# Group theory and Physics

1. Feb 1, 2012

### Hymne

Hello!
I´m currently reading 'Groups, Representations and Physics' by H.F. Jones and I have drawn some conclusions that I would like to have confirmed + I have some questions. :)

Conclusions:

1. An albelian group has always only one irrep.

2. The direct sum of two representations can be thought of as a matrix on block diagonal form, in which we have one block to each group.

Questions:
1. Jones says that the number of irreps is equal to the numer of conjugacy classes, r = k. But SO(2) is an abelian group so r = 1, but every element is its own conjugacy class since g q g^(-1) = gg^(-1) q = q, so k = infinity?

2. If conclusion 2 is right, how should i think of the matrixrepresentation for a direct product group, say SO(2)xSO(2)?

Maybe some conclusions and questions will be added in the nearby future but this is it for the moment.

Thanks helping!!

// Hymne

2. Feb 1, 2012

### chiro

Hey Hymne

I'm not sure about the first one but for the second one if you had four entries (2x2) for each representation, then you would have a 4x4 system with your first in the upper left corner and the second in the bottom right if they are completely independent from each other.

If there was some kind of dependency (which there is not in your case) then you would have mixed terms in the other areas.

3. Feb 1, 2012

### jambaugh

Your conclusion 1. is incorrect. For a finite Abelian group the number of irreps. is the order of the group. In the limit as one considers an Abelian Lie group you have a group of infinite order and so an infinite number of distinct irreps. [Edit see: http://ysharifi.wordpress.com/2011/02/08/irreducible-representations-of-finite-abelian-groups/ ]

For example with SO(2) you have each of the following irreps:
$$g(\theta) \to e^{i k \theta}, k \in \mathbb{N}$$
where $g(\theta)$ is the group element corresponding to rotating $\theta$ degrees.

If you allow projective representations then you can let the k be any real number, you have equivalence up to a scalar multiplier (since each element is a scalar).

Note that the equality of number of equivalence classes and irreps. uses the characters to be proved. Here with SO(2) the characters are the representation "matrix". Of course you can use non-trivial matrices to represent but when you diagonalize them in $\mathbb{C}$ you get the one dimensional irreps along the diagional.

Note that all Lie algebras have an infinite series of irreducible representations because their orders are likewise infinite (there is no upper bound on the order of an element).
Instead we speak of fundamental representations from which all others can be constructed as irreducible sub-representations of tensor product representations of the fundamental ones.

4. Feb 1, 2012

### Hymne

Okey, so I understood it correctly. This is however what my intuition told me that the directproduct would look like.

The xy-plane is a directprodukt of R with itself, and here we have the elements as a number pair (x, y) - i.e. we choose one element from each set. Isnt this actually what we are doing in the block matrix representation as well? When there is no dependence we choose one upper block for the first group, and one lower block for the second group...

My thinking comes from the fact that the group of 2x2 matrices can be decomposed into one subgroup of symmetric matrices with 3 dimensions, and one group of anti-symmetric matrices with dimension 1.

So we write $$3 \oplus 1$$. This example of the direct sum does not seem to be geometrically similar to the one of block matrices.

How do I represent the direct product of lets say $$SO(2) \otimes SO(2)$$ (using the field of real numbers)?

5. Feb 1, 2012

### jambaugh

Let me first show you with complex numbers:
$$g(\theta_1, \theta_2) = e^{k_1 i_1\theta_1}e^{i_2\theta_2}$$
The i's being distinct imaginary units and if you are working in a distinctly real representation must be commuting square roots of unity in a matrix algebra. So take for example the 2x2 matrix:
$$\mathbf{J}=\left(\begin{array}{cc} 0 & -1 \\ +1 & 0\end{array}\right)$$
and let...
$$i_1 = \mathbf{J}\otimes \mathbf{1}, i_2 = \mathbf{1}\otimes\mathbf{J}$$
where we can express the tensor product of two 2x2 matrices as a 4x4 matrix.

There are a couple of conventions but one is to look at the left matrix and replace each entry with its value times the right matrix. (or likewise with right and left swapped).

This will give you:
$$i_1 = \left( \begin{array}{cc} \mathbf{0} & -\mathbf{1} \\ +\mathbf{1} & \mathbf{0}\end{array}\right) = \left(\begin{array}{cccc} 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ +1 & 0 & 0 & 0\\ 0 & +1 & 0 & 0\end{array}\right)$$
and
$$i_2 = \left(\begin{array}{cccc} 0 & -1 & 0 & 0\\ +1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\end{array}\right)$$

However in the end (besides how to do this for general cases) we're not getting single irreducible representations. These real matrices still diagonalize in the complex numbers and being as they commute they diagonalize jointly. You'll get the four combinations where your two generators are +/- i, along the diagional.

