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In global coordinates the de Sitter spacetime metric is $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. A global frame of a vector bundle orthonormal for the round metric $Y_1,Y_2,Y_3$ on $S^3$ gives the displayed global pseudo-orthonormal frame. For example, identify $S^3$ with the unit quaternions and use $qi,qj,qk$. Choosing the spatial orientation and the future direction consistently gives a section of a fiber bundle for the oriented, time-oriented frame bundle. The principal bundle trivialization by a global section then proves that bundle is trivial.

All the special orthogonal groups are connected, and their second homotopy groups vanish. The rotation circle gives $SO(2)\cong S^1$. Conjugation by unit quaternions gives the twofold universal cover $S^3\to SO(3)$, so ${\pi }_{1}SO(3)=\mathbb{Z}/2$ and ${\pi }_{3}SO(3)=\mathbb{Z}$. The special orthogonal sphere fibration $SO(n-1)\to SO(n)\to S^{n-1}$ propagates the first two homotopy groups for $n\ge 4$. In dimension four its exact sequence has a zero map from the torsion group ${\pi }_4{S}^3=\mathbb{Z}/2$ to ${\pi }_{3}SO(3)=\mathbb{Z}$, giving ${\pi }_{3}SO(4)={\mathbb{Z}}^2$. For $SO(1)$ all positive groups vanish, and ${\pi }_{3}SO(2)=0$.

Every matrix in $SU(2)$ has the unique formThe four real coordinates of $(\alpha ,\beta )$ give a smooth bijection with smooth inverse onto the unit three-sphere. Matrix multiplication identifies this model with the unit quaternions. Its round metric is invariant under left and right multiplication by unit quaternions. The actionhas kernel $\{(1,1),(-1,-1)\}$: a kernel pair must have $a=b$ from $q=1$ and must commute with all quaternions. Its derivative is injective, and both dimensions are six, givingThe full isometry group is $O(4)$. Quaternion conjugation reverses orientation and supplies the additional component. Thus an arbitrary isometry has either the displayed left-right form or that form composed with conjugation.

For the Anti-de Sitter spacetime model, writeThe hyperboloid with this quadratic form equal to $-1$ is exactly $SL(2,\mathbb{R})$. Its induced Lorentzian metric is preserved by $M\mapsto AM{B}^{-1}$ for $A,B\in SL(2,\mathbb{R})$. At the identity the metric is $g(U,V)=\frac{1}{2}\mathrm{tr}(UV)$ on traceless matrices, so it is the Killing metric divided by eight. This proves the geometric identification, including the metric rather than only the underlying manifold.

A trivial left-right action again forces $A=B$ to be a scalar commuting with all determinant-one matrices, hence $A=B=\pm I$. The differential is injective and both dimensions are six, soThis is the existing left-right double cover of SO0(2,2). The full hyperboloid isometry group is $O(2,2)$. Its other components can be obtained by also using $M\mapsto M^T$ and $M\mapsto CM{C}^{-1}$, $C=\mathrm{diag}(1,-1)$. Transposition reverses one timelike coordinate; conjugation by $C$ reverses one timelike and one spacelike coordinate, so these generate the four component classes. All isometries of these constant-curvature quadrics come from ambient orthogonal transformations: a value and an orthonormal derivative at one point determine an isometry, and the ambient group realizes every such choice.

Here ${\mathrm{AdS}}_3$ means the hyperboloid with periodic time. The physically common causally unwrapped spacetime is its universal cover, modeled by the universal covering Lie group of $SL(2,\mathbb{R})$, and is not itself $SL(2,\mathbb{R})$. Its isometries lift the above local symmetries, with the global covering and central identifications changed accordingly.

One definition of a principal bundle $P\to M$ uses local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times G$ whose overlap maps arewith smooth $G$-valued transition functions of a principal bundle satisfying the cocycle condition. This definition displays the global right group action rather than assuming it: if ${\varphi }_i(p)=(x,g_i)$, define ${\varphi }_i(p\cdot h)=(x,g_{i}h)$. On an overlap,so the local definitions agree. The resulting action is smooth, free and transitive on each fibre, and respects the projection.

