Mathematics - 2025-01-24

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leslie townes

00:02

or honorific, i guess, not honorary

Martin Brandenburg

heyho

Ben Steffan

hi

Martin Brandenburg

I wonder if the category of partial orders and order-preserving maps has a cogenerator. the category of preorders has a cogenerator, namely {0 <= 1 <= 0}, but that is not a partial order, of course. Based on this I believe that there is no cogenerator, but I am not sure how to see this.

(a cogenerator is an object Q such that for every pair of distinct parallel morphisms A ->-> B there is a morphism B -> Q which doesn't coequalize them)

Thorgott

the interval {0<1} does not work?

Martin Brandenburg

if f,g : A -> B are coequalized by all B -> {0 < 1}, this means that f^(-1)(T) = g^(-1)(T) for all upsets T c B

ok take T the upset generated by b

then f(a) >= b iff g(a) >= b, so yeah, works

cool

what about the category of small categories?

with {0 <= 1 <= 0}, seen as indiscrete category, we can already separate objects

but morphisms?

Thorgott

00:20

looking at $BM$ for a monoid $M$, this would imply monoids have a cogenerator

I know groups don't have a cogenerator, do monoids?

Martin Brandenburg

monoids do neither

2

Q: Cogenerator in category of groups/monoids/rings

An object $Q$ in a category $C$ is called cogenerator (coseparator) if for any pair of arrows $f,g \in C(X,Y)$ if $f \neq g$ then there is $h \in C(Y, Q)$ such that $hf \neq hg$. For example, any set with at least two elements is a cogenerator in the category of sets, $\mathbb{Q}/\mathbb{Z}$ is a...

category-theory

but not sure why you can reduce it to that

Thorgott

ah, the same argument works, nice

Martin Brandenburg

if Q is a cogenerator of Cat, how do you produce a cogenerator for Mon?

Thorgott

if $\mathcal{C}$ were a cogenerating category, then $\prod_{x\in\mathrm{Ob}(\mathcal{C})}\mathrm{End}(x)$ should be a cogenerating monoid

cause a functor $BM\rightarrow\mathcal{C}$ is the same thing as a monoid hom $M\rightarrow\mathrm{End}(x)$ for some object $x$ of $\mathcal{C}$

Martin Brandenburg

ah, right. thanks!

what about metric spaces? (with non-expansive maps)

not sure if we can use Urysohn's lemma

Urysohn's Lemma implies that the unit interval is a cogenerator in the category of metric spaces with continuous maps, right?

Thorgott

00:31

yeah, as well as in the category of normal spaces

Martin Brandenburg

ok. I am not sure what I should think of metric spaces with non-expansive maps, tho. usually that category is much better behaved.

but my feeling is that no cogenerator exists

perhaps I should say that I mean classical metric spaces and/or Lawvere metric spaces, both are interesting

Martin Brandenburg

00:46

ok I will ask in the forum :), cya

Thorgott

00:56

yeah, I have no clue (though certainly the interval doesn't work as a cogenerator would need to have infinite diameter), cya

3 hours later…

copper.hat

03:31

@robjohn I hope that you are not directly affected by things down south.

1 hour later…

one potato two potato

05:00

arxiv.org/abs/2501.13761 "the holomorphic fibrations of a closed Kähler manifold over hyperbolic 2-orbifolds can be recovered from the profinite completion of its fundamental group."

1 hour later…

leslie townes

06:06

@copper.hat said the actress to the bishop

4 hours later…

Homesick Iguana

10:01

hi

2 hours later…

Homesick Iguana

11:31

1

Q: A $K$ Bessel function functional equation?

