Pin and Spin Groups and Their Representations

If $n>0$, $Q$ is not the zero function: otherwise polarization gives $2B(v,w)=Q(v+w)-Q(v)-Q(w)=0$ for all $v,w$, contradicting nondegeneracy. Choose anisotropic $u$. Since $Q(u)\mathrm{\ne }0$, $V=\mathbb{R}u\oplus u^{\perp }$; the restriction of $B$ to $u^{\perp }$ is nondegenerate because a vector orthogonal to $u^{\perp }$ and to $u$ is orthogonal to all of $V$. Induction on dimension gives an orthogonal anisotropic basis, with the empty basis when $n=0$.

The map $V\to V^{\ast }$ induced by $B$ is an isomorphism. Restriction $V^{\ast }\to U^{\ast }$ is onto because a linear functional on a subspace of a finite-dimensional vector space extends after extending a basis. Therefore $v\mapsto B(v,-){\mathrm{\mid }}_U$ is onto with kernel $U^{\perp }$, so rank-nullity gives $\dim U^{\perp }=n-\dim U$. If $U$ is totally isotropic then $U\subseteq U^{\perp }$, hence $\dim U\le n-\dim U$.

If $z\in \ker (A-I)$ and $y=(A-I)x$, then $B(y,z)=B(Ax,z)-B(x,z)=B(x,A^{-1}z)-B(x,z)=0$. Thus $\mathrm{im}(A-I)\subseteq \ker (A-I{)}^{\perp }$. Both subspaces have dimension $n-\dim \ker (A-I)$ by rank-nullity and step 1.2, so they are equal.

The formula for $r_u$ is linear, fixes every vector in $u^{\perp }$, and gives $r_u(u)=-u$. The orthogonal direct sum $V=\mathbb{R}u\oplus u^{\perp }$ therefore shows that $r_u$ preserves $B$ and has determinant $-1$. It is the reflection in the hyperplane $u^{\perp }$.

If $p+q=0$, there is one $0\times 0$ matrix, and $A=K=P=I_0$, $X=0$ satisfy every assertion. If $p+q=1$, write $A=(a)$; $A^{T}JA=J$ gives $a^2=1$, so $P=(1)$, $K=A$, and $X=0$, and the factorization and path assertions follow. For $p+q\ge 2$, set $M=A^{T}A$. For $x\mathrm{\ne }0$, $x^{T}Mx=\mathrm{\parallel }Ax{\mathrm{\parallel }}^2>0$, so $M$ is positive definite and symmetric. The relation $A^{T}JA=J$ gives $A^{-1}=J{A}^{T}J$ and, by transposition, $A^{-T}=JAJ$. Hence $JMJ=J{A}^{T}AJ=A^{-1}{A}^{-T}=M^{-1}$.

By [F1], diagonalize $M$ in an orthonormal basis; all its eigenvalues ${\lambda }_i$ are positive. Define $P$ to have eigenvalues $\sqrt{{\lambda }_i}$ in that basis. Then $P$ is symmetric positive definite and $P^2=M$. If $R$ is another symmetric positive definite root, it commutes with $M=R^2$ and preserves each eigenspace of $M$; on the $\lambda$-eigenspace, [F1] diagonalizes $R$ and positivity forces every eigenvalue to be $\sqrt{\lambda }$. Thus $R=P$.

Both $JPJ$ and $P^{-1}$ are symmetric positive definite roots of $M^{-1}=JMJ$. Uniqueness in step 2.1 yields $JPJ=P^{-1}$, hence $PJ=J{P}^{-1}$ and $P^{-1}J{P}^{-1}=J$.

Put $K=A{P}^{-1}$. Then $K^{T}K=P^{-1}{A}^{T}A{P}^{-1}=I$, and $K^{T}JK=P^{-1}{A}^{T}JA{P}^{-1}=P^{-1}J{P}^{-1}=J$. Thus $K$ is Euclidean orthogonal and preserves $J$; multiplying $K^{T}JK=J$ by $K$ shows $JK=KJ$, so $K$ is block diagonal in the positive and negative coordinates. The factorization is unique because its positive factor must be the positive square root of $A^{T}A$.

