prove that any central function of $SU_2$ is uniquely determined by its restriction to the following subgroup.Prove that the linear span of some functions coincide with the space of the following functions on the unit circle.$GL_2(mathbb R)$ acting on $hatmathbb R=mathbb Rcup infty$.A difficulty in understanding the solution of #2 section 1 Vinberg.Finding all subspaces invariant under F.Comparing the representation $T otimes T $ in terms of matrices and $T^2$Prove that if the dual representation is irreducible then so is the representation itself.A difficulty in understanding the universal property of modules.Prove that the image of an invariant subspace under a morphism of representations is an invariant subspace.A difficulty in understanding the definition of “Spaces of Matrix Elements.”Prove that every finite group of order larger than 2 has more than two irreducible complex representations.Calculate the characters of the left and right regular representationsof an arbitrary finite group.
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prove that any central function of $SU_2$ is uniquely determined by its restriction to the following subgroup.
Prove that the linear span of some functions coincide with the space of the following functions on the unit circle.$GL_2(mathbb R)$ acting on $hatmathbb R=mathbb Rcup infty$.A difficulty in understanding the solution of #2 section 1 Vinberg.Finding all subspaces invariant under F.Comparing the representation $T otimes T $ in terms of matrices and $T^2$Prove that if the dual representation is irreducible then so is the representation itself.A difficulty in understanding the universal property of modules.Prove that the image of an invariant subspace under a morphism of representations is an invariant subspace.A difficulty in understanding the definition of “Spaces of Matrix Elements.”Prove that every finite group of order larger than 2 has more than two irreducible complex representations.Calculate the characters of the left and right regular representationsof an arbitrary finite group.
$begingroup$
The question is given below:
But I do not know how to solve it, could anyone give me a hint please?
EDIT:
representation-theory lie-groups physics topological-groups locally-compact-groups
asked Mar 28 at 21:26
SmartSmart
717
$endgroup$
|
show 3 more comments
$begingroup$
The question is given below:
But I do not know how to solve it, could anyone give me a hint please?
EDIT:
representation-theory lie-groups physics topological-groups locally-compact-groups
asked Mar 28 at 21:26
SmartSmart
717
$endgroup$
$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
3
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
1
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52
|
show 3 more comments
$begingroup$
The question is given below:
But I do not know how to solve it, could anyone give me a hint please?
EDIT:
representation-theory lie-groups physics topological-groups locally-compact-groups
asked Mar 28 at 21:26
SmartSmart
717
$endgroup$
The question is given below:
But I do not know how to solve it, could anyone give me a hint please?
EDIT:
representation-theory lie-groups physics topological-groups locally-compact-groups
representation-theory lie-groups physics topological-groups locally-compact-groups
asked Mar 28 at 21:26
SmartSmart
717
asked Mar 28 at 21:26
SmartSmart
717
edited Mar 28 at 23:22
Smart
asked Mar 28 at 21:26
SmartSmart
717
asked Mar 28 at 21:26
SmartSmart
717
asked Mar 28 at 21:26
SmartSmart
717
717
$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
3
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
1
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52
|
show 3 more comments
$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
3
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
1
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52
$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
3
3
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
1
1
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
What does it mean to restrict a function: For a function $f:Xrightarrow Y$, the restriction of $f$ to a subset $Usubset X$ is a function $g:Urightarrow Y$ such that $g(u)=f(u)$ for all $uin U$. Extending $g$ to $X$ means that we construct a new function $g':Xrightarrow Y$ such that $g'(u)=g(u)$ for all $uin U$.
Now back to the initial problem:
Remark: Any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$.
Moreover, every matrix of this form is diagonalizable, that is, there exists $Bin SU(2)$ such that $BAB^-1$ is a diagonal matrix.
The $f(A(z))=f(A(z^-1))$ part follows from $A(z^-1)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
Uniqueness: If $f$ is a central function and $Ain SU(2)$, then
$$f(A)=f(BAB^-1)$$
where $Bin SU(2)$ such that $BAB^-1$ is diagonal.
Claim: For any matrix $Cin SU(2)$, $CAC^-1=A(z)$ is diagonal if and only if $C=B$ or $C=B^-1$, where $B$ is from above.
The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^-1$ has eigenvalues $w$ and $overline w$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^-1$ is diagonal if and only if $CAC^-1=A(w)$ or $A(w^-1)$.
