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12:00 AM
This group acts on the complex numbers (also via complex multiplication) to rotate things about.
 
@geocalc33 okay, you probably heard that groups are somehow "symmetries" of some objects. When you have an object, say for example a square, then this has some symmetries, given by rotation and reflection. You probably know that these symmetries form the dihedral group. But actually more is going on here than just having an group. In this setting, the elements of the group act on the square, i.e. they transform it in a particular way
another example (assuming you know LA): consider a vector space $\Bbb R^n$, then the invertible matrices $\operatorname{GL}_n(\Bbb R)$ act on $\Bbb R^n$ by "changing the basis" (just matrix-vector multiplication, really)
and if you just have a finite set $\{1, \dots, n\}$, then the group $S_n$ acts by permuting those elements
 
Correct me if I am wrong, @MatheinBoulomenos, but the basic idea is that you have your hands on two basic objects: a group $G$, and some kind of space $X$ (say, a topological space). In addition to this, there is a collection of functions $\varphi_g: X \to X$ (one for each element of the group) which act on the space, moving points around.
 
@XanderHenderson right
 
Moreover, these functions communicate with each other in the way that would be expected given the group structure.
 
yeah, that's completely correct
 
12:06 AM
Thanks!
 
For example, if the group operation is denoted by $+$, then $\varphi_{g} \circ \varphi_{h}(x) = \varphi_{g+h}(x)$ for all $g,h\in G$
 
and you want $\varphi_{e}(x)=x$ where $e$ is the neutral element
 
Oh, yes! That too!
 
those two things are all you need
 
I'm forcing myself to be more comfortable with group actions right now---I'm working on a precalc book with a collaborator which includes a chapter called "Group Action and Representations"
 
12:10 AM
sounds pretty advanced for precalc
 
He (my collaborator) really wants to finish that chapter with a discussion of the Galilean group
@MatheinBoulomenos It actually comes up pretty naturally.
 
I can imagine the precalc book "definition: consider a group as a one object groupoid, then an action is just a functor ..."
tbf we don't have precalc here, so I can't really comment
 
The basic approach of the book is to look at relations (primarily in the plane) and discuss how relations can be transformed.
In a precalc setting, the graph of $p(x) = a(x-h)^2 + k$ is just the graph of $f(x) = x^2$, translated right $h$ units, up $k$ units, and scaled vertically by a factor of $a$.
Some of these transformations have very natural group representations. Similarly, most of trigonometry comes down to the circle group acting on some space.
 
don't you have a time component in the Galilean group?
 
Yeah, which I why I keep telling him that the Galilean group is a bridge too far.
But once you know some trig, it isn't too hard to start thinking about dihedral groups
 
12:14 AM
that sounds like a really cool precalc book actually
 
or rigid motions of the plane
We've been teaching the class for two years now; the plan is to finish the book in the next couple of months.
I'm excited. :)
We have a chapter on derivatives that doesn't mention limits a single time. ;)
 
as a non-american I was just thinking precalc was some boring class that prepares you for calc by repeating trig and algebra etc. and maybe has you do some calculations that might come up in calc without understanding why you do them
 
@MatheinBoulomenos that is what most pre-calc classes in the US are like. They are generally pretty terrible.
We're trying to do better.
 
sounds like you're succeeding in doing better :)
 
right... my people are home... time to go make dinner
 
12:19 AM
bye @Xander!
 
what is the difference between a class of functions and a group of functions
 
12:38 AM
Could I insist that "x/0 = x/0" using the Law of Identity?
 
you would have to firstly define what x/0 means
 
Does the law of identity require that?
The law itself extends beyond mathematics as far as I know.
 
