Representation of SU(2)xSU(2) and SU(2)
Ilian Peruhov
1. SU(2)xSU(2)
It is well known that every irreducible representation of two groups
is a Kronecker product of irreducible representations of these two
groups. The "canonical" irreducible representations of SU(2) are
labeled D_J(J - index and 2J = 0,1,..). Lets for example an
irreducible representations of SU(2)xSU(2) is the Kronecker product of
the irreducible representations of J=1 and Jprim=1 (1x1).
On the other hand there is a formulae of Clebsh-Gordon which says that
the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
What is true here???
2. According to the diagrams of Young for the representations of
SU(n) for SU(2) one has m=0,1,....
I dont understand what is the representation of SU(2) with m=0. The
dimension of the space where this representation acts is 1 and
consequently is a number. Then the representation must be
commutative???
Robin Chapman
> I have some problems with the representation of SU(2)xSU(2) and SU(2)
>
> 1. SU(2)xSU(2)
>
> It is well known that every irreducible representation of two groups
You mean "of the direct product of two groups"
> is a Kronecker product of irreducible representations of these two
> groups. The "canonical" irreducible representations of SU(2) are
> labeled D_J(J - index and 2J = 0,1,..). Lets for example an
> irreducible representations of SU(2)xSU(2) is the Kronecker product of
> the irreducible representations of J=1 and Jprim=1 (1x1).
>
> On the other hand there is a formulae of Clebsh-Gordon which says that
> the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
I presume "1" means the tautological representation of SU(2).
The key to your dilemma in is your title -- you have two groups
and the notion of tensor product of representations differs between them.
Let V denote the space of 2 by 2 complex matrices.
One can regard V as a representation of SU(2) x SU(2) by the following
recipe: X.(A, B) = A^{-1} X B where X is in V and A and B are in SU(2).
One can also regard V as a representation of SU(2) by the following
recipe: X.A = A^{-1} X A where X is in V and A is in SU(2).
As a representation for SU(2) x SU(2), V is irreducible -- it has
no nontrivial [SU(2) x SU(2)]-invariant subspaces.
But as a representation for SU(2), V is reducible -- it is the direct
sum of the set of scalar matrices and the set of trace-zero matrices, both
of which are SU(2)-invariant subspaces.
> What is true here???
>
> 2. According to the diagrams of Young for the representations of
> SU(n) for SU(2) one has m=0,1,....
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
I don't follow your terminology here. How can a representation be
"commutative"? This 1-dimensional represntation is trivial.
Trivial representations W of a group G satisfy w.g = w for all w
in W and g in G.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
José Carlos Santos
> I have some problems with the representation of SU(2)xSU(2) and SU(2)
>
> 1. SU(2)xSU(2)
>
> It is well known that every irreducible representation of two groups
> is a Kronecker product of irreducible representations of these two
> groups. The "canonical" irreducible representations of SU(2) are
> labeled D_J(J - index and 2J = 0,1,..). Lets for example an
> irreducible representations of SU(2)xSU(2) is the Kronecker product of
> the irreducible representations of J=1 and Jprim=1 (1x1).
>
> the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
>
> What is true here???
The first statement. The Clebsch-Gordan formulae have to do with
the decomposition of Kronecker products as representations of
SU(2), not as representations of SU(2)xSU(2).
> 2. According to the diagrams of Young for the representations of
> SU(n) for SU(2) one has m=0,1,....
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
There's some confusion here. The representation associated with the
number k has dimension k+1.
Best regards
Jose Carlos Santos
[spelling of Clebsh-Gordon changed to Clebsch-Gordan by the moderators]
Christian Ohn
> I have some problems with the representation of SU(2)xSU(2) and SU(2)
>
> 1. SU(2)xSU(2)
>
> It is well known that every irreducible representation of two groups
> is a Kronecker product of irreducible representations of these two
> groups. The "canonical" irreducible representations of SU(2) are
> labeled D_J(J - index and 2J = 0,1,..). Lets for example an
> irreducible representations of SU(2)xSU(2) is the Kronecker product of
> the irreducible representations of J=1 and Jprim=1 (1x1).
>
> On the other hand there is a formulae of Clebsh-Gordon which says that
> the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
>
> What is true here???
If V,W are irreducible representations of SU(2), then (V tensor W) is indeed an
irreducible representation of SU(2)xSU(2), but its *restriction* to the subgroup
SU(2) sitting diagonally inside SU(2)xSU(2) is not, and decomposes according to
the Clebsch-Gordan rule.
> 2. According to the diagrams of Young for the representations of
> SU(n) for SU(2) one has m=0,1,....
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1
That is correct: actually, in your notation, m=2J, and the space of that
representation has dimension m+1.
> and
> consequently is a number. Then the representation must be
> commutative???
It's actually worse than that: m=0 gives you the *trivial* representation, i.e.
each element of SU(2) acts as the identity.
In a *faithful* representation of SU(2) (i.e. one where distinct elements of
SU(2) always act differently), such commutativity can indeed not occur. For m=0,
the representation just isn't faithful.
