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The displayed complex matrices, with real coefficients, form a four-dimensional real associative algebra. Their three imaginary basis matrices square to minus the identity and satisfy $JK=L$, $KL=J$, $LJ=K$, with the reversed products negative. For $\alpha =aI+bJ+cK+dL$, quaternionic conjugation gives $\alpha \overline{\alpha }=(a^2+b^2+c^2+d^2)I$. Every nonzero matrix therefore has inverse $\overline{\alpha }/(a^2+b^2+c^2+d^2)$ inside this algebra. This realizes the real division algebra of quaternions within $M_2(\mathbb{C})$; it is not a complex vector subspace.

Identify ${\mathbb{R}}^4$ with the quaternions. Unit quaternions act by left multiplication, embedding the SU(2) group in $SO(4)$. Under this subgroup, the complexified vector representation branches as $4\to 22$, and the Adjoint representation as $6\to 3\oplus 31$. Right multiplication supplies the commuting second factor.

Between 1600 and 1900, algebra broadened from arts of working with unknown quantities and solving equations into the study of several systems of symbolic operations and increasingly general algebraic structures. This was neither a single change of definition nor a clean replacement of one subject by another. Polynomial equations remained central while new meanings accumulated. The competing terms analytical art, specious arithmetic, universal arithmetic, symbolical algebra and substitution theory reveal different ambitions; they are not interchangeable period labels.

Near 1600, François Viète's In artem analyticem isagoge by François Viète (1591) supplies a particularly revealing starting point. His analytical art uses letters for known quantities as well as unknown ones. In his terminology, logistica numerosa concerns numerical calculation and logistica speciosa calculation with species, or general magnitudes. Vowels represent unknowns and consonants known magnitudes. This allows a single relation to represent many particular problems, but it is still tied to geometrical homogeneity: a square and a length do not simply become quantities of the same kind because they are written with letters.^[ref][ref]

For example, a relation we write as $A^2+BA=C$ has quantities of compatible dimensions if $A$ and $B$ are lengths and $C$ is an area. Letters do more than abbreviate a numerical question: they express relations between whole families of magnitudes. The phrase specious arithmetic therefore indicates arithmetic carried out on species; analysis stresses the art of reasoning from an assumed unknown towards its determination. Neither term should be reduced to the modern subject real analysis.

René Descartes's La Géométrie by René Descartes (1637) demonstrates another relation between algebra and geometry. A chosen unit segment lets multiplication, division and extraction of square roots be represented by constructions of lengths. The product of two measured lengths can itself be represented by a length relative to that unit. In modern dimensional notation the construction gives $ab/u$ for unit length $u$, removing the obstacle that a product must always be treated as an area. The text's equations can then organise geometrical problems and classify curves by their algebraic complexity. Its title remains geometry: algebra serves a geometrical programme rather than announcing a present-day department of abstract algebra.^[ref]

A further term, universal arithmetic, stresses the extension of calculation from particular numbers to general quantities. Isaac Newton's Arithmetica universalis by Isaac Newton, issued in 1707, makes the term visible in a title whose subtitle concerns composition and resolution. A title is evidence for the vocabulary, not a sufficient definition of all contemporary usage.^[ref] In Leonhard Euler's Elements of Algebra by Leonhard Euler, published in German in 1770, general numerical calculation and the solution of equations remain a substantial conception of the subject. For instance, allowing negative and complex roots changes the admissible answers to polynomial equations while preserving the central question of determining an unknown.^[ref]

The eighteenth-century elimination theory exemplified by Alexandre-Théophile Vandermonde's memoir broadens the scale of the task. Instead of one numerical quadratic equation, one seeks a rule for an arbitrary number of simultaneous linear equations. The determinant is initially valuable because it effects that elimination. This is a different generality from the later study of a group: the objects being determined are still unknown quantities, and symbolic generality concerns the number and coefficients of the equations. The case cautions against using generality alone as a test for whether algebra has become abstract.^[ref]

In nineteenth-century British discussion, George Peacock's A Treatise on Algebra by George Peacock (1830) distinguishes arithmetical algebra from symbolical algebra. If the symbols mean positive numbers, $a-(a+b)$ has no positive-number result. Yet algebraists manipulate this expression without restriction. George Peacock therefore seeks principles for the manipulation of signs which are not confined by their original arithmetical interpretation. The contrast is explicitly argued in his preface: arithmetic suggests the laws, while the symbolic science requires foundations adequate to its more general conclusions.^[ref]

The familiar extension $(-1)(-1)=1$ illustrates the stakes. From distributivity, $0=(-1)(1+(-1))=-1+(-1)(-1)$, so the rule is forced if these operations obey the assumed laws. This explains how a law can survive a change in interpretation. It does not establish that every familiar law must survive every extension: once the allowable systems themselves change, an argument must state which laws it assumes. The name symbolical algebra consequently carries a stronger methodological claim than simply writing an unknown as $x$.

