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3:23 PM
@BalarkaSen such a great album. never got around to listening to the alan parsons project, now I know what I missed
 
::classic::
 
I had not looked up Banach's life until now (Banach of Banach–Tarski paradox fame). I'm not sure I wanted to know what a "louse-feeder" was
("During World War II, feeding the lice with human subjects' blood was the only way to produce a viable typhus vaccine." - Wikipedia)
 
That's fucked up
Thanks for bringing this up right when I enter chat
 
Categories are equivalent if there are functors $F, G$ such that $F \circ G$ and $G \circ F$ are naturally equivalent to the relevant identity functors?
do you say naturally equivalent or naturally isomorphic lmao
natürliche Äquivalenz
 
naturally isomorphic
 
3:37 PM
Lol my lecturer says natürlich äquivalent
 
the isomorphisms in the category whose objects are functors and whose morphisms are natural transformations
 
yeah right
 
modulo set theory issues only nerds care about
 
literally what the lecturer said at the start too
 
nice
 
3:39 PM
"Wir ignorieren die mengentheoretischen Probleme die da auftauchen, aber falls Sie Interesse haben können Sie sich ein dickes Buch holen und die Probleme da durchlesen"
 
@BalarkaSen Turns out Polish mathematicians in WWII had it really bad
 
yeah
 
Yoneda Lemma is theorem 4.20 lmao
 
As for Tarski, he was both Polish and Jewish
> Tarski's ties to the Unity of Science movement likely saved his life, because they resulted in his being invited to address the Unity of Science Congress held in September 1939 at Harvard University. Thus he left Poland in August 1939, on the last ship to sail from Poland for the United States before the German and Soviet invasion of Poland and the outbreak of World War II.
 
@EdwardEvans lmao
 
3:44 PM
> Tarski left reluctantly, because Leśniewski had died a few months before, creating a vacancy [in Warsaw University] which Tarski hoped to fill. Oblivious to the Nazi threat, he left his wife and children in Warsaw. He did not see them again until 1946. During the war, nearly all his Jewish extended family were murdered at the hands of the German occupying authorities.
 
and to think there are people who still believe it never happened
 
@Thorgott grrrrr
 
they found a guard from one of the camps hiding in Texas the other day
 
Uhhh
I'm pretty sure he's Banach of Banach space fame
Which seems like perhaps the more important citation
 
@MikeMiller yeah I had to check whether the Banach of Banach-Tarski was a different one after Akiva's comment
By the way I wrote Assaf Shani an email and I'll start attending his course tomorrow in the end
 
3:56 PM
Nice
Friendly guy, very competent
 
I found a comment by Styxhexenhammer666 while randomly looking through some Chernobyl video some days ago lmao
Famous youtube Nazi, was retweeted by Trump sometime ago when he still had his twitter
 
That sounds like the names of most commenters on such videos
 
666 is satanic reference recaptured by online nazis
dont ask me why
its a bad name because satanists are cool people
 
that name has massive vibes lmao
 
4:00 PM
@Thorgott lol yeah
 
@BalarkaSen Styxhexenhammer666 definitely listens to NSBM
 
yeah lmao
he likes Varg
iirc
 
it sounds like the fascist twin of xX420PussySlayerXx
 
lol
 
@Thorgott lel
 
4:04 PM
i have an image to share but i dont really want to keep it up here for too long
@user2103480 bruh are you here
 
send it on dc
 
yeah sending
 
Hahn-Banach theorem is amazing
 
4:49 PM
@BalarkaSen post and delete?
Ah, you want to wait until user2103480 is here.
 
