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00:09
I mean, is there a natural way of topologizing $\frak{g}$ given that $G$ has a topology?
What about the case of infinite dimensional lie algebras?
never mind I answer my question
what is your favorite math theorem/formula and why?
00:26
At the moment, von Neumann's double commutant theorem in the theory of von Neumann Algebras; it is really illustrative of the genius of von Neumann. The theorem shows a beautiful equivalence between a topological/analytic condition and an algebraic condition.
i don't even know what is that, let me look hehe
It's somewhat technical, but very cool.
Operator algebras is the greatest branch of mathematics.
It is way over my head it seems any simple explanation?
like a little example?
hmm...there are no simple examples :P
haha I feel dum ;)
00:41
No, you shouldn't. Operator algebras is a super-technical branch of mathematics; you literally have to know every area of mathematics in order to do operator algebras, which is why a lot of people don't view it very favorably. But this is precisely why I like operator algebras.
Also, it's at the interface of fundamental, theoretical physics, which is another reason why it's so cool.
How does one tell which points in the xy-plane satisfy the inequality $xy>-1$? $y=-1/x$ forms a hyperbola in the second and forth quadrant, but since the graph’s limit is positive and negative infinity along the y-axis while simultaneously 0 along the x-axis, those points are harder to evaluate.
every time I look a term I don't know there are two news ones in the definition lool
00:53
@sch those are the points between your hyperbola and the axes, what more do you want?
plus the first and third quadrant obviously
01:14
@rapasite You’re right
 
3 hours later…
 
2 hours later…
05:59
@AkivaWeinberger hi
@AkivaWeinberger can you help.me in a problem
 
3 hours later…
08:40
I just found something online that, if $H,K$ are subgroups in $G$, then $[G:(H\bigcap K)]\leq [G:K]$, but this is nonsense, right? $H\bigcap K$ is either equal to or strictly smaller than $K$, and so it's index in $G$ must be greater than or equal to $K$'s index in $G$.
Let H=being hypnotised
Let P be a frame where the possible worlds are people. Let the thief be t in P
Let the modal operator $I_p$ to capture the notion of aware, hence $I_p \phi$ reads "p is informed that $\phi$"
Self deception is thus captured as $I_p \land I_p\neg I_p$
based on [this](https://www.irit.fr/EUMAS2013/Presentations/eumas2013_invited_2_slides.pdf) as Floridi's KTB logic is an extension of belief modal logic
typo: belief modal -> doxastic
1. Security system: People are stopped when $H \land \neg I H$
$I_t H \land I_t\neg I_t H \land I_t\neg I_t H \land I_t\neg I_t\neg I_t H$
