India has a rather "Academic Pride" thing where people(society, not recruiters) judge you by which uni you get your Bachelor's degree from.

I think this discussion is not limited to the admission procedure itself. Hierarchies of the modernity are very steadily built machines to crush people in them

@MikeMiller did you do a talk at Stony Brook a week or two ago? I was there!

Of course, that is kinda decreasing now, but the craze for some particular universities is still prevalent.

@Sid That goes for all countries, I think.

I did, thanks for coming

@Sid It's part and parcel of the rat race

I didnt understand most of it, but it seemed really interesting. What did you think that older professor who kept interrupting you? Lol

@ParthKohli Not the right response, regardless of if it's true or not. The right response is "Yes, and that's Not Good"

I liked Sullivan! I think his questions might have helped me some things clearer for other people in the room, too

I was really surprised by the attendance, both from students and from the older faculty

Sullivan corners people a lot

You mentioned something about a group action on a chain complex. Can you explain?

I was watching a talk one of his students that I personally know, and damn, he was being fried every 5 minutes in

@BalarkaSen You should put a trademark in that.

@ParthKohli not realy prevalent in Germany

@BalarkaSen Of course it isn't a good practice in any circumstance, but it isn't limited to India.

There is a field of research called group homology, and the original construction takes as input a group $G$, acting on an abelian group $M$. You can find a chain complex $C$, where $C_k = \oplus_{i_k} \Bbb Z[G]$ for each $k \geq 0$ ($i_k$ is some nonnegative integer for each $k$), and $H(C)$ is equal to $M$ in degree zero and is otherwise zero

This chain complex is called a "free resolution" of $M$; note that the group $G$ acts on basis elements $\langle g\rangle$ by sending $h \cdot \langle g \rangle = \langle hg \rangle$

Awesome!

Z[G] is the group ring?

Indeed it is

@ParthKohli Sure, but India - being a, let's be honest about this, third world backward country where capitalism has been initiated in a state of premature socio-economy - has absorbed the academic rat race as part of the country's trademark culture faster and better than any other country ever had.

So just pointing fingers at other countries who take part in this modern degeneracy isn't going to help anybody

You can "quotient by the group action" - each copy of $\Bbb Z[G]$ gets replaced by a copy of $\Bbb Z$. Algebraically, you could write this as $C \otimes_{\Bbb Z[G]} \Bbb Z$, where each element of $G$ acts as the identity on $\Bbb Z$. The homology of this quotient complex $H(C \otimes_{\Bbb Z[G]} \Bbb Z)$, is called $H_*(G;M)$, the group homology of $G$ with coefficients in $M$

Now originally $M$ carried an action by an abelian group. You can generalize this construction to the case where instead $M$ is a chain complex, with a "compatible action of $G$": precisely, if $c \in M_k$, we would want $d(gc) = g \cdot dc$ to hold in $M_{k-1}$

India has and will forever sell engineers to other countries because it's hollow in the stomach and has nothing else to sell. That's what keeps the coals burning for the rat race

Wow, sounds great. Tough to understand for me, as im just a first-year grad student. But looks like a very cool topic. This is called "Group Homology"?

group cohomology can be used to prove the most important result in 20th century number theory

Yes, it gives lots of understanding of the groups, and sometimes of the chain complexes the groups are acting on.

What's the best source for group cohomology? I need to learn some this summer

I have a few favorites, but my usual list of references isn't on this laptop, so I need a second

np

@Antonios-AlexandrosRobotis took a seminar on group cohomology, we used these notes: math.ucla.edu/~sharifi/groupcoh.pdf

thanks @MatheinBoulomenos

Curiously enough I haven't read Sharifi's notes

there's no topology in them

except for profinite groups

Eh, most references don't have that

I like Ken Brown's book

I really don't know anything about group cohomology except it's $H^\bullet(K(G, 1))$.

the springer one I suppose

boy am I glad I have so many springer pdfs :P

@BalarkaSen I think I told you the proof that a finite group with trivial group cohomology is the trivial group

Which is nice

I liked Sharifi's notes, they seem to the point and clear

Ah yeah you did

@BalarkaSen We can sit on our armchairs and rant about this academic rat-race all day, but in a country like India, this rat-race is a response to the scarcity of opportunities and resources. It's a means of survival - and we are privileged that we get to decide what we want to do. We're the minority.