Consider the function$$ R(s)=\int_{(0,1)}~ \sum_{n=1}^{\infty} e^{\frac{n^2}{\ln(x)}}x^{s-1}~dx.$$ Does $R(s)$ satisfy $$\frac{1+2R(s)}{1+2R(1/s)}=1/s^t $$ for some $1<t<2?$ Apparently $t=1.63$ is nearly exact. $R(s)$ can be seen as a Mellin transform of a kind of Jacobi theta function. Furthermo...

functional-equations analytic-number-theory bessel-functions theta-functions mellin-transform

this actually pretty neat when you think about it

the usual Jac(s) function (jacobi theta) supports this very functional equation listed for $t=1/2$

so why should the mellin transform of a Jac(s) like function satisfy the same form?

the parallels are curious: the mellin transform of a certain jacobi theta maps to the completed riemann zeta supporting the f(s)=f(1-s) type functional equation on the crit. strip.

psie

12:14

I have a question. Let $\mathcal M_\alpha=\sigma(\mathcal A_\alpha)$ be a $\sigma$-algebra generated by $\mathcal A_\alpha$, for $\alpha\in I$. Consider now $$\mathcal M=\bigcup_{\alpha\in I}\sigma(\mathcal A_\alpha).$$Is it possible to say something like the union of the $\mathcal A_\alpha$ generate $\sigma(\mathcal M)$? Seems unlikely, but I wonder what is the relationship between the $\mathcal A_\alpha$ and $\sigma(\mathcal M)$?

Binky

Hi

Homesick Iguana

hi

Ben Steffan

@psie Find a counterexample to the claim you find unlikely, and then think about whether one side is always contained in the other, say.

Binky

If I'm taking a test and I fully understand the subject of the course, but I get stuck on a technical step or mechanic that isn't part of the course content and can't move forward without solving it, is it more fair to look up the solution (like copying that step from somewhere) to complete the exercise and demonstrate my understanding of the subject, or should I stay stuck and leave the exercise incomplete, accepting the block?

psie

12:40

@BenSteffan ok 👍

Sine of the Time

@Binky copying during a test?

Binky

I mean to copy the step that is not part of the course, because maybe I don't remember how to do it, I know how to do the rest but not this step

Sine of the Time

why don't you try to understand it before the test?

Binky

I'm assuming this has never happened to me before

(the step)

Pizza

maybe you should ask your teacher about this and see what he tells you

Sine of the Time

12:54

that may happen

if it happens you usually have all the tools to arrive to the solution

or did not do enough exercises to see all possible cases

Pizza

maybe you tell him that you know how to continue but you're stuck on that step, obviously if he tells you he can't help you you have to stay stuck or try to move forward

Sine of the Time

If this has never happened to you, where will you search during the exam? Doesn't sound like a good idea

Binky

Like a calculator

Sine of the Time

I don't see how a calculator can help

Binky

online calculator on phone

Sine of the Time

13:08

I'd not cheat

Pizza

like wolfram i guess

Sine of the Time

@Pizza if you use wolfram then you know the steps

Binky

So you're telling me that if I have to do, for example, a calculus exam and I get to an exercise that I completely know how to do (those in the program) but I get stuck on one that, for example, requires a step, you accept the block and therefore do the exercise wrong, or use a calculator ONLY to move forward (I'm not telling you to see everything)

Sine of the Time

bro if you're not allowed to use a calculator then using it is cheating

are you saying it's morally acceptable?

Binky

No

It was just to know the opinions on this thing

Sine of the Time

13:17

As I said before, if you don't know how to solve an exercises you either have all the tools to find the solution or did not prepare well

Which exam do you have in mind? Calculus? Linear algebra?

In these kind of exams you don't have to "invent" anything, the exercises are standard

In more advanced exams it'll happen but that's part of the game

Binky

Do you mean that they are non -standard exercises?

Sine of the Time

yeah in more advanced exams it'll happen

For example if you have to compute the norm of an operator there's not a "road map"

You have to see each time what's the best strategy

Binky

What do you think of those who cheat all the exams?

Sine of the Time

I think it's morally wrong

Binky

Once I happened to see a person cheating with the phone during the exam and pass it also

Pizza

13:29

If someone keep cheating on all the exams, I wonder why hes still in university or college etc. Maybe he should think about whether this is really what he wants to do.

