In what sense do the categorical trace and coend count fixed points? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:29:56Z http://mathoverflow.net/feeds/question/68424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-21T19:30:45Z 2011-06-22T22:54:09Z <p>According to the nlab, the categorical trace of a 1-endomorphism $F:C\to C$ in a 2-category is the set hom$(1_C, F)$ of global elements of $F$. If $F$ is a functor in the 2-category Cat, the categorical trace is a set of natural transformations that assign to each object of $C$ a coalgebra of $F$ such that the obvious square commutes.</p> <p>Any functor can be considered a special kind of profunctor; given an endofunctor, we can compute the coend of the corresponding profunctor.</p> <p>Both of these concepts are generalizations of the trace, which for a function counts the number of fixpoints. In what sense do these "count" the fixpoints of a functor? I don't see how the categorical trace of a functor relates to fixpoints at all.</p> <p>Also, does the notion of what constitutes a fixpoint change? The coend, in particular, seems like it might count an object $c$ as a fixpoint of $F$ if it's in the same <em>endomorphism</em> class rather than the same <em>isomorphism</em> class as $Fc$.</p> http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points/68537#68537 Answer by Mike Stay for In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-22T17:21:31Z 2011-06-22T17:21:31Z <p>Here's a partial answer: in the case of an endofunctor $F$ on a discrete category $C$ (i.e. $F$ is a function), the coend of $F$ gives the <em>set</em> of fixpoints rather than the <em>number</em>: A profunctor $F:C \not\to C$ adds extra morphisms to $C$ so that the result is still a category. I'll say these morphisms are "in $F$". The coend of $F$ is the set of endomorphisms in $F$ mod conjugation by the morphisms in $C$; since the morphisms of $C$ are all identities, we just get the set of endomorphisms in $F$, i.e. fixed points of $F$.</p> <p>The categorical trace doesn't reduce to anything useful in the case of a discrete category. A natural transformation $\alpha:1_C \Rightarrow F$ chooses for each $c \in C$ a morphism $\alpha_c:c \to Fc$ in $C$. Since we're assuming all the morphisms in $C$ are identities, $\alpha_c$ can't exist unless $Fc = c$. So it looks to me like the set hom$(1_C, F)$ is empty unless $F$ is the identity functor on $C$, in which case it's the terminal set.</p> http://mathoverflow.net/questions/68424/in-what-sense-do-the-categorical-trace-and-coend-count-fixed-points/68564#68564 Answer by Mike Stay for In what sense do the categorical trace and coend count fixed points? Mike Stay 2011-06-22T22:54:09Z 2011-06-22T22:54:09Z <p>Simon Willerton explains it all very well here: <a href="http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf" rel="nofollow">http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf</a></p>