 
4 hours later…
5:14 AM
@Lispwave don't you ever write that again "x/0" :)
Hello. I have a system of equation that mathematica says the solution is the empty set {}, but in my book says something different, it says there are 4 solutions of the system. This is the system $x1 (-1 + v) + 2 x2 = 0 , x2 (-1 + v) + 2 x1 = 0, x1^2 + x2^2 = 1$
I don't know what's wrong
The solutions according to the book are $(x1,x2)=(1/\sqrt 2,1/\sqrt 2)$ and v=3, $(x1,x2)=(-1/\sqrt 2,-1/\sqrt 2)$ and v=3, $(x1,x2)=(1/\sqrt 2,-1/\sqrt 2)$ and v=-1, $(x1,x2)=(-1/\sqrt 2,1/\sqrt 2)$ and v=-1
ah forget it, I already saw the mistake, it's not just for x1,x2 it's also finding the v value
 
5:36 AM
I was meaning to reply to this sooner but hadn't had the chance. "Symplectic" is the Greek adjective form of the English word "complex", and the terminology happen to have good reasons. The symplectic group $\text{Sp}(2n) \subset \text{GL}_{2n}(\Bbb R)$ contains matrices $A$ which satisfy $A^T S A = S$ where $S = (0, I_n; -I_n, 0)$ is the "standard symplectic form on $\Bbb R^{2n}$". By Gram-Schmidt $\text{Sp}(2n)$ deformation retracts to $\text{Sp}(2n) \cap O(2n)$, which consists of matrices satisfying $A^T S A = S$ and $A^T = A^{-1}$. Equivalently, matrices which satisfy $A^T = A^{-1}$ And
The takeway from this long computation is that on a vector space of even dimension, a symplectic structure and a complex structure are "the same"
 
@Lispwave sure, but if a mathematical object is only defined via the law of identity, it is useless because it cannot otherwise be manipulated
for example, a natural thing to ask what is x/0+1 and why it is useful
 
(As an end-note, this is true linearly, but not globally I don't think. A symplectic structure on a manifold is a fiberwise symplectic structure on the tangent spaces with the integrability condition $d\omega = 0$. Similarly, there's also an integrability condition for complex structures on a manifold, which is the vanishing of Nijenhuis tensor $[JX, JY] - J[JX, Y] - J[X, JY] - [X, Y] = 0$. These conditions cause severe restriction on space of symplectic and complex structures on a manifold)
 
And to do that, you need axioms that tell us what happen when x/0 and 1 are put together
a=a, elephant=elephant is a tautology and a tautology is useless by itself
 
5:58 AM
"The sound of the color red" = "The sound of the color red"
the two statements are identical, but neither of them actually means anything
 
What's the mathematically correct way to write an expectation with respect to a posterior?
 
6:19 AM
@BalarkaSen Oh interesting, didn't know that
 
6 hours ago, by MatheinBoulomenos
Yo mama's so fat, that's not even finitely generated
Does the following exists:
sup_x (Yo mama's so far that's x)
Is there a supremum to the notion of "vastness"
e.g. vastness = {non finitely generated, non compact, countable, uncountable, not first countable, unbounded, large cardinal, proper class, incomplete,...}
Now what about:
(Yo mama's so fat that x)*
=> Yo mama's so thin that x*
Actually, inf(Yo mom's so thin that x) exists:
inf (Yo mama's s thin that x) = Yo mama's so thin that she is empty
So by the axiom of global choice, we can well order the class of all "Yo mama's so fat/thin" jokes
In mathematics, specifically in class theories, the axiom of global choice is a stronger variant of the axiom of choice that applies to proper classes of sets as well as sets of sets. Informally it states that one can simultaneously choose an element from every non-empty set. == Statement == The axiom of global choice states that there is a global choice function τ, meaning a function such that for every non-empty set z, τ(z) is an element of z. The axiom of global choice cannot be stated directly in the language of ZFC (Zermelo–Fraenkel set theory with the axiom of choice), as the choice function...
And therefore, the class of all Yo mama's so fat jokes form a lattice, known as the Yo mama's so fat lattice
 
6:40 AM
Oh my god
That escalated so damn quickly
 
lol
 
That is what happens when you try to take the supremum of any chat messages lol
 
the axiom of global choice is commonly used in category theory, even if people don't mention it explicitly
for example, one direction in the equivalent characterizations of "equivalence of categories" uses it: if you want to show that an essentially surjective and fully faithful functor has an inverse up to natural equivalence, you need to do choice on the class of objects in your category
 
no, I wasn't
 
6:53 AM
Actually, I have a joke metaconjecture:
Joke metaconjecture Let sup : Chat message -> Chat message and sup^n= sup^{n-1} sup
Then the conjecture says that: there exists a finite and very small n such that for any chat message x
sup^n(x) in category theory
 
