Local Spin lift from the matrix logarithm

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 $s(I)=1$.