$I_t H \land I_t\neg I_t H \land I_t\neg I_t\neg I_t H$
09:08
$I_t H \land (I_t\neg I_t H \land I_t\neg I_t\neg I_t H)$
$I_t H \land I_t(\neg I_t H \land \neg I_t\neg I_t H)$
$I_t H \land I_t(\neg I_tI_t H \land \neg I_t\neg I_t H)$
fail
$I_t H \land I_t\neg I_t H \land I_t\neg I_t\neg I_t H$
09:35
$I_t (H \land \neg I_t H) \land I_t\neg I_t \neg I_t H$
This reads:
"t is informed that the system have stopped him" and "t is informed that he is not informed that he is not informed that he is hypnotised".
Thus scenario 1 here suggests the thief may be hypnotised by his self deception and to him, the system can stop him
It thus left open the question on whether the system is indeed fooled
$$
\begin{array}{l}{\text { Suppose } A \text { and } B \text { are } n \times n \text { invertible matrices. Using Gaussian elimination on the }} \\ {\text { augmented matrix }[A B \quad I], \text { show that } A B \text { is invertible and }(A B)^{-1}=B^{-1} A^{-1}}\end{array}
$$
If we assume K5c is true (which is justifiable as a self deceiving thief should secretly knew well he is self decepting, then we can simplify this further:
$I_t H \land I_t\neg I_t H \land I_t\neg I_t\neg I_t H$
$I_t H \land \neg I_t H \land \neg I_t\neg I_t H$
$I_t H \land \neg I_t (H \land \neg I_t H)$
$H \land \neg I_t (H \land \neg I_t H)$
Thus scenario 2 here suggests "the thief is hypnotised and knew he is hypnotised" and "the thief is not informed that the system can stop him"
or equivalently:
"the thief is hypnotised and knew he is hypnotised" and "the thief is informed that the system cannot stop him" (assuming double negation holds as the underlying propositional logic is classical)
If double negation fails, then we have scenario 3: The thief is hyponotised, and the system may be able to stop him
Thus in all three scenarios, it is possible for the thief to bypass the security of the system as he is either aware he is hypnotised and is indeed is, or his self deception lead to him unsure whether he is being hypnotised
That there is a chance (and this is certainty for scenario 2) for the thief to still be aware he is hypnotised, means one of the two requirements for the system to stop people is foiled and hence he can bypass the system
As for how you can bypass security while being hypnotised, we will leave that question to the philosophers and worldbuilders
user280247
10:27
Is there any free university (physical or virtual) you will recommend for studying mathematics in the uk?
11:04
One could easily show that all $x\in \mathbb{R}$ is a Borel set, by simply noticing that $$x=\bigcap_{j\in\mathbb{N}} (x-\frac 1 j,x+\frac 1j)$$
is the countable intersection of Borel sets, and since the Borel $\sigma$-algebra is $\cap$-stable the statement follows.


