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The Stratonovich enhancement of Brownian motion is a geometric $p$-rough path for every $p>2$. Its second level consists of Lévy area and the prescribed symmetric tensor. Its increments satisfy the Chen identity and have finite Hölder rough-path norms at every exponent below $1/2$.

Let $\mathbf{X}$ be a weak geometric p-rough path, $N=\lfloor p\rfloor$, with increments $X_{s,t}^I$ indexed by words $I=(i_1,\dots ,i_k)$ of length at most $N$. Choose a control function for rough paths $\omega$ with $|X_{s,t}^I|\le \omega (s,t{)}^{|I|/p}$. We use the standard sufficient vector-field hypothesis $V_i\in {\mathrm{Lip}}^{\gamma }$, $\gamma >p$, with globally bounded regularity norms; smooth bounded fields are a special case. Regard $V_i$ as the differential operator $V_i\cdot \nabla$, and put $F_I=V_{i_1}\cdots V_{i_k}\mathrm{id}$.

A pathlevel solution of a rough differential equation is a continuous state path $y:[0,T]\to {\mathbb{R}}^e$, with prescribed $y_0$, satisfying a local expansionfor some $\theta >1$ on sufficiently small control intervals. One can take $\theta =\min (\gamma ,N+1)/p>1$ under the stated regularity. This definition specifies the state path, without requiring its full rough-path enhancement as additional data. For $2<p<3$, the displayed expansion readsRepeated indices are summed.

We use the following discrete form of the Davie lemma. For a finite ordered grid and a continuous superadditive control, suppose nonnegative errors $a_{s,t}$ vanish on adjacent grid points and, for every grid triple $s<u<t$, satisfyThere are $\delta >0$ and $C$, depending on $L,A,p,\theta$ but not on the grid, such that $a_{s,t}\le C\omega (s,t{)}^{\theta }$ whenever $\omega (s,t)\le \delta$. This is the discrete higher-order error estimate: splitting close to half the control size makes the two $\theta$-power errors contract; a crossing single grid interval contributes zero adjacent error.

Define the local Euler mapFor small $\omega (s,t)$ it has Lipschitz constant at most $1+L\omega (s,t{)}^{1/p}$. Taylor expansion, the Chen identity and the shuffle identity for iterated integrals give the composition estimateHere is the cancellation underlying the estimate. Assign degree $k$ to a level-$k$ increment. Expanding each $F_I(E_{s,u}(z))$ and grouping terms through degree $N$ gives the differential-operator coefficients $F_{IJ}(z)$ multiplying $X_{s,u}^I{X}_{u,t}^J$. Products of coefficients from the first increment are combined by the shuffle identity, which is the iterated Leibniz rule. Chen's identity says that the sum of these products is $X_{s,t}^{IJ}$. Thus every term of degree at most $N$ agrees with the corresponding term in $E_{s,t}$. The remaining Taylor terms have control order at least $\theta$; bounded ${\mathrm{Lip}}^{\gamma }$ norms make the estimate uniform in $z$. For smooth fields the remainder has order $(N+1)/p$.

On a partition $\pi$, construct the Euler sequence by $y_{t_{j+1}}^{\pi }=E_{t_j,t_{j+1}}(y_{t_j}^{\pi })$. For grid endpoints letIt vanishes on adjacent points. Insert $y_u^{\pi }=E_{s,u}(y_s^{\pi })+I_{s,u}^{\pi }$ into the next local map. The Lipschitz and composition estimates giveThe Davie lemma therefore yields a bound $|I_{s,t}^{\pi }|\le C\omega (s,t{)}^{\theta }$ independent of $\pi$ on small control intervals. Together with the Euler expansion this gives $|y_t^{\pi }-y_s^{\pi }|\le C^{\prime }\omega (s,t{)}^{1/p}$. A finite subdivision into small-control intervals gives a uniform global bound as well.

