From: "Jyrki Lahtonen"
Subject: Re: Questions on Lie groups
Date: 15 Jun 1999 06:48:50 GMT
Newsgroups: sci.math,sci.physics
Keywords: representations of SU(5), SO(5)
It's been almost ten years since I worked on representation theory, but
since no one else is picking it up, I give it a shot.
Valentina Rossi wrote in article
<7k2p0e$mhh$1@nnrp1.deja.com>...
>
>
> Could someone explain to me the process by which an
> irreducible 15-dimensional representation of SU(5)
> decomposes to SO(5) after restricting to real
> matrices? I think the 15 dimensional representation
> of SU(5) corresponds to the
>
> OO
>
> Young tableau, so I guess one should consider a
> ij
> symmetric tensor A ), but I'm not sure what to
> do next.
As a general recipe for questions like this I look up the weight
decomposition of the representation in question (did I tell that my
background is in Lie algebra/algebraic
groups rep theory) and then decompose using the weight decompositions of
the irreducible reps of the subalgebra. The weight decompositions of
different
irreducible representations of simple Lie algebras are tabled. (though for
small modules you can just work it out more or less immediately).
A word of warning: I have never familiarized myself with the differencies
between
real and complex Lie algebras. That may cause some errors later on, but I
doubt it.
The 15-dimensional rep of the Lie algebra A_4 is, indeed, the second
symmetric power (denote S(2,V), that 2 should really be an exponent, in
this media that is
hard to accomplish) of the natural 5-dimensional representation V. This 5-d
representation V remains irreducible, when restricted to SO(5) (= B_2 in
Lie algebra lingo). Luckily the symmetric powers of the basic module
decompose very nice in the case of orthogonal groups. Namely we have the
natural injection gotten by multiplying
symmetric tensors with the preserved quadratic form:
f: S(n,V) -> S(n+2,V)
and the quotient is known to be irreducible!
In your case we only need f: S(0,V) -> S(2,V), which simply identifies a
one-dimensional trivial representation inside S(2,V) (spanned by the
preserved quadratic form). By the above result this is complemented by a
14- dimensional irreducible representation (of highest weight 2*lambda_1,
where lambda_1 is the fundamental weight that is also the highest short
root). So this 15-d representation splits into
a sum of a 1-d trivial representation and a 14-d representation (BTW I
really hate
the physicists way of naming a representation by its dimension rather than
its
highest weight - they deserve to encounter a group with several
irreducibles of
the same dimension:)
> i
> How does a B tensor transform under the
> j
> representation
> _
> n x n
>
> of SU(n)? Is it correct to write the transformation
> law as
> i' *i' j i
> B = S S B
> j' i j' j
>
I think of the symmetric powers as homogeneous polynomials: Let V have
a basis z1, z2, z3, z4, z5, with the group acting by linear substitutions.
Then S(2,V) is the space of homogeneous quadratic polynomials, i.e. it
has a basis z1^2,z1 z2, z1 z3 ,...,z5^2. The 1-dimensional subspace stable
under SO(5) is of course the span of z1^2+z2^2+...+z5^2.
Hope this helps,
Jyrki Lahtonen, Ph.D.
==============================================================================
From: "Jyrki Lahtonen"
Subject: Re: Questions on Lie groups
Date: 17 Jun 1999 06:56:08 GMT
Newsgroups: sci.math,sci.physics
Valentina Rossi wrote in article
<7k8mfu$qa3$1@nnrp1.deja.com>...
> Thanks for the reply, but unfortunately it went right on
> top of my head! I am a physics student and I dont understand
> much of this mathematical formalism associated with group
> theory. Could you perhaps "translate" your answer in terms
> more accessible to a physicist?
