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Partition each ordered integration simplex according to the number of times before the concatenation point. The integral over each resulting region factors into an initial iterated integral and a final iterated integral. Summing these factorizations proves the identity in the truncated tensor algebra.

Inside the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the span of iterated Lie brackets of elements of $V$ of lengths at most $N$. Its exponential is the connected simply connected free nilpotent Lie group of step $N$, with tensor multiplication. The bracket is the commutator in the tensor algebra. Its grading defines dilations that multiply degree $k$ by ${\lambda }^k$.

Use the truncated tensor algebra $T^{(N)}({\mathbb{R}}^d)={⨁}_{k=0}^N({\mathbb{R}}^d{)}^{\otimes k}$, with truncated concatenation product. The step-$N$ path signature isThese iterated integrals of a path exist as Riemann-Stieltjes integrals, since a Lipschitz continuous path on a compact interval has bounded variation. The degree-zero coordinate is $1$, and the degree-one coordinate is $x_t-x_s$.

Let $S_t=S_N(x{)}_{0,t}$ and write $S_t^{(k)}$ for its degree-$k$ coordinate. Integrating first over all variables except the last givesThus $d{S}_t^{(k)}=S_t^{(k-1)}\otimes d{x}_t$ for $1\le k\le N$. If $e_i$ is a standard basis vector, define the linear vector field $W_i(g)=g\otimes e_i$, with degrees above $N$ discarded. The component identities are precisely the controlled ordinary differential equationFor a Lipschitz driver this may also be written ${\dot{S}}_t=S_t\otimes {\dot{x}}_t$ almost everywhere. Recursive integration of its levels proves uniqueness and recovers the signature formula. The Chen identity follows by splitting each integration simplex at an intermediate time.

In the truncated tensor algebra $T^{(N)}({\mathbb{R}}^d)$, multiplication isLet ${\mathfrak{g}}^N({\mathbb{R}}^d)$ be the Lie algebra generated by the degree-one space, with commutator $[u,v]=u\otimes v-v\otimes u$ and every term of degree greater than $N$ set to zero. The free step-N nilpotent Lie group isIts operation is the truncated tensor product, its identity is $e=(1,0,\dots ,0)$, and ${\exp }_N(\ell {)}^{-1}={\exp }_N(-\ell )$. Truncated multiplication is associative; the finite Lie exponential and logarithm identify this group with the free nilpotent Lie algebra as a manifold.

The Chow theorem says that on a connected manifold a smooth bracket-generating family of vector fields joins every pair of points by a piecewise smooth horizontal path. Applied to the degree-one left-invariant fields on this group, it says that every $g\in G^N({\mathbb{R}}^d)$ is $S_N(x{)}_{0,1}$ for a piecewise smooth, and hence suitably parametrized Lipschitz continuous, control path $x$ starting at zero.

For $N=d=2$, every Lie element is $v+a[e_1,e_2]$, with $v\in {\mathbb{R}}^2$ and $a\in \mathbb{R}$. The step-2 planar group coordinates giveWe realize this element explicitly. First traverse the rectangleIf $a=0$, omit the loop. Its displacement is zero, and its second-level entries areThus the loop signature is $\exp (a[e_1,e_2])$. Follow it by the straight segment from zero to $v$, whose signature is $\exp (v)=1+v+\frac{1}{2}v\otimes v$. By the Chen identity, concatenation has signaturebecause the commutator lies in the central second layer. Every group element has therefore been realized, proving this case of Chow's theorem.

The Carnot-Carathéodory distance is the infimum of lengths of horizontal controls realizing the relative group element. The construction gives the upper bound $\Vert \exp (v+a[e_1,e_2]){\Vert }_{CC}\le |v|+4\sqrt{|a|}$. Conversely, a control of length $L$ has displacement at most $L$, and its antisymmetric area satisfiesHere the control starts at zero, so its maximum Euclidean distance from zero is at most $L$. These estimates show that the homogeneous group norm is comparable to $|v|+|a{|}^{1/2}$.

