I thought it should be an orientable circle bundle over RP^2

Well, the circle bundle is orientable as a 3-manifold...

Or should be.

I don't know much about circle bundles.

It's unclear to me why you should get RP^2.

@BalarkaSen all circle bundles are orientable there, $\pi_1 S^2 = 0$

@MikeMiller Ah, rats, yes.

@BalarkaSen The idea is that the projection is $S^2 \times [-1, 1] \to S^2 \times \{0\}$, but this does not descend past the quotient action, as $\tau(x, -1) \neq (x,1)$

Instead you project to $\Bbb{RP}^2$ so that this isn't an issue

okay, I'm sorry this process was so slow. Again we cover the manifold with $M^\circ = S^2 \times (-1, 1)$, which undergoes no quotient, and $B^\circ = S^2 \times \pm 1 \times (\varepsilon, 1]$.

I thought it was geometrically clear that $\mathbb{RP}^2 = S^2/\sigma$ and $S^1 = [-1,1]/\tau\vert_{\{\pm 1\}}$... then it ought to look like a circle bundle...

@MikeMiller I don't think the construction is clear to me. From what I read, he has the mapping torus of the antipodal map of $S^2$ (<- this is unclear. Is this really what he has?). That's a nonorientable $S^2$-bundle over the circle (I had it backwards)

@BalarkaSen What is unclear? You kill the time factor.

@BalarkaSen it fibers in 2 ways

the space we are talking about could be called $S^2 \tilde{\times} S^1$

a 3d Klein bottle

Right, that's what I had in mind.

So it's nonorientable, it's double covered by S^2 x S^1

that also fibers over $\Bbb{RP}^2$ via the construction both ana and I suggested

Ah, you're fibering it over RP^2. Got it.

Different pictures, same space.

But I don't understand @anakhronizein's claim that it's orientable.

Well I haven't yet proved that it is oriented.

But that's what I am hoping to do.

yeah I dun buy it

Well, it's not orientable.

Why is it not oriented?

Think in terms of my picture. You have $S^2 \times [-1, 1]$ with the two $S^2$'s in the front and back glued by antipodal map. Consider the projection by killing the $S^2$ factor in $S^2 \times [-1, 1]$. That gives a map $p : X \to S^1$ where $X$ is this space of yours. This is a fiber bundle, with $S^2$ fibers. Notice that a lift of the identity loop in the circle doesn't complete a circle above, so $p$ is not the trivial $S^2 \times S^1$

Here is a different way of saying what your space is. $B^\circ/\varphi$ is diffeomorphic to $S^2 \times (-\varepsilon, \varepsilon)$. The point is that you're gluing two copies of $S^2 \times (-\varepsilon, 0]$ together along their boundary - the two boundary pieces are identified bijectively.

I don't see where this precludes an orientation.

The point is there is only one nontrivial $S^2$-bundle over $S^1$ (which your $X$ is from my description above), and it's not orientable as a manifold. Namely, take two solid Klein bottles (Klein bottle is a $S^1$-bundle over $S^1$, now fill in the fibers by disks $D^2$), and glue them by the boundary (the fiber becomes $D^2 \sqcup_\partial D^2 = S^2$). This is the $S^2$-bundle over circle that your space is.

@anakhronizein Carry a framing for $T_{0,x} S^2 \times [-1, 1]$ (say, $\partial/\partial t, F)$, where $F$ is a framing of $T_x S^2$, to the right, to $T_{1,x} S^2 \times [-1,1]$. This gets identified with a framing for $T_{-1, -x} S^2 \times [-1,1]$, the framing $\partial/\partial t, \tau(F)$. Walk that back to $T_{0,-x}$. The point is, then, that this orientation on the 2-sphere disagrees with the first; we are comparing an orientation $o$ to the orientation $\tau(o)$.

A neighborhood of this path is the "solid Klein bottle" that Balarka talks about

Hmmm, I don't understand either proof.

But if you two insist, then I must be mislead.

Thanks for your help, back to the drawing board I guess!

Come back at another time and we could explain this space in greater detail I'm sure.

I am really sorry for just dogpiling you with stuff that sounds like nonsense to you...

It's fine. I just don't know how to improve at geometry at this point.

Certainly no royal road, it seems.

Man you two are taking it really unconstructively. Give it some time, and come back later.

Some things take time to click.

(both in terms of understanding/explaining)

Let me try one more and then I'll stop. Your construction, phrased in the language of 'the mapping torus', is $S^2 \times [0,1]/(x, 0) \sim (\tau(x), 1)$. This is, equivalently, $S^2 \times \Bbb R$, modulo the group action by $\Bbb Z$, whose generator is $1 \cdot (x, t) = (\tau(x), t+1)$.

@KasmirKhaan Hey! I haven't forgotton about rep theory, I'm working on it.

@MatheinBoulomenos Hey :D here?

Haha

almost same second we texted each other

its okay mathein, dont be stressed , it is very important to me but I dont want you to overwork yourself because of me either :D

anyway thanks alooot again :D

Now, what is $(S^2 \times \Bbb R)/2\Bbb Z$? Note that $\tau^2$ is the identity, so that this is just the action whose generator is given by $1 \cdot (x, t) = (x, t+2)$. In particular, the quotient by that group action is just $S^2 \times (\Bbb R/2\Bbb Z)$.

I will do my best to repay you for all the help over the years :D

@KasmirKhaan I'm not overworking myself, don't worry

Good! :D

Therefore our manifold has as double cover $S^2 \times S^1$. What is the involution we quotient by to obtain our manifold? It is precisely $(x, [t]) \mapsto (\tau(x), [t+1])$.

if I'm overworking myself, then it's the 6 grad courses while also being a TA not that little rep theory assignment

Our question reduces precisely to asking whether this is or is not an orientation-preserving automorphism of $S^2 \times S^1$. But it is just a product diffeomorphism, $\tau \times -1$. The antipodal map on $S^2$ is orientation-reversing, but the antipodal map on $S^1$ is orientation preserving.

Thus our manifold is $(S^2 \times S^1)/\tau'$, where $\tau'$ is an orientation-reversing diffeomorphism.

@MatheinBoulomenos respect really for taking 6 courses >< I took once 4 and went crazy

and am sure my 4 courses arent that advanced in comparasion to ur 6 ><

one is quite basic. Some elementary projective geometry

am not sure how much time the TA takes of your time however ><

the other ones are not that basic

I wish you A - them all :D

thanks

(crappy picture of what mike is doing)

Unwrap $S^2 \times [0, 1]$ with $(x, 0) \sim (-x, 1)$ to $S^2 \times \Bbb R$ and $\Bbb Z$ acting on it by $(x, t) \sim (-x, t+1)$.

@anakhronizein feel free to get in touch with me offsite to talk about this or any other frustrations - we have chosen a frustrating road

There just seems to be an awful lot of gaps in my knowledge.

But thank you for the assistance, nonetheless. I will definitely come back to this.

is any1 here

Let $|f_n| \leq g$ be dominated, in Lesbegue's Dominated Theorem, the proof is shown via Fatou's lemma. But why can't we just apply Monotone Convergence Theorem to $(g - f_n)$?

How do you know that $g-f_n \le g-f_{n+1}$?

oh nvm i thought f_n was increasing

Taking a moment to say that this song is great

Truly think this album is one of the breakthroughs in post-rock.

Hi all, Is the center of mass of a given cube the center of the sphere inscribed in a tetrahedron? By all accounts, this seems to be so, for the ratio 1/4 still holds. However, I don't know if this can be rigorously proved, or if it's true at all.