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Start with the classification of finite-dimensional representations of SU2. Its irreducible complex group representations are the spin-$j$ spacesThe central element $-I$ acts on $V_j$ as $(-1{)}^{2j}$. These facts follow also by realizing $V_j$ as homogeneous polynomials of degree $2j$ in two variables: the raising and lowering operators connect all of their one-dimensional weight spaces, and the highest-weight classification supplies every irreducible.

Identify Euclidean four-space with the quaternions. The unit quaternions, each a copy of $SU(2)$, act byThe norm is multiplicative, so this is an orthogonal action. It preserves orientation because the acting group is connected. If it fixes every $q$, setting $q=1$ first gives $a=b$, and then this quaternion must commute with every quaternion; a real unit quaternion is $\pm 1$. The kernel is therefore $\{(1,1),(-1,-1)\}$.

For completeness, the differential is injective: if imaginary quaternions $u,v$ satisfy $uq-qv=0$ for every $q$, then $u=v$ is central and imaginary, hence zero. Both Lie algebras have dimension six. Thus the image contains a neighbourhood of the identity and is an open subgroup of the connected SO(4) group, so it is the whole group. This proves the Spin(4) double coverThe covering group is simply connected since each $SU(2)$ is a three-sphere.

The irreducible representations of a product of compact groups are tensor products of irreducibles of its two factors. One way to see this is to decompose an irreducible space into isotypic components for the first factor; the second commutes with the first, so only one isotypic component can occur. The multiplicity space must then be irreducible for the second factor. Hence the covering-group irreducibles are $V_{j_L}\otimes V_{j_R}$. By central parity on SU2 tensor products, the kernel element acts as $(-1{)}^{2j_L+2j_R}$. The group representation descends to $SO(4)$ precisely when that sign is positive. The representations of SO(4) from two SU2 spins are thereforeEvery finite-dimensional irreducible complex group representation of $SO(4)$ is obtained this way. Since the group is compact, these are also all its continuous irreducible unitary group representations, up to equivalence.

The Lie-algebra version makes the two spin labels visible locally. Choose rotation generators $J_i$ and generators $K_i$ mixing the fourth direction with the first three, normalized so thatThen $A_i=(J_i+K_i)/2$ and $B_i=(J_i-K_i)/2$ obey two commuting copies of $\mathfrak{su}(2)$. The quaternion quotient determines which Lie algebra representations integrate to the actual group, rather than only to its cover.

For example, $(0,0)$ is the scalar, $(1/2,1/2)$ is the four-vector, and $(1,0)$ and $(0,1)$ are the three-dimensional self-dual and anti-self-dual two-form group representations. The half-spin spaces $(1/2,0)$ and $(0,1/2)$ belong to the cover and do not descend to $SO(4)$. Restricting to rotations fixing the real quaternion axis gives the diagonal $SU(2)$, and the Clebsch-Gordan decomposition for SU2 yieldswith steps of one. For a descended group representation these diagonal spins are integers, as required for the spatial $SO(3)$ subgroup.

The special unitary group in two dimensions isLet $V_1={\mathbb{C}}^2$ be its defining group representation, and for $n\ge 0$ letthe homogeneous polynomial representation of SU2 on homogeneous polynomials of degree $n$ in two variables. It has degree of a representation $n+1$. The representations $V_0,V_1,V_2,\dots$ are pairwise nonisomorphic, continuous, finite-dimensional irreducible representations, and the classification of finite-dimensional representations of SU2 says that this is the complete list.

Every $A\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. The monomial basis of $V_n$ gives the character of the homogeneous polynomial representation of SU2Writing $z=e^{i\theta }$, this is $\sin ((n+1)\theta )/\sin \theta$, interpreted by continuity when $\sin \theta =0$. The Clebsch-Gordan decomposition for SU2 is

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y{)}^T$, defineSubstitution preserves degree, andso ${\rho }_n:SU(2)\to \mathrm{GL}(V_n)$ is a continuous representation of a topological group. This is the homogeneous polynomial representation of SU2.

To prove irreducibility, restrict to the diagonal circleThe monomials $x^{n-k}{y}^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operatorsRepeated applications of $E$ and $F$ connect every monomial $x^{n-k}{y}^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the character of the homogeneous polynomial representation of SU2For $z=e^{i\theta }$ this iswith the values at $\sin \theta =0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the ${\mathfrak{sl}}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weightsTheir span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the classification of finite-dimensional representations of SU2 says

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then the eigenvalues on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the character of an exterior square isThe exterior square of an SU2 irreducible representation or the Clebsch-Gordan decomposition for SU2 with flip parity givesThe dimensions $7+3=10=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ check the result.

The third elementary symmetric polynomial in the ${\lambda }_i$, together with Newton identities, givesFor the five-dimensional representation $V_4$, the canonical dualityfinishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently

The classification of finite-dimensional representations of SU2 says that every irreducible is one of the $V_n$, and compactness gives complete reducibility. The standard alternating formis $S{U}_2$-invariant and identifies $V_1$ with $V_1^{\ast }$. Taking symmetric powers gives $V_n\cong V_n^{\ast }$ for every $n$. Therefore, ifthenThis proves self-duality of finite-dimensional SU2 representations.