That seems like some victimization bullshit. Who's "we"?

What do you mean by victimisation bullshit?

I'm just saying that most people don't get to decide what they want to do, because their circumstances don't allow them to do so.

I don't understand what point you're trying to make. Yes, people don't get to decide what they want to do - that doesn't justify continuing the vacuous wheel of academic hierarchy

I've been struggling with the following problem for a long time: math.stackexchange.com/questions/65834/…

I thought you were talking about the rat-race?

I am?

"A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is the length of a radius of the sphere?"

If $G$ is finitely presented, then $H_2(G,\Bbb Z)$ and $H^2(G,\Bbb Z)$ have finite rank. Proof: since $G$ is finitely presented, we can build $K(\pi,1)$ with a finite 2-skeleton, so this follows from the construction of cellular homology. This is really neat, not sure if you can actually use it to prove that some finitely generated group is not finitely presented, but it seems possible

This was my though process: Simplify the problem into a 2D version (incircle inscribed in equilateral triangle), find the radius of the incenter in ratio to the length of the altitude, and apply said ratio to the altitude of the tetrahedron to find the radius of the inscribed sphere.

Is anyone able to help me understand what happens to a complex valued function when you raise it to a real valued power

you get another complex valued function

Thus, I used the formulas A=bh, and A=rs, and set them equal to each other, solving for r, and then found the ratio of that indaius to the length of the altitude. I then applied that ratio to the tetrahedron to get 1/3*36=12.

(note that you need to choose some branch of the complex logarithm)

@MatheinBoulomenos you should have to take a branch cut

yeah

I do not know an example of a finitely presented group with infinite rank homology

However, the answer is apparently 9, so I'm really dumbfounded.

@PathKohli In case it was unclear, the rat-race is just an extreme limit of the academic hierarchy, a form that India has successfully produced precisely because it's incompetent to manage the modern version of capitalism (well, no country is, but that's a different issue). I like to think of the hierarchy because it's more general and the root of the problem.

If anyone could help me with my problem, "A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is the length of a radius of the sphere?" that would be great.

sorry finitely generated group

@BalarkaSen would you do away entirely with the notion of hierarchy?

@Antonios-AlexandrosRobotis It's unclear how to do that, no?

Certainly a non-obvious problem

certainly, I'm just curious what alternative exists. Because one does need to celebrate great achievements.

In a healthier way, perhaps, though.

Ah, no, you're speaking of a different phenomenon.

(you should understand that I have little concept of what goes on in India)

Here's an example: Take any biological system, say trees in a tropical forest. The height of the trees vs number of trees of that height will give a distribution

If you observe it, there'll be exponentially less trees of exponentially greater heights in the distribution

That's biological hierarchy

so if I take a complex valued function, say (e^z)^(1/log(x)) where z is a complex number and x is a real number,

Actually take (sin(z))^(1/log(x))

In the modern hierarchy, if you take, say, demand vs. product, you'll get an exponentially little amount of products with exponentially high demand. That's ok so far, we can pass it off as biological hierarchy.

doesn't the exponent "squish a part of the function inside 0 and 1 for x????

the way you would define that is sin(z) = e^{log(sin(z)}, so sin(z)^{1/log(x)} = e^{log(sin(z))/log(x)}. A re-weighted exponential.

I'm following @BalarkaSen

@MikeMiller do you know any results about the cohomology of wreath products? I know $\Bbb Z \wr \Bbb Z$ is a finitely generated, not finitely presented group with a rather simple definition compared to other examples

Now take time-slices of the data. The curve not only gathers more kurtosis at the central tendency, but also becomes flatter at the end. I.e., the products which didn't have more demand gets even less demand as time flows

Somebody knows results about the cohomology of wreath products; not necessarily me

That's the tragedy of the modern hierarchy. The peak of the curve exploits the bottom.

QUillen uses it a lot

@BalarkaSen but is this problem avoidable or intrinsic to the system

I'm saying, we should find an alternative to that. People who are already crushed shouldn't be crushed exponentially more.

yeah, that I agree with

@Antonios-AlexandrosRobotis Well, this doesn't happen in the stable biological hierarchies. The curve grows steadily upwards instead of the peak growing faster and the tail end dampening down with it

So that's some evidence that it's our own making

So @MikeMiller

for x in between 0 and 1 it squishes part of the function inside there right?

