User:YohanN7/Representation theory of some important groups

Representation theory of the Euclidean group E(2)
The Euclidean group $E(2)$ in two dimensions is the group of isometries of the Euclidean plane. It is also denoted $ISO(2)$ provided reflections are excluded. The $I$ stands for inhomogeneous, referring to the translational part, and $SO$ stands for special orthogonal, referring to the rotational part. Its elements are called rigid motions. When reflections are included, the group is sometimes denoted $E^{+}(2)$ (but rarely $IO(2)$). The elements are then motions.

Notation
Generic vectors in the plane are written in boldface latin letters $$\mathbf x, \mathbf y, \ldots$$. Constant vectors in the plane use $$\mathbf a, \mathbf b, \ldots$$ or subscripted versions. In matrix notation these are taken as column vectors.

Group multiplication rule
A rigid motion can be written as


 * $$\mathbf x' = g\mathbf x = R\mathbf x + \mathbf b,\quad \mathbf x', \mathbf x, \mathbf b \in \mathbb R^2, R \in \mathrm{SO}(2), g \in \mathrm{E}(2),$$

where the vector is first rotated in the plane and then a translation is added. The group has a standard faithful three-dimensional representation. The idea is to embed $ℝ^{2}$ as the affine plane $z = 1$ in $ℝ^{3}$. Then $x ∈ ℝ^{2}$ is represented by $(x^{T}, 1)^{T} ∈ ℝ^{3}$, and

The representation by the three-dimensional matrix above for $$(-\infty < b_1,b_2 < \infty, 0 \le \theta < 2\pi)$$ is faithful.

The group multiplication rule

$$

follows by inspection of $$, and the inverse operation is then

$$

Lie algebra
The Lie algebra representation is found, using the single generator $$J$$ of $$\mathrm{SO}(2)$$ and using the series representation of the matrix exponential, from the parametric matrix form of $$g \in \mathrm{E}(2)$$ above. The Lie algebra representation in this basis is

Direct computation yields the commutation relations

where $$\varepsilon_{km}$$ is the two-dimensional Levi-Civita symbol with $$\varepsilon_{12} = 1$$.

Subalgebras
Two subalgebras can be identified, that spanned by $$J$$, isomorphic to $$\mathfrak{so}(2)$$, and that spanned by $$P_1$$ and $$P_2$$, here denoted $$\mathfrak t(2) \approx \mathbb R^2.$$ Inspection of $$ shows that $$\mathfrak t(2)$$ is an ideal in $$\mathfrak e(2).$$ It follows that $$\mathfrak e(2)$$ semi-direct sum,


 * $$\mathfrak e(2) = \mathfrak t(2) \oplus_s \mathfrak{so}(2).$$

Correspondingly, $$\mathrm E(2)$$ is a semi-direct product,


 * $$\mathrm E(2) = \mathrm T(2) \rtimes \mathrm{SO}(2),$$

in which $$\mathrm T^2$$ is a normal subgroup. The factor group is


 * $$\mathrm E/\mathrm T \approx \mathrm{SO}(2).$$

The adjoint action of $$\mathrm{SO}(2)$$ on $$\mathrm T(2)$$ is, using $$\mathrm{Ad}_{e^X} = e^{\mathrm{ad}_X}$$


 * $$e^{\theta J}P_ke^{-\theta J} = {R(\theta)^m}_k\mathrm P_m = \cos \theta P_k + \varepsilon_{km} \sin \theta P_m,$$

By the adjoint representation formula (proved here),


 * $$e^{\theta J}P_ke^{-\theta J} = e^{\mathrm{ad}_{\theta J}}P_k.$$

By $$,

Using the series expansion of the exponential map (Lie theory) and grouping terms

Substituting $$ in $$ gives


 * $$\begin{align}e^{\mathrm{ad_{\theta J}}}P_k &= \left(1 + \frac{-\theta^2}{2!} + \frac{\theta^4}{4!} + \cdots\right)P_k + \left(\theta - \frac{\theta^3}

