Awesome!

]]>I should say that I did not follow details on quiver basic here, but I got from my student Matija who was recently at a winter school on representation theory that as soon as he finishes few homeworks due soon, he plans to start doing a bit of nlabing starting with filling few entries on the role of quivers in the rep theory of fin dimensional algebras, I mean things like tilted modules, Auslander-Reiten theory and so on. He delivered a couple of seminars in Zagreb on this and I was too busy and hence missed the second one.

]]>I would think of a uniqueness of factorization into indecomposables as being that sort of thing.

]]>(PS: I'm not necessarily thinking of a small category, but of course that's immaterial for the present) ]]>

You could probably say something like “there exists a set of indecomposable morphisms such that every morphism factors uniquely as a composite of indecomposables.”

]]>We'll just leave the essential image bit alone - it's not necessary here and not what I meant at all :S

]]>I may not have meant essential image exactly. I meant the collection of objects in the image of the object component of the functor, together with all objects isomorphic to something in the image. An instructive example to think of (I hope) is vector spaces, and the functor sending to . Then the analogue of what I was talking about in the free category case is all finite dimensional vector spaces, not just the collection of s. Note that here is the category with set of objects and only identity arrows.

Sorry for confusing the issue. :)

]]>@David #56: That is a good question. A “free category” should be a category (I hope), and thinking of this example is probably what made me think the image of a functor (i.e. a diagram) should be a category. Is there something special about an adjoint functor that makes its image “essential”. I’m guessing (in the 2 minutes I have to think about this before asking my question) that “essential image” means the image is an object of the target category, so $F(G)$ should be a category.

PS: Ok! I spent an extra 60 seconds checking essential image to verify this.

PPS: If the image of an adjoint functor is always “essential”, that would be a neat bit of trivia we could add to the respective nLab pages.

PPPS: Apologies in advance if I totally botched this.

]]>A related question: is there a characterisation (algebraic, say) of a free category without saying it is in the essential image of the free category functor ? I should hope so.

]]>What is the definition of “generalised Hasse diagram” and when does a category admit one? I wasn’t quite clear from the previous discussion – does it just mean that the category is free on a quiver? If so, then I’m not sure what more of an answer one could expect than “those categories which are free on quivers.”

]]>Thank you David. That helps me a lot more than you might think. Although throwing a symbol $\pi_1$ in there doesn’t help my intuition much, but this statement makes a lot of sense:

Then non-equal arrows should correspond to holes in the space.

I’ve been thinking about non-equal parallel arrows as “paddles” with the hole filled in. So, as everyone already knew from the beginning, my idea was doomed.

Maybe now we can get back to what Toby said:

]]>I want to repeat that I agree with your remark at the beginning of comment #38 that it is interesting to ask which categories admit generalised Hasse diagrams.

Empty space between f and g. Don't forget that everything needs to work for groupoids also, if you claim it for all categories, and so should be interpretable as a space with only non-trival (for various basepoints if not connected). Then non-equal arrows should correspond to holes in the space. And adding even non-invertible higher arrows plays absolute havoc with the space (well, the homotopy type), because then these add non-trivial homotopy groups in (potentially) an infinite number of dimensions.

]]>Yep yep. That is one thing I need to think about.

I printed out some papers on computads and skimmed them on the way home. I also had another look at Street’s AOS paper again. I drew some pictures…

If you were to think of Toby’s example

$x\underset{g}{\stackrel{f}{\rightrightarrows}}y\stackrel{h}{\to} z$as a “space”, would you think of it as a loop on a handle (like the metal of a magnifying glass) with empty space between $f$ and $g$ or as a solid paddle with the space between $f$ and $g$ filled? If you know what I mean, which I would be surprised if you did since that is so vague…

]]>Are you saying that if $f\neq g$, you want to have a 2-morphism $f\to g$ but not $g\to f$? That seems problematic, given that inequality is a symmetric relation; how do you “tell” which direction the 2-morphism goes?

]]>Thanks Toby and Mike.

I need to take a step back in the lab maze because I took a wrong turn, but I still think the idea is pretty. I’m just failing to say it correctly.

