Mathematics

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ACuriousMind

@skull Please do not abuse the deletion feature to post temporary messages. Self-deletion is meant for things that you posted, then thought better of it, not to hide the content of specfic messages from the transcript.

@tatan: I think that problem does not have a unique solution. Both $f(x)=x$ and $f(x)=-x$ satisfy the given condition.

Hi prof

hi skull

Daminark

8:02 PM

Rehi Ted

rehi Demonark

how are you?

I've been mulling over a question tatan asked earlier.

desmos.com/calculator/muvqbhmf28

I can't understand it even though I've plotted it

@TedShifrin do you mind helping me?

What makes you think I can help?

tatan

8:03 PM

@TedShifrin Thanks prof

@ACuriousMind Well, that's a contradictory thing to say. Maybe he did posted it, and then thought better of that joke being part of the transcript!

5

I don't think that particular goal of the deletion feature stands out :P

Hi, by the way.

Tobias Kildetoft

@MatheinBoulomenos Hmm, I think I had more conditions that just basic for that, Possibly something like local? I recall using quivers, but I think the quivers ended up with just a single vertex, which would correspond to the algebra being local.

:-D

desmos.com/calculator/hc8vbamszw

hmm, this seems to make things clearer

@loch Oh I forgot to tell you this story relevant to the above. Say $E$ is a real line bundle on $X$. Give it a fiberwise Riemannian metric and take the unit $S^0$-bundle sitting in $E$. That's a double cover of $X$. This is a 1-1 correspondence between real line bundles on $X$ and double covers on $X$.

Now, double covers on $X$ are classified by maps $\pi_1(X) \to \Bbb Z_2$ by the covering space story. By algebra, we have $\text{Hom}(\pi_1(X), \Bbb Z_2) \cong \text{Hom}(H_1(X), \Bbb Z_2) = H^1(X; \Bbb Z/2)$.

So real line bundles on $X$ are in 1-1 correspondence with $H^1(X; \Bbb Z/2)$. I suppose you algebro-geometric people love that.

8:16 PM

I'd get that out of an exact sequence of sheaves, myself :P

I'd say you're an algebro-geometric people person

So not surprising

Lebesgue integral is way cooler than Riemann integral

But you can only calculate them using Riemann, Leaky.

So don't be too condescending about it.

true

Leaky is a Lebesgue Elitist

the very dangerous kind

8:18 PM

Why would algebraists like real line bundles? They like Chern classes.

::crickets::

@MikeMiller hm ok so im trying to figure out what the action is after I identify the unit tangent bundle with $SO(3)$ with $\mathbb{R} \mathbb{P}^3$

Unit tangent bundle of $S^2$, you mean.

Ah, you did write "with" not "of".

I can't read.

It still doesn't quite scan.

loch's writing in words that $T_1 S^2 \cong SO(3) \cong \Bbb{RP}^3$, which is why I couldn't scan the syntax

8:31 PM

Yes, I guessed what was intended.

@BalarkaSen oh yes i think i learnt this at some point in the past but have forgotten..

@loch: Think of $SO(3)$ as oriented orthonormal bases for $\Bbb R^3$.

Professor @TedShifrin have you ever talked with Robert Langlands?

Nope. Never met him that I recall.

yes - so if I identify $SO(3) \rightarrow T_1 S^2$ by mapping $\gamma \mapsto (\gamma(1,0,0), \gamma(0,1,0))$,

then this means that my action on $T_1S^2$ identifies the matrix $[v_1,v_2,v_3]$ to $[-v_1,-v_2,v_3]$ on $SO(3)$

8:35 PM

Well, you need point plus basis for tangent space, @loch

I dunno why you're putting minus signs

Certainly his conclusion is right though

I don't know why the negatives are there, Mike.

You would prefer the antipodal map on $T_1 S^2$ act as the identity?

I'm not following.

Mike asked loch to find what T_1 RP^2 is

8:38 PM

T_1 RP^2

If $e_3$ is my base point in the sphere, $e_1,e_2$ is an oriented orthonormal basis for $T_{e_3}S^2$.

So loch's figuring out what the antipodal action on S^2 extends to T_1 S^2

I missed anything about antipodal action. Then $v_3$ should be negated, too.

Disagree, this is an orientation thing

hm so my $\mathbb{Z}/2$ action on $T_1 S^2$ is given by, if I identify $T_1 S^2 = \{ (p,v) \in S^2 \times S^2 : <p,v> = 0 \}$, sending $(p,v) \mapsto (-p,-v)$.

So I think under the identification with $SO(3)$ as above, it would identify the matrix $\gamma =[v_1,v_2,v_3] \mapsto [-v_1,-v_2,v_3]$

8:39 PM

The antipodal map on $S^2$ is orientation-reversing.

Never mind. I'll just leave.

cya

hm now I'm confused.

anyway assuming what i said is right then i just have to trace through the identification $SO(3)$ to $\mathbb{R} \mathbb{P}^3$

@TedShifrin I would be glad to explain arbitrarily much (though I wanted to be careful not to do too much if it was an exercise). I wish you wouldn't try to explain why I was wrong from the start, though.