5:02 PM
tfw had to cancel my seminar talk because apparently zoom connection is too bad to simultaneously talk and draw on a whiteboard
reeeeeeeeeeeeee
 
yike
 
now I have to figure out an alternative way of presenting till next week
never have I missed the comfort of using chalk on a blackboard more
 
@BalarkaSen no
 
@Thorgott beamer
 
@Thorgott what
why not prepare a LaTeX writeup and just speak freely while pointing to things
 
5:11 PM
that's what I said
 
@Thorgott massive cringe metaller vibes lol
 
talking over a pdf is the most boring way to give a talk
the joy of commutative diagrams comes from actually drawing them
 
@Thorgott it's the easiest to follow and the most fluent
because you have time to go over it slowly
 
@Thorgott you can draw commutative diagrams on q.uiver.app
 
If you intersect three hyperboloids, the result has exactly $\ln2$ the volume of the surrounding cube
(It's like the fluffed out version of the union of two tetrahedra, the stella octangula)
 
5:22 PM
if all I'm gonna do is talk over a pdf, people might as well just read the pdf
best to follow is when something is actively derived in real time
 
I emphatically disagree
but I dont like listening anyways, if possible I read good lecture notes
 
@AkivaWeinberger cool, do you have a reference where they calculate that?
 
No
Because I calculated it myself, just now
I am so surprised that it turned out that clean
 
good job
I'm trying to see the hyperboloids but I just see the entire shape
 
A good lecture is interactive, which is why covid sucks
 
5:34 PM
I think I see them now
 
I've actually had my most interactive lectures during covid
but that's just to the credit of the prof
 
lucky you
 
The lectures I attended mostly got better during covid lol
one prof just read off a pdf but the explanations were good + a complete TeX file is great
 
One of my profs didn't have lecture notes
 
the lecturers that used electronic whiteboards were pretty well-versed with the technology
@Astyx did you ever consider just sneakily filming
*not that I ever did this*
 
5:45 PM
Oh, lectures were recorded
That's not the issue
 
okayé
can I wlog assume the wedge sum of cw complexes is glued along a zero-cell
probably right
dont know if I actually ever need this
 
@user2103480 but real slate chalkboards are so cool!
 
meh not a fan of chalkboards
 
cringe take
 
@Thorgott regarding?
 
5:50 PM
@user2103480 what does wlog mean here
@robjohn not being a fan of chalkboards
 
$\Huge W\!\!\!$ithout $\Huge L$oss $\Huge O\!$f $\Huge G\!$enerality
 
$W$ithout $L$oss $O$f $G$enerality
 
$\Huge W$ithout $\Huge L$$\Huge O$ss of $\Huge G$enerality
 
@Thorgott the question is probably better phrased as "does it matter where I glue the wedge sum of CW complexes?"
and do these "obviously" always (dim >= 1) form good pairs
 
$y=\operatorname{wlog}(x)$ iff $x=\operatorname{wexp}(y)$
 
5:54 PM
$(X,\{x_0\})$
 
matter for which purpose?
 
$y=w\log(x)\implies x=e^{y/w}$
 
the bad answer is that it obviously matters if you wedge at different connected components
 
It has to matter right? If you make a wedge of circles all at the same point you don't get the same thing as if you wedge at different points
 
you do get the same thing up to homotopy equivalence
in this specific case, at least
trying to think whether the same conclusion holds in general, but not sure
 
5:59 PM
@Thorgott Oh oh. Thor and Ted agree again.
Howdy @Astyx, @Thor, @robjohn, @user2
 
hi Ted
 
@TedShifrin G'day
 
@Thorgott that is probably the only thing that matters since its all about (co)homology
 
@AkivaWeinberger Congruent hyperboloids? So this is a variant of my favorite three-intersecting-cylinders problem?
 
6:01 PM
I wonder if Archimedes knew how to do this one.
 
$x^2+y^2-z^2=1$ and the two other permutations
It sits snugly in a 2x2x2 cube (those "x"s in the picture are planar)
and its volume at that scale seems to be $8\ln2$
 
I think if the complexes are path connected you're fine
 
Presumably this has the same symmetries as the cylindrical case.
 
@TedShifrin Finding the volume of the intersection of three congruent cylinders along each axis?
 
Yes, @robjohn. One of my favorite multivariable calc assignments, but in an MAA talk I showed how Archimedes did it.
 
6:07 PM
@TedShifrin so you're question was rhetorical since you knew Archimedes? ;-p
 
Is that just computing a hideous integral or is there a trick?
 
No hideous integral.
 
@Astyx sounds sensible thanks
 
@robjohn Only in passing. He was busy.
 
@user2103480 I'm by not means familiar enough on this topic for you to take my word for it, but that's what my guts tell me
 
6:09 PM
@TedShifrin taking a bath?
 