The given path is the pure-area rough path ${\mathbf{y}}_t=\exp (t[e_1,e_2])$. Centrality givesThe graded dilations ${\delta }_r(v,a)=(rv,r^{2}a)$ scale horizontal lengths by $r$. Inversion preserves the norm. Hence, with $c=d_{CC}(e,\exp ([e_1,e_2]))>0$,Thus the path is exactly $1/2$-Hölder continuous: on a compact time interval it has every exponent at most $1/2$, and none greater.

Its p-variation is finite precisely when $p\ge 2$. Indeed, the $p$th power of the variation sum is $c^p{\sum }_j(t_{j+1}-t_j{)}^{p/2}$. For $p\ge 2$ it is bounded by $c^p{T}^{p/2}$; for $p<2$, the partition into $m$ equal intervals gives $c^p{T}^{p/2}{m}^{1-p/2}\to \infty$. In the usual step-two rough-path range it is weak geometric for $2\le p<3$, including $p=2$. For $p\ge 3$, the same formula $\exp (t[e_1,e_2])$ in $G^{\lfloor p\rfloor }$ gives the additional levels and a weak geometric $p$-rough path there. The original $G^2$ data alone are a truncated object when those additional levels are required.

In the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the linear span of all iterated Lie brackets of elements of $V$ of length at most $N$, with bracket $[a,b]=a\otimes b-b\otimes a$. Brackets of length greater than $N$ vanish. The free step-N nilpotent Lie group isThe exponential and logarithm are the finite seriesIts multiplication is the tensor multiplication above, its identity is $e=(1,0,\dots ,0)$, and its inverse isThe truncated Baker--Campbell--Hausdorff formula shows that this multiplication stays in the exponential of the Lie algebra. This is the connected simply connected Lie group of the free step-$N$ nilpotent Lie algebra, rather than a discrete free nilpotent group.

Give the first layer $V$ its usual Euclidean norm. An absolutely continuous curve $\gamma$ in $G^N(V)$ is horizontal iffor an absolutely continuous control $x$ in $V$. In the truncated tensor algebra this equation is ${\dot{\gamma }}_t={\gamma }_t\otimes {\dot{x}}_t$, truncated at degree $N$. Its horizontal length is $\int |{\dot{x}}_t|dt$. The Carnot-Carathéodory distance isThe first-layer Lie brackets span the full Lie algebra, so horizontal curves reach every endpoint. Thus this infimum is finite; it is a left-invariant distance inducing the usual manifold topology. Dilations ${\delta }_{\lambda }(a{)}^{(k)}={\lambda }^k{a}^{(k)}$ give $d_{CC}({\delta }_{\lambda }g,{\delta }_{\lambda }h)=\lambda d_{CC}(g,h)$ for $\lambda >0$. Equivalently $d_{CC}(e,g)$ is the least length of a bounded variation control whose step-$N$ path signature is $g$.

For $p\ge 1$ set $N=\lfloor p\rfloor$. A weak geometric p-rough path is a continuous path $\mathbf{X}:[0,T]\to G^N(V)$, usually based at $e$, withIts increments ${\mathbf{X}}_{s,t}={\mathbf{X}}_s^{-1}\otimes {\mathbf{X}}_t$ obey ${\mathbf{X}}_{s,u}={\mathbf{X}}_{s,t}\otimes {\mathbf{X}}_{t,u}$. Equivalently their degree-$k$ tensor components have finite $p/k$-variation and satisfy the path signature algebraic constraints. This is a group-valued definition; being a limit of smooth path signatures in the same $p$-variation topology is the stronger definition of a geometric p-rough path.

Put $V={\mathbb{R}}^d$ and write $a^{(k)}$ for the degree-$k$ component of a tensor. We use the truncated tensor algebraHere $V^{\otimes 0}=\mathbb{R}$, and products of degree above $N$ are discarded. The answers to the three printed parts follow with these conventions; the middle part is missing from the converted TeX.

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.