@MikeMiller do you think that $H^2(\Bbb Z \wr \Bbb Z, \Bbb Z)$ might have infinite rank? This is equivalent to a statement about central exensions $1 \to \Bbb Z \to E \to \Bbb Z \wr \Bbb Z \to 1$

it's like an extra little piece

I don't know what that means @geocalc33

@BalarkaSen So, in a stable system, both parts grow at the same rate?

I might not be helpful

@Sid In a stable system growth is not the goal. Exponential growth is itself unstable.

@Sid Not necessarily at the same rate, I don't think. But they grow nonetheless. I'll have to find a reference with the math in it.

@MatheinBoulomenos Hmm, I don't know. There is a theorem by Nakaoka about the cohomology rings of wreath products but it seems to be for wreath products of finite groups

@MikeMiller I know for a stable system, The population grows as a Sigmoid curve. I don't understand what the Carrying Capacity of this world actually is

Hahahahaha probably not much

Stupid question; how do I rewrite this so that it moves in the opposite direction? i.imgur.com/3El9L8R.png

id imagine it's probably not very different from 7 billion

we may have long passed the point where we have a population that can sustain itself; the exponential growth might be too late

@EricSilva I think human carrying capacity is a function of our technological development

just not something that can be curtailed or well-understood beforehand

indeed @MikeMiller

@Antonios-AlexandrosRobotis So perhaps we should not have developed agriculture :)

@MikeMiller well, that is kinda the source of the whole problem LOL

division of labor fucked us forever

of course agriculture increased the carrying capacity but it also increased the population growth rate even more lol.

something like that anyway

certain explicit upper bounds were done away

It was a mistake to move the Malthusian point farther, @MikeMiller

I agree with that completely

I have a philosophical semi-babble that we should hype up technology and warfare like the Futurist did to monger the first world war. I see no other way to reduce population than an explicit warfare (of course, to make that idea into reality is a morally reprehensible action, so there's a morality vs. need conflict in that direction).

@MatheinBoulomenos projecteuclid.org/download/pdf_1/euclid.ijm/1256047257

is @Balarka an accelerationist

I mean the utilitarian thing to do is to round up the "worst" x% of the population and kill them all.

Doesn't mean it's the right thing to do.

o..o

lol

can I take a moment to say that mercio's reaction emojies are the best

@EricSilva A part of me is, I think

yeah that is pretty spot on lol

@Balarka war is a way less effective way to kill people than the free market

there's a way to reduce the population growth and Japan is working on it: sex robots

hahahahah @Eric

@BalarkaSen I guess the slower and surer way is to just stop having kids. Of course, that's a personal decision though

@EricSilva LOL

just turn everyone gay

it will be fabulous

LOLOLLOLOL

@Sid Yeah I can't say I don't think it's a morally reprehensible action to reproduce.

there was ths guy that got hit on the head and then his sexuality changed

we need to research that

@EricSilva Depends on the day but yes

@mercio did he develop a fetish for being hit on the head?

Hey, has anybody a good knowledge in random walk theory?

can i attach a graph here

How can I write this so it increases in the opposite direction? i.imgur.com/3El9L8R.png

I don't remember that much but probably not ...

@EricSilva eh. not sure how easy it is to disentangle modern warfare from modern capitalism

@MatheinBoulomenos The first Corollary of that paper says that $$H_2(\Bbb Z \wr \Bbb Z) \cong H_2 \Bbb Z \oplus H_2 \Bbb Z \oplus_{i=1}^\infty (H_1 \Bbb Z \otimes H_1 \Bbb Z)/(x \otimes y + y \otimes x).$$

@Semiclassical I was told there would eb no politics =P

@Semiclassical dont read super far into my hot take

eb=be

lol

the first two terms are zero, the last term is $(\Bbb Z/2)^\infty$

lmao

we can debate for hours about the kernels of truth hidden in what i said but i also havent had coffee today

some stronk covfefe needed to discuss that shite

@BalarkaSen golf clap

@MatheinBoulomenos That's a really cool result, I gotta say

This group is definitely not finitely generated though right?

@BalarkaSen I dislike that you equate covfefe to Coffee when the President said it's a real scrabble-legal(I think) word.

@Sid are you t r i g g e r e d by my usage of that word?

sounds like you're a radical SJW left

the SJW left isnt radical u take that back

lol

@MikeMiller wreath products of finitely generated groups are finitely generated, right?