I added the second 2-morphism $\beta:g\to f$ at the last minute as I was typing. I should probably spend some time on my wikiweb to think this through. I understand now that having both would mean $f$ and $g$ are equivalent, so my immediate thought is that we should not keep both. We should fill the diagram with just 1 2-morphism $\alpha:f\to g$ (as I originally had in mind).

I think what I’m trying to do should somehow be related to computads. I think I mentioned somewhere the idea could probably be described as a “Hasse computad”.

]]>Toby said what I was going to say: if you have a unique identity 2-morphism between equal arrows, and a unique nonidentity morphism between unequal arrows, then any two arrows are uniquely isomorphic, so your resulting 2-category is equivalent to a poset. So at least if you want to treat this 2-category 2-categorically, then it is not encoding the difference between equality and inequality of arrows.

]]>@ Eric

I’m trying to understand the $2$-category that you propose in response to my example of a category with no generalised Hasse diagram in #39. You would put in $2$-morphisms $\alpha\colon f \Rightarrow g$ and $\beta\colon g \Rightarrow f$, yes? And based on your comment #45, I think that you would also have $\beta \bullet \alpha = 1_f$ (since $f = f$) and $\alpha \bullet \beta = 1_g$ (since $g = g$). And I guess that the whiskerings $h \triangleright \alpha$ and $h \triangleright \beta$ should be the identity on $h \circ f = h \circ g$.

OK, now we have a strict $2$-category, that’s fine. But this strict $2$-category is *equivalent* (as a strict $2$-category) to $a \to b \to c$ (the $1$-category produced from my example by identifying $f$ and $g$, interpreted as a $2$-category). This is because $\alpha$ and $\beta$ are isomorphisms (since they have inverses: each other), so they serve to identify $f$ and $g$ up to isomorphism. Yet in my original $1$-category, $f$ and $g$ are as distinct as distinct can be; they should not be made isomorphic. So your $2$-category (if I understand it) doesn’t really have much to do with my $1$-category.

Even if it is relevant, I don’t see how you can expect to get this $2$-category from some sort of $2$-quiver. After all, it *has* loops: $f \overset{\alpha}\Rightarrow g \overset{\beta}\Rightarrow f$ and $g \overset{\beta}\Rightarrow f \overset{\alpha}\Rightarrow g$.

I want to repeat that I agree with your remark at the beginning of comment #38 that it is interesting to ask which categories admit generalised Hasse diagrams. I just don’t agree with your answer to that question at the end of comment #38.

]]>What???

Eric, you should give the first chapter of HTT a shot before you try to formulate this definition. It appears that you're trying to formulate the definition of a quasicategory, but you aren't really doing it right. It would save you and us a lot of time if you paged through HTT.

]]>PS: I am going to reproduce Domenico’s comment here since it is so cool

I guess this is well known, but let me try writing it here and see what happens.

If a connection on a principal $G$-bundle is locally represented by the 1-form $\omega$ with values in $\mathfrak{g}$, then the connection is flat if and only if the curvature 2-form $F=d\omega+\frac{1}{2}[\omega,\omega]$ vanishes, that is, if $\omega$ is a solution of the Maurer-Cartan equation.

Now, what is interesting is that one can see curvature only on 2-dimensional paths (it is a 2-form): if one restricts $\omega$ to a 1-dimensional submanifold, then there $\omega$ is clearly a solution to the Maurer-Cartan equation. so, if I think of an infinitesimal 1-simplex and look at my connection there, I could say that my connection is 1-flat. then, moving to an infinitesimal 2-simplex I see that the connection is (generally) not 2-flat: holonomy along two sides of the 2-simplex is not the same thing as holonomy along the third side. not the same, but in a very precise way: the curvature $FF$ exactly measures the gap to go from a horn of the 2-simplex to the third edge. this is very 2-categorical, and suggests I could cure the lack of flatness of my original connection by adding a copy of $\mathfrak{g}$ in degree -1 and cooking up a 2-Lie algebra $\mathfrak{l}$. Maurer-Cartan equation for this 2-Lie algebra would coincide with the original one one the 1-simplex (since only degree 1 elements are 1-forms with values in $\mathfrak{l}_0$. but on the 2-simplex we would have, in addition to these elements, also 2-forms with coefficients in $\mathfrak{l}_{-1}$, and the Maurer-Cartan equation on teh simplex would look like $d\omega+\frac{1}{2}[\omega,\omega]+\delta F=0$ where $\delta\colon \mathfrak{l}_{-1}\to \mathfrak{l}_{0}$ is the differential of the $L_\infty$-algebra $\mathfrak{l}$ (and it should be induced by the identity of $\mathfrak{g}$, thought as a degree 1 map from $\mathfrak{g}[-1]$ to $\mathfrak{g}$). So the original equation telling that $\omega$ had curvature $F$ is now equivalent to say that $\omega+F$ is flat.