Thorgott

13:57

@Ben by the way, it turns out that the claim in Moishe's answer linked in Achim's answer is quite a bit subtle

can you think of an argument that a countable, finite-dimensional CW-complex is equivalent to a locally finite, countable CW-complex of the same dimension?

(it's definitely true, don't worry, but not so easy to see; it's also easy if you allow the dimension to increase by 1, which suffices if you don't want dimension estimates)

Xander Henderson

14:13

While I don't think that cheating is a moral issue (it is the result of a nexus between incentives, pressure, and opportunities), it is not something that I would encourage.

When a student cheats, they deprive themselves of an education, and they negatively impact other students, as well (e.g. by throwing off the curve and changing expectations for what a "good" grade should look like). In an environment where cheating is common, the students who don't cheat are punished.

And, frankly, if you can't work your way through a derivative because you don't know how to add two fractions, you should not be in a calculus class.

(You didn't describe what kind of "block" you were talking about, but that is the kind of thing that I commonly see---students who don't know the prereqs struggling with basic arithmetic).

Binky

15:12

for example I am evaluating a limit using Taylor expansions, I know what to do but I get stuck because I don't know if a - or + is needed for an expansion. I'm assuming I have to remember everything by heart

So I could try to see from my notes to be sure to write the right sign or risk writing the wrong sign and making mistakes all the exercise

Even if I know how to continue the exercise

Xander Henderson

@Binky I mean, there are principles involved. You should know those principles.

So you either figure it out, or you make a choice and continue with the problem, or you throw up your hands and give up.

Vladimir Lysikov

If you know the theory behind Taylor expansion and derivatives, then you can check what the sign should be

Sine of the Time

It's not that hard to memorize the Taylor series of the most important functions

Binky

it was just an example

Sine of the Time

Some teachers allow you to keep the notes during the exam

handan_toddler

15:24

ok, let's hear about another example

Sine of the Time

the only topic that may be challenging from this point of view is ODE

Thorgott

@SineoftheTime this is pretty standard for math here

Sine of the Time

We usually can keep one handwritten sheet with the formulas

In other exams you're allowed to keep everything but that won't help you :D

Binky

@SineoftheTime also the phone ?

Xander Henderson

@SineoftheTime And even if you don't memorize them, you should understand how they work. It doesn't take that long to integrate by parts once or twice to see the first couple of terms of the general Taylor series expansion, and then generalize.

Or to work out the first couple of derivatives.

Sine of the Time

15:29

@Binky no not the phone

Xander Henderson

If you actually understand the material that you are being tested on, you should not need to cheat.

Binky

@XanderHenderson How do you understand if someone used cheats in a test?

Xander Henderson

@Binky Are you asking how not to get caught?

Sine of the Time

Plus if you do a lot of exercises you'll automatically internalize the first terms of the expansion

Xander Henderson

Why should I explain that to you?

@SineoftheTime Indeed. Things become automatic.

Binky

15:32

I mean how do you understand from his work if he has cheated or not

Xander Henderson

@SineoftheTime I prefer the term "internalize", though.

Over "memorize".

@Binky I'm not answering that.

Sine of the Time

thumb up emoji

Binky

because I have seen people use the phone and pass the test without ever being caught

Sine of the Time

you'll always encounter cheaters in life

Binky

So I think that sometimes you can't notice it

Xander Henderson

15:35

Oh, well, if everyone else cheats, it must be okay. [THIS IS SARCASM]

Binky

So if there is a person next to me who cheats, can I do it too?

I think no

Xander Henderson

@Binky No!

Thorgott

I took one exam once where we were allowed to use the internet

Sine of the Time

bro...

Xander Henderson

Do you know what sarcasm is?

Jeebus...

I guess you are just not allowed to be sardonic on the internet...

Sine of the Time

15:38

Study instead of wasting time elaborating strategies :D

Xander Henderson

Honestly... it takes more time to figure out elaborate ways to cheat than it does to just learn the bloody material.