I have a proof in my head and seems not too hard, but maybe something is wrong

Is that true? I trust you if so, I would worry about the fact that we add infinitely many copies of $\Bbb Z$ in the wreath process

I haven't thought at all about it though

@MikeMiller raising a number such as 3 to the 1/lnx+1/ln(1-x) makes a bell shaped curve confined to the unit square

isn't that cool

I suppose, I don't think that's the sort of thing that usually gets me excited

@EricSilva SJW annoys me. It's basically just a meaningless pejorative

Mhm.

i.e. "you're a SJW, therefore I don't have to think about anything you say"

sure, I was riffing, it's basically just used as a mean word for liberals on the internet

fair

relevant:

@MatheinBoulomenos If you want to, I would be interested in knowing what square-0 extensions these generators correspond to

@MikeMiller I'm working with the restricted wreath product where one takes the direct sum and not the direct product

this is one of my favorite meta-memes of this month

not sure what definition is used for that cohomology result

Sure, it's still really big though

it's $\oplus^\infty \Bbb Z \rtimes \Bbb Z$ right?

yeah

they're using the restricted one like you

i guess finitely generated groups can have infinite rank subgroups but I'm just scare

what's the finite generating set?

F_2 has F_infty as subgroup

oh

of course I see the finite generating set

:(

2-generated, even

yeah

and the construction works in general

@Semiclassical i dont like to use it personally cause it would put me in a weird horseshoe territory, my complaint w what one might call the "social justice left" is that they're not radical enough which is a weird positioning for someone who would say something like 'SJW'

well, SJW is both too extreme and not extreme enough

PC-supporters is the right alternative I think

PC = Political Correctness

yeah, no

@Balarka does PC = NPC though?

does yeah, no or no, yeah mean yes

idk which

@EricSilva Yes

@Daminark Good question

@Daminark lmao

It's the last word that matters

@Daminark Damn, I was about to do that...

Pew pew

you should post a paper on vixra or r/shittymathematics or something

@MikeMiller I had a question about quotient manifolds. I have a manifold M and then a map $\phi\colon M\to M$ which I want to take the quotient of $M$ by. That is, mod out by the equivalence relation $(x,\phi(x))$

@Semiclassical idk what this means

Phi has finite order?

In what sense of the word?

finite order in the endomorphism monoid

There is some n for which phi^n(x) = x for all x

This follows I think from such a fact for each x, with no bound on n(x), but I think that's hard

Yes.

Okay, so I can think of this as an action of Z/n on M

The question though is I do not know how to show that I get an orientable manifold.

I have showed

is $\phi$ continuous ?

I think at least, that the map phi is orientation preserving.

I am confused. You're identifying $x$ and $\phi(x)$?

If a group G acts freely on M and M/G is oriented, then G acted by orientation preserving maps upstairs on M

Yes, @BalarkaSen

The converse is also seen to be true

The problem here is that I am only doing the identification of $\phi$ on part of the manifold.

That is, the boundary.

So you just need to see that Z/n acts orientably, and in particular that the generator phi does

@MikeMiller I would also like to know which extensions these generators corresponds to

going to play some soccer/football, be back later, all.

Happy mathing.

@anakhronizein You're quotienting by a group action on the boundary?

By $\phi$ on the boundary

I don't see clearly how this would be seen as a group action.

Oh $\phi$ has finite order. I was confused by how $\langle \phi^k \rangle$ would be a group, hah.

I believe the result would be a manifold without boundary if phi is order 2 and a non-manifold if phi has larger order

phi is indeed order 2

Then the action of phi is an action by Z/2 on the boundary

Let me think now

@anakhronizein We start with a manifold $M$ with a boundary component $\partial M$. $M$ is oriented, and this induces an orientation on the boundary, via outward-normal-first conventions (or whatever convention is right, I never remember).

It is best to think of the normal bundle to $\partial M$ as being identified with $\det(T\partial M)$. The outward-pointing normal vector is identified with a positive orientation on $\partial M$.

Now $\partial M$ carries a fixed point free action of $\Bbb Z/2$. The normal line bundle to this on the quotient manifold is $N(\partial M)/(\Bbb Z/2)$. If $\phi$ acts in an orientation-preserving way, it fixes some positively-oriented section, and thus this quotient is a trivial line bundle. Otherwise, if it was not orientation preserving, you get the nontrivial line bundle corresponding to that double cover $\partial M \to (\partial M)/(\Bbb Z/2)$.

