Unitary if and only if $U = 2P - I$, for some orthogonal projection $P$. I'm not sure if this helps.
Any ideas?
linear-algebra matrices group-theory
linear-algebra matrices group-theory
asked Nov 29 '18 at 3:21
user334137user334137
515210
asked Nov 29 '18 at 3:21
user334137user334137
515210
asked Nov 29 '18 at 3:21
user334137user334137
515210
asked Nov 29 '18 at 3:21
user334137user334137
515210
asked Nov 29 '18 at 3:21
user334137user334137
515210
515210
5
You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22
add a comment |
5
You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22
5
5
You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22
You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22
add a comment |
1 Answer
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Almost all pairs of involutions satisfy this property.
Unitary involutions are in correspondence with subspaces, so the space of pairs $(U,V)$ (or representations of $mathbb Z/2 *mathbb Z/2$) is a disjoint union of products of Grassmannians, of course.
The condition that $U,V$ determine a faithful representation of $mathbb Z/2*mathbb Z/2$ is equivalent to $langle UVrangle cong mathbb Z$. Another way to see this condition is to remove the pairs $(U,V)$ such that $(UV)^n = I$ for each $n$. Since the matrix equation is just a set of polynomial equations in the coefficients of the matrices, for fixed $n$ it defines a subvariety. On the $G_delta$ set away from the countable union of these subvarieties, the condition is true. This is what algebraic geometers call a "very generic property."
We're over an uncountable base field, so a very generic property always holds on a non-empty set. In more analytic language, a subvariety has measure 0, so by subadditivity their countable union also has measure zero. Thus the property holds almost everywhere.
answered Nov 30 '18 at 15:09
BenBen
2,608616
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
add a comment |
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Almost all pairs of involutions satisfy this property.
Unitary involutions are in correspondence with subspaces, so the space of pairs $(U,V)$ (or representations of $mathbb Z/2 *mathbb Z/2$) is a disjoint union of products of Grassmannians, of course.
The condition that $U,V$ determine a faithful representation of $mathbb Z/2*mathbb Z/2$ is equivalent to $langle UVrangle cong mathbb Z$. Another way to see this condition is to remove the pairs $(U,V)$ such that $(UV)^n = I$ for each $n$. Since the matrix equation is just a set of polynomial equations in the coefficients of the matrices, for fixed $n$ it defines a subvariety. On the $G_delta$ set away from the countable union of these subvarieties, the condition is true. This is what algebraic geometers call a "very generic property."
We're over an uncountable base field, so a very generic property always holds on a non-empty set. In more analytic language, a subvariety has measure 0, so by subadditivity their countable union also has measure zero. Thus the property holds almost everywhere.
answered Nov 30 '18 at 15:09
BenBen
2,608616
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
add a comment |
Almost all pairs of involutions satisfy this property.
Unitary involutions are in correspondence with subspaces, so the space of pairs $(U,V)$ (or representations of $mathbb Z/2 *mathbb Z/2$) is a disjoint union of products of Grassmannians, of course.
The condition that $U,V$ determine a faithful representation of $mathbb Z/2*mathbb Z/2$ is equivalent to $langle UVrangle cong mathbb Z$. Another way to see this condition is to remove the pairs $(U,V)$ such that $(UV)^n = I$ for each $n$. Since the matrix equation is just a set of polynomial equations in the coefficients of the matrices, for fixed $n$ it defines a subvariety. On the $G_delta$ set away from the countable union of these subvarieties, the condition is true. This is what algebraic geometers call a "very generic property."
We're over an uncountable base field, so a very generic property always holds on a non-empty set. In more analytic language, a subvariety has measure 0, so by subadditivity their countable union also has measure zero. Thus the property holds almost everywhere.
answered Nov 30 '18 at 15:09
BenBen
2,608616
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
add a comment |
Almost all pairs of involutions satisfy this property.
Unitary involutions are in correspondence with subspaces, so the space of pairs $(U,V)$ (or representations of $mathbb Z/2 *mathbb Z/2$) is a disjoint union of products of Grassmannians, of course.
The condition that $U,V$ determine a faithful representation of $mathbb Z/2*mathbb Z/2$ is equivalent to $langle UVrangle cong mathbb Z$. Another way to see this condition is to remove the pairs $(U,V)$ such that $(UV)^n = I$ for each $n$. Since the matrix equation is just a set of polynomial equations in the coefficients of the matrices, for fixed $n$ it defines a subvariety. On the $G_delta$ set away from the countable union of these subvarieties, the condition is true. This is what algebraic geometers call a "very generic property."
We're over an uncountable base field, so a very generic property always holds on a non-empty set. In more analytic language, a subvariety has measure 0, so by subadditivity their countable union also has measure zero. Thus the property holds almost everywhere.
answered Nov 30 '18 at 15:09
BenBen
2,608616
Almost all pairs of involutions satisfy this property.
Unitary involutions are in correspondence with subspaces, so the space of pairs $(U,V)$ (or representations of $mathbb Z/2 *mathbb Z/2$) is a disjoint union of products of Grassmannians, of course.
The condition that $U,V$ determine a faithful representation of $mathbb Z/2*mathbb Z/2$ is equivalent to $langle UVrangle cong mathbb Z$. Another way to see this condition is to remove the pairs $(U,V)$ such that $(UV)^n = I$ for each $n$. Since the matrix equation is just a set of polynomial equations in the coefficients of the matrices, for fixed $n$ it defines a subvariety. On the $G_delta$ set away from the countable union of these subvarieties, the condition is true. This is what algebraic geometers call a "very generic property."
We're over an uncountable base field, so a very generic property always holds on a non-empty set. In more analytic language, a subvariety has measure 0, so by subadditivity their countable union also has measure zero. Thus the property holds almost everywhere.
answered Nov 30 '18 at 15:09
BenBen
2,608616
edited Nov 30 '18 at 15:32
answered Nov 30 '18 at 15:09
BenBen
2,608616
answered Nov 30 '18 at 15:09
BenBen
2,608616
answered Nov 30 '18 at 15:09
BenBen
2,608616
2,608616
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
add a comment |
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
Are you taking the correspondence between unitary involutions and subspaces to be via the $U = 2P - I$ decomposition I mentioned above?
– user334137
Nov 30 '18 at 15:52
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
@user334137 Yes, $U$ corresponds to the image of $P$.
– Ben
Nov 30 '18 at 15:56
add a comment |
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You could just take $U$ and $V$ to be two real orthogonal reflections about lines through the origin such that the angle between them is not a rational multiple of $pi$. Then $UV$ has infinite order.
– Derek Holt
Nov 29 '18 at 9:22