{3!} + \frac{\theta^5}{5!} + \cdots\right)\varepsilon_{km}P_m.\\ \end{align}$$

Recognizing the series expansion of the sine and the cosine, this is


 * $$e^{\mathrm{ad_{\theta J}}}P_k = \cos\theta P_k + \varepsilon_{km}\sin \theta P_m.$$

In matrix notation this becomes with


 * $$R(\theta) = \left(\begin{matrix}\cos \theta&-\sin\theta\\\sin\theta&\cos\theta \end{matrix}\right).$$

in component notation of matrices


 * $$e^{\mathrm{ad_{\theta J}}}P_k = {R(\theta)^m}_{k}P_m,$$

and in pure matrix form, this is


 * $$\mathbf P'^{\mathrm T} = \mathbf P^{\mathrm T}R(\theta).$$

The effect on $$\mathbf b \cdot \mathbf P = b^kP_k$$ is seen to be


 * $$e^{\mathrm{ad_{\theta J}}}b^kP_k = {R(\theta)^m}_{k}b^kP_m = (R(\theta)\mathbf b)^mP_m = R(\theta)\mathbf b \cdot \mathbf P.$$

-

leading to

$$

and hence

$$

Casimir operator
The operator $$P^2 = P_1^2 + P_2^2$$ commutes with all Lie algebra elements since


 * $$[J, P_1^2 + P_2^2] = P_1[J, P_1] + [J, P_1]P_1 + P_2[J, P_2] + [J, P_2]P_2 = 0,$$

where $$ was used in the last step.

When unitary representations are assumed, The $$P_k$$ will be anti-Hermitian, meaning $$P_k^\dagger = -P_k$$, and hence $$P^2$$ will be positive-semidefinite. Its eigenvalues serve to partly classify the unitary representations.

Representation theory from the method of induced representations

 * For each $$m \in \mathbb Z$$ there is a one-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$(0, m)$$, where the first coordinate refers to the eigenvalue of the Casimir operator $$P^2$$, and the second coordinate is a further label referring to the eigenvalue of the Casimir operator $$J$$ of the little group $$\mathrm{SO}(2)$$. The action of the Lie algebra is given by

$$
 * At the group level,

$$
 * is obtained.


 * For each $$p > 0 \in \mathbb Z$$ there is an infinite-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$p$$, the square root of eigenvalue of the Casimir operator. The action of the Lie algebra is given by

$$


 * At the group level,

$$

To derive these results, the standard representation on $$\mathbb R^2$$ is examined for subgroups leaving invariant a vector $$\mathbf p \in \mathbb R^2$$.

Little groups of Euclidean group E(2)
There are only two cases. Either $$(p_1, p_2) = (0, 0)$$ in which case the little group is $$\mathrm{SO}(2)$$, or $$(p_1, p_2) \ne (0, 0)$$ in which case the little group is the trivial group $$1.$$ The basis is chosen such that the the Hermitean representatives the commuting $$P_1, P_2$$ are simultaneously diagonalized. This is called the linear momentum basis.

Nonzero vector: The one-element group
Here the labeling of states $$\left|p_1, p_2, \ldots\right\rangle$$ is introduced. By definition per above, the operators $$\pi(P_i)$$ act by


 * $$\begin{align} \pi(P_1)\left|p_1, p_2, \ldots\right\rangle &= p_1\left|p_1, p_2, \ldots\right\rangle,\\ \pi(P_2)\left|p_1, p_2, \ldots\right\rangle &= p_2\left|p_1, p_2, \ldots\right\rangle.\end{align}$$

The dots indicate possible further labels.

At the group level this is


 * $$e^{-i\mathbf a \cdot \mathbf P}\left|p_1, p_2, \ldots\right\rangle = \left|p_1, p_2, \ldots\right\rangle e^{-i\mathbf a \cdot \mathbf p}.$$

The Lie algebra $$\mathfrak 1$$ of the little group $$1$$ is trivial, $$\mathfrak 1 = \{\emptyset\},$$ and $$1$$ has only one irreducible unitary representation, the trivial one.

