# Non-uniqueness of loop-erasure for continuous-time curves

Question. Is there a continuous curve in the plane that has a non-unique loop-erasure?

Here is the definition of a loop-erasure. A continuous curve $$Y:[c,d]\to\mathbb R^2$$ is a loop-erasure of a curve $$X:[a,b]\to\mathbb R^2$$ if there exists an increasing and right-continuous function $$w:[c,d]\to [a,b]$$ such that:

• $$w(c)=a$$,
• $$X(w(d))=X(b)$$,
• $$Y(t)=X(w(t)), \forall t$$,
• For every $$T$$, the image of the curve $$Y(t), c\leq t\leq T$$ does not intersect the image of the curve $$X(s), w(T).

Note that if $$w$$ has a jump at time $$t$$, then one should have $$X(w(t^-))=X(w(t))$$. This corresponds to erasing a loop in $$X$$.

Update. Two loop-erasures are equivalent if they have the same image.

• Couldn't two erasers of $X$ erase different loops of $X$? – Jochen Wengenroth Sep 24 at 12:02
• @JochenWengenroth No, the definition implies that $Y$ should not intersect itself. So, for instance, if $X$ has finitely many self-intersections, then one should erase all of the loops and there is a unique loop-erasure. – Ali Khezeli Sep 24 at 15:40
• Okay, I see now. – Jochen Wengenroth Sep 24 at 15:52

The paper that Iosif Pinelis mentioned in his answer has an example to this problem: Consider the compact space obtained by adding $$\pm\infty$$ to the strip $$\{ z\in \mathbb C: 0\leq \mathrm{Im}(z)\leq 1\}$$. Consider the curve that connects the integer points of this set (by segments) in the following order: $$\ldots, n, n+1, n+i, n+i+1, n+1, n+2,n+i+1,n+i+2,n+2,\ldots$$. Then there are two non-equivalent loop-erasures of this curve: One has image $$\mathbb R\cup\{\pm\infty\}$$ and one has image $$(\mathbb R+i)\cup\{\pm\infty\}$$.

Zhan, Dapeng, Loop-erasure of planar Brownian motion, Commun. Math. Phys. 303, No. 3, 709-720 (2011).

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – András Bátkai Sep 24 at 7:46
• @AndrásBátkai According to your comment, I edited my answer and included the example here. – Ali Khezeli Sep 24 at 15:39

$$\newcommand\tt{\tilde t}\newcommand\tY{\tilde Y}\newcommand\tw{\tilde w}$$A loop-erasure is never unique whenever it exists (see e.g. this paper for a nontrivial example of such an existence):

Zhan, Dapeng, Loop-erasure of planar Brownian motion, Commun. Math. Phys. 303, No. 3, 709-720 (2011). ZBL1217.60076.

Indeed, if a continuous curve $$Y\colon[c,d]\to\mathbb R^2$$ is a loop-erasure of a curve $$X\colon[a,b]\to\mathbb R^2$$, witnessed by a function $$w\colon[c,d]\to[a,b]$$, then for any real $$s$$ the curve $$[c-s,d-s]\ni\tt\mapsto\tY(\tt):=Y(s+\tt)$$ is also a loop-erasure of the curve $$X$$, witnessed by the function $$[c-s,d-s]\ni\tt\mapsto\tw(\tt):=w(s+\tt).$$

• I forgot to say that such examples are considered equivalent. I will update the question. But the paper that you mentioned has already an example for my question! Thanks. – Ali Khezeli Sep 24 at 5:12