If $s:M\to P$ is a global section of a fiber bundle, set $F(x,h)=s(x)\cdot h$. The free and transitive fibre action makes $F:M\times G\to P$ bijective. In a local trivialization, write $s(x)=(x,a_i(x))$; then $F(x,h)=(x,a_i(x)h)$ and its inverse is $(x,g)\mapsto (x,a_i(x{)}^{-1}g)$. Both are smooth and respect the action and projection. Conversely a principal bundle isomorphism $M\times G\to P$ sends $x\mapsto (x,e)$ to a global section. Hence

For a pseudo-Riemannian manifold of metric signature $(p,q)$, its orthonormal frame bundle has fibre at $x$ consisting of linear isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$. Its structure group is $O(p,q)$ and its right action is $u\cdot h=u\circ h$. Two such frames differ by a unique element of $O(p,q)$, proving the required free and transitive fibre action. A global section is exactly a global pseudo-orthonormal frame. Orientation and time orientation, when specified, restrict the structure group and the allowed frames.

Realize unit-radius four-dimensional de Sitter spacetime asIts tangent space is $X^{\perp }$, with Lorentzian metric signature $(3,1)$. If $e_0,e_1,e_2,e_3$ is a pseudo-orthonormal frame at $X$, append $X$ as the final column. The resulting five-by-five matrix preserves $\eta$, and conversely every such matrix supplies a point $X$ and a frame at that point. Thus the full frame bundle is $O(4,1)$, with projection $g\mapsto g{e}_4$ and fibre group $O(3,1)$.

The printed $SO(4,1)$ statement requires the orientation restriction. Choose the tangent orientation by requiring the appended matrix to have determinant one. Thenis precisely the oriented frame bundle of de Sitter spacetime, with right structure group $SO(3,1)$ embedded as $\mathrm{diag}(h,1)$. If future time orientation is also required, use $S{O}_0(4,1)$ and $S{O}_0(3,1)$. Calling $SO(4,1)$ the full bundle without specifying orientation would exclude valid frames of the opposite orientation.

Finally write the global coordinatesDirect differentiation gives the de Sitter spacetime metric $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. The assumed global frame of a vector bundle on $S^3$ can be orthonormalized smoothly by Gram-Schmidt orthogonalization for the round positive metric. More explicitly, view $S^3$ as the unit quaternions: $Y_1(q)=qi$, $Y_2(q)=qj$, $Y_3(q)=qk$ form a smooth global orthonormal tangent basis. SetThese have inner products $g(E_0,E_0)=-1$, $g(E_i,E_j)={\delta }_{ij}$ and $g(E_0,E_i)=0$. They are smooth everywhere because $\cosh \tau$ never vanishes. Choose the spatial orientation consistently, reversing one $Y_i$ if necessary, and choose $E_0$ future-directed. This global frame on four-dimensional de Sitter spacetime yields a global section of the full frame bundle and also of either restricted bundle. Yes: the de Sitter frame bundle admits a global section and is trivial.

Sending a matrix to its last column gives the special orthogonal sphere fibrationThe action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by Gram-Schmidt process near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the long exact sequence of homotopy groups of a fibration to be applied.

At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify ${\mathbb{R}}^3$ with imaginary quaternions. A unit quaternion acts by $v\mapsto qv{q}^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos (\theta /2)+u\sin (\theta /2)$. The kernel is precisely $\{1,-1\}$, since a quaternion commuting with all imaginary quaternions is real. Thus this gives the twofold covering space $S^3\to SO(3)$. Its simply connected total space is the universal cover, yielding ${\pi }_{1}SO(3)=\mathbb{Z}/2$. Coverings induce isomorphisms on higher homotopy groups, so ${\pi }_{2}SO(3)={\pi }_2{S}^3=0$ and ${\pi }_{3}SO(3)={\pi }_3{S}^3=\mathbb{Z}$.