Interpolate each Euler sequence linearly in time, and take partitions with mesh tending to zero. Continuity of the control makes these paths uniformly equicontinuous: the grid estimates apply to the interval enlarged to its neighboring grid endpoints, and the enlargement tends to zero uniformly with the mesh. The paths are uniformly bounded. The Arzelà-Ascoli theorem supplies a uniformly convergent subsequence with continuous limit $y$.

For arbitrary $s<t$ in a small control interval, choose neighboring grid endpoints converging to $s,t$. Continuity of $\mathbf{X}$, the coefficient functions and the control lets us pass to the limit in the grid error estimate. We obtain the defining expansion with $|R_{s,t}|\le C\omega (s,t{)}^{\theta }$, and $y(0)=y_0$. This constructs a pathlevel RDE solution. No continuity theorem for the RDE solution map was used to obtain existence.

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.

For Brownian motion $B$ starting at zero, define its enhanced Brownian motion by the Stratonovich second level:The associated group-valued path is ${\mathbf{B}}_t={\mathbf{B}}_{0,t}$. The Stratonovich chain rule givesConsequently its symmetric second level is $\frac{1}{2}{B}_{s,t}\otimes B_{s,t}$ and the remaining antisymmetric part $A_{s,t}=\frac{1}{2}({\mathbb{B}}_{s,t}-{\mathbb{B}}_{s,t}^{\mathsf{T}})$ is Lévy area. Splitting the integral at $u$ proves the Chen identity, so these increments lie in $G^2({\mathbb{R}}^d)$ and are the relative increments of ${\mathbf{B}}_t$.

The group norm for the Carnot-Carathéodory distance is comparable to $|B_{s,t}|+|A_{s,t}{|}^{1/2}$. The planar construction in Question 2 extends to dimension $d$: prescribe every coordinate-plane area with a closed rectangle, concatenate these central-area loops, and finish with the displacement segment. The converse bound follows from length bounds on displacement and on each area integral. Thus the comparison constants depend only on $d$.

Brownian scaling and stationarity of increments imply that $(B_{s,t},A_{s,t})$ has the distribution of $(\sqrt{t-s}{B}_{0,1},(t-s)A_{0,1})$. The permitted moment estimates for Lévy area, together with Gaussian moments of Brownian displacement, therefore give, for every finite $q\ge 1$,This also verifies the hypothesis of the metric-valued Kolmogorov continuity theorem. We give the dyadic argument to obtain a simultaneous version explicitly.

Fix $0<\beta <1/2$ and choose $q$ so large that $q(1/2-\beta )>1$. The Markov inequality and a union bound giveThe right-hand side is summable. By the Borel-Cantelli first lemma, almost surely there is a finite random constant $K_{\beta }$ bounding every adjacent dyadic increment by $K_{\beta }{2}^{-n\beta }$, including the finitely many exceptional levels.

Decompose any interval with dyadic endpoints into successive dyadic intervals, with at most two intervals of each size and no size exceeding its length. The metric triangle inequality and the sum of the geometric series givefor all dyadic $s,t$. The finite-dimensional Carnot group is complete, so the path extends uniquely from dyadic times to a continuous $\beta$-Hölder path. At each fixed time the original stochastic enhancement and its values at approaching dyadic times converge in probability by the moment estimate. The extension has the same limit almost surely, hence is a modification of the original enhancement.

Perform this construction on one countable sequence of exponents increasing to $1/2$. The extensions agree on the dense set of dyadic times, so they give one modification that is Hölder at every exponent below $1/2$. On $[0,1]$, a $\beta$-Hölder estimate implies the same estimate at every $0\le \alpha \le \beta$. Therefore

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.

Fix $T<\infty$ and a standard $d$-dimensional Brownian motion $B$ starting at zero. Let $D_n$ be deterministic nested finite partitions of $[0,T]$ with mesh tending to zero; dyadic partitions are the basic example. Let $B^n$ be the polygonal interpolation through the partition values and put ${\mathbf{B}}^n=S_2(B^n)$. The limit is enhanced Brownian motion, with incrementsHere the integral is a Stratonovich integral. In terms of Itô integrals,Its symmetric part is $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, while its antisymmetric part is Lévy area,These identities and splitting the integral at an intermediate time give the Chen identity, so $\mathbf{B}$ has values in $G^2({\mathbb{R}}^d)$.

