Fundamental ring of a circle

MathOverflow Asked by Tegiri Nenashi on January 1, 2022

Starting with fundamental group, say of a circle, let’s reflect back to path groupoid a little. The path concatenation operation is partial, but this can be remedied by focusing on the sets of paths, rather than individual paths. Then, if the two sets of paths of the multiplier have no end points coinciding with start points of the multiplicand, then the result is the empty set, which is the additive identity of the ring. Therefore, Boolean algebra of sets provides the additive structure of the ring.

The ring multiplication is inherited from groupoid, that is path concatenation, and it is non commutative. There is a multiplicative identity, which is the set of trivial loops around all points in the space.

To prove the distributivity property, let $P$,$Q$,$R$ be sets of paths. Then

$P (Q cup R) = {p cdot x | p in P wedge x in Q cup R } = $

$= {p cdot x | (p in P wedge x in Q) vee (p in P wedge x in R) } = $

$= {p cdot x | (p in P wedge x in Q) } cup {p cdot x | (p in P wedge x in R) } $

At this point, it is not immediately obvious if any congruence is needed to be introduced. Yet, I’m puzzled why there appears nothing on the web matching the "fundamental ring" search term. Or there is? And what is the ring of path sets of the circle, is it a familiar mathematical object?

Add your own answers!

Ask a Question

Get help from others!

© 2024 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP