Proposition. Isometries transport the Spin vector action [fcap-0029]
Proposition. Isometries transport the Spin vector action [fcap-0029]
Let \(R\) be a commutative ring, let \(M_1,M_2\) be \(R\)-modules with
quadratic forms \(Q_1,Q_2\), and let \(f:M_1\to M_2\) be an \(R\)-linear map
satisfying
\[Q_2(f(m))=Q_1(m).\]
The Clifford universal property gives an algebra homomorphism
\(\operatorname {Cl}(f):\mathcal {C}\kern -2pt\ell (Q_1)\to \mathcal {C}\kern -2pt\ell (Q_2)\) characterized on generators by
\[\operatorname {Cl}(f)(\iota _{Q_1}(m))=\iota _{Q_2}(f(m))\]
CliffordAlgebra.map_apply_ι.
It preserves the Spin carrier and therefore restricts to a group homomorphism
\(\operatorname {Spin}(f):\operatorname {Spin}(Q_1)\to \operatorname {Spin}(Q_2)\)
QuadraticMap.Isometry.spinGroupMap. Concretely, the
Clifford map preserves Lipschitz membership
QuadraticMap.Isometry.map_mem_lipschitzGroup, evenness
CliffordAlgebra.map_mem_even, and Clifford conjugation
CliffordAlgebra.map_star; these are the carrier conditions
separated in Remark [fcap-0028]. Neither construction requires \(f\) to have an
inverse.
Now suppose that \(2\) is invertible in \(R\). If \(a_i\) denotes the Spin
vector action on \(M_i\), then
\[a_2\bigl (\operatorname {Spin}(f)(s),f(m)\bigr )
=f\bigl (a_1(s,m)\bigr ).\]
This identity says that the square with horizontal maps \(a_1,a_2\) and
vertical maps \(\operatorname {Spin}(f)\times f,f\) commutes. It is the vector-action
naturality equation, with no finite-dimensionality or nondegeneracy
hypothesis. If \(f\) is instead an isometry equivalence, it induces a Spin
equivalence
QuadraticMap.IsometryEquiv.spinGroupEquiv, and the same
square packages as naturality of the Spin-to-special-orthogonal homomorphism
source theorem.
A changed Gram matrix shows why the target form is part of the data. Over \(\mathbb Q\), set \(Q_1(x,y)=x^2+y^2\) and \(T(x,y)=(2x,y)\). Keeping the same form on the target does not give an isometry: at \(e_1\), \[Q_1(Te_1)=4\ne 1=Q_1(e_1).\] Correspondingly, the source generator \(\iota _{Q_1}(e_1)\) squares to \(1\), while its proposed image \(\iota _{Q_1}(2e_1)\) squares to \(4\); that assignment cannot define the claimed Clifford map.
For the transported target form \(Q_2(x,y)=x^2/4+y^2\), one has \(Q_2(Tv)=Q_1(v)\). In column coordinates, \[G_1=I,\qquad G_2=(T^{-1})^\top G_1T^{-1} =\begin {pmatrix}1/4&0\\0&1\end {pmatrix},\qquad T^\top G_2T=G_1.\] Now \(\iota _{Q_2}(2e_1)^2=Q_2(2e_1)=1\), so the generator relation is preserved. Thus an arbitrary basis scaling is not a fixed-form isometry; it becomes an isometry only after the domain and target forms are related in the correct direction. This differs from the module equivalence \(T_Q:\mathcal {C}\kern -2pt\ell (Q)\simeq \bigwedge M\) in Lemma [fcap-0004], which is not an algebra map.
Meinrenken derives \(\operatorname {Cl}(f)\) from the same universal property for a bilinear-form-preserving linear map [meinrenken2013clifford, Section 2.2.2, Proposition 2.2, p. 29]. Chevalley's tensor-quotient construction gives the corresponding extension from generator images whose squares are \(Q(x)\) [chevalley1954algebraic, II.1.1, pp. 38--39]. In Meinrenken's convention \(Q(v)=B(v,v)\), so his \(B\) is the half-polar form \(b_Q\) of Convention [fcap-0027]. These sources work over finite-dimensional vector spaces over fields; Meinrenken assumes characteristic different from two. The formal action square above has the commutative-ring hypotheses stated here. This card makes no claim about scalar extension, orthogonal sums, or composition of extension maps.