While the antipodal map on $S^2$ is orientation-reversing, this implies that the induced map on $T^1 S^2$ is orientation-preserving. (It is a nice trick of luck that this is always true, no matter what. We reverse the orientation in the vector-factor as well.)

This is related to the statement that $TM$ is always an orientable manifold.

The reason I am confident with no work that @loch's calculation is correct is that this involution should correspond to an element of SO(3), which can be diagonalized to the element stated.

In terms of orthonormal oriented bases, note that these can be equivalently described by two vectors, the second orthonormal to the first. One sees from this that the action is by $-1$ on these. However, when you identify with oriented orthonormal bases (forgetting the third vector), you see that $(v,w) \mapsto (-v, -w)$ is sent to $(v,w,u) \mapsto (-v,-w,u)$ (so that this is oriented the same way).

So now the remaining task is to understand $SO(3)/(\Bbb Z/2)$. This may be easier to do inside of the double cover, $SU(2)$.

philmcole

Hi, I can somebody help me with the proof that a bounded function on a Jordan null-set is Riemann integrable with integral zero? I have the feeling it's not that difficult though I'm confused with the definitions and can't get it right...

These are my definitions:

- $N \subseteq \Bbb R$ is a Lebesgue null-set, if for all $\varepsilon \gt 0$ there exists a sequence of open cuboids $(Q_l)_l$ such that $N \subseteq \bigcup_l Q_l$ and $\sum_l \text{vol}(Q_l) \lt \varepsilon$.

- The Lebesque theorem states that a function $f: Q \to \Bbb R$ is Riemann integrable iff the set of points where it's not continuous is a Lebesgue null-set.

- $J \subseteq \Bbb R^n$ is Jordan measurable, if there exists a closed cuboid $Q \subseteq \Bbb R^n$ such that $\mathbb 1_J: Q \to \Bbb R$ is Riemann integrable. We define $\text{vol}(J) = \int_Q 1…

---

So let $J \subseteq \Bbb R^n$ be a Jordan null-set. Then $J$ is Jordan measurable which means there exists a closed cuboid $Q \subseteq \Bbb R^n$ such that $\mathbb 1_J: Q \to \Bbb R$ is Riemann integrable.

Now how do I conclude that also $\mathbb 1_Jf: Q \to \Bbb R$ is Riemann integrable?

right - so i think the action there is also right multiplication by $[-e_1, -e_2, e_3]$ ($e_i$ being my standard basis)

and i think the preimage of this in $SU(2)$ is $\pm [ (i,0); (0,-i)]$

8:53 PM

Ayup

so looking at $SU(2) = S^3$, multiplying this matrix on the right is the action sending $(x,y,z,w) \mapsto (-y,x,w,-z)$. So I'm modding out this action and $\{ \pm 1 \}$ hmm

Have you heard of lens spaces?

9:09 PM

ah i've heard of them but somehow we never spent much time on them in my classes

but at least i remember an example would be modding out $S^1$ by $\mathbb{Z}/n\mathbb{Z}$

So here I'm modding out $S^3$ by $\mathbb{Z}/4\mathbb{Z}$

(Check out, say, the wikipedia page)

it is also Hatcher 2.43

hm ok - so the upshot is they are orientable e.g. by Hatcher's 2.43 computation of its homology, knowing that we are modding out $S^3$ by a free action so we still get a closed manifold

and so this is related to what you were saying about the unit bundle of a vector bundle is orientable as a manifold govenrns whether the total space is orientable or not + the fact that the total space of the tangent bundle of a manifold is always orientable (as a manifold)

9:26 PM

Yeah, specifically you have shown that $T^1 \Bbb{RP}^2$ is a specific, rather concrete oriented 3-manifold, $L(4,1)$

yeah

I like that quite a bit :)

yeah that's pretty neat - i also never spent too much time on these calculations so it was a good exercise too + I guess I now know some 3 -manifolds that is not $S^3$ or $\mathbb{R} \mathbb{P}^3$ that I should keep in mind :)

thanks!

is LaTex still worth learning? is it out of date?

You probably don't have time for this, but I calculated all of the $S^1$ bundles over $\Bbb{RP}^2$ in a similar way with Balarka recently; they all must come as quotients of the circle bundles $L(n,1) \to S^2$ by some involution which is antipodal downstairs

9:30 PM

@Bob yes and no

(Except for precisely three, which come as quotients of $S^2 \times S^1$)

So you can end up analyzing everything in terms of quotients of $SU(2)$ by subgroups, or $S^2 \times S^1$

@loch are you saying it is worth learning but it is out of date?

it is worth learning and it is not out of date

recently I was trying to help a friend learn it and it did not go well

he did not like the fact that it was not wizwig

That's what I like about markdown

WYSIWYG as fuck boii

9:33 PM

is markdown a new tool?

I believe it is

thanks and have a nice day

bye