you need to use the axiom of global choice to show that a category is equivalent to its skeleton, which can be pretty important. Basically the skeleton of a category removes all "unnecessary" isomorphic objects: if two objects in the skeleton are isomorphic, they are equal. (So cardinals are the skeleton of sets, ordinals of well-ordered sets, for vector spaces, you still have cardinals, corresponding to dimension, but different morphisms etc.)
You can construct that without choice, by modding out by isomorphisms basically, but you can't show that it is equivalent without choosing a represe
"We will not concern ourselves with subtle
foundational issues (set-theoretic issues, universes, etc.). It is true that some people
should be careful about these issues. But is that really how you want to live
your life? (If you are one of these rare people, a good start is [KS2, §1.1].)" - Vakil
 
lol
What's KS2? I don't recognize the initials but I'm awful at remembering names
 
Kashiwara and Schapira - Categories and Sheaves
 
What kind of isomorphisms are we considering in order to make a skeleton from a category?
 
@MatheinBoulomenos Ah, I was expecting something more set theoretical
The isomorphisms of the category we want to quotient is my guess
 
7:05 AM
@Secret every category naturally comes with its own notion of an isomorphism: bijections for sets, linear isomorphism for vector spaces, homeomorphism for toplogical spaces etc.
The general definition is that an isomorphism is a morphism which has a two-sided inverse. Now what a "morphism" is is and how you compose them is part of the datum of a category
 
Now that I'm done with differential geometry I only have the algebraic number theory exam left , expect a lot of stupid questions from me :P
 
Ah I see, so it's kinda like a generalisation of the notion of bijective maps
 
Bijective maps are precisely the isomorphisms in the category of sets
 
@Secret: How did you get to abstract nonsense?
 
(If only I would read before writing I would have noticed Mathei already wrote that)
 
7:09 AM
@Rudi_Birnbaum I don't really know. I have a habit of explosive generalisation in order to explore more mind bending concepts from familiar ones. Somehow via the metaconjecture it almost always end me up in category theoretic territory without realising
 
"Now that I am done with differential geometry"

I congratulate you on this glorious milestone
5
 
@AlessandroCodenotti it's not really set-theoretic, but it works consequently in Tarski–Grothendieck set theory and doesn't leave out any details in that direction. There are sections on universes and on cardinals (in the context of Tarski-Grothendieck set theory, of course) and there are some categorical notions that depend on a cardinal, e.g. there are $\kappa$-filtrant categories or $\kappa$-accessible objects for each cardinal $\kappa$
@AlessandroCodenotti answering ANT questions sounds great
 
@Daminark lol, well I meant with the exam, I'll probably learn more diffgeo in the future
 
In the past, there are some users said I almost rediscover most of concrete category theory just from the way I speak and at that time I don't even know such field of maths exists
I guess my only explanation I had so far is thinking about categories may have something in come with holistic thinking styles
 
categories are all about holistic thinking!
 
7:17 AM
@Secret: "habit of explosive generalisation" good!
I once had a try with Alouffi. Was a attempt to pimp up my algebra a bit and at the same time start approximating the big things (Grothendieck et al) but I am not sure if it made me happy. My problem is that I need to discuss stuff while I am learning...
 
I don't know if I can understood grothendieck topology stuff yet. Right now it still sound a bit far from me
 
@Secret you want to have a clear understanding of sheaves on a topological space before you can get the motivation behind grothendieck topologies
Hi @loch
 
Yes, I meant going from 0 to 1 approximating 100 ...
 
Lol yeah I prob bash diff geo a bit too much
 
Yeah, and I will need to be much more confident with my topology foundations before touching sheaves (it's purpose of prescribing data is quite attractive to me though, thus that might help me to understand it quicker when I am ready)
 
7:26 AM
@Rudi_Birnbaum I can only imagine that it's quite difficult to begin with learning that abstract mathematics without having classmates, lecturers, TAs, math friends etc. to discuss things
@Daminark nah, you don't bash diff geo enough
 
Someone told me he saw the work on the board of the guy who taught manifolds this year since had a class right after (that guy definitely had a hand in my previous encounter with DG that made me dislike the subject so much), and felt that he approached the subject in a particular (and kinda bad) way
@Mathein tru actually, worst subject tbh
 