Like, you've wasted an hour here that you could have used to study...

Binky

But if someone is cheating, can I tell the teacher of this or not?

Xander Henderson

Personally, I think it is totally appropriate to let the instructor know.

Binky

Ah okay , however I have never cheated, but I saw people do it.

Xander Henderson

I don't really care. You are asking a lot about cheating. You are asking questions which sound like you are trying to figure out how to get away with cheating, questions which suggest that you are looking for the line regarding what is acceptable, and questions which suggest that you would really like to figure out how to cheat better.

Stop.

You have better things to be thinking about.

Ben Steffan

15:51

@Thorgott no, I'm not good with these sorts of things

Thorgott

16:06

fair, I think this one is genuinely difficult to boot

you have to artificially enlarge the $0$-skeleton and then ???

Ben Steffan

I have an intuitive idea for how it should go up to 2-skeleton but beyond that I don't know

and even that idea seems, well, unfun to formalize

say you have countably many 2-cells $e_2^i$, $i \in \mathbb{Z}$ attached via multiples of some $\gamma$ representing a 1-cell, for simplicities sake. How do you pull these apart? You could replace $\gamma$ by an (infinite cylinder) $\mathbb{R} \times S^1$ and glue in $e_2^i$ on the 1-cell $\{i\} \times S^1$

if the attaching words are more complicated you'll have to do more tinkering

Thorgott

16:23

yeah, it's not at all clear

Ben Steffan

actually hold on maybe this works: Inductively assume the $k$-skeleton is locally finite (let's take $k = 1$ as our induction base). To "locally finitize" the $(k + 1)$-skeleton, replace $X_k$ by $X_k \times \mathbb{R}$ and pick some ordering $e^i_{k + 1}$, $i \in \mathbb{Z}$ on the $(k + 1)$-cells you're attaching, then attach $e^i_{k + 1}$ to $X_k \times \mathbb{R}$ via the composite $S^k \to X_k \cong X_k \times \{i\} \hookrightarrow X_k \times \mathbb{R}$.

These attaching maps are cellular, so this gives a CW-complex, and at each stage the new complex $X_{k + 1}'$ obtained via this procedure is homotopy equivalent to $X_{k + 1}$ via the obvious linear homotopy equivalence $X_k \times \mathbb{R} \simeq X_k$

Since $X_k \times \mathbb{R} \hookrightarrow X'_{k + 1}$ is a cofibration, this should yield a locally finite CW-complex $X'$ homotopy equivalent to $X$

$k = 0$ also works

1 hour later…

psie

17:39

@BenSteffan are you interested in seeing an argument for my claim earlier today? I actually don't have any use of it anymore I think, but in case you are interested, I think my "unlikely" claim was true :) i.e. the $\sigma$-algebra generated by an arbitrary union of $\sigma$-algebras (which in turn have their generating sets), is generated by the union of the individual generating sets of the sigma algebras in the arbitrary union.

Thorgott

@Ben I'll think about that in a second, meanwhile I think I've proven (and am massively shocked by its generality): If $K(G,n)$ has homology concentrated in only finitely many degrees, then either $n=1$ and $G$ is a subgroup of $\mathbb{Q}^d$ for some $d\ge1$ or $n\ge2$ and $G=\mathbb{Q}^d$ for some $d\ge1$.

Ben Steffan

well your unlikely claim is false so let's hear it @psie

@Thorgott !!

that's huge

Thorgott

I claim: For $n\ge2$, the reduction mod $p$ morphism $K(\mathbb{Z},n)\rightarrow K(\mathbb{F}_p,n)$ induces a non-zero map on mod $p$ cohomology in infinitely many degrees. (Proof: Rely on the algebra description in Tamanoi's paper.)
Then, this map obviously cannot factor through a $K(G,n)$ with only homology (and hence mod $p$ cohomology) in finitely many degrees (this is generalizing Will Sawin's idea), so $G$ has no $p$-torsion. Thus, $G$ is divisible, but we already know Prüfer groups have homology in infinitely many degrees and an infinite product of $\mathbb{Q}$s does too by Künneth, …

psie

Ok :) well recall we had $$\mathcal M=\bigcup_{\alpha\in I}\sigma(\mathcal A_\alpha),$$and I was wondering about whether $\sigma(\mathcal M)$ is generated by $\bigcup_{\alpha\in I}\mathcal A_\alpha$.

