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1. • CommentRowNumber1.
   • CommentAuthorKarol Szumiło
   • CommentTimeJul 20th 2013
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   Author: Karol Szumiło
   Format: MarkdownItexLet's say that I have a pointed endofunctor $(S, \sigma)$ of a category $\mathcal{C}$ (i.e $S : \mathcal{C} \to \mathcal{C}$ and $\sigma : \mathrm{id}_\mathcal{C} \to S$). I don't want to assume that $(S, \sigma)$ is well-pointed (i.e. that $S \sigma = \sigma S$). I can form certain colimits that look like naive attempts at constructing free algebras over this functor. Namely, for any $X \in \mathcal{C}$ I could take the colimit $X \to S X \to S^2 X \to \ldots$ where there morphisms are $\sigma_X, \sigma_{S X}, \sigma_{S^2 X} \ldots$ or a similar one with morphisms $\sigma_X, S \sigma_X, S^2 \sigma_X, \ldots$. These are different sequences, but are the colimits also different in general? Note that I don't care whether I obtain free algebras or even algebras at all, I'm only interested in comparison between these colimits. I'm happy to assume that $\mathcal{C}$ is a category of presheaves and that $S$ preserves filtered colimits (and I'm iterating only $\omega$ times).

   Let’s say that I have a pointed endofunctor $(S, \sigma)$ of a category $\mathcal{C}$ (i.e $S : \mathcal{C} \to \mathcal{C}$ and $\sigma : \mathrm{id}_\mathcal{C} \to S$). I don’t want to assume that $(S, \sigma)$ is well-pointed (i.e. that $S \sigma = \sigma S$). I can form certain colimits that look like naive attempts at constructing free algebras over this functor. Namely, for any $X \in \mathcal{C}$ I could take the colimit $X \to S X \to S^2 X \to \ldots$ where there morphisms are $\sigma_X, \sigma_{S X}, \sigma_{S^2 X} \ldots$ or a similar one with morphisms $\sigma_X, S \sigma_X, S^2 \sigma_X, \ldots$. These are different sequences, but are the colimits also different in general? Note that I don’t care whether I obtain free algebras or even algebras at all, I’m only interested in comparison between these colimits. I’m happy to assume that $\mathcal{C}$ is a category of presheaves and that $S$ preserves filtered colimits (and I’m iterating only $\omega$ times).

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJul 20th 2013
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   Author: Mike Shulman
   Format: MarkdownItexInteresting. My initial reaction is "surely they are different" but it seems to be a bit trickier than I expected to come up with a counterexample.

   Interesting. My initial reaction is “surely they are different” but it seems to be a bit trickier than I expected to come up with a counterexample.

3. • CommentRowNumber3.
   • CommentAuthorKarol Szumiło
   • CommentTimeJul 20th 2013
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   Author: Karol Szumiło
   Format: MarkdownItexThat's my experience too. I think that Kan's $\mathrm{Ex}$ might be a counterexample but I find it difficult to analyze. I was also thinking about the case when $S$ is the free monoid functor on the category of sets. Then $S^m \varnothing$ is the set of finite planar trees of height less than $m$. If we take colimit over $S^m \sigma_\varnothing$, then we get the set of all finite planar trees with its standard filtration by height. If we take colimit over $\sigma_{S^m \varnothing}$, we get something that looks different. It's a set whose elements are trees, but some of them are identified and some have multiple copies. However, both sets are countable so even though they "feel different" it's difficult to quantify what exactly it means.

   That’s my experience too. I think that Kan’s $\mathrm{Ex}$ might be a counterexample but I find it difficult to analyze. I was also thinking about the case when $S$ is the free monoid functor on the category of sets. Then $S^m \varnothing$ is the set of finite planar trees of height less than $m$. If we take colimit over $S^m \sigma_\varnothing$, then we get the set of all finite planar trees with its standard filtration by height. If we take colimit over $\sigma_{S^m \varnothing}$, we get something that looks different. It’s a set whose elements are trees, but some of them are identified and some have multiple copies. However, both sets are countable so even though they “feel different” it’s difficult to quantify what exactly it means.

4. • CommentRowNumber4.
   • CommentAuthorjim_stasheff
   • CommentTimeJul 20th 2013
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   Author: jim_stasheff
   Format: TextIs it just my reader or are there all sorts of weird symbols overlapping Karel's text?

   Is it just my reader or are there all sorts of weird symbols overlapping Karel's text?
5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeJul 20th 2013
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   Author: Todd_Trimble
   Format: MarkdownItexLooks fine to me, Jim.

