Identifying irreps of $SU(2)$ How does one verify that, the representations of $SU(2)$ corresponding to $j=1/2$ or $j=1$ is irreducible? I think showing the irreducibility (taking the representative matrices into a block-diagonal form is not always trivial.) Is there a theorem which can tell us whether a $SU(2)$-representation is irreducible or not?
 A: Schur's lemma asserts that a representation is irreducible if its commutant is trivial, that is it only contains multiples of the identity.
Let $G$ be your group and let $\pi$ be any representation of $G$ over the representation vector space $V$ (usually a Hilbert space). Then $\pi$ is a group homomorphism between $G$ and $\text{GL}_n(\mathbb C)$. The commutant of the representation is the set
$$\pi(G)' = \{S\in\text{GL}_n(\mathbb C)\ |\ ST = TS\quad\forall T\in \pi(G)\}$$
i.e. all the elements in $\text{GL}_n(\mathbb C)$ that commute with every element in the image of the representation. If $\pi$ is irreducible, by the above mentioned Schur's lemma, this commutant reduces to
$$\pi(G)' = \{\lambda\cdot 1_{\text{GL}_n(\mathbb C)}\ |\ \lambda\in\mathbb C\}$$
which means that the only elements that commute with the image of the representation are just multiples of the identity matrix.
The centre of the representation $\pi$ is the intersection $\pi(G)\cap\pi'(G)$, which contains all the elements of the representation $\pi$ that commute with $\pi$ itself. By Schur's lemma, such operators are scalars, i.e. just numbers. An example of such an element is the Casimir operator, which in the case of the group $SU(2)$ defines the total angular momentum, and takes the form
$$J^2 = j(j+1)\cdot 1_{\text{GL}_n(\mathbb C)}.$$
Since there is no way of linking two irreducible representations that have a different representation of the Casimir operator (i.e. a different value of $j$) by an intertwiner, every value of $j$ gives you a different (i.e. unitarily inequivalent) irreducible representation.
For the second question, the above argument shows that a test for the irreducibility of a representation is just to check that its commutant is trivial.
A: An alternative method to Schur's lemma is to use the orthogonality relations of the characters. Let $\chi_R(g)$ be the trace of the matrix of $g$ in the representation $R$. If you have a finite group $G$, with $|G|=n$, there is an inner product between representations
$$\langle \chi_R,\chi_{R'} \rangle = \frac{1}{n}\sum_{g\in G} \chi_R(g)\chi_{R'}(g)^*$$
Now when $R$ and $R'$ are irreducible, this gives unity if $R=R'$ and zero otherwise. This implies that if $R$ is reducible, $\langle \chi_R,\chi_{R} \rangle $ will give some number larger than 1: the sum of the squares of the number of times each irreducible rep appears in $R$.
Of course, $SU(2)$ is not finite, so this is no good. But it works with a simple modification: replace the sum $\frac{1}{n}\sum_{g\in G}$ with an integral over the group, using the invariant Haar measure, normalised so that the group volume is unity:
$$\langle \chi_R,\chi_{R'} \rangle =\int_{G} \chi_R(g)\chi_{R'}(g)^* dg$$
This works for compact Lie groups $G$. The Haar measure for $SU(2)$ is the usual measure on $S^3$, parametrized be Euler angles, say, and divided by the volume of the sphere to normalize.
So you need to work out the measure on the group, find the traces of the representation, and compute the integral $\int_{G} |\chi_R(g)|^2 dg$. If you get one, then $R$ is irreducible; otherwise, yo will get some integer greater than one.