Diagonalize $P$ by [F1] and define $X=\log P$ by replacing each positive eigenvalue $\lambda$ by $\log \lambda$. Then $X$ is symmetric. Conjugation by $J$ preserves the matrix logarithm, so $JXJ=\log (JPJ)=\log (P^{-1})=-X$. Consequently $J\exp (sX)J=\exp (-sX)=\exp (sX{)}^{-1}$; because $K$ commutes with $J$ and is orthogonal, $K\exp (sX)$ preserves $J$. At $s=0$ this is $K$ and at $s=1$ it is $A$. Its determinant is constant and equals $\det A$, proving the path assertion.

If $n=0$, $O(V,B)=\{I\}$ and the empty product proves the claim. If $n=1$, choose an anisotropic basis vector $u$ by [F1]. Every $A\in O(V,B)$ sends $u$ to $\pm u$, so $A=I$ or $A=r_u$; these have reflection lengths zero and one, respectively, and the parity agrees with the determinant. We now prove the induction step for $n\ge 2$.

Suppose $n\ge 2$ and inductively assume the theorem in smaller dimensions. If $K$ contains an anisotropic $u$, then $A$ fixes $u$ and preserves the nondegenerate space $u^{\perp }$. The induction hypothesis factors $A{\mathrm{\mid }}_{u^{\perp }}$ into at most $n-1$ reflections with normals in $u^{\perp }$; extending these reflections by the identity on $\mathbb{R}u$ gives the same reflections in $V$ and their product equals $A$.

Suppose instead $W$ is not totally isotropic. Then $w\mapsto Q(w)$ and $w\mapsto Q((A-I)w)$ are nonzero real polynomials: the first by nondegeneracy and the second because some vector in $W$ is anisotropic. Their product is a nonzero polynomial because the real polynomial ring is an integral domain. A nonzero polynomial on ${\mathbb{R}}^n$ does not vanish everywhere: for $n=1$ it has finitely many roots; for $n>1$, write $P(x,t)={\sum }_j{p}_j(x)t^j$ with a nonzero coefficient $p_j$, use induction to choose $x$ with $p_j(x)\mathrm{\ne }0$, and then choose $t$ outside the finite root set of the nonzero one-variable polynomial $P(x,t)$. Choose $w$ where the product is nonzero and put $u=(A-I)w$. Then both $w$ and $u$ are anisotropic.

It remains that both $K$ and $W$ are totally isotropic. By [F1], $W=K^{\perp }$. Since $K$ and $W$ are both totally isotropic, [F1] gives $\dim K\le n\mathrm{/}2$ and $\dim W\le n\mathrm{/}2$. But $\dim W=n-\dim K$, so both dimensions equal $n\mathrm{/}2$; the inclusion $K\subseteq K^{\perp }=W$ then gives $K=W=K^{\perp }$ and $n=2\dim K$. The map $A$ is the identity on $K$, and because $(A-I)V=W=K$ it is also the identity on $V\mathrm{/}K$. In a basis extending a basis of $K$, its matrix is block triangular with identity diagonal blocks, so $\det A=1$.

Symmetry and orthogonality give $B((A+I)w,u)=B(Aw,Aw)-B(Aw,w)+B(w,Aw)-B(w,w)=0$. Thus $(A+I)w$ is fixed by $r_u$ and $u=(A-I)w$ is negated, so $r_{u}Aw=w$. The orthogonal map $r_{u}A$ has the anisotropic fixed vector $w$, and the argument of step 1.2 factors it with at most $n-1$ reflections. Prepending $r_u$ factors $A$ with at most $n$.

Since $n$ is even in step 1.4, choose anisotropic $u$ and put $A_1=r_{u}A$. Then $\det A_1=-1$, so $A_1$ cannot be in the isotropic case of step 1.4. Steps 1.2 and 2.1 give a factorization of $A_1$ of length at most $n$. Each reflection has determinant $-1$, so this length is odd; because $n$ is even it is at most $n-1$. The identity $A=r_u{A}_1$ now gives at most $n$ reflections.

