Spin group

The spin group $\mathrm{Spin}(n)$ is the connected double cover of $SO(n)$ constructed inside the units of a Clifford algebra.

For $\mathrm{Spin}(2m)$, $m\ge 3$, the two complex half-spin irreducible representations have dimension $2^{m-1}$. Their weights are $\frac{1}{2}(\pm e_1\pm \cdots \pm e_m)$ with respectively even and odd numbers of minus signs, after a chirality convention is chosen; every weight multiplicity is one. A model is the even or odd part of the exterior algebra of ${\mathbb{C}}^m$, where an occupied mode reverses the corresponding sign. Each is a minuscule representation. These are representations of the spin group and its Lie algebra, but do not descend through the double covering to the special orthogonal group.

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

On two-by-two Hermitian matrices, the determinant of $X=tI+x\cdot \sigma$ is $t^2-|x{|}^2$. The action $X\mapsto AX{A}^{†}$ preserves it for $A\in SL(2,\mathbb{C})$. A kernel element is unitary and commutes with every Hermitian matrix, hence is $\pm I$. The derivative is injective and both real dimensions are six, giving the full proper orthochronous Lorentz group as the connected image.

Congruence $X\mapsto AX{A}^T$ on real symmetric two-by-two matrices preserves the quadratic form $-\det X$ of signature $(2,1)$ when $\det A=1$. Its kernel is $\{\pm I\}$ and its derivative is injective: $aX+X{a}^T=0$ for every symmetric $X$ implies $a=0$. Both Lie groups have dimension three, so the connected image is the full identity component. Also $A^{T}JA=(\det A)J$ identifies $Sp(2,\mathbb{R})$ with $SL(2,\mathbb{R})$.

A fixed volume form makes wedge pairing on ${\Lambda }^2{\mathbb{R}}^4$ a symmetric form of signature $(3,3)$. The exterior-square action of special linear group matrices preserves it. Its kernel is $\{\pm I\}$: a transformation acting trivially on every bivector preserves every two-plane and hence every line, so is scalar, with scalar square one. The derivative is injective and both groups have dimension fifteen, so the connected image is the full identity component. This is a concrete real spin-group cover.

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.

A spinor representation is a representation of a Spin group, or infinitesimally of a Special orthogonal Lie algebra, constructed from a Clifford algebra. In even dimensions it can split into two chiral spinor representations.

The Spin group $\mathrm{Spin}(7)$ double covers SO(7). Its central kernel element $-1$ acts trivially in the vector representation and as $-I$ in the spinor representation. Thus a spinor Lie algebra representation does not define an ordinary SO(7) group representation. Under $\mathrm{Spin}(6)\cong SU(4)$ the same obstruction is visible because $-I_4$ acts trivially on ${\Lambda }^{2}4$ but negatively on $4\oplus \overline{4}$.

Choose five Hermitian four-by-four matrices obeying the Euclidean Clifford algebra relations ${\gamma }_a{\gamma }_b+{\gamma }_b{\gamma }_a=2{\delta }_{ab}I$. For $M_{ab}=E_{ab}-E_{ba}$, the displayed anti-Hermitian matrices satisfy the Special orthogonal Lie algebra Lie bracket relations. A direct proof first establishes $[\rho (M_{ab}),{\gamma }_c]={\delta }_{bc}{\gamma }_a-{\delta }_{ac}{\gamma }_b$, then applies the commutator product rule to $[{\gamma }_c,{\gamma }_d]$. One explicit choice is ${\gamma }_1={\sigma }_1\otimes I$, ${\gamma }_2={\sigma }_2\otimes I$, ${\gamma }_3={\sigma }_3\otimes {\sigma }_1$, ${\gamma }_4={\sigma }_3\otimes {\sigma }_2$, ${\gamma }_5={\sigma }_3\otimes {\sigma }_3$, with the Pauli matrices. The Cartan subalgebra operators $-i\rho (M_{12})$ and $-i\rho (M_{34})$ have simultaneous eigenvalues $(\pm 1/2,\pm 1/2)$, identifying the four-dimensional spinor representation. A full $2\pi$ rotation acts as $-I_4$, proving it does not descend to $SO(5)$.

For the B2 root system with long simple root first, the four-dimensional spinor representation has highest weight $(0,1)$. Its symmetric square has highest weight $(0,2)$ and dimension ten, so it is the Adjoint representation. The exterior square has the four nonzero defining weights and two zero weights; it splits into the five-dimensional vector representation of highest weight $(1,0)$ and a trivial representation. Thus the tensor square decomposes as displayed. The invariant alternating bilinear form reflects the pseudoreal representation structure. The spinor representation is a group representation of the Spin group $\mathrm{Spin}(5)$, while its tensor square descends to $SO(5)$.

An irreducible complex Clifford algebra module in even spacetime dimension $2k$ has dimension $2^k$ and splits under the connected Spin group into two Weyl spinor modules of dimension $2^{k-1}$. In odd dimension $2k+1$, an irreducible spinor still has dimension $2^k$ but has no Weyl splitting. Reality conditions depend on signature and are additional to these complex counts.