@Daminark numerical analysis is worse
 
^
 
But yeah in any event I've been deciding on whether to take DG next year with a prof who will have a very different approach apparently, so we'll see if it's worth it or not. Though honestly I'm more inclined if anything to approach the stuff from the standpoint of algebra
 
the right way to introduce manifolds is as locally ringed spaces, i.e. topological spaces with a sheaf of $\Bbb R$ (or $\Bbb C$ or $\Bbb Q_p$)-algebras satisfying some conditions
 
7:29 AM
Lol numerical analysis is something I don't have to engage with so it doesn't register as a thing in my mind
 
@Daminark I envy you
 
:P
 
@Daminark if you want to learn manifolds in a way that differential geometers probably find horrible I recommend Wedhorn - "Manifolds, Sheaves and Cohomology"
 
Hmm, I might try that out at some point
For now though, I think I'll be focusing on other things. My mentor recommended this book called "Rational Points on Elliptic Curves" by Silverman so I'm prob gonna be reading that
 
It takes manifolds more as a "toy model" as the spaces where sheafy stuff is easier than say on schemes
 
7:33 AM
I see
 
but it does do some serious cohomology theory on manifolds, too
 
Actually I remember him talking briefly about schemes, he basically said that one of the early things where you can see that it's important is how it's somewhat better at distinguishing certain things than varieties. If you consider the curve $y^2 = x^3$, at has a singular point
And you can take the map from the y-axis there by $t\mapsto (t^2,t^3)$, somehow variety business wrt this map can't really single out the singular point, while schemes can. Seemed p cool
 
yeah, schemes get particularly important when you want to do stuff in char p
 
What is a good book for an introduction to category theory? I am doing tangent bundles and vector bundles at the moment and there are sections of category theory which I don't understand?
 
>sections
:P
 
7:38 AM
Is MacLaine too outdated now?
 
It's more than a week old
Practically Euclid's Elements at this point
 
MacLane is still great
but it's not particularly easy if you don't know algebra and/or (ideally and) algebraic topology to understand the motivations
the problem with category texts in general is that the theory is self-contained, but you have to know some serious mathematics to understand the (interesting) examples.
if you don't have the background, then it's all just empty formalism, that's like learning a language without having anything to talk about
 
I think that was my problem the first time I tried to read MacLane.
I didn't have sufficient knowledge to motivate it. Plus, the language itself I found a bit confusing
 
8:44 AM
@Albas You don't need to know category theory to understand vector bundles. What gives you that idea?
 
@BalarkaSen isn't it better
vector bundles is all about maps
that's the philosophy of category theory
 
@MatheinBoulomenos Well, when I was 17,18 I did that one scribd.com/document/317454424/… including all exercises. Its nicely axiomatical and I guess was some classics German undergraduate Analysis textbook. Later on I attacked the Scheja/Storch (Algebra I) several times, but always got stuck somewhere in the first third. And I feel when you don't use and work with the concepts all the time they don't get "incarnated", if you know what I mean.
 
@LeakyNun That is indeed the philosophy, but it is unfortunately continental philosophy
 
9:03 AM
@BalarkaSen Tu and Lee both have sections on functors, dual functors etc which they say are important for tangent and vector bundles
 
@Albas Bleh. Don't read that stuff. Grab Guillemin-Pollack.
 
What's that book about?
 
Differential topology
Same thing Tu and Lee are about
 
Okay cool, thanks.
 
All of you completely ignored Mathein's book rec above for manifolds
 
9:06 AM
Was it sheaves on manifolds my Kashiwara?
 
It is your Kashiwara but not mine
 
Oops
 
Throw away your keyboard, you're wrecked bro
 
Now I have to type with my mouse and a virtual keyboard :(
 
Maybe I don't want to get flagged
 
9:08 AM
Now I have to use voice to text
 
It's painfully early
 
Where is the text recommendation?
 
All time zones are the same and you should go to sleep too
 
Identify all time zones
 
The Wedhorn thing above @Alex
 
9:09 AM
Oh I see
It does have sheaves in the title, so that's a good sign
 
9:19 AM
The set of reals is uncountable. But if we were to restrict the domain to say [0,1] even then that set of reals would be uncountable right?
 