But this is equivalent to $C=B$ or $C=B^-1$.
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
$endgroup$
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
1
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
|
show 4 more comments
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1 Answer
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1 Answer
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$begingroup$
What does it mean to restrict a function: For a function $f:Xrightarrow Y$, the restriction of $f$ to a subset $Usubset X$ is a function $g:Urightarrow Y$ such that $g(u)=f(u)$ for all $uin U$. Extending $g$ to $X$ means that we construct a new function $g':Xrightarrow Y$ such that $g'(u)=g(u)$ for all $uin U$.
Now back to the initial problem:
Remark: Any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$.
Moreover, every matrix of this form is diagonalizable, that is, there exists $Bin SU(2)$ such that $BAB^-1$ is a diagonal matrix.
The $f(A(z))=f(A(z^-1))$ part follows from $A(z^-1)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
Uniqueness: If $f$ is a central function and $Ain SU(2)$, then
$$f(A)=f(BAB^-1)$$
where $Bin SU(2)$ such that $BAB^-1$ is diagonal.
Claim: For any matrix $Cin SU(2)$, $CAC^-1=A(z)$ is diagonal if and only if $C=B$ or $C=B^-1$, where $B$ is from above.
The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^-1$ has eigenvalues $w$ and $overline w$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^-1$ is diagonal if and only if $CAC^-1=A(w)$ or $A(w^-1)$.
But this is equivalent to $C=B$ or $C=B^-1$.
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
$endgroup$
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
1
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
|
show 4 more comments
$begingroup$
What does it mean to restrict a function: For a function $f:Xrightarrow Y$, the restriction of $f$ to a subset $Usubset X$ is a function $g:Urightarrow Y$ such that $g(u)=f(u)$ for all $uin U$. Extending $g$ to $X$ means that we construct a new function $g':Xrightarrow Y$ such that $g'(u)=g(u)$ for all $uin U$.
Now back to the initial problem:
Remark: Any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$.
Moreover, every matrix of this form is diagonalizable, that is, there exists $Bin SU(2)$ such that $BAB^-1$ is a diagonal matrix.
The $f(A(z))=f(A(z^-1))$ part follows from $A(z^-1)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
Uniqueness: If $f$ is a central function and $Ain SU(2)$, then
$$f(A)=f(BAB^-1)$$
where $Bin SU(2)$ such that $BAB^-1$ is diagonal.
Claim: For any matrix $Cin SU(2)$, $CAC^-1=A(z)$ is diagonal if and only if $C=B$ or $C=B^-1$, where $B$ is from above.
The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^-1$ has eigenvalues $w$ and $overline w$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^-1$ is diagonal if and only if $CAC^-1=A(w)$ or $A(w^-1)$.
But this is equivalent to $C=B$ or $C=B^-1$.
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
$endgroup$
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
1
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
|
show 4 more comments
$begingroup$
What does it mean to restrict a function: For a function $f:Xrightarrow Y$, the restriction of $f$ to a subset $Usubset X$ is a function $g:Urightarrow Y$ such that $g(u)=f(u)$ for all $uin U$. Extending $g$ to $X$ means that we construct a new function $g':Xrightarrow Y$ such that $g'(u)=g(u)$ for all $uin U$.
Now back to the initial problem:
Remark: Any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$.
Moreover, every matrix of this form is diagonalizable, that is, there exists $Bin SU(2)$ such that $BAB^-1$ is a diagonal matrix.
The $f(A(z))=f(A(z^-1))$ part follows from $A(z^-1)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
Uniqueness: If $f$ is a central function and $Ain SU(2)$, then
$$f(A)=f(BAB^-1)$$
where $Bin SU(2)$ such that $BAB^-1$ is diagonal.
Claim: For any matrix $Cin SU(2)$, $CAC^-1=A(z)$ is diagonal if and only if $C=B$ or $C=B^-1$, where $B$ is from above.
The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^-1$ has eigenvalues $w$ and $overline w$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^-1$ is diagonal if and only if $CAC^-1=A(w)$ or $A(w^-1)$.
But this is equivalent to $C=B$ or $C=B^-1$.
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
$endgroup$
What does it mean to restrict a function: For a function $f:Xrightarrow Y$, the restriction of $f$ to a subset $Usubset X$ is a function $g:Urightarrow Y$ such that $g(u)=f(u)$ for all $uin U$. Extending $g$ to $X$ means that we construct a new function $g':Xrightarrow Y$ such that $g'(u)=g(u)$ for all $uin U$.