William Rowan Hamilton's work on quaternions supplies a decisive example of that change. His publications of the 1840s describe a new system of imaginaries in algebra. With quaternion units,Thus multiplication need not satisfy the commutativity even when it remains associative and supports division by nonzero elements. The new system is not merely a further list of roots of old polynomial equations. It has its own operations, interpretation and laws. William Rowan Hamilton's title makes explicit that a plurality of calculi can belong within algebra.^[ref]

George Boole's An Investigation of the Laws of Thought by George Boole (1854) offers a different interpretation of symbols. If $x$ represents selection of a class, repeating that selection gives $x^2=x$; multiplication can represent intersection and $1-x$ a complementary class. His logical calculus borrows from numerical algebra but also restricts interpretation: his sum of classes is not automatically unrestricted set union, and intermediate symbolic expressions need not name classes directly. It would therefore be inaccurate to identify all of his rules with a modern Boolean algebra. The example nevertheless shows why algebraic reasoning could extend beyond magnitudes.^[ref]

The shift can also take a concrete linear form. Arthur Cayley's A Memoir on the Theory of Matrices by Arthur Cayley (1858) treats matrices with operations of their own. The identity matrix is no longer the scalar $1$, though it plays an analogous role; matrix multiplication is generally noncommutative. For example,produce $AB=\mathrm{diag}(1,0)$ and $BA=\mathrm{diag}(0,1)$. This is a checkable reason why universal arithmetic can no longer mean that every rule of ordinary numbers applies unchanged.^[ref]

Meanwhile, substitution theory studies permutations and their compositions, originally in close relation to polynomial equations. Évariste Galois's work connects the solvability by radicals of an equation with its associated group. Augustin-Louis Cauchy's work on substitutions and Heinrich Weber's 1882 definition illustrate successive forms of generalisation: from particular permutations towards arbitrary elements satisfying operational conditions. This is not a claim that one man supplied the complete modern hierarchy of groups, rings and fields at once. It identifies the change in the question: one can now investigate what follows from the laws of an operation, independently of whether the elements are numbers, substitutions or classes.^[ref]

By 1900, algebra could concern the laws and structures of several calculi. Earlier meanings persisted in teaching and in the theory of equations. The strongest account of the change therefore combines continuity with enlargement: letters for general quantities, symbolic operations with explicit laws, nonnumerical interpretations, and structural questions about those laws. Analytical art foregrounded a method of discovery, universal arithmetic foregrounded general calculation, symbolical algebra foregrounded formal operations, and substitution theory foregrounded a particular class of transformations. Their differing uses help explain the expansion more precisely than a single date for the birth of abstraction.

A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form $T(x)=Rx+b$, with translation $b\in {\mathbb{R}}^3$ and an orthogonal matrix $R$. For a proper physical rotation, the coefficient constraints areEquivalently the columns of $R$ are an orthonormal basis with positive orientation. A reflection also preserves distance but has $\det R=-1$ and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.

Computationally use the homogeneous coordinates representationThe last row is $(0,0,0,1)$; matrix multiplication composes transforms. Alternatively store a unit quaternion together with $b$, maintaining its unit norm. A transpose gives the rotational inverse: $T^{-1}(y)=R^T(y-b)$.

Write the cubic Bézier curve as $P(t)={\sum }_{i=0}^3{B}_i^3(t)p_i$, where $B_i^3(t)=\left(\genfrac{}{}{0ex}{}{3}{i}\right)(1-t{)}^{3-i}{t}^i$. The binomial theorem gives ${\sum }_i{B}_i^3(t)=1$. Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.

Orthogonality was not used in the calculation. The identical proof applies to every affine map $T(x)=Ax+b$, including scales, shears and even singular affine maps. It applies to every representation $P(u)={\sum }_i{\varphi }_i(u)p_i$ with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.

The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all $x,y$ and $0\le t\le 1$. This segment identity, followed by continuity, forces $T$ to be affine. Nonlinear transforms such as perspective division generally fail it.

There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls $(w_i{p}_i,w_i)$, then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.

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.

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.

The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.

The quaternionic projective space ${\mathbb{HP}}^n$ is the space of one-dimensional right quaternion subspaces of ${\mathbb{H}}^{n+1}$. Equivalently it is the quotient of the unit sphere $S^{4n+3}$ by simultaneous right multiplication by unit quaternions. Its coordinate filtration has one open cell ${\mathbb{H}}^k\cong {\mathbb{R}}^{4k}$ in each dimension $4k$, for $0\le k\le n$. Hence its cellular cohomology is $\mathbb{Z}$ in those dimensions and zero otherwise.