Absolutely.
 
(Meanwhile in Rambles, still trying to "count" the reals by monitoring how fast the number of new element function blows up)
 
9:38 AM
@BalarkaSen vector bundles are just continuous cocontinuous endofunctors of $C^\infty(M)$-mod
 
yo mom is a continuous cocontinuous endofunctors of C^infty(M)-mod
 
that's Serre-Swan-Eilenberg-Watts basically
 
Sorry for the playground insult, had to do it
Meh, they are projective modules over C^infty(M) aren't they?
 
finitely generated projectives, yes
 
Thanks
So why meddle that beautiful theorem with categorical nonsense? Please stop
I like Serre-Swan. I do not like Eilenberg-Watts
 
9:41 AM
and finitely generated projectives turn out to be precisely that modules such that tensoring with them preserves limits (of course it always preserves colimits)
 
Yeah who cares
who care my m8
 
idk, my memories were that you liked Eilenberg-Watts when you asked me why one would care about bimodules
 
Hmm, did I? Maybe because you gave me a good answer to the question :)
I don't like the meddling in this context though
 
[Random diagram]
 
@BalarkaSen sanity check: if $p:E \to B$ is a vector bundle and $\pi: B \to X$ is a covering with finite fibers, then $\pi \circ p$ is a vector bundle, right?
 
9:44 AM
Ok I failed, I was trying to draw an impression of this:
 
@MatheinBoulomenos Fiber over a point seems to be disjoint union of a bunch of k-planes to me
 
What mathematics characterise that noodle pattern of tubules that is seen in a glomerulus?
I always found that impossible to draw freehand
 
@BalarkaSen oh right
hmm, I was just pondering if there is a relation between the $\pi_1$ action on fibers of coverings and holonomy on a flat bunde
 
It's not quite knot theory because as you can see there are branching points at certain locations
 
@MatheinBoulomenos There is!
A flat bundle's structure group reduces to a discrete group
There is an associated covering space you can extract out of it
 
9:48 AM
> holonomy
> homology
> homotopy
 
@BalarkaSen ah! very interesting stuff
 
salami
 
@MatheinBoulomenos What's your definition of flat?
 
@BalarkaSen let's just take that holonomy is homotopy-invariant (this should be equivalent to actually having vanishing curvature tensor I think)
 
Yeah that's right.
 
9:52 AM
So if you take the $\pi_1$-action on the fibers on the associated covering space and look at the permutation representation induced by that, this should give us the representation of $\pi_1$ from holonomy?
 
So you have a well-defined representation $\pi_1(B, x) \to GL(T_x B)$ by the holonomy, whose image is going to be a discrete group $G \subset GL(T_x B)$. The associated covering space is indeed, like you say, the covering space associated with this permutation repn
But how do you see it, is the question
 
Thanks! this makes a lot of sense
 
Here is the answer. If you have the vector bundle $p : E \to B$, consider $dp : TE \to TB$. There is a vertical subbundle $V = \ker(dp)$ of $TE$ coming from this projection, and once you have a fiberwise Riemannian metric on $E$ you can take it's orthogonal complement, the horizontal subbundle $H$
$TE = V \oplus H$
Flatness is equivalent to requiring that $H \subset TE$ is an integrable distribution, i.e., there is a foliation $\mathcal{F}$ of $E$ by submanifolds ("leaves" of the foliation) which are always tangential to $H$
 
I don't know distributions or foliations (I can make sense of foliations of a torus induced by lines of irrationals slopes in $\Bbb R^2$, but that's about it)
 
I have forgotten the right visual reason for this but it should be that the holonomy representation $\Omega(B, x) \to GL(T_x B)$ comes a lifting procedure: If you have a connection $\nabla$ on $E$, and a loop $\gamma \in \Omega(B, x)$, you take a vector $v \in T_x B$ and using $\nabla$ when one parallel transports $v$ along $\gamma$ one is really lifting the vector $v$ to the horizontal subbundle $H$
@MatheinBoulomenos $H$ is just a subbundle of $TE$ which projects to $TB$ by $dp$. The vertical $V$ gets killed.
Think of the foliation as a partition of $E$ into submanifolds $\mathcal{F}_i$ such that they are always tangent to $H$
I.e., for any $x \in \mathcal{F}_i$, $T_x \mathcal{F}_i$ is a fiber of $H$
What happens is that $p$ gets the structure of a foliated vector bundle, so that $E$ admits a foliation $\mathcal{F}$ by submanifolds which are always transverse to the fibers.
 
