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For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

A Lie algebra representation is a linear map $\rho :\mathfrak{g}\to \mathrm{End}(V)$ preserving brackets: $\rho ([X,Y])=[\rho (X),\rho (Y)]$. Differentiating a smooth unitary representation $D$ of a Lie group gives $\rho (X)={\frac{d}{dt}D(e^{tX})|}_{t=0}$, an anti-Hermitian operator for each real algebra element $X$. In the physics convention one uses Hermitian generators $t_X=i\rho (X)$ and writes $D(e^{tX})=e^{-it{t}_X}$. Conversely, integration of a Lie-algebra representation gives a representation of the connected simply connected group. A representation of another group with the same algebra additionally has to be trivial on the kernel of its covering map.

For the angular momentum commutation relations, take $J_3$ Hermitian and $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space. A highest-weight vector has $J_3|j,j⟩=j|j,j⟩$ and $J_{+}|j,j⟩=0$. Set $v_{\ell }=(J_{-}{)}^{\ell }|j,j⟩$. Commuting $J_3$ through each lowering operator gives $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. Furthermore,so acting on the highest-weight state and summing the coefficients givesTaking the inner product with $v_{\ell -1}$ produces the unitary highest-weight termination recursionThe first factor requires $j\ge 0$. If $2j$ is not an integer, the first integer $\ell >2j+1$ would make the norm negative, while every preceding norm is positive. ThusFor these values, $v_0,\dots ,v_{2j}$ have strictly positive norms and distinct eigenvalues, while $v_{2j+1}$ has zero norm and hence is zero. They therefore span a space of dimension $2j+1$, invariant under all three generators. Its weights are $j,j-1,\dots ,-j$. This proves integer or half-integer spin and finite dimension $2j+1$. The positive-definite unitary assumption matters: an unrestricted algebraic highest-weight module need not terminate. Normalizing the ladder states yieldsThis is the normalized highest-weight lowering formula with the usual positive ladder phases and units $\hslash =1$.

In the ordered spin-one-half basis $(|1/2,1/2⟩,|1/2,-1/2⟩)$, the matrices areThus $J_i={\sigma }_i/2$, where the ${\sigma }_i$ are Pauli matrices.

Choose the active-rotation convention. The SU(2) lift of an axis-angle rotation isThe Pauli matrix multiplication law shows that conjugating $\mathbf{x}\cdot \mathbf{\sigma }$ by this matrix produces $(R\mathbf{x})\cdot \mathbf{\sigma }$, withThis is the ordinary three-dimensional rotation. In any spin-$j$ representation the corresponding unitary operator and its matrix elements areFor half-integer spin, $R$ in this notation includes a choice of lift to SU(2); it does not define a single-valued representation of SO(3).

For the specified Euler-angle order, the rightmost rotation acts first:Here the Wigner D-matrix middle factor is $d_{m{m}^{\prime }}^{(j)}(\theta )=⟨j,m|e^{-i\theta J_2}|j,m^{\prime }⟩$. It can be calculated explicitly by realizing spin $j$ as the symmetric product of $2j$ spin-one-half factors. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The fundamental rotation sends $|+⟩$ to $c|+⟩+s|-⟩$ and $|-⟩$ to $-s|+⟩+c|-⟩$. Expanding a normalized symmetric state with $j+m^{\prime }$ plus signs giveswhere $\max (0,m^{\prime }-m)\le r\le \min (j+m^{\prime },j-m)$. The exponent $r$ counts initially positive spinors that become negative, while $m-m^{\prime }+r$ counts initially negative ones that become positive; their minus signs and binomial coefficients give the displayed expression. Every factorial argument is an integer even for half-integer spin.

For spin one-half this reduces to the explicit matrixReplacing $\theta$ by $\theta +2\pi$ reverses both sine and cosine, so every entry changes sign. In particular, a pure $2\pi$ rotation gives $D^{(1/2)}=-I$, while a $4\pi$ rotation gives $I$.

Finally, the Adjoint double cover from SU(2) to SO(3) is surjective by the axis-angle construction. Its kernel consists of matrices commuting with all Pauli matrices, hence scalar matrices; within SU(2) these are $\pm I$. ThusTheir Lie algebras are isomorphic, but their global topology and representations differ. In spin $j$, the central element $-I$ acts as $(-1{)}^{2j}I$, so exactly the integer-spin representations descend to ordinary SO(3) representations. This is the descent of an SU(2) representation to SO(3) criterion.

The SU(2) group consists of complex $2\times 2$ matrices $U$ with $U^{†}U=I$ and $\det U=1$. The SO(3) group consists of real $3\times 3$ matrices $R$ with $R^{T}R=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This givesConversely the matrix exponential of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the Lie algebras, with bracket the matrix commutator.

