0
$\begingroup$

There is this operation you learn in algebraic topology when working with homotopy groups and loops i.e. paths on a topological space $X$, $p:[0,1]\rightarrow X$ with $p(0) = p(1)$. Basically it is an operation $*$ between two loops $p,q:[0,1]\rightarrow X$ defined by

$$ p*q(t):= \left\{ \begin{array}{ll} p(2t) & t\in[0,\frac{1}{2}]\\ q(2t-1) & t\in(\frac{1}{2}, 1]\\ \end{array} \right. $$

It so happened that I work a lot now with paths on a specific topological space (mostly with real manifolds). And I tend to work a lot with this operation not on loops but on paths in general (i.e. $p(0)$ is not necessarily the same as $p(1)$ but...). I redefined this operation on generic paths in such a way that you get the same equality as above if $p(1) = q(0)$ and otherwise we define $p*q = p$ (so in general this is nothing but a concatenation of paths if this can be done otherwise we leave it to be the first path). With this it is easy to see that the set of paths on $X$ and the operation makes a noncommutative semigroup and I am not sure if there is anything more algebraic about it.

I am not an algebraic topologist so obviously I am just going by intuition here and not any past experience. So my question is: is this something of any relevance to people in algebraic topology (or any other field)?

$\endgroup$
3
  • $\begingroup$ Your operation is only associative up to homotopy (perhaps you want to talk about end-point preserving homotopy classes of paths, in which case the fundamental groupoid may be a more natural object to study). The lack of symmetry here is a bit unsettling, too. You could alternatively get a partial monoid by declaring $p\ast q$ to be defined only when $p(1)=q(0)$. $\endgroup$
    – Mark Grant
    Apr 1 '14 at 13:04
  • $\begingroup$ Yes, indeed I don't see much to $*$ here. you can't assume associativity so it isn't strictly a semigroup. Though you can force associativity if you for instance require that $*$ concatenates the two paths by attaching the first path to the second path if the first path endpoint intersects the second path, so halfway thru you have the first path and the other halfway is a section of the second part starting from the first point of intersection of the first path endpoint to the second path (if this intersection exists). $\endgroup$
    – Jose Capco
    Apr 1 '14 at 16:10
  • $\begingroup$ See ncatlab.org/nlab/show/fundamental+groupoid and ncatlab.org/nlab/show/path+groupoid. $\endgroup$ Apr 1 '14 at 22:57
2
$\begingroup$

Check out my answer to https://math.stackexchange.com/questions/706093/ which explains something about Moore paths; concatenation of these gives a category, i.e. is strictly associative with strict identities. The general setting for homotopy classes rel end points of paths is that of groupoids.

$\endgroup$
1
  • $\begingroup$ wow thanks for the link. Something new learned! Basically Moore paths uses the same idea for forcing associativity as proposed by Mark in his comment (to define $p*q$ only when $p(1)=q(0)$.). $\endgroup$
    – Jose Capco
    Apr 1 '14 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.