So the orientability of $M/\phi$ hinges on the orientation preserving property of $\phi$?

Ah yeah ok. So in either case you end up with an oriented total space, and if the quotient is called $N$, it sits inside the total space with normal bundle isomorphic to $\det(TN)$.

It's just that because the normal bundle and tangent bundle have the same determinant bundle, they sum to something with trivial determinant bundle; hence $T(M/\phi) \big|_N$ is orientable. And $T(M/\phi) \big|_{M/\phi \setminus N}$ is orientable.

Model cases: $D^n$, quotient map on the boundary induces $\Bbb RP^n$

The point is the collar neighborhood of $\partial M$ is like $\partial M \times [0, 1]$. If $\Bbb Z_2$ acts orientation-preservingly, you quotient to the trivial line bundle on $\partial M/\Bbb Z_2$

@MikeMiller oh btw what's the paper with the result on the cohomology of the wreath product?

I thought I linked it but might have forgotten to, here i tis

projecteuclid.org/download/pdf_1/euclid.ijm/1256047257

thanks!

So does this work similarly if we are not discussing the boundary of M, and its on all of M?

group cohomology is really cool. Some of my peers said they found it too technical, but I like it. (I guess algebraists like technical things)

@anakhronizein I think there is a very different story going on between quotienting on all of M and doing quotients on the boundary

The boundary stuff is tricky, the whole-manifold stuff is not

What's the specific situation, @anakhronizein?

If there is any

Well the specific situation is even more complicated than what I am asking about

hi peopel

Yikes

@MatheinBoulomenos I really like group cohomology; one of the main novelties of the paper I've been writing forever is to phrase a certain equivariant cohomology construction in my field, over a compact Lie group, as coming from "dga cohomology", and properties thereof

which I enjoyed

^this was the link sorry :P

@MikeMiller that sounds really cool!

$M := S^2\times [-1,1]$ with the map $\varphi := \sigma\times\tau$ where $\sigma$ is the antipode map on $S^2$ and $\tau\colon \pm1\mapsto\mp1$.

So you get a circle bundle on RP^2

I don't really know anything about dgas other than that they're monoid objects wrt to the graded tensor product on the category of graded abelian groups (or $R$-modules) and very few examples

You quotient out by $\varphi\vert_{S^2\times\{\pm 1\}}$

That's a great example @anakhronizein

@anakhronizein If you quotient at the boundary... yeah.

The quotient map gives you a copy of $\Bbb{RP}^2$ sitting inside, as well as its normal bundle being the nontrivial bundle over $\Bbb{RP}^2$

Thus a neighborhood of that $\Bbb{RP}^2$ has boundary $S^2$

The issue I am finding is I don't see how to show it is orientable. The quotient is modded out irregularly on $M$ but yet $\phi$ is orientation preserving.

I was definitely uncareful to ignore the group action on the boundary like I did

Do it in two steps. Call $M^\circ = S^2 \times (-1, 1)$ and $B^\circ = S^2 \times \{[-1, -1 + \varepsilon) \cup (1-\varepsilon, 1]\}$ (lazy notation, small neighborhood of the boundary).

Then $M^\circ/\varphi$ is orientable (quotient of open smooth manifold by an orientation-preserving involution is orientable), as is $B^\circ/\varphi$ is orientable because it's just $S^2 \times (1-\varepsilon, 1]$

I think there must be some mistake here because the quotient shouldn't have boundary

How is $B^\circ/\varphi$ just $S^2\times (1-\varepsilon,1]$?

Isn't there a unique representative of each $\varphi$-orbit of $B^\circ$ in the $(1-\varepsilon, 1]$ bit?

Shouldn't it be RP^2 cross (1-e,1]?

That would be quotienting by an order 4 action

Note that S^2 x 2 intervals -> RP^2 x 1 interval has fibers of cardinality 4

Which part of the line integral notation gives direction of increase of t?

I actually think your $M/\varphi$ has one boundary component, and that the thing you're describing is the unit disc bundle of the tautological line bundle over $\Bbb{RP}^2$

I am still not seeing how it's $S^2\times(1-\varepsilon,1]$ and not $\mathbb{RP}^2\times(1-\varepsilon,1]$

Is there another way of seeing that?

Do you understand my point why it cannot be the latter?

