Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
I have touched the first paragraphs, adding hyperlinks (for instance to permutative monoidal category) and making some slight changes. Please check if you can live with it.
I also looked for, found and added a reference. It turns out to be by you! :-)
By the way, since the nLab doesn’t go by a Wikipedia-style “neutral point of view”-paradigm, but instead by the idea that if you are the expert on a subject, then we want you to not shy away from sharing your insights, you are invited to add references to your own articles. In fact you should not put that burden on other shoulders! Like mine in this case.
That’s just the evident consequence of you editing an entry on a given subject in the first place. If you have more to say on that subject, and if you did say it in a preprint or publication, then let us know where to find this. In converse, if your publications are inappropriate for linking on the nLab, then your edits to the nLab must by extension be inappropriate, too, and then we you will hear from the steering committee anyway. :-)
Re #2, possibly the reference is Doctrines of Algebraic Geometry? (Is there a way to link directly to a “published” web from here?)
John, I haven’t thought about this question, but where is it written down the claim about the category of line bundles? I would think that might give inspiration.
Regarding the conjecture, suppose we consider the groupal case. Section 2 of Johnson-Orsono shows that a symmetric groupal groupoid (i.e. Picard groupoid) is equivalent to a skeletal and strictly-associative one, and that the symmetry is determined by a cocycle $c : G\times G \to M$ (where $G$ is the group of objects and $M$ the group of automorphisms of the unit) such that $c(x,y+z) = c(x,y) + c(x,z)$ and $c(x,y) = - c(y,x)$. Two such cocycles $c,c'$ determine equivalent Picard groupoids if they are “cohomologous” in the sense that there is a $k:G\times G\to M$ such that $k(x,0)=k(0,y)=0$ and $k(x+y,z) + k(x,y) = k(x,y+z)+k(y,z)$ and $c(x,y) - c'(x,y) = k(x,y)-k(y,x)$. Now a commutative groupal groupoid would have $c(x,y)=0$ for all $x,y$, while the condition given in the conjecture says $c(x,x)=0$ for all $x$. So the question in that case is, if $c(x,x)=0$ for all $x$, is $c$ necessarily cohomologous to $0$? To say that $c$ is cohomologous to $0$ means there is a $k$ satisfying the cocycle conditions and with $c(x,y) = k(x,y)-k(y,x)$. This last condition certainly implies that $c(x,x)=0$, but we knew that already. It’s not at all clear to me how assuming that $c(x,x)=0$ helps in constructing such a $k$, but I haven’t thought about it for more than a minute.
I changed the syntax of the hyperlink to James Dolan’s page to the more robust double square bracket syntax.
I also changed “we conjecture” to “one might conjecture”.
Maybe you want to have it say: “In Baez-Master 18 it is conjectured…”, which would work. But first person pronouns don’t really work on public wiki pages.
But first person pronouns don’t really work on public wiki pages.
I tend to agree, and yet the nLab has quite a few of them (many from Toby’s hand).
Sometimes I write something like “I (Mike Shulman) conjecture that…”. I think it’s reasonable to write down our conjectures in our lab book, as long as we’re clear about who we are.
I do that too, and also think that’s okay. But I’ve noticed that Toby doesn’t identify himself in many cases.
1 to 16 of 16