Suppose I have a (homotopy) fiber sequence of spaces $\Omega A \to B \to C$, can I always rotate this to the right once ? i.e., can I extend it to a fiber sequence $\Omega A \to B \to C \to A$ ?

Oh, I'm dumb, I guess yes iff the map on the left $\Omega C \to \Omega A$ is a loop map, is that right ? Is there an "easy" way to check that this map is indeed a loop map, given just the initial fiber sequence $\Omega A \to B \to C$ ?

That's exactly the condition that $B\to C$ is a principal fibration

oh awesome! Thanks. I'll go read on that

To everyone: suppose I have a cosimplicial $E_\infty$-algebra and I understand it completely. I want to take the limit and compute its homotopy groups: those are given by the standard Bousfield-Kan spectral sequence, so no problem there. Now I want to understand the action of the Dyer-Lashof algebra over those (say that my E_\infty-algebra is over F_2). What can I do? That is, does anyone have a nice example where this stuff is worked out?

@DenisNardin maybe Theorem 1.1 in arxiv.org/pdf/1102.0020.pdf looks interesting

Thanks! I'll try to have a look at that, though it talks about spaces rather than E_\infty-rings. However I'm more looking for examples of computations rather than theoretical framework

The H-infinity book I think has lots of computations of DL (and other!) operations on things

ah right yeah, I actually just got curious by your question cuz I'm also using the BK spectral sequence so I googled a few minutes. Do the differentials respect the action by Dyer-Lashof algebra ?

That's sort of ignoring your setup questions though, possibly

Is it "H_infinity-ring spectra and their applications"?

yes

Thank you very much :)

arxiv.org/pdf/1503.08456v1.pdf

@DenisNardin I was going to suggest Phils stuff as well.

The Hinfty book has computations but not of the kind you seem to be asking about. I would try using Phil's thesis. What he does is work out how the operations and the differentials interact. I think this shouldn't be so special as to not work for $E_{\infty}$ ring spectra. They are algebras over the same monad. I guess you might have to mess around a bit with the universal example stuff, but really it should all be fine if you stick to working with ordinary homology.

Oh, he has another paper where he looks at the infinite loop space case, which is closer to what you want.

@SeanTilson Thanks! I still can't believe that there isn't a standard reference on the operation I'm trying to do...

@DenisNardin Did you figure out what was going on with Day convolution in the end?

Uh no sorry. I haven't had time to work on it these days (stuff has come up), but I promise I want to see this through. I think that the multicategory I was describing may exists by chance in the general case as the multiassociativity is trickier than I expected but the proof in Saul's paper still looks good to me, so there must be some kind of trick

Does anyone know what the effect of the Todd genus is on coefficients? Specifically, what map of rings does it induces $MU_* \to KU_*$ or $ku_*$?

@Sean I may be mistaken, but shouldn't it have values more in the periodic rational cohomology?

Hello

no worries

I might ask Tom Leinster later, since he's here for a conference

cool

This may be a long shot, but do ppl know a reference for the following fact:

Let $f : X \to Y$ be a morphism of schemes, locally of finite presentation, finite and flat. Then the locus $\{y \in Y: X_y\to y is \'{e}tale\}$ is open in $Y$.

I looked at Milne's etale cohomology property 2.14 but that proposition is about an open set in $X$.

Should I ask on mathoverflow?

You mean the locus $\{y\in Y : X_y\to y is étale\}$?

yea sorry.

I'm not sure I would expect that to be true

In that case isn't it just the complement of the support of $\Omega^1_{X/Y}$?

Well it better be, because I need it to show that the diagonal of $\mathscr{M}_g$ is representable @ZhenLin

That's what I guessed too

Well we have to be careful, I think we want it to be the complement of the support of $\Omega^1_{X/Y}$ as an $\mathcal{O}_Y$-module

because Kahler differentials are a sheaf on $X$.

Yep, so it's the same thing as the complement of the pushforward, which I suppose isn't always coherent

well under a finite morphism it will be

anyway the problem is I don't know what coherent means here

?

The schemes are possibly non-noetherian

Sure, no problem there

Do you know what coherent means for a non-noeterian ring?

well IIRC there is one but in most cases the category is like empty or something right

It means finitely generated + that every finitely generated submodule is finitely presented

Ok.

