= Solution
In the <truncated tensor algebra> $T^{(N)}(\mathbb R^d)$, multiplication is
$$
(a\otimes b)^{(k)}=\sum_{i+j=k}a^{(i)}\otimes b^{(j)}.
$$
Let $\mathfrak g^N(\mathbb R^d)$ 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> is
$$
G^N(\mathbb R^d)=\exp_N\bigl(\mathfrak g^N(\mathbb R^d)\bigr),\qquad \exp_N(\ell)=\sum_{k=0}^N\frac{\ell^{\otimes k}}{k!}.
$$
Its operation is the truncated tensor product, its identity is $e=(1,0,\ldots,0)$, 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> give
$$
\exp(v+a[e_1,e_2])=1+v+\tfrac12v\otimes v+a(e_1\otimes e_2-e_2\otimes e_1).
$$
We realize this element explicitly. First traverse the rectangle
$$
(0,0)\to(r,0)\to(r,\sigma r)\to(0,\sigma r)\to(0,0),\qquad r=\sqrt{|a|},\quad\sigma=\operatorname{sgn}(a).
$$
If $a=0$, omit the loop. Its displacement is zero, and its second-level entries are
$$
\int x^1\,dx^2=\sigma r^2=a,\qquad\int x^2\,dx^1=-a,\qquad\int x^i\,dx^i=0.
$$
Thus the loop signature is $\exp(a[e_1,e_2])$. Follow it by the straight segment from zero to $v$, whose signature is $\exp(v)=1+v+\tfrac12v\otimes v$. By the <Chen identity>, concatenation has signature
$$
\exp(a[e_1,e_2])\otimes\exp(v)=\exp(v+a[e_1,e_2]),
$$
because the commutator lies in the central second layer. Every group element has therefore been realized, proving this case of Chow's theorem.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-31-chow-loop.png]
{title=A rectangular signed-area loop followed by a straight endpoint segment}
The <Carnot-Carathéodory distance> is the infimum of lengths of horizontal controls realizing the relative group element. The construction gives the upper bound $\|\exp(v+a[e_1,e_2])\|_{CC}\leq|v|+4\sqrt{|a|}$. Conversely, a control of length $L$ has displacement at most $L$, and its antisymmetric area satisfies
$$
|a|=\left|\tfrac12\int(x^1\,dx^2-x^2\,dx^1)\right|\leq\tfrac12 L^2.
$$
Here 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 gives
$$
\boxed{\mathbf y_s^{-1}\otimes\mathbf y_t=\exp((t-s)[e_1,e_2])=\left(1,0,\begin{pmatrix}0&t-s\\-(t-s)&0\end{pmatrix}\right).}
$$
The graded dilations $\delta_r(v,a)=(rv,r^2a)$ scale horizontal lengths by $r$. Inversion preserves the norm. Hence, with $c=d_{CC}(e,\exp([e_1,e_2]))>0$,
$$
\boxed{d_{CC}(\mathbf y_s,\mathbf y_t)=c|t-s|^{1/2}.}
$$
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 $p\geq2$. Indeed, the $p$th power of the variation sum is $c^p\sum_j(t_{j+1}-t_j)^{p/2}$. For $p\geq2$ it is bounded by $c^pT^{p/2}$; for $p<2$, the partition into $m$ equal intervals gives $c^pT^{p/2}m^{1-p/2}\to\infty$. \b[In the usual step-two rough-path range it is weak geometric for $2\leq p<3$, including $p=2$.] For $p\geq3$, 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.
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