Use the theory of characters to derive the following relation for the representations of $SU_2.$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Question 4, chapter III, section 7 in Vinberg “Linear representations of groups. ”Show that the characters of the representations $phi_n$ of $SU(2)$ constitute a complete orthogonal set.A discrepancy in understanding a solution given to me here for a problem of Vinberg section 8.How to find a representation with given order?Sum of squares of dimensions of irreducible characters.Irreducible characters form orthonormal basis of set of class functionsCharacters of permutation representations for $S_4$Reducible modules and representationsDefining induced representation with motivationRepresentations of the generalized quaternionsIndex of subgroup generated by characters induced from $p$-elementary subgroups in the ring of virtual charactersConstruct a basis of the matrix elements in the space $mathbbC[S_3]$A discrepancy in understanding a solution given to me here for a problem of Vinberg section 8.
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Use the theory of characters to derive the following relation for the representations of $SU_2.$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Question 4, chapter III, section 7 in Vinberg “Linear representations of groups. ”Show that the characters of the representations $phi_n$ of $SU(2)$ constitute a complete orthogonal set.A discrepancy in understanding a solution given to me here for a problem of Vinberg section 8.How to find a representation with given order?Sum of squares of dimensions of irreducible characters.Irreducible characters form orthonormal basis of set of class functionsCharacters of permutation representations for $S_4$Reducible modules and representationsDefining induced representation with motivationRepresentations of the generalized quaternionsIndex of subgroup generated by characters induced from $p$-elementary subgroups in the ring of virtual charactersConstruct a basis of the matrix elements in the space $mathbbC[S_3]$A discrepancy in understanding a solution given to me here for a problem of Vinberg section 8.
$begingroup$
The question is given below:
And the hint at the back of the book says:
Establish the corresponding equality for characters.
And this was a question I was helped on it, which establish the relation between characters and $Phi_n$: Question 4, chapter III, section 7 in Vinberg "Linear representations of groups. "
So, I end up with having $$ tr Phi_m trPhi_n = fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1$$ but then what, could anyone help me in establishing the above relation mentioned in the question?
representation-theory lie-groups lie-algebras physics topological-groups
asked Apr 8 at 13:15
SmartSmart
948
$endgroup$
add a comment |
$begingroup$
The question is given below:
And the hint at the back of the book says:
Establish the corresponding equality for characters.
And this was a question I was helped on it, which establish the relation between characters and $Phi_n$: Question 4, chapter III, section 7 in Vinberg "Linear representations of groups. "
So, I end up with having $$ tr Phi_m trPhi_n = fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1$$ but then what, could anyone help me in establishing the above relation mentioned in the question?
representation-theory lie-groups lie-algebras physics topological-groups
asked Apr 8 at 13:15
SmartSmart
948
$endgroup$
2
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago
add a comment |
$begingroup$
The question is given below:
And the hint at the back of the book says:
Establish the corresponding equality for characters.
And this was a question I was helped on it, which establish the relation between characters and $Phi_n$: Question 4, chapter III, section 7 in Vinberg "Linear representations of groups. "
So, I end up with having $$ tr Phi_m trPhi_n = fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1$$ but then what, could anyone help me in establishing the above relation mentioned in the question?
representation-theory lie-groups lie-algebras physics topological-groups
asked Apr 8 at 13:15
SmartSmart
948
$endgroup$
The question is given below:
And the hint at the back of the book says:
Establish the corresponding equality for characters.
And this was a question I was helped on it, which establish the relation between characters and $Phi_n$: Question 4, chapter III, section 7 in Vinberg "Linear representations of groups. "
So, I end up with having $$ tr Phi_m trPhi_n = fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1$$ but then what, could anyone help me in establishing the above relation mentioned in the question?