So, for every $\beta\in I$, we have $\sigma(\mathcal{A}_\beta)\subset \sigma(\bigcup_{\alpha\in I}\mathcal{A}_\alpha)$, therefore $\mathcal{M}\subset \sigma(\bigcup_{\alpha\in I}\mathcal{A}_\alpha)$, so also $\sigma(\mathcal{M}) \subset \sigma(\bigcup_{\alpha\in I}\mathcal{A}_\alpha)$.

On the other hand, since $\mathcal{A}_\alpha \subset \sigma(\mathcal{A}_\alpha)$, we have $\bigcup_{\alpha\in I}\mathcal{A}_\alpha \subset \mathcal{M}$, therefore $\sigma(\bigcup_{\alpha\in I}\mathcal{A}_\alpha) \subset \sigma(\mathcal{M})$.

Thorgott

(The $n=1$ case is a separate argument.)

To see the claim, we have to do it by cases. Note that the map clearly pulls back the fundamental class of $K(\mathbb{F}_p,n)$ to the mod $p$ fundamental class of $K(\mathbb{Z},n)$.
1. If $p=2$, the mod $2$ cohomology ring in either case is a polynomial algebra and has the (mod $2$) fundamental class as a generator, so their powers are non-trivial and the pullback map is non-trivial in their degrees.
2. If $p$ is odd, the mod $p$ cohomology ring is a tensor product of a polynomial algebra on the even-dimensional and an exterior algebra on the odd-dimensional generators.

Ben Steffan

17:54

@psie are the $\mathcal{A}_\alpha$ subsets of the set on which you are defining your $\sigma$-algebras or of its powerset?

psie

I would say the power set. They are all subsets of the power set of some set $X$.

Ben Steffan

then I've misunderstood your intent, sry :)

psie

no worries :) I'm not really sure how this question arose in the end

Ben Steffan

@Thorgott right, that looks good

NoSingularities

18:18

Suppose one finds the standard deviation of a sample called "No aging" to be 4.86 and the standard deviation of a sample called "Aging" to be 6.49. My text says these numbers show "the variation in 'Aging' is smaller than the variation in 'No Aging'" but isn't it the opposite here?

Thorgott

18:32

@Ben my claim for $n=1$ was obviously nonsense, though

there are plenty of non-trivial $K(G,1)$ with only homology in finitely many degrees, e.g. any aspherical manifold

but I still think I can prove (by a separate argument) that $K(G,1)\simeq M(G,1)$ only happens for subgroups of $\mathbb{Q}$

Ben Steffan

yeah, the case $n = 1$ is different

Thorgott

cause if that is the case, it's clearly finite-dimensional (construct an $M(G,1)$ explicitly), hence so is its universal cover, which is also the universal cover for any $K(H,1)$ with $H\le G$. thus, $G$ is torsion-free since $K(\mathbb{Z}/p\mathbb{Z},1)$ is not finite-dimensional. this implies $G$ embeds into $G\otimes\mathbb{Q}$, which is a $\mathbb{Q}$-vector space. it can only have dimension $1$ cause you otherwise collect homology in higher degrees by Künneth.