10:03 AM
A foliation is a decomposition of a manifold into submanifolds so that locally (at any point) you can pick a chart so that it's the decomposition of R^n into parallel R^k s
 
The transition functions then not only has to be linear automorphisms of $\Bbb R^n$ but preserve a foliation of $\Bbb R^n$. That makes the structure group discrete
 
(not locally near a leaf, as your example of irrational slope on a torus shows)
 
Also why am I writing $ \to GL(T_x B)$? I should write $\to GL_x(E_x)$. We're working with the general vector bundle not the tangent bundle
Fiber over that point
 
@MatheinBoulomenos finite extension of local fields is local fields right
is it easy to show if i define local field as a locally compact non-discrete topological field
 
@LeakyNun yes
 
10:05 AM
But the associated cover is now simply projection of a leaf $\mathcal{F}_i \to B$
 
@LeakyNun if you define local field as locally compact non-discrete topological field, then a finite extension will just be a finite-dimensional vector space over that, i.e. topologically and in terms of the additive structure it's just $K^n$
 
oh
oh
wait it isn't just the product
how do we prove that multiplication is continuous
 
multiplication will be $K$-linear in each argument,i.e. it's a linear map $L \otimes_K L \to L$ Linear maps on finite-dimensional topological vector spaces over a topological field are continuous
(the thing as over $\Bbb R$)
maybe I should add that the vector space should be Hausdorff
yeah that works just over local fields, I think
I'll check my ANT notes
I don't know if you can prove that multiplication is continuous without running through the same kind of arguments that give you the structure theory for local fields
but anyway, local fields have a norm that induces the topology and finite-dimensional Hausdorff topological vector spaces always have have a norm that induces the topology, as well (here I'm using norm in two different, but related meanings)
and the proof that linear maps on finite-dimensional normed spaces are bounded, thus continuous is easy
you can use the same proof you would use for $\Bbb R$
 
10:26 AM
@MatheinBoulomenos hi
 
@MatheinBoulomenos can i not use the norm
hi @loch
 
Hi @LeakyNun
Based on you guys hi-ing me at random times apparently I come online from time to time when I’m not actually around lol
 
@loch yeah
@LeakyNun you really cannot do a lot with the definition as a locally compact field, honestly
 
:(
 
you can avoid the norm if you use the Haar measure, but the Haar measure is what gives you the norm anyway, so that's just a reformulation
 
10:29 AM
sad
 
you can use "complete with respect to a discrete valuation with finite residue field", that's enough to do most things
and it's not difficult to show the equivalence to locally compact
(if you know Haar measures, of course)
 
let's say I don't want to deal with Haar measure at all
for now
@MatheinBoulomenos that's non-arch...
 
What are you trying to do?
 
@LeakyNun ah, sure
@LeakyNun if you don't use Haar measures, then the definition as locally compact field is completely useless, sorry
 
:(
 
10:33 AM
@MatheinBoulomenos When you say 'that's enough to do most things' are you saying there is a different definition with a more broad classification or something?
 
@loch if you have MSE on phone, sometimes the page refreshes on its own, and if you just like open but inmediately tab away it does so too
 
@Alex yeah I could've said that it's enough to do everything, but I wanted to be on the safe side.
The other defintion is just Laurent series over a finite field or a finite extension of $\Bbb Q_p$
 
non-arch!!!!!!!!
 
or $\Bbb R$ or $\Bbb C$, fine
 
:P
 
10:35 AM
@Alex under the assumption that you absolutely avoid to show that this stuff is equivalent, which is what Leaky seems to do
 
That would be very naughty
 
@MikeMiller ohh I see
 
@MatheinBoulomenos is T1 assumed?
locally compact non-discrete topological field?
where is t1
 