--
Christian Ohn
e-mail: christian|dot|ohn|at|univ|hyphen|valenciennes|dot|fr
Squark
> I have some problems with the representation of SU(2)xSU(2) and SU(2)
>
> 1. SU(2)xSU(2)
>
> It is well known that every irreducible representation of two groups
> is a Kronecker product of irreducible representations of these two
> groups.
> On the other hand there is a formulae of Clebsh-Gordon which says that
> the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
>
> What is true here???
In the first case, you had an irrep of the _right_ SU(2) and an irrep
of the _left_ SU(2) and their tensor product is an irrep of the product
of the two copies of SU(2). In the second case you had _the same SU(2)_
acting on both. The second case is obtained from the first if consider
your tensor product as a representation of the diagonal SU(2) subgroup
of SU(2) X SU(2), naturally. As a representation of this subgroup it is
no longer irreducible and splits accordingly.
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
Simply enough, it's the trivial representation. I.e., all elements
of SU(2) go, not only to a number, but to the specific number 1!
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and change the
extension in the obvious way)
Andrew Swann
> I have some problems with the representation of SU(2)xSU(2) and SU(2)
>
> 1. SU(2)xSU(2)
>
> It is well known that every irreducible representation of two groups
> is a Kronecker product of irreducible representations of these two
> groups. The "canonical" irreducible representations of SU(2) are
> labeled D_J(J - index and 2J = 0,1,..). Lets for example an
> irreducible representations of SU(2)xSU(2) is the Kronecker product of
> the irreducible representations of J=1 and Jprim=1 (1x1).
You need to distinguish the factors in your product group, e.g. call it
SU(2)_+ x SU(2)_-, and you need to distinguish the representations of each
factor.
Let V=C^2=D_{1/2} be the standard 2-dimensional representation of SU(2).
Write V_+=D_{1/2} and V_-=D_{(1/2)'} for the standard representations of
SU(2)_+ and SU(2)_- on C^2.
The irreducible representations of SU(2) are the kth symmetric powers S^kV
= C^{k+1}=D_{k/2}. The irreducible representations of SU(2)_+ x SU(2)_-
are the tensor products (S^aV_+) \otimes (S^bV_-) = C^{(a+1)(b+1)} =
D_{a/2}D_{(b/2)'}.
> On the other hand there is a formulae of Clebsh-Gordon which says that
> the Kronecker product 1x1 is reducible and splits like 1x1=0+1+2.
In this notation 1x1=0+1+2 is correct, and we also have 1'x1'=0'+1'+2'. But
1x1' is irreducible. The representations 1 and 1' are of different factors
in your product group and are inequivalent.
> 2. According to the diagrams of Young for the representations of
> SU(n) for SU(2) one has m=0,1,....
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
It is C as the trivial representation (every element of SU(2) acts as the
identity).
I hope this helps
Andrew
- --
Andrew Swann sw...@imada.sdu.dk http://www.imada.sdu.dk/~swann
Department of Mathematics and Computer Science, Tel +45 6550 2354
University of Southern Denmark, Campusvej 55, Dept +45 6550 2387
DK-5230 Odense M, Denmark Fax +45 6593 2691
Alfred Einstead
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
I don't see any problem. Take the Lie algebra
[X,Y] = Z, [Y,Z] = X, [Z,X] = Y
and map X, Y and Z to 0. Obviously the mapping preserves
the Lie brackets, mapping them all to 0. Therefore, it's
commutative.
Alfred Einstead
> I dont understand what is the representation of SU(2) with m=0. The
> dimension of the space where this representation acts is 1 and
> consequently is a number. Then the representation must be
> commutative???
I don't see any problem. Take the Lie algebra
Ilian Peruhov
As Christian Ohn writes this rep is not faithful. Everything would be
nice if one has only this rep. One can ignore it.
But than exactly the Clebsch - Gordan formulae says that if we have a
higher rep of SU(2) it can decompose to m=0 + other rep.
Than are these higher reps faithful?
Or for SU(2)(direct) x SU(2) one may want to use the Kr. pr. of m=0
and an other rep. (j,0) and (0,j)
What about these reps : are they faithful???
And is there a problem when a representation is not faithful?
Ilian Peruhov
1. SU(2)xSU(2) and SU(2) representation
I suppose I've got the idea. Hereafter I will outline my
understanding.
I will be thankful if someone takes a look at this to reaffirm is it
right.
===========================================================================
If one has a rep of SU(2) by say (2x2) matrices
( a b )
( -b* a* )
and wants to receive a new rep using the Kronecker product one has to
take the same matrix again:
( a b ) ( a b )
( ) Kronecker X ( )
( -b* a* ) ( -b* a* )
Whereas if he wants to receive the rep of SU(2) direct X SU(2) one
takes all
( a b ) ( c d )
( ) Kr. X ( )
( -b* a* ) ( -d* c* )
SU(2) x SU(2) has (much more elements) - say n^2, that SU(2) (say n).
In fact the matrix rep of SU(2) by the Kronecker product uses a small
(diagonal) part of the matrices representing SU(2)dir xSU(2); it is
restriction of the rep of SU(2)dir xSU(2) to SU(2).