Ben Steffan

but non-simply connected spaces are bad etc. :)

@Thorgott I assume you're assuming $G$ is abelian here

Thorgott

yes, otherwise the question is not really well-formed

though we could ask when $K(G,1)\simeq M(G^{ab},1)$ happens, hmm

Ben Steffan

perhaps it is best to leave the non-abelian groups be

Thorgott

18:42

the first half of the above argument still goes through and $G$ is torsion-free, but then I don't know

anyway, I'll try to type up the argument for MO and then hope nobody comes around and tells me I completely misunderstood the description of the cohomology rings

Ben Steffan

always a risk, but to the best of my knowledge this checks out

although I haven't verified the excess calculation

Thorgott

excess for odd $p$ is super weird (and apparently there's multiple inconsistent conventions lol)

but I stuck with what Tamanoi gives and it should work out

Ben Steffan

yeah I saw the description in Tamanoi's paper and did not bother to parse it

I guess the situation at odd primes is generally considered annoying, enough so that textbooks don't treat it

Thorgott

right, the $p=2$ case is in Hatcher (though I never read it)

it's kind of weird that, in the odd case, $P^1$ has excess $2$, but it checks out cause you don't want $P^1\iota$ to be a generator for the cohomology of $K(\mathbb{Z},2)$

there's a divergence cause the Bockstein becomes a separate entity in the odd case, whereas it's just $Sq^1$ for $p=2$

Ben Steffan

that, and you now also have signs

I guess these two things are related

Thorgott

18:52

right, signs are why everything for $p=2$ is polynomial, whereas for odd $p$ it's polynomial in even and exterior in odd degrees

but for some reason, for $p=2$ and $\mathbb{Z}/2^k$, the $k=1$ and $k>1$ case also have an asymmetry, I can't really explain that one

Ben Steffan

Yes, that one is weird

Thorgott

also, the mod $p$ cohomology algebra of $K(\mathbb{F}_p,1)$ is already surprisingly complicated compared to the $p=2$ analogue

it's the tensor product of an exterior factor generated by the fundamental class and a polynomial algebra generated by its Bockstein, so weird

the homology looks analogous to $p=2$, but the ring structure is completely different

Ben Steffan

Actually, on that one I disagree somewhat

It's the closest that ring can be to a polynomial ring on a generator in degree 1

you could also note that $H^*(K(\mathbb{F}_p, 1); \mathbb{Z}) \cong \mathbb{Z}[\tilde{\iota_1}] / (p\tilde{\iota}_1)$ for all $p$ and then ask what ring structures with $\mathbb{F}_p$-coefficients are compatible with this

PM 2Ring

19:10

All primes are odd, except for 2, which is the oddest prime of all.

10

Q: Why are even primes notable?

There are much-discussed theorems like Fermat's theorem on sums of two squares which make statements about odd primes only. This makes $2$ seem to be a "special" prime. In their book The book of numbers, Conway and Guy accordingly state that "Two is celebrated as the only even prime, which in som...

number-theory prime-numbers soft-question

Ben Steffan

indeed

what's strange in algebraic topology is that in most of the "basic" stuff you'd learn from lectures the situation at $p = 2$ is particularly nice, like it is here

but at some point the situation inverts and then $p = 2$ is the case where things don't work out all of a sudden

especially stably

Thorgott

19:24

@BenSteffan oh cool, I never calculated this

yeah, it's not too strange, but it's still a far cry from a polynomial algebra (because it has to be for graded-commutativity reasons)

PM 2Ring

@BenSteffan Right. More on that at the linked question:

78

Q: What's so special about characteristic 2?

I've often read about things which do not work in a field with a characteristic $2$, mainly things which have to do with factoring, or similar things. I'm not exactly sure why, but the only example of such a field I could think of is $\mathbb{Z}/2\mathbb{Z}$, which itself is an interesting field ...

soft-question field-theory finite-fields big-list

copper.hat

@leslietownes i am such a prude, always speaking in euphemisms...