representation-theory lie-groups lie-algebras physics topological-groups
representation-theory lie-groups lie-algebras physics topological-groups
asked Apr 8 at 13:15
SmartSmart
948
asked Apr 8 at 13:15
SmartSmart
948
asked Apr 8 at 13:15
SmartSmart
948
asked Apr 8 at 13:15
SmartSmart
948
asked Apr 8 at 13:15
SmartSmart
948
948
2
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago
add a comment |
2
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago
2
2
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Expand out one of the terms your expression:
$$fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1 = fracz^m+1 - z^-m-1z-z^-1 (z^n+z^n-2+dots +z^-n)$$
Do the multiplication:
$$= fracz^m+n+1+z^m+n-1+dots+z^m-n+1 - z^-m+n-1 - z^-m+n-3 dots - z^-m-n-1z-z^-1$$
Rearrange and collect pairs of terms from outside to inside:
$$= frac(z^m+n+1- z^-m-n-1) + (z^m+n-1 - z^-m-n+1)+dots+(z^m-n+1 - z^-m+n-1)z-z^-1$$
answered 2 days ago
NateNate
7,17511122
$endgroup$
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Expand out one of the terms your expression:
$$fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1 = fracz^m+1 - z^-m-1z-z^-1 (z^n+z^n-2+dots +z^-n)$$
Do the multiplication:
$$= fracz^m+n+1+z^m+n-1+dots+z^m-n+1 - z^-m+n-1 - z^-m+n-3 dots - z^-m-n-1z-z^-1$$
Rearrange and collect pairs of terms from outside to inside:
$$= frac(z^m+n+1- z^-m-n-1) + (z^m+n-1 - z^-m-n+1)+dots+(z^m-n+1 - z^-m+n-1)z-z^-1$$
answered 2 days ago
NateNate
7,17511122
$endgroup$
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
add a comment |
$begingroup$
Expand out one of the terms your expression:
$$fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1 = fracz^m+1 - z^-m-1z-z^-1 (z^n+z^n-2+dots +z^-n)$$
Do the multiplication:
$$= fracz^m+n+1+z^m+n-1+dots+z^m-n+1 - z^-m+n-1 - z^-m+n-3 dots - z^-m-n-1z-z^-1$$
Rearrange and collect pairs of terms from outside to inside:
$$= frac(z^m+n+1- z^-m-n-1) + (z^m+n-1 - z^-m-n+1)+dots+(z^m-n+1 - z^-m+n-1)z-z^-1$$
answered 2 days ago
NateNate
7,17511122
$endgroup$
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
add a comment |
$begingroup$
Expand out one of the terms your expression:
$$fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1 = fracz^m+1 - z^-m-1z-z^-1 (z^n+z^n-2+dots +z^-n)$$
Do the multiplication:
$$= fracz^m+n+1+z^m+n-1+dots+z^m-n+1 - z^-m+n-1 - z^-m+n-3 dots - z^-m-n-1z-z^-1$$
Rearrange and collect pairs of terms from outside to inside:
$$= frac(z^m+n+1- z^-m-n-1) + (z^m+n-1 - z^-m-n+1)+dots+(z^m-n+1 - z^-m+n-1)z-z^-1$$
answered 2 days ago
NateNate
7,17511122
$endgroup$
Expand out one of the terms your expression:
$$fracz^m+1 - z^-m-1z-z^-1 .fracz^n+1 - z^-n-1z-z^-1 = fracz^m+1 - z^-m-1z-z^-1 (z^n+z^n-2+dots +z^-n)$$
Do the multiplication:
$$= fracz^m+n+1+z^m+n-1+dots+z^m-n+1 - z^-m+n-1 - z^-m+n-3 dots - z^-m-n-1z-z^-1$$
Rearrange and collect pairs of terms from outside to inside:
$$= frac(z^m+n+1- z^-m-n-1) + (z^m+n-1 - z^-m-n+1)+dots+(z^m-n+1 - z^-m+n-1)z-z^-1$$
answered 2 days ago
NateNate
7,17511122
answered 2 days ago
NateNate
7,17511122
answered 2 days ago
NateNate
7,17511122
answered 2 days ago
NateNate
7,17511122
7,17511122
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
add a comment |
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
2
2
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
Since we have unitary representations of compact group it should be possible in a few lines to explain why equality of character guarantees isomorphism of representation
$endgroup$
– reuns
2 days ago
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
$begingroup$
what about the isomorphism?
$endgroup$
– Smart
yesterday
1
1
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
Since $SU(2)$ is a compact Lie group, the theory of characters tells us that equality of characters implies the existence of an isomorphism of representations. This is really the point of looking at characters, it let's us reduce problems about isomorphisms of spaces with compatible group actions to algebraic computations with Laurent polynomials.
$endgroup$
– Nate
yesterday
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
math.stackexchange.com/questions/3177056/… @Nate could you please look at this question (if you do not mind), when you have time?
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
$begingroup$
and also this question (if you do not mind), when you have time please? math.stackexchange.com/questions/3178900/…
$endgroup$
– hopefully
6 hours ago
add a comment |
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2
$begingroup$
$Phi_n$ is the representation on homogeneous polynomials of degree $n$, sending $f(x,y)$ to $f(ax+by,cx+dy)$. The LHS is tensor product of representations, the RHS is direct sum. $sum_j=0^n tr Phi_m-n+2j = ?$
$endgroup$
– reuns
Apr 8 at 13:26
$begingroup$
Could you provide more details please?@reuns why you added $2j$?
$endgroup$
– Smart
Apr 8 at 15:13
$begingroup$
How $phi_n$ sends $f(x,y)$ to $f(ax + by + dy)$?@reuns
$endgroup$
– Smart
2 days ago
$begingroup$
Do I have to use orthogonality relations for characters?@reuns
$endgroup$
– Smart
2 days ago