> Thanks,
>
> Valentina
>
Sorry, I did get carried away a tad. I think that in order to have a
general algorithm that finds the decomposition pattern, you do need a
little Lie algebra/ Lie group representation theory. The idea is that
you account for all the highest "weight" spaces and go from there. You
have probably seen this mechanism in action, when adding two angular
momenta/spins together - there you really use representation theory of
SO(3) or SU(2). If you tensor two irreducible reps of SU(2), say those
of dimensions m and n, corresponding to highest weights (here "weight"
is a scalar multiple of "the eigenvalue of the operator L_z") (m-1)
and (n-1), you can immediately find an eigenvector corresponding to a
maximal weight (m-1)+(n-1)=m+n-2 inside the tensor product. This gives
you a decomposition factor of dimension m+n-1. But inside the tensor
product you have two independent vectors belonging to the weight m+n-2
and the first decomposition factor only used up one of them, so you
get another decomposition factor of dimension m+n-3 et cetera.
No doubt you have seen similar things done with representations of
SU(3) (like 3d-rep tensored with its conjugate decomposes into 8+1),
where the weight lattice (combinations of eigenvalues of diagonal
matrices) is now a two-dimensional pattern (such as a hexagon in the
case of octet or a triangle in the case of the 10-d
representation). For SU(5) the weight lattice is a 4-dimensional
pattern (if only I had a piece of 4d paper, I would draw it for
you:). For SO(5) it is again a 2-dimensional pattern. A problem (in
the general setting) is to figure out how the weights in the
4-dimensional pattern (SU(5)) are mapped into weights in the
2-dimensional pattern of the subgroup (SO(5)). While that is not
difficult, I'm not going to spend time working it out now. The upshot
here is that once you know that mapping (and have access to a table
containing the dimensions of the different weight spaces in different
irreducible representations) you do the same thing as when adding two
angular momenta: exhaust the weight spaces beginning from the highest.
The easiest way to construct the 15-dimensional representation you
have in mind is to view it as symmetric tensors of rank 2. Both SU(5)
and SO(5) have a basic 5-d representation. Let z_1,z_2,z_3,z_4 and z_5
be a basis for this representation, so the groups act by linear
substitutions:
g (sum_i a_i z_i) = sum_j (sum_i g_ji a_i) z_j
or on the basis elements
g z_k = sum_j g_jk z_j (*).
Tensoring this 5-dimensional representation with itself gives us a
25-dimensional representation that immediately splits into the sum of
symmetric and antisymmetric tensors (of dimensions 15 and 10
respectively). The symmetric tensors may be viewed as second degree
polynomials in the unknowns z_i, i=1,2,3,4,5. The group action is
gotten then naturally by multiplying (*) with itself:
g (z_k z_l) = sum_j sum_i g_jk g_il z_i z_j.
But SO(5) should preserve the quadratic form Q=sum_k z_k^2. Indeed,
g (Q)= sum_k sum _j sum_i g_jk g_ik z_i z_j
= sum_i sum_j (sum_k g_jk g_ik) z_i z_j
here the sum over k is delta_ij, when g is an orthogonal matrix, so this is
equal to
= Q.
Thus we have a trivial 1-dimensional representation of SO(5) inside.
(Note that this isn't a stable subspace under the bigger group SU(5)!).
It turns out that the remaining 14-dimensional summand is an
irreducible representation of SO(5) (to identify this representation,
see my earlier post). Sorry, I don't have the time to find a nice
basis for this representation in terms of the z_k z_l (k less than or
equal to l). I do have a guess, but hope you can figure it out
yourself.
Hope this helps,
Jyrki
[text reformatted -- djr]
==============================================================================
From: "Jyrki Lahtonen"
Subject: Re: Questions on Lie groups
Date: 17 Jun 1999 06:57:51 GMT
Newsgroups: sci.math,sci.physics
Valentina Rossi wrote in article
<7k8mfu$qa3$1@nnrp1.deja.com>...
> Thanks for the reply, but unfortunately it went right on
> top of my head! I am a physics student and I dont understand
> much of this mathematical formalism associated with group
> theory. Could you perhaps "translate" your answer in terms
> more accessible to a physicist?
> Thanks,
>
> Valentina
>
I forgot to add that Young tableaus only work for SU(n), not SO(n).
The weight space diagrams really are (in a way) the general subsitute
for Young tableaus.
Jyrki