If your ring is coherent as a module over itself it is equivalent to finite presentation

Right.

So here's a question

If we have a map of rings A \to B which is finitely presented

then $\Omega^1_{B/A}$ is also finitely presented as a $B$-module yes

@DenisNardin Ah ok if I want to prove my proposition can I reduce to the case of $Y,X$ being affine?

I think so, because $f$ is finite

If f is finite I think you're basically done for what we have said

ah ok I think it's ok

everything is ok now :D

$\Omega^1_{B/A}$ is still a finite $A$-module since $A \to B$ is finite

@DenisNardin I am surprised for a homotopy theory guy you seem to have a good command of schemes

Maybe that's what 11 courses at MIT does to you :D

Well I started as an algebraic geometer

Moreover a certain brand of homotopy theory guys use a lot of schemes

Ok, that explains how your previous life is affecting your current life :D

@DenisNardin Actually we don't even need loc. finite presentation

All I used was the fact that the morphism is finite, which makes sense given that the relevant exercise in Olsson doesn't assume loc. finite presentation

doesn't this follow from a result of Thomason, since it is the complement of the support of Omega^1

@Adeel which result of thomason?

let me try to find the reference

Isn't the proof just as I have written above?

i haven't read what you wrote

but Thomason proved that if you have a perfect complex, the complement of its support is a quasi-compact open subscheme

Since $f : X \to Y$ is finite, we can reduce to the case that $Y,X$ are affine.

but yes this is probably overkill

Then we are looking at a ring map $A \to B$ that is finite flat

and then $\Omega^1_{B/A}$ is finite as an $A$-module. Its support as an $A$-module is closed, and we now just take the complement of that in Spec A @Adeel

@Adeel As much as I like perfect complexes why can't you just use coherent modules? :)

:D

I don't know anything about perfect complexes

by coherent modules do you mean finitely generated modules?

No, I mean coherent modules

(of course they coincide if the ring is Noetherian)

by Thomason's result it should hold without the finite and flat hypotheses, since by finite presentation, Omega^1 should be compact

@DenisNardin From what I've heard this category, even though nice is most often empty or something?

I don't know what a coherent module is

I don't believe that, at least if the ring is coherent

@Adeel finitely generated + every finitely generated submodule is finitely presented

stacks.math.columbia.edu/tag/05CU

If the ring is coherent it should be the smallest abelian subcategory containing fg free modules or something like that

i see

@Adeel Anyway my original motivation for the question was to show that a certain subfunctor of a certain hilbert scheme is open

They're used mostly in complex geometry, where asking for your rings to be noetherian is a bit too much

anyway, I think in the derived setting, where the cotangent complex is not concentrated in degree zero, you have to use this Thomason-style result, or the analogue of it. at least that's the only proof i know

That's way beyond me atm.

yeah, just trying to explain why this overkill approach is the first thing that came to mind for me

I was trying to use it to show that the diagonal of M_g is representable

Staying in the topic of perfect complexes, does anyone knows when the restriction of a perfect complex to a closed subscheme (say an effective Cartier divisor) is perfect?

very rarely, probably?

@BenLim since you are working with M_g you don't need to bother about non-noetherian things. It's a general fact that, for X locally of finite type over some base S, the R-points of X, for arbitrary R, are determined by the A-points of X, for A of finite type over S.

Hmm in my case is just the inclusion at zero $X\to X\times A^1$, but I suppose this won't work even there

@DenisNardin I am confused: the pullback of a vector bundle is always a vector bundle.

@bananastack You're right, I was confusing perfect complexes and pseudocoherent complexes. Thanks a lot :)

i was also confused, it's always true

Well, we should never complain when things are way easier than we thought they would be. :)

@bananastack Are you sure that you don't need X locally of finite presentation?

@AndrewSenger for what? I think if he wants to prove something for M_g over Z it doesn't matter as the integers are noetherian. (but I think you are right, I should be saying finite presentation, I think I was confusing myself with the case of modules)

Nice! arxiv.org/abs/1503.08456

Been wondering about that for a while...

yeah, looks like a big step

although surely not the end of the story

Certainly not. I haven't looked at it yet really, but I'm interested in the methods actually used.