Then
( a b ) ( a b )
( ) Kr. X ( ) = KS --> (4x4 matrix)
( -b* a* ) ( -b* a* )
can be brought to block diagonal form by A.KS.A^-1 (A is digital 4x4
matrix), whereas for all
( a b ) ( c d )
( ) Kr. X ( )
( -b* a* ) ( -d* c* )
this is impossible, while c is not.eq. to a, and b to d.
I managed to bring KS (by operating (+,-) with the columns and rows of
KS)in the form
(X 0 )
(0 Y )
where X is digit 2, and Y is (3x3), which is in accordance with Clebsh
- Gordon formulae Ω x Ω = 0+1.
I suppose this proves that there is such digit matrix A (independent
from the elements of SU(2)).
I would be grateful if one can describe a procedure for receiving the
matrix A that turns KS in block diagonal form.
Unfortunately my Y is not traceless: this problem arose from Mr. Robin
Chapman reply - in his post he says:
{
But as a representation for SU(2), V (that is the Kronecker product)
is reducible -- it is the direct sum of the set of scalar matrices and
the set of trace-zero matrices, both of which are SU(2)-invariant
subspaces.
}
I thought that only the generators of the rep must be traceless??
Thanks all in advance.
Ilian Peruhov
Robin Chapman
>
> Unfortunately my Y is not traceless: this problem arose from Mr. Robin
> Chapman reply - in his post he says:
> {
> But as a representation for SU(2), V (that is the Kronecker product)
> is reducible -- it is the direct sum of the set of scalar matrices and
> the set of trace-zero matrices, both of which are SU(2)-invariant
> subspaces.
> }
I'm afraid that I have been unable to follow this posting of yours.
If you are having difficulty with my earlier reply, all I can do is
to reiterate it in more detail.
A representation of a group G is a vector space V, together with a map
V x G - > G which I'll denote as (v,g)|-> v.g
with the following properties:
(v+w).g = v.g + w.g,
(a v).g = a(v.g) (where a is a scalar)
and
a.(gh) = (a.g).h.
For example, let G = SU(2), the group of 2 by 2 unitary matrices.
There is a "tautological" representation of G, with
V = C^2 = {(z,w): z,w in C}
and where (z w).A (for A in SU(2)) is simply the ordinary matrix product.
We can create a new representation of SU(2) by tensoring
the tautological representation with itself. I'll spare the details,
but the new representation is isomorphic to that with space W = M_2(C),
the set of 2 by 2 matrices over C, and action
M.A = A^{-1} M A (where M is in W and A is in SU(2)).
Now this representation is reducible. Let W_0 = {M in W: trace(M)=0}.
Then A^{-1} M A is in W_0 whenever M is in W_0. So W_0 is
a three-dimensional subrepresentation of W. Also if M = aI, with a in C,
then onviously A^{-1} (aI) A = aI. Hence C I = {aI: a in C}
is a (trivial) one-dimensional subrepresentation of W.
Now W is the direct sum of W_0 and C I : if M is in W then
we can write M = M_0 + aI for unique M_0 in W_0 and aI in C I,
by taking to be 1/2 trace(M).
> I thought that only the generators of the rep must be traceless??
>
What are "generators of a representation"?
Robin Chapman
> 2. About m=0 rep of SU(2)
>
> As Christian Ohn writes this rep is not faithful. Everything would be
> nice if one has only this rep. One can ignore it.
>
> But than exactly the Clebsch - Gordan formulae says that if we have a
> higher rep of SU(2) it can decompose to m=0 + other rep.
> Than are these higher reps faithful?
All irreducible representations of SU(2) apart from the
trivial 1-dimensional representation are faithful.
> Or for SU(2)(direct) x SU(2) one may want to use the Kr. pr. of m=0
> and an other rep. (j,0) and (0,j)
> What about these reps : are they faithful???
If we have irreducible representations phi and psi of groups
G and H respectively, then the tensor product phi (x) psi is an irreducible
representation of G x H. The kernel of phi (x) psi is
the product of the kernels of phi and of psi. In short, phi (x) psi
is faithfull iff both phi and psi are.
> And is there a problem when a representation is not faithful?
Non-faithful representations are not problematical
José Carlos Santos
> Ilian Peruhov wrote:
>
> > 2. About m=0 rep of SU(2)
> >
> > As Christian Ohn writes this rep is not faithful. Everything would be
> > nice if one has only this rep. One can ignore it.
> >
> > But than exactly the Clebsch - Gordan formulae says that if we have a
> > higher rep of SU(2) it can decompose to m=0 + other rep.
> > Than are these higher reps faithful?
>
> All irreducible representations of SU(2) apart from the
> trivial 1-dimensional representation are faithful.
Not quite. The even-dimensional representations of SU(2) are faithful
but the odd-dimensional ones are not; their kernel is {Id, -Id}.
Ilian Peruhov
(X 0 )
(0 Y )
where X is digit 2, and Y is (3x3), which is in accordance with Clebsh
- Gordon formulae Ω x Ω = 0+1.
This is to be read Gordon formulae 1/2 x 1/2 = 0+1