In the end your irreps for SO(2)xSO(2) are:
$$g(\theta_1,\theta_2) = e^{i(n\theta_1 + m\theta_2)}$$
a 1 dim x 1 dim = 1 dim irrep.

6. Feb 6, 2012

### Hymne

Thanks so much for the answers! You guys are a Redwood tree to hold on to when it gets to windy!

Here comes some more questions that I struck upon:

1) In trying to get an geometric intuition for the group U(1) I wonder how the determinant of e^i\theta can be interpreted in terms of rotations and reflection... :/
For O(n) and SO(n) their relation is obvious.. but U(n) is not disconnected as O(n) right? How do you think about it?

2) Is there a quick way to prove that a innerproduct that is not positivedefinite will yield a non-compact group to keep it invariant?

3) One of the authors Im reading is taking two one-dimensional, non-equivalent representations of U(2) as an example:

U_1(1) = diag( e^i\theta, e^i\theta ) = e^i\theta I and U_2(1) = diag( e^i\phi, e^-i\phi ).

First I thought this was rotation without reflection and rotation with one reflection but the determinants will be e^i2\theta and 1...

So U_2(1) takes all elements i U(2) and force them together into SU(2) (maybe?).
But what does U_1(2) represent?

Can this be related to the lefthanded and righthanded spinors and how we build the Dirac spinor? :)

Interesting questions I think! Dont feel the need to answer them all, Im thankful for all possible help I can get.

All the best!
/Hymne

7. Feb 7, 2012

### jambaugh

Many coincidence happen in low dimensions i.e. $sp(2)\simeq so(3) \simeq su(2)$. So beware of generalizing form small dimensional cases.

In trying to get an intuitive handle on U(n) you might begin with an embedding such as U(n)$\subset$ SO(2n). Treat an n-dimensional complex vector space as a 2n dimensional real space and consider rotations. Likewise consider the embedding U(n)$\subset$ Sp(2n).

(You can understand Sp(2n) as canonical transformations of n P's and n Q's, canonical momenta and coordinates, generated by the quadratic functions $Q_iQ_j, P^iP^j,P^iQ_j$, via Poisson brackets. While not compact on its face, one can complexify and then select the compact real subalgebra. Imagine O(n) rotations, plus evolutions of harmonic oscillator systems which rotate in phase space and mixtures of such.)

I think of U(n) as the intersection of SO(2n) and Sp(2n) groups of transformations on $\mathbb{R}^{2n}\simeq \mathbb{C}^n$. The real and imaginary part of the preserved Hermitian form of Hilbert space are respectively the orthogonal and symplectic metrics.

I find further insight by not restricting myself to the compact cases. For example one can embed GL(n) in SO(n,n). Consider the vectors and dual vectors of GL(n) as forming a single n+n dimensional indefinite orthogonal space each a set of null and dual null vectors. (GL(n) vectors are elements of a totally null subspace, a subspace of SO(n,n) vectors on which the metric reduces to the zero form. The SO(n,n) metric is the "evaluation form" of dual vector applied to vector to yield scalar.)

This in turn relates to Fermi-Dirac quantification ("2nd quantization") in how the GL(n) generators are mapped to fermionic creator annihilator products which are in turn generators of the Clifford algebra for O(n,n).

Similarly GL(n,C) embeds in Sp(2n,C) as the group generated by quadratic creator annihilator products in Bose-Einstein quantification. (I complexified here just to avoid having to digress further about which symplectic group I'm defining. The pre-complex embedding also works with the non-compact Sp(2n,R) group.)

The quantification procedure is to map the GL(n,C) generator $|i\rangle\otimes \langle j|$ to the operator $\mathbf{a}^i \mathbf{a}^\dagger_j$.
When one considers the fact that the set: $\{\mathbf{a}^i\mathbf{a}^j, \mathbf{a}^i\mathbf{a}^\dagger_j, \mathbf{a}^{\dagger}_i\mathbf{a}^\dagger_j\}$ closes under commutators, they generate a Lie group which will be SO(n,n) in the case of fermionic creators/annihilators and will be Sp(2n,R) in the bosonic case.

In this setting one sees the orthogonal or symplectic groups respectively as the groups preserving the creator annihilator anti-commutation/commutation relations which are accordingly symmetric/anti-symmetric forms.

There are other ways to to embed and relate as well and I don't think any single one provides a final picture. One's best bet is to consider the meaning of the groups in question within the physical setting and how they relate. The relevant groups are transformations preserving certain structures and the type of group relates to the form of the structure. The meaning of the structure comes from the application, e.g. the quantum unitary groups preserve the inner product which establishes the transition amplitudes of the physical systems.