I wasn't there for the Eureka moment, @robjohn, but it was the talk of the city.
 
Archimedes was actually grumping while taking that bath. He only took that bath because his students said, "You reekah!"
They were from Rome
 
@robjohn 👎
 
I think robjohn has too much free time on his hands.
 
@Astyx interestingly this is the only other way you can find WLOG in that expression
 
6:14 PM
 
Indeed, and there are many more instances WOL, but only one of WOLG
 
6:58 PM
Howdy, @loch. Long time!
 
long time indeed
 
@Leaky: Did I miss an answer to my query yesterday?
 
hello @TedShifrin
long time indeed!
 
7:20 PM
You doing OK, @loch?
 
@TedShifrin I don't know yet
 
OK. I thought it was too early. Keep me posted.
 
sure
 
@Ted: did your answer to the triple tube intersection involve no integration whatsoever? I had to compute $\int_{1/\sqrt3}^1\left(1-z^2\right)\,\mathrm{d}z$.
 
not too bad
what about you?
 
7:33 PM
I just used Burnside's lemma to compute the number of non-congruent ways to distinctly color all 6 faces of a cube by 6 different colors.
Lol
 
I did a different integral for the integration approach (natural region of symmetry is easy in cylindrical coordinates), but it's morally equivalent to yours. Yes, a completely integral-free approach, as Archimedes figured the volume of a sphere. If you're interested, I'll give you the link to my MAA slides.
 
@TedShifrin sure
 
@loch: Well, much to the chagrin of the room, I haven't disappeared yet. :) Hanging in there. Antique enough — and lucky enough — that I've actually been vaccinated.
 
@BalarkaSen sounds reasonable; not sure how else you would do it
once?
 
Twice.
@Balarka: I love Burnside. I only learned it when I wrote my algebra book :P
 
7:35 PM
hmm, it's quick in USA
 
@LeakyNun Burnside's lemma is overkill, it's only useful when you need not distinctly color
I immediately think of Burnside in these kind of situations
 
@TedShifrin It look like it's going to be a while until my wife and I get vaccinated.
 
there is no kill like overkill
 
Truth
 
@Leaky: Far from quick. We're at about 10%, I think.
 
7:36 PM
It's less slow than in France
At least last time I checked
 
The point is the action is free here.
 
@robjohn: Here's the MAA link.
 
So the answer is just 6!/4! = 30
 
@TedShifrin thanks!
 
@Astyx: Well, it's only because the former president who thinks he won in a landslide is gone that things are moving. But still way behind on people of color, teachers, and slogging through the over-65 folks.
 
7:38 PM
wow, there's really discrimination?
Don't people realize the virus leaves only when everyone is vaccinated?
 
Oh, our society (including the military and medical personnel) is full of people who refuse to be vaccinated. I suspect most of them are doing this in solidarity with Trompolini. We require all sorts of vaccinations to allow kids to go to school or young adults to go to university. I think things have to be mandatory.
But, yeah, people without computers or cell phones (and sometimes cars) are disadvantaged for sure.
 
@Astyx I'm tryingn to make more visualizations like yours, how did you get the text display which also included numerical expressions involving the variables?
 
7:53 PM
There's a section called "Measure" in the geometric tools
It lets you measure angles, lengths and area
 
Got it
 
Ah, like Geometers Sketchpad.
 
Alternatively, there's a "Text" object you can add, and you can use variables by clicking on "Advanced"
 
Last Q, how did you name it?
The applet
 
What do you mean?
 
7:58 PM
It's named "transpose" (you can see it in the tab of the webpage) and I'm trying to use your thing as a base to make other illustrations so I'd like to be able to change that
 
Ah ok, I thought you were wondering what the name was. Upon trying to save it, they prompt for a name
 
OK, I made one to illustrate what column addition == shears look like and why they preserve area :)
 
cool! :D
 
Hey
The Jewish holiday of Purim ("the silly one") is coming up
and someone I know, Isaac Mayer, recorded Steamed Hams (that skit from the Simpsons) in the style of the Scroll of Esther
(The original, if you somehow hadn't seen it before:
)
 
Hysterical
I love it
 
8:10 PM
Hamentaschen aren't silly! They can be yummy.
 