Now back to the initial problem:
Remark: Any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$.
Moreover, every matrix of this form is diagonalizable, that is, there exists $Bin SU(2)$ such that $BAB^-1$ is a diagonal matrix.
The $f(A(z))=f(A(z^-1))$ part follows from $A(z^-1)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
Uniqueness: If $f$ is a central function and $Ain SU(2)$, then
$$f(A)=f(BAB^-1)$$
where $Bin SU(2)$ such that $BAB^-1$ is diagonal.
Claim: For any matrix $Cin SU(2)$, $CAC^-1=A(z)$ is diagonal if and only if $C=B$ or $C=B^-1$, where $B$ is from above.
The eigenvalues of a matrix do not change under unitary change of basis, and since $BAB^-1$ has eigenvalues $w$ and $overline w$ (because diagonal), $A$ has the same eigenvalues. Hence $CAC^-1$ is diagonal if and only if $CAC^-1=A(w)$ or $A(w^-1)$.
But this is equivalent to $C=B$ or $C=B^-1$.
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
edited Mar 30 at 0:25
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
answered Mar 28 at 23:30
Pink PantherPink Panther
382113
382113
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
1
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
|
show 4 more comments
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
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– Pink Panther
Mar 28 at 23:47
1
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In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
1
1
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
$begingroup$
@Smart $SO(2)$ is the group of all orthogonal matrices with real entries, i.e. all matrices $Ainmathbb R^2times 2$ such that $AA^top=I$ and $det A=1$. SU(2)$ is the group of unitary matrices with complex, i.e. all matrices $Ainmathbb C^2times 2$ such that $AA^dagger=I$ and $det A=1$, where $I$ denotes the identity matrix and $A^dagger$ denotes the complex conjugate transpose.
$endgroup$
– Pink Panther
Mar 28 at 23:47
1
1
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
$begingroup$
In particular $SO(2)$ is a proper subgroup of $SU(2)$. @Smart
$endgroup$
– Pink Panther
Mar 29 at 11:25
1
1
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
$begingroup$
why $A(z^-1) = A(barz )$?
$endgroup$
– Smart
Mar 29 at 17:39
1
1
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
$begingroup$
@Smart You are right, this is not correct. In my head it was like $|z|^2=zoverline z$ and then something... But it is not relevant for the argument, so i will edit that.
$endgroup$
– Pink Panther
Mar 29 at 19:55
1
1
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
$begingroup$
@Smart no wait, it is correct that if $|z|=1$ then $z^-1=overline z$, because $$z^-1=1/z=fracoverline zzoverline z=fracoverline z^2=overline z$$ And yes, I meant to write $Bin SU(2)$, I guess I made a typo there...
$endgroup$
– Pink Panther
Mar 30 at 0:22
|
show 4 more comments
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$begingroup$
what is your definition for "central function"?
$endgroup$
– Pink Panther
Mar 28 at 22:21
$begingroup$
okay I will edit my question to include the definition.@PinkPanther
$endgroup$
– Smart
Mar 28 at 22:22
3
$begingroup$
Consider the fact that any matrix $Ain SU(2)$ can be written as $beginpmatrixz & -overline w \ w & overline zendpmatrix$ for some $w,zinmathbb C$ with $|w|^2+|z|^2=1$ and that every matrix of this form is diagonalizable, that is, there exists $Bin SO(2)$ such that $BAB^-1$ is a diagonal matrix. The $f(A(z))=f(A(z^-1)$ part will then follow from $A(z^-1)=A(overline z)=BA(z)B^-1$, where $B=beginpmatrix0 & 1 \ -1 & 0endpmatrix$.
$endgroup$
– Pink Panther
Mar 28 at 22:42
$begingroup$
and why the determination is unique? also the first equality in the last line follows from the fact that A is unitary? and why in the second equality there is no baar on z? @PinkPanther
$endgroup$
– Smart
Mar 28 at 22:49
1
$begingroup$
yes, unitary is crucial here. Since $SU(2)$ is a group, $BAB^-1$ is again in $SU(2) $ and any diagonal matrix in $SU(2)$ is of the form $A(z)$ for some $z$ with $|z|=1$.
$endgroup$
– Pink Panther
Mar 28 at 22:52