Clebsch--Gordan decomposition for $mathrmSU(2)$, in indices

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Let $pi_m$, $m geq 0$, be the unitary irreps of $mathrmSU(2)$. The Clebsch--Gordan decomposition then gives that
$$ pi_m otimes pi_n = bigoplus_k=0^min(m,n)pi_m+n-2k.$$
But suppose I want to think of this decomposition as matrices. Evaluating at a point $x in mathrmSU(2)$, on the left I have
$$ (pi_m(x))_ij (pi_n(x))_pq.$$
How do the indices $i,j$ and $p,q$ correspond to the indices on the big matrix on the right?



Some motivation for the question. Expressions involving products of Fourier coefficients arise naturally in nonlinear harmonic analysis on $mathrmSU(2)$. In this specific instance I am interested in calculating $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n)pi_n(x)) = hatf(pi_m)^ji hatf(pi_n)^qp pi_m(x)_ij pi_n(x)_pq$ using the Clebsch--Gordan expansion above. Here $hatf(pi) = int_mathrmSU(2) f(x) pi(x^-1) mathrmd x. $




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  • The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
    – LSpice
    Aug 10 at 14:54






  • 1




    Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
    – onamoonlessnight
    Aug 10 at 15:21







  • 1




    The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
    – onamoonlessnight
    Aug 10 at 15:33







  • 1




    In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
    – onamoonlessnight
    Aug 10 at 16:06






  • 1




    Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
    – onamoonlessnight
    Aug 10 at 23:03














up vote
4
down vote

favorite












Let $pi_m$, $m geq 0$, be the unitary irreps of $mathrmSU(2)$. The Clebsch--Gordan decomposition then gives that
$$ pi_m otimes pi_n = bigoplus_k=0^min(m,n)pi_m+n-2k.$$
But suppose I want to think of this decomposition as matrices. Evaluating at a point $x in mathrmSU(2)$, on the left I have
$$ (pi_m(x))_ij (pi_n(x))_pq.$$
How do the indices $i,j$ and $p,q$ correspond to the indices on the big matrix on the right?



Some motivation for the question. Expressions involving products of Fourier coefficients arise naturally in nonlinear harmonic analysis on $mathrmSU(2)$. In this specific instance I am interested in calculating $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n)pi_n(x)) = hatf(pi_m)^ji hatf(pi_n)^qp pi_m(x)_ij pi_n(x)_pq$ using the Clebsch--Gordan expansion above. Here $hatf(pi) = int_mathrmSU(2) f(x) pi(x^-1) mathrmd x. $




matrices representation-theory lie-groups




share|cite|improve this question






















  • The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
    – LSpice
    Aug 10 at 14:54






  • 1




    Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
    – onamoonlessnight
    Aug 10 at 15:21







  • 1




    The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
    – onamoonlessnight
    Aug 10 at 15:33







  • 1




    In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
    – onamoonlessnight
    Aug 10 at 16:06






  • 1




    Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
    – onamoonlessnight
    Aug 10 at 23:03












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Let $pi_m$, $m geq 0$, be the unitary irreps of $mathrmSU(2)$. The Clebsch--Gordan decomposition then gives that
$$ pi_m otimes pi_n = bigoplus_k=0^min(m,n)pi_m+n-2k.$$
But suppose I want to think of this decomposition as matrices. Evaluating at a point $x in mathrmSU(2)$, on the left I have
$$ (pi_m(x))_ij (pi_n(x))_pq.$$
How do the indices $i,j$ and $p,q$ correspond to the indices on the big matrix on the right?



Some motivation for the question. Expressions involving products of Fourier coefficients arise naturally in nonlinear harmonic analysis on $mathrmSU(2)$. In this specific instance I am interested in calculating $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n)pi_n(x)) = hatf(pi_m)^ji hatf(pi_n)^qp pi_m(x)_ij pi_n(x)_pq$ using the Clebsch--Gordan expansion above. Here $hatf(pi) = int_mathrmSU(2) f(x) pi(x^-1) mathrmd x. $




matrices representation-theory lie-groups




share|cite|improve this question














Let $pi_m$, $m geq 0$, be the unitary irreps of $mathrmSU(2)$. The Clebsch--Gordan decomposition then gives that
$$ pi_m otimes pi_n = bigoplus_k=0^min(m,n)pi_m+n-2k.$$
But suppose I want to think of this decomposition as matrices. Evaluating at a point $x in mathrmSU(2)$, on the left I have
$$ (pi_m(x))_ij (pi_n(x))_pq.$$
How do the indices $i,j$ and $p,q$ correspond to the indices on the big matrix on the right?