@bananastack For the general statement of the result that you quoted above. And I'm not sure it make sense to show that the diagonal of M_g is representable using the fact that M_g is locally of finite type over Z: how do you make sense of local finite type-ness of a stack without already knowing that its diagonal is representable?

@DenisNardin are you coming to the UVA conference?

Instead, one should reduce to the Noetherian case using the finite presentedness of genus g curves. (I guess that one could define locally of finite presentation for stacks as a 2-version of the definition for contravariant functors on the category of schemes and declare that a stack over a Noetherian base is locally of finite type iff it is locally of finite presentation, but this seems backwards...)

@AndrewSenger you are probably right

@DenisNardin Well, what is the $E_{\infty}$ complex orientation of $ku$ then? It isn't one that kills all the higher $x_i$'s, that would violate Johnson-Noel.

@Jon No I'm not unfortunately.

Ah too bad!

@Sean What complex orientation of ku are you talking about? The one induced by the complex orientation of KU?

If that's the one it is given by the multiplicative formal group law, so in principle you should be able to explicitely compute the image of all x_i's (but it may be a bit painful)

@DenisNardin There must be multiple orientations that induce the multiplicative formal group. I am curious about the one that is $E_{\infty}$. Maybe this is in something of @TylerLawson's

suppose I have a category with direct sums

what more do I need to assume to ensure that any retraction is the projection onto a direct summand?

what does "category with direct sums" mean? does it mean Ab-enriched category with finite biproducts?

if so, nothing

if not, i don't know what a direct summand is

no, it means a category with finite coproducts and finite products such that the morphism from the former to the latter is iso

in general, it's enriched in commutative monoids, not Ab

okay. well, certainly Ab-enrichment is enough. if m is the corresponding idempotent, then m and 1 - m respectively are the projections onto the direct summand and its complement

sure

but I don't know I can subtract

(and don't you need to first say that the category has a zero object before you talk about "the morphism from the former to the latter"?)

maybe idempotent completeness is the right condition for me - it's certainly enough

right, sorry

yeah, idempotent completeness is a pretty dank condition

Empty coproduct ---> Empty product exists and is an iso, so zero object :P

i guess it's necessary and sufficient that if m is the corresponding idempotent then there's some other idempotent m' such that 1 = m + m'...?

@Jeremy: right, that works, but it seems like you have to say that separately first

oh, that's not right - idempotent completr means that every idempotent comes from a retraction, not that it comes from a direct summand

oh, yeah, i was being silly. i guess commutative monoids itself is a counterexample

I'm worried I might have to enforce the existence of some kind of fibers or cofibers

wouldn't commutative monoids continue to be a counterexample? I bet you can find a small commutative monoid with a retract that isn't a direct summand

probably take something as far away from an abelian group as possible... maybe a semilattice

e.g. the commutative monoid {0, 1, 2} with operation min(-, -) appears to have {0, 1} as a retract but not a direct summand

If A-->B-->A is your retraction, does saying the A splits off imply that the other summand is fib(B--->A)? Is this property equivalent to the existent of fibers for the morphisms of interest?

it does if you're Ab-enriched

the main reason i think commutative monoids is a counterexample is that i bet you can find lots of commutative monoids with idempotent endomorphisms that don't have complements

like endomorphism semirings of commutative monoids should be as badly behaved as arbitrary semirings

e.g. in the above example, the projection {0, 1, 2} -> {0, 1} (given by rounding down) has fiber {0}

that example is quite funny in a way – the obstruction to being a direct summand is number-theoretic!

anyway, other than not being Ab-enriched I bet CMon has a lot of lovely properties as a category (e.g. it's certainly complete and cocomplete) so I don't think completeness / cocompleteness-type conditions are the way to go

yeah ok

stable is definitely enough though

well, if you are thinking about kernels, then it might be worth thinking about the Mal'cev condition

what's that?

category-theoretically, the Mal'cev condition says that every reflexive binary relation is an equivalence relation

For my own curiosity, is it not the case that a retraction A--->B--->A in an additive category splits if and only if fib(B-->A) exists?

but in practical terms, it ensures that kernels are useful

@Jeremy: I don't follow. what you've written down is a thing that is already split...

sorry by splits I mean Saul's B--->A is a projection onto a direct summand

@Jeremy: then yes, I believe so

Ok, thank you. I was trying to get intuition about why Saul hoped a condition other than the existence of fibers in general might suffice.