And since we know that the Borel $\sigma$-algebra is stable under countable unions (intersections), I don't seem to understand how the Cantor set can have subsets that are not Borel sets.
I've seen numerous answers in the site proving that statement by using the fact that the cardinality of the Cantor set is $2^{\aleph_0}$ (a
11:41
@Rithaniel I think it follows from third isomorphism theorem for groups
or second
whichever
one of them
Hi all, what is the conventional notation for a large number of summations?
So, for example $N\rightarrow\infty$ times $\sum$?
12:06
@RScrlli A simpler way to see that $\{x\}$ is Borel for every $x\in\Bbb R$ is to observe that it is a closed set
You're right that all singletons are Borel, but not all of their unions are, only the countable unions
A $\sigma$-algebra is not closed under arbitrary unions
@ÍgjøgnumMeg Second, but we don't know that $K$ or $H\bigcap K$ are normal, so it's not applicable here. I actually think it might be a typo, and that they're either trying to say $[G:(H\bigcap K)]\geq [G:K]$ or $[H:(H\bigcap K)]\leq [G:K]$
12:53
Lets say I have a commutative ring $A$. Is the localization $A_{\mathfrak{p}} $of a prime ideal $\mathfrak{p} \in \operatorname{Spec}(A)$ to $\operatorname{Spec}(A)$ what a tangent space $T_pM$ is to a smooth manifold $M$?
@Perturbative I don't think so
the cotangent space is $\mathfrak m/\mathfrak m^2$, which is a $A_\mathfrak p/\mathfrak m$-vector space
here $\mathfrak m = \mathfrak p A_\mathfrak p$
the tangent space is the dual
there is some functorial construction involving $A[X]/(X^2)$ that I forgot
maybe you can figure that out geometrically
13:15
For a field $k$, and a scheme $X$ over $k$ and a $k$-rational point $p \in X(k)$, the tangent space at $p$ is the fiber over $p$ of the map $X(k[\varepsilon]/(\varepsilon^2)) \to X(k)$
@RScrlli Borel sets are closed only under countable unions not arbitrary unions.
14:07
@AlessandroCodenotti @SayanChattopadhyay You are right, I absolutely forgot that detail, thanks guys!
@Perturbative the analogue is really the germ of (smooth/holomorphic etc.) functions at a point - but even then geometrically there's some difference because open sets in the Zariski topology are very big
14:47
Is there a polynomial f:C->C such that f vanishes at all points of the real line but f is not a zero polynomial?
@famesyasd a non-zero polynomial has only finitely many roots, so no
thanks
@Mathein a new Veranstaltung has come up on the Vorlesungsverzeichnis
Konstruktion von L-Funktionen
ooh, interesting
Yeah looks cool :)
Only mündliche Prüfung
Might take that as well lol
14:56
sure why not
Then I have Modular forms, ANT1 and L-Functions in one semester
hahaha
nice
100% NT
yeah, and then topology proseminar
and Iwasawa theory of elliptic curves at the start yey
I already saw L-functions for modular forms and for representations of $\mathrm{GL}_2(K)$ for a local field $K$, so it would be nice to see some general theory
Nice :)
it mentions L-functions for representations of GL(n,K)
for number fields K
and their "local factors" whatever that means
15:07
Question I had this morning: is there such a thing as hyper-spherical trigonometry?
I think it's a factor in an Euler product, like for example $\zeta(s)=\prod_{p \in \Bbb P} \frac{1}{1-p^{-s}}$
ah I see
Apparently there is, and according to someone’s thesis it’s related to elliptic functions and integrability
So...that’s neat
adopt naively the definition of the Dedekind zeta function to the completion of a number field $K_v$, then as the valuation ring $\mathcal{O}_v$ is a DVR, we get that all ideals are powers of the maximal ideal so we have $\zeta(K_v,s)=\sum_{n \geq 1} q^{-ns}=\frac{1}{1-q^{-s}}$
@famesyasd not only there isn't such a polynomial, there's no holomorphic $f$ vanishing on $\Bbb R$ except for the zero function
15:12
where $k$ is the cardinality of the residue field
by multiplicativity of norms and unique factorization into prime ideals, we even have $\zeta(K,s)=\prod_{v \in \Sigma_K} \zeta(K_v,s)$
What's the intuition behind nilpotent groups (and central series in general)?
where $\Sigma_K$ is the set of places, i.e. equivalence classes of valuations
this explains the terminology "local factor" since the Euler factor is attached to a local field (or in general, an object defined over a local field)
by $k$ you meant $q$ I assume
yeah, oops
that's cool af tho
love dedekind magic
that makes sense, I read something like this getting contributions from locally defined zeta functions to get something global
what about at the infinite places?
15:27
@ÍgjøgnumMeg I think the infinite place contribution is that factor with $\pi$ and $\Gamma$ that appears in the functional equation
but I don't know the theory behind it
Guess we'll see lol
dedekind magic, hyperbolic madness, flow induced by generators, call it consequential sadness, cause the rapture'll capture the batches of evil rats and cats consuming the poison, the wrath of the mad match, those hats that attach wack, pop cats with their snap-backs, consuming media in fat stacks, and missing all the fun of the dedekind magic, now that's wack
 