   Looks fine to me, Jim.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJul 21st 2013
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   Author: Mike Shulman
   Format: MarkdownItexOkay, I think this is a case of not thinking trivially enough. Let $S:Set \to Set$ be constant at the 2-element set $2 =\{a,b\}$ and let $\sigma_X:X\to S X$ be constant at $a$. Then the colimit of $\sigma_X, \sigma_{S X}, \sigma_{S^2 X} \ldots$ has 1 element, while the colimit of $\sigma_X, S \sigma_X, S^2 \sigma_X, \ldots$ has 2 elements.

   Okay, I think this is a case of not thinking trivially enough. Let $S:Set \to Set$ be constant at the 2-element set $2 =\{a,b\}$ and let $\sigma_X:X\to S X$ be constant at $a$. Then the colimit of $\sigma_X, \sigma_{S X}, \sigma_{S^2 X} \ldots$ has 1 element, while the colimit of $\sigma_X, S \sigma_X, S^2 \sigma_X, \ldots$ has 2 elements.

7. • CommentRowNumber7.
   • CommentAuthorKarol Szumiło
   • CommentTimeJul 21st 2013
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   Author: Karol Szumiło
   Format: MarkdownItexThat's definitely a counterexample, thanks! Still, I have a feeling that perhaps something positive could be said under some assumptions. If we assume that $\sigma_X$ is always a monomorphism and $S$ preserves monomorphisms, is there still a counterexample? In particular, I would be curious to know whether two ways of iterating $\mathrm{Ex}$ give the same answer.

   That’s definitely a counterexample, thanks!

   Still, I have a feeling that perhaps something positive could be said under some assumptions. If we assume that $\sigma_X$ is always a monomorphism and $S$ preserves monomorphisms, is there still a counterexample? In particular, I would be curious to know whether two ways of iterating $\mathrm{Ex}$ give the same answer.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJul 21st 2013
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   Author: Mike Shulman
   Format: MarkdownItexWell, if we assume that $\sigma_X$ is always a monomorphism and $S$ preserves monomorphisms, then in the category of *sets*, the results are always isomorphic, because a set is determined up to (unnatural) isomorphism by its cardinality, and in this case the cardinality of both colimits will be the supremum of the cardinalities of the sets $S^n X$. I suspect that this will not be the case in other categories, though.

   Well, if we assume that $\sigma_X$ is always a monomorphism and $S$ preserves monomorphisms, then in the category of sets, the results are always isomorphic, because a set is determined up to (unnatural) isomorphism by its cardinality, and in this case the cardinality of both colimits will be the supremum of the cardinalities of the sets $S^n X$. I suspect that this will not be the case in other categories, though.

9. • CommentRowNumber9.
   • CommentAuthorKarol Szumiło
   • CommentTimeAug 6th 2013
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   Author: Karol Szumiło
   Format: MarkdownItexI didn't have time to think about it for the past two weeks, but I am still interested. Of course in the category of sets we can do some cardinality estimates, but this gives us no hint about generalizations to other categories. I was trying to see whether we can at least write down some natural maps between these colimits. I have two ideas, but so far I wasn't able to do anything useful with them. 1. We can write down a big lattice diagram (indexed over $\mathbb{N} \times \mathbb{N}$) so that one of our sequences is its zeroth row and the other one is the zeroth column. Thus we have natural morphisms from colimits of both sequences to the colimit of the big diagram. Are there some reasonable conditions under which these morphisms (or at least one of them) are isomorphisms? 2. It seems that there are no interesting natural transformations between the two sequences, but I have a feeling that there might be some interesting ind-morphisms between them (though I haven't quite succeeded in writing them down) that could be ind-isomorphisms under some conditions. If anyone has any thoughts about these remarks I would very curious to hear them.

   I didn’t have time to think about it for the past two weeks, but I am still interested.

   Of course in the category of sets we can do some cardinality estimates, but this gives us no hint about generalizations to other categories. I was trying to see whether we can at least write down some natural maps between these colimits. I have two ideas, but so far I wasn’t able to do anything useful with them.

   1. We can write down a big lattice diagram (indexed over $\mathbb{N} \times \mathbb{N}$) so that one of our sequences is its zeroth row and the other one is the zeroth column. Thus we have natural morphisms from colimits of both sequences to the colimit of the big diagram. Are there some reasonable conditions under which these morphisms (or at least one of them) are isomorphisms?

   2. It seems that there are no interesting natural transformations between the two sequences, but I have a feeling that there might be some interesting ind-morphisms between them (though I haven’t quite succeeded in writing them down) that could be ind-isomorphisms under some conditions.

   If anyone has any thoughts about these remarks I would very curious to hear them.