A precise nested polygonal approximation of Brownian rough paths statement is: for every $0<\alpha <1/2$,where the rough path metric used here isIn particular, for each $2<p<3$, choosing $1/p<\alpha <1/2$ gives convergence in the inhomogeneous $p$-variation rough path metric. The limiting signed area is the Lévy area, and the limiting full second level is Stratonovich, not Itô. This distinction is necessary: a polygonal lift has the prescribed symmetric second level, whereas an Itô lift would lack half the quadratic-variation correction.

Here is a proof sketch emphasizing the nested martingale structure. SetA Brownian bridge between two prescribed endpoints has the affine interpolation as conditional mean. ConsequentlyFor $i\ne j$ the coordinate processes remain independent when their respective partition values are conditioned on. Approximating the Itô integral by its left sums, conditioning each product, and then taking its $L^2$ limit givesIndeed each conditioned factor is its polygonal interpolation, and their sums converge to the ordinary integral of those piecewise linear functions. For a fixed pair $s,t$, the right side is therefore a martingale in the refinement index $n$. It is not being asserted to be a martingale in physical time; polygonal interpolation uses the right endpoint of its current cell.

The union of the partitions is dense. Continuity of $B$ implies that ${\mathcal{G}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{G}}_n)$ contains the full Brownian path and its stochastic integrals, up to completion. The martingale convergence theorem for conditional expectations therefore yields convergence of $B_t^n$ and all off-diagonal ${\mathbb{B}}_{s,t}^{n,ij}$ at every fixed pair, almost surely and in $L^q$ for finite $q$. Simultaneously it holds for a countable dense collection of pairs. The diagonal terms must be handled separately:They are not conditional expectations of the diagonal Brownian second level: the latter would also contain a conditional variance.

Pointwise convergence alone is insufficient for the rough path metric. The Burkholder-Davis-Gundy inequalities, Brownian scaling and Gaussian increment moments give, for arbitrarily large $q$,Equivalently, using the homogeneous norm on $G^2$, the $q$-th moment of the Carnot-Carathéodory distance of an increment is bounded by $C_q|t-s{|}^{q/2}$. The metric form of the Kolmogorov continuity theorem, with $q$ chosen large, gives for every $\beta <1/2$ a random $K_{\beta }$ having all finite moments such thatConditioning these bounds givesThe diagonal bounds follow by squaring the first bound. The two conditional-expectation processes are nonnegative martingales; the Doob Lp maximal inequality makes their suprema over $n$ finite almost surely, with all needed moments. Thus all the polygonal lifts have a common random bound at exponent $\beta$.

These bounds and the Chen identity give joint equicontinuity of the increments as functions of $(s,t)$. Convergence on the dense collection of pairs consequently gives uniform convergence of both levels on the time triangle. Take $\alpha <\beta <1/2$. For either level $k=1,2$, with $M$ a common bound and ${\epsilon }_n$ its uniform difference, interpolation givesThis proves the stated Hölder rough path metric convergence. Finally, for a partition $D$, the $k$-th level difference satisfiesbecause $\alpha p>1$. Taking the appropriate powers and suprema proves $p$-variation convergence too. Nesting is what permits the conditional-expectation martingale proof; without nesting one instead needs direct approximation estimates.

The path signature collects the ordered iterated Riemann-Stieltjes integrals of a continuous path of bounded variation. Its truncation belongs to the free step-N nilpotent Lie group. It multiplies under concatenation by the Chen identity, is independent of orientation-preserving reparametrisation of the whole path, and records more than its endpoint displacement.

A weak geometric $p$-rough path is a continuous path in the free step-N nilpotent Lie group, with $N=\lfloor p\rfloor$, whose increments have finite $p$-variation for the Carnot-Carathéodory distance. Its multiplicative increments are ${\mathbf{X}}_{s,t}={\mathbf{X}}_s^{-1}\otimes {\mathbf{X}}_t$. They obey the Chen identity and the algebraic symmetries of path signatures.