Some motivation for the question. Expressions involving products of Fourier coefficients arise naturally in nonlinear harmonic analysis on $mathrmSU(2)$. In this specific instance I am interested in calculating $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n)pi_n(x)) = hatf(pi_m)^ji hatf(pi_n)^qp pi_m(x)_ij pi_n(x)_pq$ using the Clebsch--Gordan expansion above. Here $hatf(pi) = int_mathrmSU(2) f(x) pi(x^-1) mathrmd x. $





matrices representation-theory lie-groups





share|cite|improve this question













share|cite|improve this question




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edited Aug 10 at 23:11

























asked Aug 9 at 17:38









onamoonlessnight

692414




692414











  • The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
    – LSpice
    Aug 10 at 14:54






  • 1




    Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
    – onamoonlessnight
    Aug 10 at 15:21







  • 1




    The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
    – onamoonlessnight
    Aug 10 at 15:33







  • 1




    In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
    – onamoonlessnight
    Aug 10 at 16:06






  • 1




    Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
    – onamoonlessnight
    Aug 10 at 23:03
















  • The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
    – LSpice
    Aug 10 at 14:54






  • 1




    Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
    – onamoonlessnight
    Aug 10 at 15:21







  • 1




    The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
    – onamoonlessnight
    Aug 10 at 15:33







  • 1




    In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
    – onamoonlessnight
    Aug 10 at 16:06






  • 1




    Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
    – onamoonlessnight
    Aug 10 at 23:03















The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
– LSpice
Aug 10 at 14:54




The only possible answer is "it depends". The statement is that there is an isomorphism $pi_m otimes pi_n cong bigoplus_k = 0^min(m, n) pi_m + n - 2k$; once you choose bases of the left- and right-hand sides, you'll get a matrix $A$ describing this isomorphism, and then the answer is that $A(pi_m(x))_i j(pi_n(x))_p qA^-1$ is the matrix of the direct-sum representation in terms of your chosen basis for the right-hand side. Do you have a preferred choice of bases?
– LSpice
Aug 10 at 14:54




1




1




Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
– onamoonlessnight
Aug 10 at 15:21





Yes, I suppose that would be it. Ultimately, the reason for my question is because I want to know how to contract the indices $(pi_m(x))_ij (pi_n(x))_pq$ on the left with a tensor $f^ijpq$, so the answer will be basis independent. But I want to do it in terms of the sum on the right!
– onamoonlessnight
Aug 10 at 15:21





1




1




The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
– onamoonlessnight
Aug 10 at 15:33





The problem comes from harmonic analysis on $mathrmSU(2)$, and I am dealing with the product $operatornameTr(hatf(pi_m) pi_m(x)) operatornameTr(hatf(pi_n) pi_n(x))$ for some function $f$ on $mathrmSU(2)$. There is more detail to the problem that leads me to want to use the Clebsch--Gordan decomposition as above as a means to compute $hatf(pi_m)^ji pi_m(x)_ij hatf(pi_n)^qp pi_n(x)_pq$, but it might entirely be that I am missing something obvious!
– onamoonlessnight
Aug 10 at 15:33





1




1




In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
– onamoonlessnight
Aug 10 at 16:06




In my convention $hatf(pi)$ is $int f(g) pi(g^-1) , mathrmd g$, but yes, and I think I agree that $operatornameTr(hatf(pi_m) pi_m(x)) = ( f * operatornameTr(pi_m))(x)$. Does this help? (Ideally I want the end result in terms of $hatf$, not $f$, but if this lends itself nicely to the decomposition then that would be useful too).
– onamoonlessnight
Aug 10 at 16:06




1




1




Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
– onamoonlessnight
Aug 10 at 23:03




Thanks a lot. This has given me a few ideas to try too, and I will edit the question to include the motivation as well.
– onamoonlessnight
Aug 10 at 23:03















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