@robjohn what if you did this along all lines through the origin?
 
${4\over 3}\pi r^3$
 
yeah that was my guess
 
not too hard to prove
 
only reasonable answer
idk how to formally offhand
 
8:16 PM
hm how do I get that an odd-dimensional constant singular simplex is the boundary of some higher-dimensional chain
 
what does constant mean?
 
$\Delta^n \rightarrow \{\mathrm{pt}\} \subset X$
 
what is a map from a simplex to a point?
 
a map that... maps every point in a simplex to one single point?
something something terminal object
 
oh ok kinda obvious. does the point have to be different from the vertices of the simplex?
@user2103480 mmm that map always exists and is unique?
but it's not always in $X$ I take it
 
8:23 PM
unique up to choice of terminal object; that wasn't serious
 
it's the boundary of the constant simplex one dimension higher
 
I mean yeah that is ok
 
I just mean any map from the n-simplex to some point in the space X and I want that to be zero in homology except in dimension 0 ofc
@Thorgott what's the boundary of the constant simplex in dimension 1, thor
 
0
 
bc then it is a boundary & homology is cycles/boundaries?
 
8:25 PM
uhh I wanted to make a point but that backfired I think lol. I switched up parity I think. One case is trivial, the other leads to 0. Yeah I meant even dimensions
boundary of odd-dimensional simplex is 0 so I don't immediately get that the even-dimensional constant simplex is null in homology
Probably if one takes some smart chain?
 
@user2103480 What exactly are you trying to do here?
 
no, even-dimensional constant simplex is boundary of odd-dimensional constant simplex one dim higher
no wait, now I'm off
it's not a cycle in even dimensions
you have either no non-trivial cycles or that all cycles are boundaries, depending on parity
 
@Thorgott ah damn should have checked that before trying to prove its 0
 
Well, I'm not needed. Lunchtime.
 
The point of all this is proving something completely different anyways. I want to show that for spaces $X,Y$ with chosen points $x_0,y_0$ such that $(X,x_0)$ and $(Y,y_0)$ form good pairs, $p_{X}^{*} \oplus p_{Y}^{*}: H^{k}(X ; R) \oplus H^{k}(Y ; R) \rightarrow H^{k}(X \vee Y ; R)$ is an isomorphism, where $p_X$, $p_Y$ are the collapse maps from $X \vee Y$ to $X$ resp. $Y$.

The problem is basically that an isomorphism in the other direction is easy to get via the LES of a pair, and in principle I just gotta concatenate this with the sum of the collapses and see that it yields the identit
MV is also a possibility to get from $X \vee Y$ to the sum, maybe easier to compute
Our lecturer gave the following "proof" for connected CW complexes
I should probably also assume connectedness. Also, $k \geq 1$
 
8:38 PM
yeah, anything but MV seems needlessly complicated tbh
 
yeh
 
once you have that, it's easy to confirm that the collapse maps induce isomorphisms too
 
8:50 PM
I'm sorry, does X + Y mean disjoint-union?????
 
yes
 
I refuse
@Thorgott Amusingly this completely fails for cubes!
 
@MikeMiller K(G,n)-bundles over B is classified by H^(n+1)(B;G)?
 
9:07 PM
comments you expect to make when grading a physics quiz problem: "You assumed the acceleration was zero when it's not." "You didn't account for the weight." "You included a spurious force in your diagram."
comment you don't expect to make: "There are 60 minutes in an hour, not 60 seconds."
 
we've all been there
 
the fact that it's not just one person is a bit distressing
but it is funny when the only point you dock for someone is "an hour has sixty minutes"
 
everyone makes that sort of mistake. difference between a surgeon and a student.
 
that's why cubical homology sucks
don't let Brown know I said this
 
9:12 PM
@LeakyNun What's a bundle?
 
principal bundle
K(G,n) is a group
 
Then yes and you understand the argument
 
great
thanks
so you can define H^2 for G nonabelian? @MikeMiller
 
No, K(G,1) is not a group for G nonabelian
I take back saying you understand the argument
 
ok
 
9:24 PM
@Semiclassical may I suggest a quick review of unit conversion in the form of a worksheet for those that need it?
 