CroCo

19:40

Hi everyone, is there anyone here who is good at geometric algebra (GA)? I am seeking advice on how GA can enhance collision detection. I did ask this question in Computer Graphics but I didn't get any solid answer. The question is

I have a strong background in computer graphics mathematics, particularly linear algebra, including rotation and translation, which I gained from my background in robotics. I have started learning modern OpenGL (i.e., shaders), and I feel I'm moving relatively fast due to my strong background in C/C++ coding. I want to move toward computer haptics (CH) (i.e., so…

--------------------------
At present, I am employing dual quaternion algebra, which allows me to effortlessly represent points, lines, and planes. Distances can be easily obtained; however, they can also be obtained using vector algebra. I'm attempting to find a gap in terms of collision detection, but to no avail.

Ben Steffan

@Thorgott this is also in hatcher, somewhere. It's not too hard: A model for $K(\mathbb{F}_p, 1)$ is given by the infinite-dimensional lens space $L^\infty_p := S^\infty / (\mathbb{Z} / p)$

@PM2Ring I see no topology there :^)

a prime (ha) example of what I'm talking about would be that the mod 2 moore spectrum $\mathbb{S} / p$ does not support the structure of a homotopy ring spectrum at $p = 2$ but does so for odd primes

or the fact that the height 1 morava $K$-theory spectrum $K(1)$ at $p = 2$ somehow has two non-equivalent $A_1$-ring structures, neither of which is obviously preferable to the other :^)

Thorgott

@BenSteffan yeah, I just don't know (at least without putting effort into it) how to compute the ring structure

Ben Steffan

but perhaps one should expect things like this, in particular in the chromatic theory, given that it's at least partially arithmetically geometric in nature

@Thorgott I wonder if it's actually easier to compute $H^*_{\mathrm{grp}}(\mathbb{Z} / p, \mathbb{Z})$

I don't have a lot of experience with group cohomology

Thorgott

20:05

I don't think so

best you can do is write down an explicit resolution and compute, which is the same as doing a cellular computation for an appropriate CW-model

Ben Steffan

come to think of it, this is a straightforward computation with the Serre specseq

take $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} / p \to 0$, apply $K({{-}}, 1)$ and then just compute

if all you have is a hammer...

Thorgott

20:25

@Ben I suppose now we wait mathoverflow.net/a/486563/517429

@BenSteffan is it clear this is orientable for $p=2$

Ben Steffan

yes

The fibration you obtain from an exact sequence of groups $1 \to A \to B \to C \to 1$ by applying $K({{-}}, 1)$ is orientable iff $A$ lies in the center of $B$

according to an exercise in Hatcher I haven't done, admittedly

...but also do you really need the case $p = 2$ :^)

Thorgott

20:42

no, but I felt like being a pedant

but that's a good Exercise, I suppose it makes sense

Ben Steffan

one would very much hope it's true

Thorgott

21:06

@Ben The point is this, I think: The conjugation by an element $b$ induces a map $K(B,1)\rightarrow K(B,1)$ that is homotopic to the identity. This restricts to a map on the fiber $K(A,1)\rightarrow K(A,1)$ corresponding to the restricted conjugation on $A$ (this is a self-map of $A$ since it's a normal subgroup). The restricted nullhomotopy witnesses that this map on the fiber is the one induced by the image of $b$ in $C=\pi_1(K(C,1))$.

by the UP of EML-spaces, the base acts trivially on the fiber iff the conjugation by any element is the identity on $A$ iff $A$ is central

(I suppose acting trivially on the homology of the fiber is a strictly weaker condition than acting trivially on the fiber itself, but the sufficient condition is enough for us)

Ben Steffan

yes, something like that

I remember that these questions about local systems and actions in fibre sequences were the bane of my existence when I first learnt this material

Thorgott

@BenSteffan also, going back, I do think this construction works. the issue, of course, is that what we really want is not a locally finite, countable, finite-dimensional CW-complex, but a locally finite, countable, finite-dimensional simplicial complex of the right homotopy type, but I really don't wanna think about that

@BenSteffan repeat after me: every homotopy type is 1-connected

Ben Steffan

every homotopy type is 1-connected

2 hours later…

pie

23:42

How do you guys write $\zata$?

mine looks like ح

and some times I write it like ع

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