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The Proper orthochronous Lorentz group is the connected Lorentz group in the question. Its double cover is $SL(2,\mathbb{C})$, viewed as a real Lie group. Identify a spacetime vector with the Hermitian matrixThe action $X\mapsto SX{S}^{†}$ for $S\in SL(2,\mathbb{C})$ preserves this determinant and hence the Minkowski metric. The group is connected, so its image is proper and orthochronous. Its kernel consists of $\pm I$: a matrix in the kernel first preserves $I$, hence is unitary, and then commutes with every Hermitian matrix, so is scalar; determinant one forces the two signs. Matrices in $SU(2)$ generate spatial rotations, and positive Hermitian determinant-one matrices generate boosts. Rotations and boosts generate the connected Lorentz group, so the action is onto. Polar decomposition also gives $SL(2,\mathbb{C})\cong SU(2)\times {\mathbb{R}}^3$ as a manifold, proving that it is simply connected. This establishes the Lorentz spinor double cover.

In an anti-Hermitian rotation-generator convention, the Lorentz algebra brackets areThe negative sign in the last bracket distinguishes boosts from Euclidean four-dimensional rotations. After complexification, setA direct bracket calculation gives $[A_i,A_j]={ϵ}_{ijk}{A}_k$, $[B_i,B_j]={ϵ}_{ijk}{B}_k$ and $[A_i,B_j]=0$. Thus the chiral decomposition of the complex Lorentz algebra isComplexification matters: the real Lorentz algebra is not the compact real algebra $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$.

Each spin-$j$ homogeneous polynomial representation of SU2 extends from $SU(2)$ to $SL(2,\mathbb{C})$ as $D_j(S)={\mathrm{Sym}}^{2j}S$. Its complex-conjugate extension uses $\overline{S}$. The finite-dimensional irreducible complex group representations of the covering group are thereforeThe two separate complexified Lie-algebra factors act irreducibly on the two spin spaces, so their tensor product is irreducible. Conversely an invariant complex subspace for the real group is invariant under its complexified Lie algebra; the highest-weight classification for the two ${\mathfrak{sl}}_2$ factors gives exactly these tensor products. This constructs all finite-dimensional complex Lorentz representations.

Again $-I\in SL(2,\mathbb{C})$ acts by $(-1{)}^{2j_L+2j_R}$. Therefore the irreducible representations of the connected Lorentz group itself, in this finite-dimensional complex category, areIf the sum is a half-integer, the group representation is a group representation of the spin cover, or a projective group representation of the Lorentz group, and is not an ordinary single-valued group representation of the group named in the question.

The scalar $(0,0)$ and four-vector $(1/2,1/2)$ descend. The left and right Weyl spinors, $(1/2,0)$ and $(0,1/2)$, do not. Their direct sum is a Dirac spinor, reducible under the connected group; parity exchanges its two chiral summands. The group representations $(1,0)$ and $(0,1)$ describe the two complex chiral parts of an antisymmetric tensor. On the rotation subgroup, self-duality of $SU(2)$ irreducibles identifies the conjugate spin space with the usual spin space, so the Clebsch-Gordan decomposition for SU2 gives the same spin range $|j_L-j_R|,\dots ,j_L+j_R$ as in part (i).

These are group representations used for fields, and they need not be unitary for a positive-definite inner product. Indeed no nontrivial finite-dimensional group representation of this group is unitary: if it were, its differential would embed the simple real Lorentz algebra into an algebra of skew-Hermitian matrices. The trace form $-\mathrm{tr}(XY)$ would give an invariant positive-definite form $b$ on that algebra. Invariance and the boost brackets would then force $-b(J_3,J_3)=b([K_1,K_2],J_3)=b(K_1,[K_2,J_3])=b(K_1,K_1)$, impossible for positive-definite $b$. Equivalently nontrivial boosts in these polynomial group representations have real exponential rather than phase eigenvalues.

The qualification about dimension is necessary because a noncompact group also has infinite-dimensional unitary group representations. They too can be built using $SU(2)$ spin spaces, but not by a single finite pair. For example, the rotation content of induced Lorentz representations is obtained from normalized induced representations of the upper triangular subgroup of $SL(2,\mathbb{C})$, with $n\in \mathbb{Z}$, $\rho \in \mathbb{R}$ and unitary characters $(a/|a|{)}^n|a{|}^{2i\rho }$ on its diagonal $(a,a^{-1})$, trivial on its unipotent part. In the compact picture this uses functions on $SU(2)$ satisfyingExpanding functions into $SU(2)$ matrix coefficients selects a single right-torus weight from each spin space, giving the rotation contentNormalized induction supplies the boost action, coupling these infinitely many rotation spaces. The central sign is $(-1{)}^n$, so the even-$n$ family descends to the connected Lorentz group and has integer rotation spins. This explains both the finite-dimensional field construction and why a classification of unitary group representations cannot simply be identified with the finite two-spin labels.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

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

WriteThe homogeneous polynomial representation of SU2 is the symmetric-power action determined byand extended multiplicatively to $V_n$.

In the ordered basis $(x^3,x^{2}y,x{y}^2,y^3)$, the matrix for $g$ is

The character of a representation $(V,\rho )$ is ${\chi }_V(g)=\mathrm{tr}\rho (g)$. Every $g\in S{U}_2$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$, and hence the character of the homogeneous polynomial representation of SU2 isFor $z=e^{i\theta }$ this is $\sin ((n+1)\theta )/\sin \theta$, interpreted by continuity when $\sin \theta =0$.