For $n\ge 4$, both ${\pi }_1{S}^{n-1}$ and ${\pi }_2{S}^{n-1}$ vanish. The Serre fibration sequence therefore identifies ${\pi }_{1}SO(n-1)$ with ${\pi }_{1}SO(n)$ and propagates vanishing of ${\pi }_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer isHere ${\pi }_0$ records the single connected component.

For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth homotopy group, isThe first homomorphism is zero since a torsion group cannot map nontrivially to $\mathbb{Z}$. Thus it reduces to a short exact sequence $0\to \mathbb{Z}\to {\pi }_{3}SO(4)\to \mathbb{Z}\to 0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. ThereforeThese are the low homotopy groups of special orthogonal groups.

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.

Choose a complex projective line in each Complex projective plane summand. Make the connected sum at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The intersection form is diagonal with entries one, henceThe oriented normal bundle of $S$ is a rank-two real bundle with Euler number two, because its Euler number equals the self-intersection number. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $T{S}^2$. The boundary of a closed tubular neighborhood $N$ is consequently the unit tangent bundle of $S^2$.

A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in ${\mathbb{R}}^3$. Thus $UT{S}^2\cong SO(3)\cong {\mathbb{RP}}^3$, the latter identification coming from the double covering by unit quaternions. Therefore $\partial N$ is the required smoothly embedded ${\mathbb{RP}}^3$. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus \partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of separating projective three-space in a positive connected sum.

For a connected base with a chosen frame at a base point, flat connections modulo the based unitary gauge group correspond to flat holonomy representations ${\pi }_1(X,x)\to G$. Removing the frame, or allowing the full unitary bundle gauge group, also divides by conjugation in $G$. On a prescribed underlying bundle, retain only those representations whose associated flat bundle has that bundle type.

The correspondence can be seen directly. Zero vector-bundle curvature makes parallel transport invariant under homotopies of paths with fixed endpoints, giving a group homomorphism on loops. Conversely, a representation $\rho$ gives the flat bundle $(\tilde{X}\times G)/{\pi }_1(X)$, with the deck action on the first factor and $\rho$ on the second, and the product horizontal distribution descends. If two framed connections have identical holonomies, compare their parallel transports along a path from the base point to each point. The comparison is path independent and defines a bundle gauge transformation equal to the identity at the base point. These constructions are inverse. This is framed flat connections and holonomy.

The punctured torus retracts onto a wedge of two circles, so its fundamental group is the free group $F_2$. A representation into $SU(2)$ is specified freely by the two generator images $(A,B)$, with no relation. Every resulting principal $SU(2)$ bundle is trivial, since a bundle with connected structure group over a graph is trivial. Thus no representation is excluded by the fixed trivial-bundle condition. Holonomy depends continuously on a connection, and the flat-bundle construction gives locally continuous choices of representatives as the two matrices vary; local choices of paths in $SU(2)$ give this continuity without requiring a global logarithm. ConsequentlyThe last identification is the description of $SU(2)$ by unit quaternions. The unbased quotient used next is simultaneous conjugation, not independent conjugation of the two generator images.

By part (b), the moduli space is $M(T)=SU(2{)}^2/SU(2)$ under simultaneous conjugation. Regard $SU(2)$ as the unit quaternions and writewith $u,v\in {\mathbb{R}}^3$. Conjugation rotates both imaginary vectors by the same element of $SO(3)$. Their lengths and inner product determine an orbit, even when one vector vanishes or the vectors are collinear: an isometry between the two spanned planes can be extended to an orientation-preserving rotation of ${\mathbb{R}}^3$. Thus complete invariants areEvery allowed triple is realized by a pair of vectors with the prescribed lengths and dot product. The induced continuous bijection from the compact quotient to this closed subset of ${\mathbb{R}}^3$ is a homeomorphism.