in other words, what seemed a non-flat connection was so since I was not seeing the 2-bundle, but only a 1-bundle approximation. and on a 1-bundle I can only clearly see up to 1-simplices, wher my connection was actually flat.

once curvature has come in, we can repeat the argument: now we have a 2-flat connection and test it on the 3-simplex. if it has 3-curvature, that will presumibly be because we are not seeing the 3-bundle, yet. so I find it natural to wonder (to conjecture) whether any connection on a principal bundle (and more generally any $n$-flat connection on an $n$-bundle) can be seen as a flat connection on an $\infty$-bundle.

What I’m talking about is *trying* to be a simplified version of this where instead of talking about trivial holonomies, i.e. flat connections, we are talking about commuting diagrams (which I think are closely related)

We can “test” all 1-diagrams to see if they 1-commute. If one does not, we can “fill” it with 2-morphisms to make the individual filled 2-diagram 2-commute. Then we can test all subsequent 2-diagrams to see if they 2-commute. If one does not, we can fill it with 3-morphisms to make it 3-commute. Etc etc.

Something like that anyway…

PS: The prescription chosen for “filling” a category like this could be called something like the “$\infty$-curvature of an $\infty$-category”, where the category itself is essentially an $\infty$-connection.

]]>Hi Mike,

I think (hope!) I am saying something different. The composition of two non-identity 2-morphisms can be an identity 2-morphism. So $f\ne g$ implies a non-identity 2-morphism $\alpha:f\to g$ and $g\ne h$ implies a non-identity 2-morphism $\beta:g\to h$, but if $f = h$, I don’t think there is a contradiction. If $f = h$, it means the composite is the identity 2-morphism, i.e. $\beta\circ\alpha = 1$.

Then, in case $f\ne h$, this implies a non-identity 2-morphism $\gamma:f\to h$ AND a non-identity 3-morphism $\beta\circ\alpha \to \gamma$.

The picture here is very pretty. I would be sad if there was no truth to it.

From what I can tell, this is a pure analogy to Urs’ and Domenico’s discussion of flat connections. A non-flat 1-connection is a projection of a flat 2-connection. A noncommuting 1-diagram is a projection of a commuting 2-diagram. If a 2-connection is not flat, you can fill it to obtain a flat 3-connection. If a 2-diagram does not commute, you can fill it to obtain a commuting 3-diagram. It is very pretty. I wish I could express myself better…

]]>Eric, I feel like we’ve told you a bunch of times that you can’t represent *inequalities* by morphisms in a higher category, because inequalities can’t be composed: if $f\neq g$ and $g\neq h$, it doesn’t follow that $f\neq h$. My apologies if you’re saying something different now, but it seems like the same thing to me.

Just as every connection is flat when viewed in a higher category, every diagram commutes when viewed in some higher category.

This is palpably false.

]]>Right right. Sorry. I stated that very poorly.

What I’m trying to say is that for a 1-category with noncommuting diagrams, there is an $n$-category with “fillers” that fill any noncommutating diagram whose $n$-diagrams commute such that the original category is a projection of this one.

This idea is inspired by Urs’ and Domenico’s discussion and making an analogy between “commutative diagrams” and “flat connections”.

Just as every connection is flat when viewed in a higher category, every diagram commutes when viewed in some higher category.

So there is a 2-category with 2-morphism fillers such that all 2-diagrams commute and your category is a projection. This means there are 2-morphisms $\alpha:f\to g$ and $\beta:g\to f$, but since this 2-diagram commutes, it means $\beta = \alpha^{-1}$. There is a Hasse 2-diagram for this 2-category which projects down to your 1-category.

I hope that is a little better, but I imagine it still is pretty vague (maybe incomprehensibly so). Worse, what I said could be totally wrong…

]]>