Using the Pauli matrices, put $t_a=-i{\sigma }_a/2$. They form a real basis of the SU(2) Lie algebra. The Pauli matrix commutator identity gives $[t_a,t_b]={ϵ}_{abc}{t}_c$. Define $J_a$ on ${\mathbb{R}}^3$ by $J_{a}v=e_a\times v$. These are a basis of the SO(3) Lie algebra; the vector triple-product identity gives $[J_a,J_b]={ϵ}_{abc}{J}_c$. Thusis a real linear bijection preserving the bracket. This is the SU(2)-SO(3) Lie algebra isomorphism. It is a Lie-algebra isomorphism, not a group isomorphism: the Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$.

The SU(3) group is defined similarly by $U^{†}U=I$ and $\det U=1$ on complex $3\times 3$ matrices. One SU(2) subgroup is $\{\mathrm{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an SO(3) group subgroup, since real orthogonality is also complex unitarity.

Under the defining group action, the group orbit of $e_1$ is exactly the unit sphere in ${\mathbb{C}}^3$:Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The isotropy group of $e_1$ must also preserve its orthogonal complement, and therefore isThis proves the unit sphere orbit of the defining special unitary action, including $S^5\cong SU(3)/SU(2)$.

For the complex quadric, write $z=x+iy$. Its defining relation separates intoThe Hermitian norm is then $z^{†}z=1+2|y{|}^2$, which is not constant on this set. For example, $e_1$ and $\sqrt{2}{e}_1+i{e}_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the SU(3) group preserves that norm, the quadric cannot be a single SU(3) group orbit. In fact it is not even invariant under the whole group: $\mathrm{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne 1$.

The real SO(3) group does preserve the quadric, acting simultaneously on $x$ and $y$. Its group orbits are classified completely by $r=|y|\ge 0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the group orbit $S^2$ and stabilizer subgroup $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the stabilizer subgroup is trivial. Thus the real rotation orbits on a complex unit quadric areIn particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y{|}^2}n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the tangent bundle of $S^2$.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

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 scalar potential can be written asIts vacuum manifold is the sphere ${\varphi }_a{\varphi }_a=v^2$. The Adjoint double cover from SU(2) to SO(3) rotates any nonzero vacuum expectation value to $(0,0,v)$, with $v>0$. A local unitary gauge removes the two angular fluctuations, leaving $\varphi =(0,0,v+\eta )$. This choice holds in a neighbourhood of the nonzero vacuum; a single global rotation cannot align arbitrary position-dependent fields, and topologically nontrivial configurations can obstruct a global unitary gauge.

The generator $t^3$ fixes the vacuum expectation value, while $t^1,t^2$ do not. Thus the Higgs mechanism in this SU(2) gauge theory with an adjoint Higgs field givesThe two angular Goldstone bosons become the longitudinal polarizations of two massive gauge bosons. Define physical fieldsThese labels describe the fields of this model; they do not identify it with the Standard Model. With the stated adjoint covariant derivative,To express all Yang-Mills theory terms in physical fields, introduce the gauge field strengthsThen $F_{\mu \nu }^3=f_{\mu \nu }+ig{Q}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}={\mathcal{W}}_{\mu \nu }^{\pm }$. Dropping the constant vacuum energy, the Lagrangian density isTaking the positive gauge coupling convention $g>0$, the canonically normalized mass terms giveThe residual U(1) gauge symmetry makes $W^{\pm }$ oppositely charged. The Yang-Mills theory terms supply interactions of $A$ with $W^{\pm }$ and four-vector interactions. The Higgs mode has cubic and quartic self-interactions, together with $2g^{2}v\eta W^{+}\cdot W^{-}$ and $g^2{\eta }^2{W}^{+}\cdot W^{-}$ couplings; it is neutral under the surviving U(1) gauge symmetry.

Coupling fermions permits a massless electromagnetic gauge boson and massive charged mediators of the weak interaction; fermion couplings with chirality can produce parity-violating weak charged currents. However, this model has no massive neutral Z boson, and its only charge generator is $t^3$. A fundamental doublet has opposite charges, so it cannot reproduce the observed doublet charge assignments through an independent hypercharge. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_Q$, a complex Higgs doublet, three absorbed Goldstone bosons, a massive Z boson and a weak mixing angle.

Because the bilinear form from part (ii) is invariant, $\rho (SU(2))\subseteq O(3)$. The mapis continuous. Since $SU(2)$ is connected and $\det \rho (I_2)=1$, it is identically one. Hence

To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation withforces $U$ to be diagonal. Commutation withthen forces its two diagonal entries to agree. Thus $U=z{I}_2$. Since $U\in SU(2)$,and therefore

It remains to prove surjectivity. The derived representation at the identity isIf this vanishes for every $A$ in the SU(2) Lie algebra, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d{\rho }_{I_2}$ is injective. Its domain and codomain are both three-dimensional real Lie algebras, so it is an isomorphismThe inverse function theorem now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The special orthogonal group $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, soTogether with the kernel calculation, this is the Adjoint double cover from SU(2) to SO(3).

The Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$. Under SU(2) as the three-sphere, multiplication by $-I$ is the antipodal map, so the quotient is Real projective space ${\mathbb{RP}}^3$. Equivalently, an axis-angle ball of radius $\pi$ has opposite boundary points identified. The fundamental group is ${\mathbb{Z}}_2$.

The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.