2 hours later…
user131753
17:13
Is it true that the notion of Logical Consequence (denoted by the symbol $\vdash$) is a meta-level notion? Of course, in usual discourses on Logic we define it as such, but my question is: How can we really be sure that it really is a meta level notion and not something that is convenient to be represented as a meta level notion?
17:49
Current mood: Writing $\pi$ as $\Gamma(\frac{1}{2})^2$
tau over two
alternatively, write everything as it x1 ad infinitum
Of note: $a=a+0-0$
big if true
@Rithaniel: You are in a bizarre mood today :)
That I am. Too much work. Going insane, more than likely.
(I joke, of course)
18:03
Well, you were halfway there to start :)
I take that as a compliment
18:19
Howdy, Mr. @Eric. How goes the new adventure?
Philosophers love their terms, its annoying.
Just to say that a society has power structure there's no need to introduce terms like hegemony, discourse, consent and what not
Philosophers are entitled to their terms, just as mathematicians are entitled to theirs!
But it gets confusing. I mean even mathematics with excess of language becomes weird.
You'll complain that to speak English (or French or Irdu) you must learn English (or French or Irdu) vocabulary.
never learn what numbers are
18:27
The effect of literature comes largely from the use of words and sentence structure. Can't have that, either.
But that vocabulary is constant throughout mostly. Taking normal meaning terms and giving them twisted meaning depending on various contexts and using them in various other contexts is some twisty stuff
I don't believe the philosophers are changing the meanings of the words. They're just using words you're not comfortable with.
Hmm yeah, maybe.
philosophers make me uncomfortable
but i respect them
"Philology recapitulates ontogeny."
18:30
:scared:
I guess it's supposed to be "Ontogeny recapitulates philogeny."
So I totally messed up :)
Language bad. Me talk simple. Understood.
^^
everyone should talk like they are writing for simple english wikipedia
You mean everyone should use a third-grade vocabulary ... not that I'm thinking of anyone with power at the moment.
18:34
@SayanChattopadhyay But each of those words mean more than "power structure". They capture specific aspects of that power structure and how it operates.
No, third-grade is too high. Only grunts and pointing.
Good for @MikeM for actually engaging in the point.
I wonder why Eric's icon keeps entering and departing.
bc ive been restarting my device
@MikeMiller Yeah, but it gets difficult to keep track of what they mean sometimes.
I agree that it is sometimes hard to read philosophy because the terminology can be overwhelming sometimes. This is why it's valuable to be able to take these ideas and explain them without that language. (I think any philosopher and anyone interested in philosophy should learn this skill.) But at the same time, to build on old ideas, you need to actually use those old ideas!
18:36
@TedShifrin i like it here way better than i thought i would tbh
That is to say: the best sort of philosopher engages in a back-and-forth of simplification (hopefully without vulgarization) and deepened understanding of their own subject.
@TedShifrin Bored grad student syndrome.
@MikeMiller Yeah, rightly said.
I'm very glad for you, @Eric. Keep it up.
Observation: If $X$ is a smooth $G$-manifold whose stabilizers $\Gamma_x$ are all conjugate to a fixed closed subgroup $H \subset G$, then $X/G$ is a smooth manifold and $X \to X/G$ is a smooth $G/H$-fiber bundle. (Easy enough to see.)
@Mike I think this is almost completely analogous to what one does within mathematics right?
18:38
A little harder: the transition functions of this fiber bundle lie in $G$.
@MikeM: This seems tangentially related to a question I pseudo-addressed on main a few days ago. When you have a smooth group action, when can you say that an invariant function must be smooth?
This implies that we can construct a principal $G$-bundle $\tilde X \to X/G$ so that $X = \tilde X/H$.
Actually, I think maybe I have made an error. What would the $G$-action be on $\tilde X/H$?
This seems like it might only work if $H$ is central which is a terribly restrictive ask.
Questions like this that seem "obvious" usually turn out to be wrong.
I need a concrete example of what you're doing ... :)
18:41
What I want to know is: in the situation with $X$ as above, can I construct a $G$-vector bundle over $X$ just from the data of an $H$-vector bundle over $X/G$?
@TedShifrin I have one, but I'd have to extract it from my head, so I think that might be difficult to do in the span of a couple minutes.
I am thinking $G = SO(3), H = SO(2)$, and a certain $G$-space that I would have to think for a bit to describe carefully.
So you're trying to do some sort of fibered product construction.
Basically. This is why I wanted $\tilde X$.
I do feel that $X$ as above should be "very well-behaved", since they seem analagous to principal bundles...
Something about this reminds me of things that go wrong, but I'd have to work on it and I'm rusty.