Every reflection has determinant $-1$, hence any reflection factorization of $A$ has parity determined by $\det A$. The constructed length is at most $n$ in all cases, and its parity is even precisely for determinant $1$. This proves the assertion, including the isotropic kernel and moved-space case.

On the tensor algebra, send a word $v_1\cdots v_m$ to $(-1{)}^m{v}_1\cdots v_m$ for $\alpha$, and to $v_m\cdots v_1$ for $t$. The first is an algebra automorphism and the second reverses products. Each defining relation $v^2-Q(v)1$ is fixed by both maps. The ideal generated by these relations is therefore invariant under each map: applying either map to a generator of the ideal gives another generator, with left and right factors exchanged for $t$. Thus both maps descend to the quotient and square to the identity.

Their composite $\bar{x}=t(\alpha (x))$ is an order-reversing involution and satisfies $\bar{u}=-u$ for $u\in V$. Hence $N(u)=u\bar{u}=-u^2=-Q(u)1$.

For $k=0$, the empty product is $1$ and $N(1)=1$. If $N(y)$ is scalar, then $N(xy)=xy\overline{xy}=xy\bar{y}\bar{x}=xN(y)\bar{x}=N(y)N(x)$. Induction on $k\ge 1$, starting with step 2.1, therefore gives $N(u_1\cdots u_k)={\prod }_{j=1}^k(-Q(u_j))1$.

For anisotropic $u$, $u^{-1}=u\mathrm{/}Q(u)$ by [F1]. Hence ${\rho }_u(v)=-uvu\mathrm{/}Q(u)$. The polarized relation gives $uv+vu=2B(u,v)1$, so $-uvu\mathrm{/}Q(u)=v-2B(v,u)u\mathrm{/}Q(u)=r_u(v)$.

Let $x\in \mathrm{\Gamma }$ and $y_v={\rho }_x(v)\in V$. Since $\alpha (y_v)=-y_v$, applying $\alpha$ to $y_v=\alpha (x)v{x}^{-1}$ gives $xv\alpha (x{)}^{-1}=\alpha (x)v{x}^{-1}$, that is, ${\rho }_{\alpha (x)}(v)={\rho }_x(v)$. Using this equality in one factor and [F1] gives $2B(y_v,y_w)1=y_v{y}_w+y_w{y}_v=\alpha (x)(vw+wv)\alpha (x{)}^{-1}=2B(v,w)1$. Thus ${\rho }_x$ preserves $B$; it is injective because multiplication by invertible elements is injective, and finite dimensionality makes its image all of $V$.

If $n=0$, [F2] gives $\mathrm{C}\mathrm{l}(V,Q)=\mathbb{R}$, so every kernel element is scalar. If $n>0$ and ${\rho }_x$ is the identity, then $\alpha (x)e_i=e_{i}x$ for every vector of an orthogonal anisotropic basis $e_1,\dots ,e_n$. Write $x=x_0+x_1$ with even and odd parts. The grading gives $x_0{e}_i=e_i{x}_0$ and $-x_1{e}_i=e_i{x}_1$. For a square-free monomial $e_I$ of length $r$, moving $e_i$ past it gives commutation sign $(-1{)}^r$ when $i\mathrm{\notin }I$ and $(-1{)}^{r-1}$ when $i\in I$. For each nonempty even $I$, choose $i\in I$: the required commutation fails; for each odd $I$, choose $i\in I$: the required anticommutation fails. Distinct $I$ containing a fixed $i$ give distinct deleted basis monomials, so no cancellation can remove these coefficients. By [F2], only the empty even monomial remains and $x$ is scalar. Conversely every nonzero scalar acts trivially.