To deduce the action of the full group $$\mathrm E(2)$$, the action of $$\mathrm {SO}(2)$$ is examined by examining the effect of $$P_k, k = 1, 2$$ on rotated states. To facilitate notation, write $$(p_1, p_2)$$ as $$\mathbf p.$$


 * $$\begin{align}

P_kR(\theta)\left|\mathbf p, \ldots\right\rangle &= \left[R(\theta)R(\theta)^{-1}\right]P_kR(\theta)\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[R(\theta)^{-1}P_kR(\theta)\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[{R(-\theta)^m}_kP_m\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[{R(-\theta)^1}_kP_1 + {R(-\theta)^2}_kP_2\right]\left|p_1, p_2, \ldots\right\rangle,\\ \end{align}$$

or


 * $$\begin{align}

P_1R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[{R(-\theta)^1}_1P_1 + {R(-\theta)^2}_1P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1\cos \theta -p_2\sin \theta  \right],\\ P_2R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[{R(-\theta)^1}_2P_1 + {R(-\theta)^2}_2P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1\sin \theta + p_2 \cos \theta \right]. \end{align}$$

On infinitesimal form, this is


 * $$\begin{align}

P_1R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \cos \delta\theta - p_2 \sin \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 - p_2 \delta\theta \right],\\ P_2R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \sin \delta\theta + p_2 \cos \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \delta\theta + p_2 \right]. \end{align}$$

Since this is deduced from the postulated behavior of $$P_1$$ and $$P_2$$ on a single vector, and the result are eigenvalues different from the postulated ones for the single vector, the only reasonable conclusion is that $$R(\theta)\left|\mathbf p\right\rangle$$ is a new eigenvector of $$P_1$$ and $$P_2$$ orthogonal to $$\mathbf p$$. Evidently,


 * $$\begin{align}R(\theta)\left|\mathbf p\right\rangle &= \left|R(\theta)\mathbf p\right\rangle\\J\left|\mathbf p\right\rangle &= \left|J\mathbf p\right\rangle.\end{align}$$

Since elements are orthogonal matrices, the norm of $$\left|R(\theta)\mathbf p\right\rangle$$ is the same as the norm of $$\mathbf p.$$ Thus the eigenvalue of the Casimir operator remains the same, and an infinite-dimensional unitary representation of $$\mathrm E(2)$$ is characterized by this eigenvalue.

Zero vector: Rotation group SO(2)
Here the labeling is $$\left|0, 0\, \ldots\right\rangle$$ is introduced. The first two zeros refer to the eigenvalues of the $$P_i$$. They act by definition according to


 * $$\begin{align}P_1\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset,\\ P_2\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset.\end{align}$$

It remains to work out how the little group acts. If the representation is to be irreducible, it must be one-dimensional, since only one-dimensional irreducible representations of $$\mathbf{so}(2)$$ exist. In these representations, labeled by $$m \in \mathbb Z$$ the generator $$J$$ of $$\mathbf{so}(2)$$ acts by


 * $$J|0, 0, ...\rangle = |0, 0, \cdots\rangle m.$$

This suggests the labeling $$\left|0, 0, m\right\rangle, m \in \mathbb Z$$ for the basis vector.

At the group level, one obtains


 * $$\begin{align}e^{-i\mathbf a \cdot \mathbf P}|0, 0, m\rangle &= |0, 0, m\rangle,\\ e^{-i\theta J}|0, 0, m\rangle &= e^{-im\theta}|0, 0, m\rangle,\end{align}$$

As it happens, the actions of the abelian subgroup $$\mathrm T(2)$$ and of the little group $$\mathrm{SO}(2)$$ described so far exhausts the action of all of $$\mathrm E(2)$$.