You need some sort of triviality of $G\to G/H$ or something.
This should be the "spacifying" of the fact that $G$-vector bundles over $G/H$ are $H$-vector spaces.'
Um ...
18:51
You object?
I'm not sure I understand. Are you just talking an associated $H$ bundle structure?
Hey guys
Hey there, Demonark.
How's everything going?
Decently. How's your new life?
18:58
Doing well, getting back into the swing of things
Ah, the demons are all here. Hi, demonic @Alessandro.
Things are taking a little while to pick up but I guess this gives some breathing room for sure
@TedShifrin I mean there is a canonical bijection between vector bundles with $G$-action over $G/H$ and vector spaces with $H$-action. One direction of the bijection is given by taking the fiber above $eH$, the other direction sends $V$ to $G \times_H V$.
I might have handedness wrong somewhere.
19:00
What's everyone thinking about mathematically nowadays? I haven't been here much lately
This semester I'm doing AG, rep theory, and possibly a reading course on something related to number theory
(Topology ended up conflicting with my TA section)
I try not to think about mathematics but I don't do very well.
I'd like to understand this paper. arxiv.org/pdf/1812.02448.pdf
I think I'm being really dumb, but I don't see why $G\times_H V$ is a vector space rather than a bundle.
@TedShifrin It's a vector bundle with global $G$-action given by acting on the left. There is some break in communication here, I am not sure where.
@Daminark sounds like a cool schedule
19:03
No, I'm just really dumb. You're using language in a way that I don't.
The bijection is between (vector bundles over $G/H$, which are equipped with a compatible action of $G$, in the sense that it is linear on each fiber and the projection map to the base is equivariant) and (vector spaces equipped with $H$-action).
We're back to that language thing again.
"Using language in different ways" is precisely a break in communication
Should be fun. Rep theory will have a project which is hype but it's going rather slow. AG is nice, proved Nullstellensatz last class
I'd like to solve it
19:03
How about you?
I don't call something a vector space unless it's actually a vector space.
I never once did. Please point to me where you see that so I can help clarify.
Demonark, so they're not actually assuming solid commutative algebra and going with it?
As I said, my brain is not working, but I don't see how to add two elements of $G\times_H V$.
@Daminark is it going to be classical AG or schemes?
It depends year to year, I think this guy is trying to ease in the commutative algebra
19:05
I never called that a vector space. That is what I called a $G$-vector bundle over $G/H$.
Classic first semester, schemes second
The projection $G \times_H V \to G/H$ is induced by the projection $V \to \{0\}$.
math.wisc.edu/wiki/index.php/Math_763_--Algebraic_Geometry_I--_Detailed_list_of_topics
Oh, OK, I see where I missed it. Sorry, @MikeM.
19:06
I am not bothered, I hope my terseness doesn't give that impression. I just figured less language makes it easier to pin down the issue.
I think I should go eat lunch and turn off my brain.
You're OK.
The link is a bit of a mess...
Demonark: That link is empty.
He copied an abbreviated link instead of the actual link
So it's broken
19:07
What I pasted in here has more
I'll hyperlink
I googled and got it. So he's using Hartshorne. Surprise surprise.
I am surprised more people don't use Vakil's notes, since they seem so admired. I don't know them at all.
Good that he's going to do linear systems and Riemann-Roch and not be entirely abstract.
Damn you suck at this
It's []()
LOL, yup.
He's not using Hartshorne actually, there's no official book
I sucked at it for my first few days here, but, as I recall, Demonark has been around a while.
Oh, the page I got by googling listed Hartshorne and referred to Milne and Harris and Shafarevich.
19:09
I'm trying to get some grad students here to organize a seminar for themselves on early mathematical gauge theory, starting with Atiyah-Bott
Yeah this is hard to Google, the guy first semester recommends Shafarevich and Milne
Oh, I found the fall 2018 syllabus.
Caldararu
Anyhow, I'm taking a lunch break.
Oh yeah that guy is more meme AG
So who's teaching this year?
Dima Arinkin, he does geometric rep theory
19:11
Aha.
Nobody knows what 'meme AG' means that's older than around 21 or so.
After being memed on here years ago, I think I can extrapolate, @MikeM.
I dunno like the sorta heavy/derived category theory, "homological mirror symmetry"
funny you think geom. rep theory isn't like that :p
19:13
for what it's worth mirror symmetry is not derived category theory, whether or not it uses derived categories in its statement...
i'll let loch say whether or not geometric representation theory is higher category theory
Ah I see. Really I dunno much about either lmao
But people say he's the guy who likes both, also string theory apparently
mirror symmetry is a mathematical incarnation of part of string theory
i know 0 geom. rep theory