To prove homogeneity, let $A={\rho }_x\in O(V,B)$. By [F4], write $A=r_{u_1}\cdots r_{u_k}$ and put $y=u_1\cdots u_k$. Step 1.1 and homogeneity of each vector give ${\rho }_y=A$. Equality ${\rho }_x={\rho }_y$ implies, after multiplying by $\alpha (y{)}^{-1}$ on the left and $x$ on the right, that $\alpha (y^{-1}x)v=v(y^{-1}x)$ for every $v$. The PBW kernel argument of step 1.3 applies to the even and odd parts of $y^{-1}x$ and makes it a nonzero scalar. Therefore $x$ is a scalar multiple of the homogeneous product $y$.

Products of homogeneous normalizers act by composition: ${\rho }_{xy}={\rho }_x\circ {\rho }_y$. Step 2.1 shows every unrestricted normalizer is homogeneous, and step 1.2 shows its action maps $V$ onto itself. These facts make the set closed under products and inverses, so $\mathrm{\Gamma }$ is a group. Step 1.3 gives its scalar kernel, and step 2.1 proves equality with the homogeneous normalizer.

For a vector product, [F1] makes $N(x)$ a real scalar. By [F3], every $x\in \mathrm{\Gamma }$ is $\lambda u_1\cdots u_k$ for $\lambda \in {\mathbb{R}}^{\times }$ and anisotropic $u_j$; $N(\lambda )={\lambda }^2$, so $N(x)$ is scalar. For scalar-valued norms, $N(xy)=xy\bar{y}\bar{x}=xN(y)\bar{x}=N(y)N(x)$, which proves multiplicativity.

The empty product has norm $N(1)=1$. Every normalized vector belongs to $\mathrm{\Gamma }$ by [F3], has inverse $u\mathrm{/}Q(u)=\pm u$, and has norm $N(u)=-Q(u)\in \{1,-1\}$. Thus subgroup words are normalized-vector products, and multiplicativity in step 1.1 shows that they have norm $\pm 1$. In dimension zero Pin is $\{1,-1\}$, and both scalars belong to $\mathrm{\Gamma }$ and have norm $1$. Hence Pin is contained in the asserted norm locus in every dimension.

Conversely, take $x\in \mathrm{\Gamma }$ with $N(x)=\pm 1$. By [F4], factor ${\rho }_x$ into reflections and lift them by normalized vectors to a product $y$ with ${\rho }_y={\rho }_x$. By [F3], $x{y}^{-1}=\lambda$ is a nonzero scalar, and step 1.1 gives $N(x)={\lambda }^{2}N(y)$. Since both norms are $\pm 1$ and ${\lambda }^2>0$, ${\lambda }^2=1$, so $\lambda =\pm 1$. If $\dim V>0$, $-1$ is itself a product of normalized vectors: for $Q(u)=1$, $u(-u)=-1$, and for $Q(u)=-1$, $u^2=-1$. If $V=0$, it is included by definition. Thus every norm-$\pm 1$ element lies in Pin.

The grading parity of a vector product is its number of factors modulo two, and nonzero scalars are even. In positive dimension, the even elements of Pin are therefore exactly the even normalized products, which are ${\mathrm{Spin}}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ by [F2]. In dimension zero every Clifford element is scalar by [F3], so both signs are even and have norm $1$, giving all three groups $\{1,-1\}$. In every dimension the definition of ${\mathrm{Spin}}_{N=1}$ selects the norm-one subgroup; multiplicativity in step 1.1 ensures that it is a subgroup.

Let $A\in O(V,B)$. By [F1], write $A=r_{u_1}\cdots r_{u_k}$. Replacing each anisotropic $u_j$ by $u_j\mathrm{/}\sqrt{\mathrm{\mid }Q(u_j)\mathrm{\mid }}$ preserves its reflection and gives a Pin generator; [F2] shows their product maps to $A$. Thus $\rho$ maps Pin onto $O(V,B)$.

Each normalized vector has determinant $-1$ under $\rho$. Therefore an even product maps into $SO(V,B)$, and every $A\in SO(V,B)$ has an even reflection factorization by [F1]; its normalized lift lies in full Spin. Hence $\rho$ maps full Spin onto $SO(V,B)$.

