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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.