T0 at least yeah
or you can say non-discrete non-indiscrete
a topological field is Hausdorff iff it doesn't have the trivial topology
 
:o
where does non-indiscrete come
 
10:40 AM
I'm using non-indiscrete for "doesn't have the trivial topology"
 
really
 
@loch That happens to me a lot, since I usually have old copies of this tab open
 
11:02 AM
@LeakyNun let's show this, it's not that hard (I don't want to do all the details, though)
first: a topological group is Hausdorff iff $\{e\}$ is closed where $e$ is the neutral element
 
that i know
 
okay, now if you have a topological ring, then the connected component of the identity is actually an ideal
two-sided ideal (if the ring is not commutative)
so if you have a topological ring that doesn't have a nontrivial two-sided ideal, then it's either Hausdorff or the topology is trivial
 
interesting
 
(so this argument doesn't need continuity of inversion, even)
 
11:04 AM
i have a question
 
I learned this from the book "Locally compact groups" by Stroppel
 
@BalarkaSen Maybe this'll make me sad after mirror reaper (rip)
 
Mirror Reaper was harrowing. I'm going through their discography atm
 
how do you call the formula for: $P_{X|Y}(x | y) =\frac{P_{XY}(x,y}}{p_Y(y)}$?
is it conditional probability function?
wondering about the correct terminology
 
locally compact groups are such a wide class that it's amazing that you can say so much nontrivial stuff about them: they include Lie groups (p-adic, real or complex), compact groups, in particular profinite groups, discrete groups etc.
 
11:24 AM
@loch You were right that the pinch point fails condition A by the way
I just checked
The tangent line to the 1-dimensional singularity set at the origin is not contained in a limiting tangent plane to that point
 
11:57 AM
@MatheinBoulomenos If I have a field equipped with a topological ring structure, can the inverse fail to be continuous?
is there any example?
 
12:53 PM
What is actually the dot product of two vectors. I thought it was the distance of two vectors until I learned otherwise
 
@LeakyNun Thanks XD
 
1:25 PM
It has a bounty
 
.
0
Q: Deduction of eigenvalues of the matrix $A$ which satisfies $x^t A A^t x = \alpha x^t x$?

BAYMAXLet $A$ be a $m \times n$ matrix of rank $m$ with $n > m$. If for some non-zero real number $\alpha$, we have $x^t A A^t x = \alpha x^t x$, for all $x \in \Bbb{R}^m$ then $A^tA$ has exactly two distinct eigenvalues where $0$ is an eigenvalue with multiplicity $n-m$ and $\alpha$ is a non-zero eige...

 
-3
Q: Tommy’s integral $\int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 }{36}\pi ^2 $

mickConsider Tommy’s integrals: $$a) \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4}\, dx = \frac{5 }{36}\pi ^2 $$ $$ b) \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^5}\, dx = \frac{5}{72} \pi^2 - \frac{ ln^{2}(3) }{4} - \frac{ \text{Li}_2(1/3) }{2} $$ $$ c) \int_0^1 \fr...

 
What we can say after $AA^t = \alpha I$
 
1:39 PM
Hello, i have two sets $$A=\{(x,y)\in\mathbb{R}^2, y> a_1 x+b_1; y>-a_1 x+c_1\}$$ and $$B=\{(x,y)\in\mathbb{R}^2, y> a_2 x+b_2; y>-a_2 x+c_2\}$$
where $a_1,a_2>0$
$b_1,b_2.c_1,c_2\in\mathbb{R}$
how to prove that $A\capB\neq\emptyset$
i suppose that $A\cap B=\emptyset$
that is $\forall (x,y)\in A, (x,y)\notin B$
that is : $$y>a_1 x+b_1~\text{and}~y>-a_1 x+c_1$$ and $$y\leq a_2 x+b_2~\text{or}~ y\leq -a_2 x+c_2$$
i can't find a contradiction
 
1:57 PM
Someone have an idea?
 