The kernel of either restriction is contained in the scalar kernel ${\mathbb{R}}^{\times }$ by [F2]. If a scalar $\lambda$ lies in Pin, [F3] gives $N(\lambda )={\lambda }^2\in \{1,-1\}$. Since $\lambda$ is real, ${\lambda }^2>0$, so ${\lambda }^2=1$ and $\lambda =\pm 1$. Conversely both scalars belong to Pin and full Spin and act trivially. This proves exactness with kernel $\{1,-1\}$.

Scalars are central in the Clifford algebra, so the two kernels are central. For a definite form with $n>0$, the normalized-vector set $U=\{u:\mathrm{\mid }Q(u)\mathrm{\mid }=1\}$ is a compact sphere. Given $x\in \mathrm{Pin}$, factor ${\rho }_x$ by [F1] into at most $n$ reflections and lift them to a normalized-vector product $y$. Then $x{y}^{-1}$ is scalar by [F2], and step 1.3 makes it $\pm 1$; hence every Pin element is $\pm$ a product of at most $n$ vectors. If $x$ is full-even Spin, its image has determinant $1$, so [F1] gives an even reflection factorization and the same argument gives an even product of at most $n$ vectors. Each fixed-length product is the continuous image of a finite product of $U$, hence compact; the groups are finite unions of these images. In dimension zero both are the finite group $\{1,-1\}$.

If $n=0$, set $U=\{I_0\}$, $L=0$, and $s(I_0)=1$; this is a continuous section with norm one. For $n>0$, choose a submultiplicative matrix norm and shrink a neighborhood $U$ of $I$ in $SO(V,B)$ until $\mathrm{\parallel }A-I\mathrm{\parallel }$, $\mathrm{\parallel }{A}^T-I\mathrm{\parallel }$, and $\mathrm{\parallel }{A}^{-1}-I\mathrm{\parallel }$ are all less than $1$. Then the series $L=\log A={\sum }_{m\ge 1}(-1{)}^{m+1}(A-I{)}^m\mathrm{/}m$ and the corresponding series for $A^T$ and $A^{-1}$ converge absolutely and continuously. The scalar power-series identities, by absolute Cauchy products, give $\exp L=A$, $\log (A^T)=L^T$, and $\log (A^{-1})=-L$. From $A^{T}JA=J$ one gets $JAJ=A^{-T}$, so conjugating the logarithm series gives $JLJ=-L^T$, equivalently $L^{T}J+JL=0$.

For $i<j$, direct use of [F1] gives $[e_i{e}_j,e_k]=2(B(e_j,e_k)e_i-B(e_i,e_k)e_j)$. The $n(n-1)\mathrm{/}2$ bivectors therefore map under $b\mapsto {\mathrm{ad}}_b{\mathrm{\mid }}_V$ to a basis of the matrices $L$ satisfying $L^{T}J+JL=0$: each coefficient $c_{ij}$ independently gives entries $L_{ij}=2c_{ij}{ϵ}_j$ and $L_{ji}=-2c_{ij}{ϵ}_i$. Thus there is a unique bivector $b(L)={\sum }_{i<j}{c}_{ij}{e}_i{e}_j$ with $[b(L),v]=Lv$ for every $v\in V$, and it depends linearly and continuously on $L$.

Set $s(A)=\exp (b(L))$ in the finite-dimensional Clifford algebra. Since $b(L)$ is even, $\alpha (s(A))=s(A)$. The convergent commutator series gives $s(A)vs(A{)}^{-1}=\exp ({\mathrm{ad}}_{b(L)})(v)=\exp (L)v=Av$, so $s(A)\in \mathrm{\Gamma }$ and $\rho (s(A))=A$.

Reversion negates each bivector, hence Clifford conjugation sends $b(L)$ to $-b(L)$ and $\overline{\exp (b(L))}=\exp (-b(L))$. Therefore $N(s(A))=\exp (b(L))\exp (-b(L))=1$. By [F4], $s(A)\in {\mathrm{Spin}}_{N=1}$. Continuity of matrix logarithm on $U$, of $b(L)$, and of the finite-dimensional algebra exponential proves continuity of $s$, with $s(I)=1$.