No, not quite.

I would have thought the former would have been quotienting by $1\times\tau$

Okay. First, the quotient is a manifold, with one boundary component. Thus there is a covering map of order 2 $\partial M \to \partial (M/\phi)$. In the first case, the boundary is $S^2 \sqcup S^2$, and in the second case, $\Bbb{RP}^2$. From Euler characteristic considerations - the first has $\chi = 4$, the second $\chi = 1$, - a covering map from the first to the second has degree 4.

Now suppose we have $S^2 \sqcup S^2$, mod the action $(x, 1) \sim (-x, 2)$. That's what you're talking about, on the boundary.

Then the quotient $(S^2 \sqcup S^2)/\sim \cong S^2$. The map $S^2 \to (S^2 \sqcup S^2)/\sim$ is given by $x \mapsto [(x,1)]$.

I feel bad because I am getting hung up on almost everything you are saying.

This is surjective, because given any $[y,2]$, then it is equivalent to $[-y, 1]$, and hence in the image of this map. It is injective, because no $[x, 1]$ is equivalent to itself.

I don't want you to feel bad. There just must be some place we're coming at this with different pictures in our heads.

Here, I have a more general way of phrasing this that might be helpful.

Let $X$ be a space with $G$-action. Consider $X \times G$, with the diagonal $G$-action ($G$ acts on itself by the group multiplication).

Is this taking into account that we are quotienting out by $\varphi\vert_{S^2\times\{\pm1\}}$?

Then $X \times G$ is always isomorphic, as a $G$-space, to $X \times G$ where the first component has the trivial action and the second component the standard action.

In our case, we have $S^2$ with antipodal action, and pass to $S^2 \times \pm 1$.

Yeah, I think so. I'm actually only talking about $S^2 \times \{\pm 1\}$. My notation here is bad: I write $(x,1)$ to mean an arbitrary element of $S^2 \times \{-1\}$ (the first of the two spheres) and $(y,2)$ for an arbitrary element of $S^2 \times \{1\}$.

Lol

Sorry that's just funny

yeah it's really bad

I lol'd irl

indiana.edu/~jfdavis/notes/duh.pdf

#2

@anakhronizein So I am attempting to understand what's happening. We have $S^2 \times [0, 1]$ and you quotient at the boundary by $\Bbb Z/2$ (which in reality acts on the full space by the antipodal map on $S^2$ and reversing homeomorphism on $[0, 1]$)

That's a nontrivial $S^1$-bundle over $\Bbb{RP}^2$. What's the question?

Is it "why is this a nontrivial bundle?"

@MikeMiller Okay so I see now where the $S^2$ arises instead of $\mathbb{RP}^2$ (From your first paragraph there). However, $B^\circ \neq \partial(S^2\times[-1,1])$?

@BalarkaSen the goal is I want to show that this quotient is orientable.

I got lazy

@anakhronizein The full quotient?

@BalarkaSen $(S^2\times[-1,1])/\varphi\vert_{S^2\times\{\pm 1\}}$

@anakhronizein Here is how I am organizing my thinking: There is an S^2 of fixed points, and away from that the action is free. I am going to try to understand the quotient in a neighborhood of the fixed set and then in the complement of the fixed set

@MikeMiller where is this RP^2 coming from?

A neighborhood of the fixed set is modeled on the trivial line bundle over $S^2$, with $\Bbb Z/2$ action on $S^2 \times \Bbb R$ by $\tau(x,t) = (-x,-t)$. This quotients to a nontrivial line bundle over $\Bbb{RP}^2$, the tautological bundle. This is my local model in a neighborhood of the fixed set, after we quotient

Brain-typo

sorry

Well question still stands, what exactly is the S^2 of fixed points?

@anakhronizein Let's call $\phi = \varphi|_{S^2 \times \{\pm 1\}}$. Then $\phi(x, 1) = (-x, -1)$, yes?

Yes, @BalarkaSen

I must have been confused. I thought you were quotienting by a global group action. I'm sorry.

And you're taking the mapping torus of the antipodal map of $S^2$, to put it succinctly.

It's fine, the more I talk about this, the more it doesn't make me scratch my head.

@BalarkaSen I don't know what is meant by that.

Okay, I still like the language of local models.

Yes, that's basically it, @BalarkaSen

So that's a nonorientable circle bundle over $S^2$, is it not?