Let $\mathcal{H}$ be the quaternionic tautological line bundle. Its unit sphere bundle is $S^{4n+3}\to {\mathbb{HP}}^n$, with fibre $S^3$. The Gysin sequence of a sphere bundle shows that multiplication by its Euler class $e\in H^4$ is an isomorphism from $H^{4k}$ to $H^{4k+4}$ for $k<n$. Choose the generator $u=-e$. Its powers generate every nonzero positive degree, giving the cohomology ring of quaternionic projective spaceThis also accounts for $n=0$.

First take $n=2$. Under the coordinate inclusion $i:{\mathbb{CP}}^2\to {\mathbb{HP}}^2$, the pulled-back quaternionic line is the quaternionic extension of the complex tautological line $L$. As a complex rank-two bundle it is $L\oplus \overline{L}$: a transition scalar $c\in \mathbb{C}$ acts on the two complex coordinates of a quaternion by $c$ and $\overline{c}$. Put $x=c_1(L^{\ast })$, the degree-two generator of the cohomology ring of complex projective space. The Whitney sum formula for Chern classes givesHere the Euler class of a complex vector bundle is its top Chern class, using the complex orientation. In particular this degree-four pullback has coefficient one; it is not a multiple of larger absolute value. The compatible tautological bundles on the projective filtrations give the same equality for every $n$, and multiplicativity then determines the whole ring map:It is zero whenever $2k>n$. These facts are the complex inclusion into quaternionic projective space.

For an odd prime $p$, the Steenrod reduced powers are natural stable cohomology operationsThey satisfy $P^0=1$, the Cartan formula, $P^{i}a=0$ when $2i>|a|$, and $P^{i}a=a^p$ when $|a|=2i$. In particular, on ${\mathbb{CP}}^{\infty }$ one has $P^{1}x=x^p$ and $P^{i}x=0$ for $i>1$. The Cartan formula and the binomial theorem givePass to the infinite projective spaces, where $i^{\ast }:H^{\ast }({\mathbb{HP}}^{\infty };{\mathbb{F}}_p)\to H^{\ast }({\mathbb{CP}}^{\infty };{\mathbb{F}}_p)$, $u\mapsto x^2$, is injective. The equality just obtained determines the Steenrod powers on quaternionic projective space; restricting to the finite spaces giveswith coefficients modulo $p$ and powers above $n$ set to zero. In particular $P^{1}u=2u^{(p+1)/2}$, $P^{2}u=u^p$, and $P^{i}u=0$ for $i>2$. The infinite-space argument matters: the finite inclusion cannot detect those degrees for which $x^{2k}=0$.

Finally put $Q={\mathbb{HP}}^{15}/{\mathbb{HP}}^{12}$ and $S={\Sigma }^{48}{\mathbb{HP}}^3$. Both have reduced cohomology $\mathbb{Z}$ in degrees $52,56,60$ and zero otherwise. Choose integral generators $a_{13},a_{14},a_{15}$ for $Q$ whose pullbacks under the quotient map are $u^{13},u^{14},u^{15}$, and suspended integral generators $b_1,b_2,b_3$ for $S$ from $u,u^2,u^3$.

Use $p=5$. Naturality for the quotient and the formula above giveStability under the suspension isomorphism instead givesAny homotopy equivalence $h:Q\to S$ would induce isomorphisms on the rank-one integral groups, so $h^{\ast }{b}_1=\epsilon a_{13}$ and $h^{\ast }{b}_3=\delta a_{15}$ with $\epsilon ,\delta \in \{1,-1\}$. Reducing modulo five and commuting $h^{\ast }$ with $P^1$ would require $\epsilon =2\delta$ in ${\mathbb{F}}_5$. Neither $1$ nor $-1$ equals $2$ or $-2$ modulo five. ThereforeThe essential point is that integral generator signs constrain Steenrod comparisons. Arbitrary changes of basis over ${\mathbb{F}}_5$ could rescale these two nonzero coefficients into agreement; a genuine equivalence must also preserve the integral lattices, where only the two signs are available.

No. The quaternion groupis another non-abelian group of order eight: for example $ij=k$ while $ji=-k$. Its multiplication comes from the associative quaternion algebra, so it does define a group.

It is not isomorphic to the dihedral group of the square. In $Q_8$, only $-1$ has order two; the six elements $\pm i,\pm j,\pm k$ have order four. The square's dihedral group has five elements of order two: four reflections and the half-turn. Group isomorphisms preserve element orders, so these groups are distinct. The hypothesis in part (c) excludes $Q_8$, since its only element of order two is already in each cyclic subgroup of order four.

Write a matrix in the set as $M(a,b)$. The zero matrix $M(0,0)$ has no multiplicative inverse matrix, so $S$ is not a group. Its determinant iswhich vanishes only at that zero matrix. It is therefore the only element that needs to be removed.

For closure, direct matrix multiplication givesThe product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The identity matrix is $M(1,0)$, associativity is inherited from matrix multiplication, andThus $\boxed{S\setminus \{0_{2\times 2}\}\text{ is a group}}$. It is the multiplicative group of nonzero quaternions in the complex matrix representation of quaternions.