but my impression of geom. rep theory (maybe specifically in stuff related to langlands) consists of a lot of statements along the lines of some derived category = some other derived category
well, that's also what mirror symmetry is
so I guess we both get to list them as applications of derived blah
and then at some point people come in and say 'look at mah infinite categories' and now you get people using a lot of higher categorical language in the subject
19:20
Oh, somehow in my mind geometric rep theory was more... I mean I guess I see rep theory and I'm like alright this is something I can work with
I'm going to go clean my floor
See ya
waves
i should get out to your city again at some point
i actually read one of the introductions of one of caldararu's papers (categorical GW invariants)

let me pretend i know something about this and essentially regurtitate his intro

mirror symmetry gives predictions in math - one incarnation of that is computing gromov witten invariants of some space is 'related' to variation of hodge structures or something like that in a different space (its 'mirror') -- think of this as a prediction giving some equality of numbers

another incarnation of that is this 'homological mirror symmetry' thing - which is some equivalence of categories between sh
@MikeMiller you should!
19:30
@AkivaWeinberger on that note, recently I got my hands on a simple 4-wheel combo lock and I’ve gotten decently good at unlocking it just by feel
Ah, that's interesting
also if you want a feel of how meme-y geom. rep theory can get - take a look at gatisgory's work
Oh my guy has actually collaborated a couple times with Gaitsgory it seems
And apparently with Caldararu
@loch I don't think the Fukaya category is too hard to understand, but it probably depends on how you mean "understand"
oh
it just means i dont know a lot of symplectic geometry (if at all)
19:45
Then understand it for Kahler manifolds :)
After all, you need both a 2-form and a complex structure on the tangent bundle for define Fuk(M)
hm
i always assumed i should know a decent amount of sympletic geometry first to understand it but maybe i'll look at it again later
the objects are lagrangians, which is certainly a symplectic notion but a very basic one, and the "morphism $A_\infty$-algebras" are defined by counting holomorphic curves w/r/t a compatible almost complex structure $J$, and sometimes weighted depending on the area of the curve
that is to say for a Kahler manifold $(\omega, J)$ the hard part is really the part coming from $J$ :)
 
2 hours later…
ABC
ABC
22:16
Why momentum of inertia of mass $m$ in that system: ibb.co/C7SzR71 is $I=mr^2$.

My text doesn't give me the length of the thread.
It is as if he assumed that the thread has length $0$
22:28
@ABC: moment of inertia about an axis through the center of the spool, perpendicular, of course.
ABC
ABC
yes sorry moment of inert about center of pulley
@Ted hey
ABC
ABC
should it not be $mD ^ 2$ where with $D$ I indicate the distance of $m$ from the center of the pulley?
@ABC: $r$ should be the distance from the axis of rotation. I need to know the exact question, but remember that if the mass is moving downward, it is not rotating about the axis.
hi @Leaky
The motion will decompose into a translation and a rotation.
ABC
ABC
mass is going up. It's like a fishing rod reel :')
22:36
Right, so it's not rotating about the center.
ABC
ABC
So how can I calculate $I$?
So angular momentum is a cross product.
Why are you computing $I$?
You trying to do total energy?
ABC
ABC
because i need to write this : $I_{tot}\alpha=\frac{Mgl}{2}-mgr$
where $I_{tot}:=I_m+I_M$
I don't know what that means.
ABC
ABC
It is the results of moments of external forces
22:39
There's rotational energy for the rotating pulley and there's translational kinetic energy for the mass. The moment of inertia only works if it's a rigid body rotating about an axis. This is not that.
I think you want to think about angular momentum and not moment of inertia.
ABC
ABC
moment of inertia about a point of mass definition
I know what moment of inertia is. You're not interpreting it correctly.
You need a rigid body rotating about an axis. Your mass is not doing that.
ABC
ABC
I know I'm wrong, I don't want to say you're wrong. I'm trying to explain myself.
This stuff is even in the multivariable book/course I taught.
On the RHS of your equation you have what looks like potential energy. The left is not energy.
Have you worked with your teacher on this?
Anyhow, I need to leave for a while.
ABC
ABC
On the right of my equation I have the sum of momentum of forces
ok thanks for your patience, good night!

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