# Is path lifting property is utilized in Rough path theory?

I am trying to learn rough path theory as described in https://en.wikipedia.org/wiki/Rough_path. In this wikipedia page "rough path lift" is mentioned. I have gone though the following link which talks about path lifting property https://www.youtube.com/watch?v=snHIkGVlmkM .

Is there a connection between path lifting property mentioned in the youtube video and "rough path lift" mentioned in the wikipedia page.

An explanation how they are related or unrelated would be highly appreciated.

My intuition is that can we assume the d-dimensional rough path space as X as mentioned in the video. Further, is Y of video is the space of iterated integrals or something like that? In such case what are $$\phi$$ of the video or $$p$$ of the video?

• @verret, thanks for the edit. The rough path theory talks about group like element so my intuition was that a principle bundle may be associated in the question so I added the group theory. Oct 26, 2018 at 17:58
• Crossposted to MO: mathoverflow.net/questions/314807/…. In future, when you do so, update your post with a link to the other one. Nov 8, 2018 at 3:02
• @NateEldredge I wrote in the mathoverflow that I have asked the question here, but did not give the link, sorry. Nov 8, 2018 at 3:13

They're both just two examples of the generic notion of "lifting": given some spaces $$X,Y$$ with a "projection" map $$p : Y \to X$$ (typically very far from injective), for each $$x \in X$$, identify, in a "canonical" way, some $$y \in Y$$ with $$p(y) = x$$.
In the video, $$X,Y$$ are topological spaces and $$p$$ is a covering map.
In rough paths theory, $$X$$ could be the space of ordinary paths, $$Y$$ the space of "enhanced" rough paths (where an element consists of a path together with some higher-order objects that can be seen as iterated integrals of the path), and the map $$p$$ is "forget the higher-order stuff". So "lifting" just means making a "good" choice for what the higher-order stuff ought to be, in a particular context.
• May I ask where $\phi$ stands in rough path? Is it just inverse of $p$? Nov 8, 2018 at 3:47
• The map $\varphi$ from the video doesn't have an analogue in rough paths. There, the idea was that $p$ is a covering map and the set $X_i$ is evenly covered, so $p^{-1}(X_i)$ is homeomorphic to the product of $X_i$ with a discrete set $D_i$, and $\phi$ is defined as that homeomorphism. The projection $p$ in rough paths isn't a covering map and the preimage of a set doesn't have the same kind of structure. Nov 8, 2018 at 3:57