Let $U$ be the subset where $A\ne \pm 1$, equivalently $-1<a<1$. It is open and dense because the excluded first holonomies can be approximated by noncentral ones. Rotate $u$ to the positive $i$ axis, so $A=a+\sqrt{1-a^2}i$. Its remaining stabilizer rotates the $j,k$ plane. If $B=b+ti+yj+zk$, the orbit is determined by $(b,t)$, andwith the remaining radius $\sqrt{1-b^2-t^2}$. These coordinates give the corrected resultContinuity of the inverse follows by taking $y=\sqrt{1-b^2-t^2}$ and $z=0$ as a representative. The complement consists of $A=1$ and $A=-1$, separately. In either case conjugation classes of $B$ are parametrized by its real part $b\in [-1,1]$. Therefore $M(T)\setminus U$ is two disjoint closed intervals, as requested.

The printed $S^2$ factor is incorrect: the factor must be a closed disk. The stabilizer acts by conjugation, fixing the circle of quaternions $b+ti$; it is not the free Hopf circle action on $S^3$. To rule out a different choice of dense open set with the printed topology, we can also identify the whole space. Replace $c$ by $d=ab+c$. The invariant region becomesThese are the Gram matrices of the three unit vectors $1,A,B$ in ${\mathbb{R}}^4$. Conversely, a positive-semidefinite matrix of this form can be realized by such vectors, hence by the original quaternion data. The set $\mathcal{C}$ is compact and convex and contains the identity matrix as an interior point in its three coordinates. Radial parametrization of a convex body therefore identifies it with a closed three-ball. ThusThis also agrees with Theorem 6.5 of the primary character-variety paper.

An open subset homeomorphic to the boundaryless three-manifold $S^2\times (0,1)$ cannot contain a boundary point of this closed ball. Otherwise a local chart, followed by inclusion into ${\mathbb{R}}^3$, would be a continuous injection from an open subset of ${\mathbb{R}}^3$ whose image contains a ball-boundary point while staying inside the closed ball, contradicting invariance of domain. Its complement would therefore contain the entire boundary sphere. But a sphere cannot embed in two disjoint intervals: its connected image would lie in one interval and be a nondegenerate compact interval, whose interior points disconnect it, unlike $S^2$. This proves that the literal printed pair of claims is impossible, and supplies the complete corrected SU2 character variety of the punctured torus description.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

Identify ${\mathbb{R}}^4$ with the quaternions, and use the identification of unit quaternions with $SU(2)$ proved in the root Solution. The actionpreserves the Euclidean norm by multiplicativity of quaternion norms. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has determinant $+1$.

An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all quaternions, so $q$ is real; since it has unit norm, $q=\pm 1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.

The differential acts as $v\mapsto av-vb$ for imaginary quaternions $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both Lie algebras have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thuswhere ${\mathbb{Z}}_2$ acts diagonally, not separately on the two factors.

The imaginary quaternions form Euclidean ${\mathbb{R}}^3$. Conjugation by a unit quaternion acts byIt preserves the imaginary subspace and its norm, and continuity from $q=1$ gives determinant $+1$. The kernel consists of quaternions commuting with all imaginary quaternions, hence is $\{\pm 1\}$.

For a unit imaginary quaternion $n$, take $q=\cos (\theta /2)+n\sin (\theta /2)$. If $v$ is perpendicular to $n$, quaternion multiplication giveswhile the component parallel to $n$ is unchanged. This produces the rotation through angle $\theta$ about axis $n$, and every element of $SO(3)$ has such an axis-angle description. The map is surjective, yieldingThe half-angle also explains why $q$ and $-q$ give the same rotation.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

For $n\ge 1$, let $\#$ denote concatenation in the first cube coordinate and let $\cdot$ denote pointwise multiplication of maps into the topological group. Both operations descend to the homotopy group, and the constant map is their common identity. They satisfy the interchange rulewhich follows by considering the two halves of the first coordinate. Using the identity class $0$ givesTherefore $[\alpha \beta ]=[\alpha ]+[\beta ]$. The equality is on based homotopy classes; the usual reparametrization homotopies justify the unit identities for concatenation. This is the Eckmann-Hilton argument, and proves homotopy-group addition in a topological group even for $n=1$.