Hey guys
What is the difference between a class of functions and a group of functions?
 
group is an algebraic structure on top of a class or set
So a group of function is richer than an equivalence class of functions, and certainly much richer than a class of functions
 
oh okay
 
as for that polyp (something) class mentioned in that MSE of yours, I have no knowledge of that thus I cannot comment further
 
2:17 PM
Yeah it's kind of odd notation
with the 1/log(x) and the zeta function
 
2:31 PM
What is the "restriction topology"?
 
what do you mean
 
I mean that I'm looking at $\Bbb Z_p \cong \varprojlim \Bbb Z/p^n \Bbb Z$ and the document says that this is a topological group with the restriction topology
but I don't know what this topology is lol
 
@ÍgjøgnumMeg subspace topology
it lives in $\prod_{n} \Bbb Z/p^n \Bbb Z$
 
I see
Just confused because this says that each $\Bbb Z/p^n \Bbb Z$ has the discrete topology, the product has the product toplogy, and the projective limit has the restriction toplogy
 
maybe it's the smallest topology that makes the projections $\Bbb Z_p \to \Bbb Z /p^n \Bbb Z$ continuous ?
or maybe it sees $\Bbb Z_p$ as a subset of the product of all the $\Bbb Z/p\Bbb Z$ and it is the induced topology
 
2:44 PM
I think that's what @Leaky was saying
 
ah indeed
 
@ÍgjøgnumMeg well it's also with the profinite topology
 
Reading some fractional calculus
 
they're the same
 
@Leaky I think it means the "restricted product topology" which I googled about 5 seconds ago
lol
 
2:54 PM
aha
 
In mathematics, the restricted product is a construction in the theory of topological groups. Let I {\displaystyle I} be an indexing set; S {\displaystyle S} a finite subset of I {\displaystyle I} . If for each i ∈ I {\displaystyle i\in I} , G i {\displaystyle G_{i}} is a locally compact group, and for each ...
 
yeah
need to learn some more topology then
 
$A_{m \times n}, n>m, AA^t =\alpha I , \alpha> 0$, and rank A = m , what can we conclude about eigenvalues of $A^tA$ ?
any idea
 
3:04 PM
hey guys I have a question, permutation of type (2,2) means (abc) = (ab)(de)(de)(bc) so (abc)=(ab)(bc) so (ab) has length 2 and (bc) has length, so we have permutation of type (2,2)?
 
3:15 PM
$$ \sum_{n=1}^{\infty} n^{-n} = \int_0^1 x^{-x} dx $$
 
Good morning all. I have a silly notation question: what's the mathematically correct way to write an expectation with respect to a posterior distribution?
 
In mathematics, sophomore's dream is the pair of identities (especially the first) ∫ 0 1 x − x d x = ...
:P
 
I see posterior distribution called $X|\Theta$ and its expectation $E[X|\Theta]$
 
I am actually wondering what governs the set of all Sophomore's dream
Generalised Sophomore's dream problem
 
@M.Nestor I mean, the expectation of another quantity, with respect to a posterior.
 
3:25 PM
Let $P, Q$ be possibly unbounded operators in the reals, then what is the criteria for the following to be true in terms of P, Q:
$$P(x^{-x}) = Q(x^{-x})$$
::once again explosive generalisation send me into abstract nonsense again, ooops::
 
ok it smells like coincidence that they will meet because of three equations
 
$\nexists$ "coincidence"
 
@M.Nestor Very helpful, thanks!
 
3:43 PM
hi @loch
 
 
1 hour later…
4:56 PM
Given the function $ \Phi(s,x)=\zeta(s)^{1/\text{log}(x)} $ how do you solve for $s$
 
5:18 PM
anyone?
 
5:54 PM
in The h Bar, 9 secs ago, by Secret
in Mathematics, 28 secs ago, by Secret
in The h Bar, 1 min ago, by Secret
vampires dies not exist

 Inner sphere

A room where users constructs and speaks in cryptic, a mixed c...
 
Does this expression make sense: $E[E[f(X)|X=x]X\sim p] = E(f(X))$, where $p$ is a prior distribution for random variable $X$.
Oops that should read E[E[f(X) | X=x] | X\sim p]
 
what does E[ number | X ~ p ] mean, I never saw a conditional probability with a pure number in one of the entries?
 
p is a prior on X
Oh, I see what you're saying
 
The stuff on the left of the | is just a number (because all expected values maps to numbers, thus I am not sure what you mean there)
anyway I am going to sleep, so you may need to find someone else to help you on your question
 

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