Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 32 / 2 / Solution

For a Lipschitz continuous path $x:[0,T]\to V$, define its signature of a bounded variation path in the truncated tensor algebra byThe Riemann-Stieltjes integrals exist since $x$ has bounded variation. Equivalently, since $x$ is absolutely continuous, the integrand is $x^{\prime }(u_1)\otimes \cdots \otimes x^{\prime }(u_k)$ with Lebesgue integration on the ordered simplex. The degree-zero component is $1$, andThe lift solves $dS=S\otimes dx$, starting at the identity. Since its velocities belong to the first layer, its endpoint lies in the free step-N nilpotent Lie group.

For $s\le t\le u$, partition the degree-$k$ integration simplex according to how many integration variables lie before $t$. If exactly $j$ do, the integral factors as the first $j$-fold integral on $[s,t]$ tensor the last $(k-j)$-fold integral on $[t,u]$. The boundaries with a time equal to $t$ have zero measure. Thuswhich is precisely the Chen identity:This proves the identity at every level, including $k=0$, rather than just checking the first two levels.

For a strictly increasing continuously differentiable map $\psi :[a,b]\to [0,T]$, the actual change of variables formula isTo prove it, substitute $u_j=\psi (v_j)$ in each ordered integral. Strict increase preserves the order, and the almost-everywhere chain rule gives $d(x\circ \psi )(v)=x^{\prime }(\psi (v)){\psi }^{\prime }(v)dv$. The resulting integral is over $\psi (a)<u_1<\cdots <u_k<\psi (b)$. A vanishing derivative at some points causes no difficulty: monotone continuously differentiable substitution still applies, and the zero derivative contributes no variation.

Therefore orientation-preserving reparametrisation of the whole segment preserves its path signature. On $[0,1]$ the printed equality needs $\psi (0)=0$ and $\psi (1)=1$, equivalently that $\psi$ be onto. These endpoints are not forced by the stated assumptions. For example $x(t)=t{e}_1$ and $\psi (t)=t/2$ satisfy those assumptions, butwhich already differ at level one for every $N\ge 1$. The endpoint-preserving interpretation proves the intended invariance, while the general formula states exactly what remains true without it.