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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 $V={\mathbb{R}}^2$ with basis $e_1,e_2$, the step-two Lie algebra has basis $e_1,e_2,[e_1,e_2]$ and its last element is central. WriteIts first level is $v=(x,y)$ and its second level isThe step-two multiplication givesConsequently the mapis an isomorphism onto the real Heisenberg group. The shift by $xy/2$ converts exponential coordinates into the matrix coordinates used for that group. In exponential coordinates the left-invariant horizontal vector fields areTaking $X,Y$ orthonormal gives exactly the Carnot-Carathéodory distance inherited from $G^2({\mathbb{R}}^2)$.

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.

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 first level of this group path is zero but its second level grows linearly. Its Carnot-Carathéodory distance increments are a constant times $|t-s{|}^{1/2}$. It therefore has finite $p$-variation precisely when $p\ge 2$, and is a step-two weak geometric rough path at $p=2$, even though its first-level trace is constant. For higher truncations the same Lie exponential defines a compatible lift.

For $p\ge 1$, the displayed quantity measures the accumulated oscillation of a path in a metric space. For $p=1$ it is total variation. A Hölder continuous path of exponent $\alpha$ has finite $p$-variation on a compact interval if $p\alpha \ge 1$. Applying this construction to the Carnot-Carathéodory distance gives the defining regularity of a weak geometric p-rough path.

Write a group element as $\exp (v+a[e_1,e_2])$. The step-two Baker-Campbell-Hausdorff formula gives the displayed multiplication and inverse $(-v,-a)$. The scalar $a$ is the antisymmetric second signature coordinate, which equals the signed area for a closed planar path. The Carnot-Carathéodory distance is comparable to $|v|+|a{|}^{1/2}$ near the identity, with the same homogeneous scaling globally.

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.