I am studying rough paths from the 2007 St Flour lecture notes and I came across the example at the end of chapter one of the sequence of paths $X(n):[0,2\pi]\to \mathbb R^2$ given by $X_t(n) = \frac{1}{n} (\cos(n^2t),\sin(n^2t))$, which for each $n$ enclose an area of $\pi$ but converge to the zero path $\mathbf{0}$ in the uniform norm $\|\cdot\|_{\infty,[0,2\pi]}$. To compute the area in the sense I'm referring to, the area of a loop ($\frac{\pi}{n^2}$) must be multiplied by the winding number (number of loops, which is $n^2$).
It can also be shown that the $2$-variation $\|\cdot\|_{2,[0,2\pi]}$ of $X(n)$ is bounded. By using Lemma 1.13 and the fact that $X(n)\to \mathbf{0}$ in the uniform norm, it can be shown that $X(n)$ in fact converges to $\mathbf{0}$ in the $p$-variation norm $\|\cdot\|_{\mathcal{V}^p([0,2\pi])}:= \|\cdot\|_{p,[0,2\pi]} + $$ \|\cdot\|_{\infty,[0,2\pi]}$ for any $p>2$. Hence, the book concludes that the map sending $X\in \mathcal{V}^p([0,2\pi])$ (where $\mathcal{V}^p([0,2\pi])$ is the space of paths with finite $p$-variation) to $A^X_{2\pi}:=\frac{1}{2}[\int_{0<u_1<u_2<2\pi} (\dot{X_{u_1}}\otimes \dot{X_{u_2}})du_1 du_2-\int_{0<u_1<u_2<2\pi} (\dot{X_{u_2}}\otimes \dot{X_{u_1}})du_2 du_1]$ is not continuous in the $p$-variation norm for any $p>2$. The object $A^X_t$ gives the area enclosed by the curve described by $X$ until time $t$, if we complete the curve to a close curve by a straight line and provided the winding number is taken into account if $X$ loops into itself.
I have heard from an expert that "the limit of the paths in the $2$-variation topology is a path with zero increments and positive area". This sounds confusing. I think this is what he meant:
The $2$-variation topology is a topology on $2$-rough paths, and is introduced in Chapter 3. A $2$-rough path $\mathbb X_{s,t}:=(1, X^1_{s,t}, X^2_{s,t})$ is a mapping from the set $\triangle_T :=\{(s,t)| s,t \in[0,T] \:\text{ and } s\leq t\}$ into $V^{\otimes 0}\oplus V^{\otimes 1}\oplus V^{\otimes 2}$, where $V$ is a vector space, whose entries satisfy certain condition. Take a smooth and finite one-variation path $X:[0,T]\to V$, then $X_{s,t} :=(1, X_t-X_s,$$ \int_{s<u_1<u_2<t} \dot{X_{u_1}}\otimes \dot{X_{u_2}}du_1 du_2)$ turns out to be a $2$-rough path (the entries in fact correspond to the iterated integrals (i.e. $X^k_{s,t} := \int^t_{s}X^{k-1}_{s,u}\otimes \dot{X}_u du$). The second entry $X^2_{s,t}:=\int_{s<u_1<u_2<t} \dot{X_{u_1}}\otimes \dot{X_{u_2}}du_1 du_2$ can be decomposed as $X^2_{s,t}:=S^X_{s,t}+A^X_{s,t}$, where $A^X_{s,t}$ has the same interpretation as the "enclosed area" as before (now starting at $s$, not at $0$). It can be shown that $S^X_{s,t}\to 0$ in the uniform norm in the above example, so that $X^2_{s,t}$ tends to $\pi\:(\:= A^{X(n)}_{s,t},\: \forall n)$.
I think that what the expert meant is that, in the $2$-variation topology, the $2$-rough paths associated with each $X_n$ converge to $(1,0,\pi)$, whereas the $2$-rough path associated with the zero path is $(1,0,0)$.
Can someone shed light on the matter? If they don't understand rough paths, maybe they can try to answer the question independently of rough path theory considerations. Can a path have zero increments but enclose a positive area? Can the limit of the sequence of paths I have described in a relevant topology be considered an example of such a pathological path? Or am I right and what Lyons meant is that we can construct a ($2$-) rough path with zero increments and positive area, i.e. $(1,0,\pi)$?