uh
i think "knows the difference between hours, minutes, and seconds" shouldn't require a worksheet
i don't think this is a unit conversion problem, it's a "zomg have to finish the exam" problem
 
Just setting up the units to "cancel"
 
yeah but
you have to know what the units are first
i mean, i had people writing 100 kmh * 1000m/km * 1hr/60s
they knew how to convert units, they just forgot them in the rush
 
in an exam, I once multiplied a solution to an equation by a constant to get another "linear independent" solution
last minute scramble does weird things to your brain
 
exactly
"this is your brain. this is your brain on exam time."
 
9:31 PM
@Semiclassical huh?
 
context: the problem had multiple parts, and this person did almost everything correct
however, they were told the speed in 100kmh
and to compute with that, you need to convert that to m/s
 
@TedShifrin glad that you got vaccinated!
 
which they did try to do....but unfortunately treated it as 100kmh = 100km/60 s
i mean, it doesn't affect anything except the final answer, so all I docked was one point
 
100km/h?
 
9:33 PM
Yes. I get it. I got it. But look what you typed to which I said huh
 
not 100kmh
 
ah. I meant that as "an hour has sixty minutes" is all I'd have to leave as a comment
 
OK.
 
hmm. i guess i'm thinking kph
 
right
 
9:35 PM
ye
 
Interesting. We write mph but km/h.
 
Still, this issue needs attention.
rushed or not
 
@TedShifrin I have seen kph, however.
 
9:48 PM
but "kph" implies "mph" for a meter per hour :-/
 
$kh^{-1}m$
 
is that a kilo-hour :P
 
It's a kilo (1/hour x meter)
 
10:35 PM
@user85795 Not if you specify it beforehand.
 
What are ways to prove that the euler-characteristic of the boundary of a 2n-dimensional manifold is zero, other than this one: Use that the LES of the pair $(M, \partial M)$ with $\Bbb Z_2$ coefficients is a sequence of vector spaces so alternating sum of dimensions is zero, then use that $\dim_{\Bbb Z_2}(H_k(M;\Bbb Z_2)) = \dim_{\Bbb Z_2}(H_{2n-k}(M, \partial M;\Bbb Z_2))$
I mean, it's sensible since the LES relates boundary and manifold itself, and the last result helps us get rid of the relative homologies
still feels kinda random
ah I guess this needs compactness to ensure finiteness of dimensions
 
You're using Poincare duality in an essential way
Not super random
Just exploring consequences of it
 
@MikeMiller of course this isnt like actually random
 
11:07 PM
Also every (2n-1) manifold has euler characteristic zero
 
this does feel random, because "being a boundary" is irrelevant
every odd-dimensional.... what Mike said
 
Closed
 
right
 
11:21 PM
Is there a continuous involution on the projective plane without fixed points
I want to answer some form of "Can the number of the points on the sphere be a multiple of 4, requiring continuity"
Not sure how to ask it rigorously
Perhaps: Can I embed the sphere $S^2$ into the space $\{X\subseteq S^4 : |X|=4\}$ such that $x\in f(x)$ and the image of $f$ is a partition of $S^2$
 
I Admit I havent opened math in a while, I know how to take partial derivatives but regarding this video youtu.be/4b4MUYve_U8?t=1749 why the partial derivative of theta-j Xj will be zero except to the theta-J, Can you point me what do I need to read
 
or alternatively, can I partition $S^2$ into sets of cardinality 4 such that the associated map $S^2\to\{X\subseteq S^4:|X|=4\}$ is continuous
 
any map RP^2->RP^2 has a fixed point by Lefschetz, no?
 
11:44 PM
2
Q: What volume is enclosed by $k$ evenly-spaced, overlapping American footballs whose axes are diameters of a unit sphere?

geocalc33Take $k \in \Bbb N$ intersecting American footballs and configure them inside a unit sphere such that each football touches two opposite ends of the sphere. Each of the shapes are spaced evenly apart. In the case $k\to \infty$ the volume should be equal to the volume of a sphere. Related Problem ...

related to the 3 intersecting cylinder problem
 
11:54 PM
@Thorgott Oh yeah
 
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