For the unit quaternions, the explicit inverse isBoth compositions cancel in the displayed order, without commuting the quaternions. Multiplication, inversion and all integer powers are continuous, so this proves that ${\Phi }_{k,l}$ is a homeomorphism for all integers $k,l$.

Put $m=k+l$. Let $a,b$ be the standard generators of $H_3(S^3\times S^3;\mathbb{Z})$, represented by the first and second factors. The previous pointwise-multiplication result and the Hurewicz theorem imply that the mapping degree of $q\mapsto q^m$ on $S^3$ is $m$, including negative integers. Restricting $\Phi$ to the two factors therefore givesThe map on $H_0$ is the identity. For the top homology group, take the dual degree-three cohomology classes $x,y$. We have ${\Phi }^{\ast }x=mx+y$, ${\Phi }^{\ast }y=x$. Since these classes have odd degree, graded commutativity of the cup product givesThus ${\Phi }_{\ast }$ is multiplication by $-1$ on $H_6$, and all other homology groups of the product are zero.

For the gluing, regard $\Phi$ as the attaching identification from $\partial (D^4\times S^3)$ to $\partial (S^3\times D^4)$. In these coordinates its first input is the boundary coordinate of $D^4$, its second input the fibre coordinate; its second output is the boundary coordinate of the other $D^4$. This makes the two pieces the usual two trivializations of a three-sphere bundle over the four-sphere.

Let $A=S^3\times D^4$ and $B=D^4\times S^3$, and parametrize their common boundary $T=S^3\times S^3$ using the $B$ coordinates. Collar neighbourhoods give an open cover with the same homotopy types, so the Mayer–Vietoris sequence applies. Both pieces retract onto $S^3$. In degree three, the map into the homology of the pieces isIndeed, inclusion into $A$ retains the first output coordinate, while inclusion into $B$ retains the second input coordinate. Exactness now givesThe second relation eliminates the second generator of the cokernel, leaving a single generator with relation $m$ times that generator equal to zero. Thus $\mathrm{coker}L=\mathbb{Z}/m\mathbb{Z}$. The kernel is zero if $m\ne 0$ and is generated by $(1,0)$ if $m=0$.

The degree-six boundary class gives $H_7(X_{k,l})\cong H_6(T)=\mathbb{Z}$. All remaining positive-degree groups outside degrees $3,4,7$ vanish by the same Mayer–Vietoris sequence; connectedness gives $H_0=\mathbb{Z}$. Consequently the complete integral answer isHere $\mathbb{Z}/0\mathbb{Z}$ means $\mathbb{Z}$, and a negative $m$ gives the same cyclic group as $|m|$. In particular, $m=\pm 1$ gives the integral homology of a seven-sphere, while $m=0$ gives the integral homology of $S^3\times S^4$.

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.

Two unit quaternions act on Euclidean four-space by $q\mapsto aq{b}^{-1}$. This norm-preserving action maps onto the SO(4) group. Its kernel is precisely $(1,1)$ and $(-1,-1)$: a kernel pair must have equal entries commuting with every quaternion. The two simply connected SU(2) factors therefore form the universal spin cover.

For an oriented rank-four real vector bundle $E\to S^4$, its sphere bundle has fibre $S^3$. Let $m=⟨e(E),[S^4]⟩$ be its Euler class evaluated on the orientation class. The Gysin sequence gives integral homology $\mathbb{Z}$ in degrees $0,7$, $\mathbb{Z}/m\mathbb{Z}$ in degree $3$, and a copy of $\mathbb{Z}$ in degree $4$ exactly when $m=0$, with all other groups zero. Here $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$. A transition function $v\mapsto q^{k}v{q}^l$ on the unit quaternions has Euler number $k+l$, up to an overall orientation sign: evaluating it at a fixed unit vector gives the power map $q\mapsto q^{k+l}$, whose mapping degree is $k+l$. The same homology calculation follows by applying the Mayer–Vietoris sequence to the two trivializations over the hemispheres.