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1:47 AM
why are monoids useful?
2:09 AM
The natural numbers are a monoid. Understand monoids and you understand the natural numbers
2:46 AM
@Rithaniel vast overstatement
Well it's true, isn't it?
@Thorgott cheat proof that a^n/n! converges to 0: by the quotient criterion (for large enough n), the sum of a^n/n! converges, so the terms must go to zero
I'm pretty sure I actually used that argument in an analysis class once
@user2103480 There's nothing wrong with that proof
@Rithaniel No, it's not. The natural numbers have an addition, multiplication and order on top of the successor function
@MikeMiller i know, it feels cheaty though
2:52 AM
it's not much of a cheat though, since once you unwind the proof of the quotient criterion and forget all the summation parts, it just goes like:
Pick $N$ large enough such that $N>a$. Then $\frac{a^n}{n!}=\frac{a^N}{N!}\frac{a^{n-N}}{(N+1)\cdot\dots\cdot n}\le\frac{a^N}{N!}\frac{a^{n-N}}{N^{n-N}}=\frac{a^N}{N!}\left(\frac{a}{N}\right)^{n-N}$ for $n>N$ and the RHS goes to $0$ as the bracketed term is $<1$.
Although I do find it more satisfying to prove concergence of the geometric series instead of proving that a^n converges to 0
Ah, well, okay I should therefore qualify my statement. "You understand the natural numbers as a monoid, which is a nontrivial part of their structure."
The natural numbers are pretty much the most mystifying structure in math
And the most central, logically
You can code up everything with em
Generate evermore complexity
I can agree with that. I was about to argue that polynomial rings might take that title, but then I realized they're just monoid rings where the monoid is the natural numbers
did you know that you can just code everything up with arrows instead
3:03 AM
Ah wait no
Forget that
We need that for the prood to go through lmao
So yeah it all boils down to showing a^n goes to 0 and then the proofs are both analogous
@Thorgott arrows?
You mean like in deduction laws?
The implications in Hilbert axiom systems?
Lindenbaum-Tarski algebras?
i mean like categories man
3:25 AM
Oh, you were definitely right that epsilons and deltas can be wholely circumvented in that proof
a^n is monotonely decreasing so has a limit which must naturally be 0, and if one goes by the geometric series proof, one can argue that it's monotonely increasing and bounded above by 1/(1-a), without proving that a^n goes to 0 first
For a > 0
1 hour later…
4:34 AM
in The h Bar, 12 mins ago, by Nihar Karve
Anybody know software to plot 3D plane waves?
Any Comtupar Science Engynar here?
5:27 AM
@Astyx: I lied. The answer to that geometry problem was $\arctan(1/2)$. I made a dumb, dumb mistake. Nevertheless, I contend that a math multiple choice question should say "which is these is approximately the correct answer?" ...
3 hours later…
8:37 AM
@TedShifrin Here you go
3 hours later…
12:02 PM
what structure normally comes with a function space?
I think it should come with a norm, and should be modeled over a vector space
12:20 PM
does anyone like $\Bbb R^2_+$ as a vector space?
12:45 PM
can anyone help me with this;
Let X be a non-empty set and n ∈ N. Then X n is the Cartesian product of n copies of X. A relation can be defined on Xn by (a1,a2,...,an) ∼ (b1,b2,...,bn) if and only if every x ∈ X appears the same the number of times in the first list as it does in the second.

Question:Let X=R^3. For the relation above, list all elements which are related to (0,1,2).
What have you managed so far?
I thought it'd be all ordered triples of 0,1 and 2 but I'm not sure
the definition isn't clear for me
That's my understanding as well
$(0,1)^2$ is not a vector space right?
I feel like that would be quite a long list
12:50 PM
@geocalc33 Not unless you give it a vector space structure
@Emmamath Only 6 of them I think
Which would be?
@Astyx but how can you give it a vector space structure? say you add elements .8+.9, this would not respect the closure condition?
(0,1,2), (0,2,1), (1,0,2), etc
@geocalc33 You have a bijective map $(0,1)^2 \to \Bbb R^2$, so you can use the vector space structure on $R^2$. Of course the operations are not going to be the standard real operations on $(0,1)$
A vector space is no just a set, it's a set and its operations
@Astyx so the operation would have to be the dot product in order to combine vectors and secure closure?
in $(0,1)^2$
I can't see addition working
12:57 PM
Which operation?
dot product?
I have no clue what you're trying to say
@Astyx $f:\Bbb R^2_- \to (0,1)^2$ is not a linear map and $\Bbb R^2_-$ is a semiring so I was wrong to think of vector spaces
$\Bbb R^2_{-}:=\Bbb R^2_{\le 0}$
1:33 PM
I don't remember if you listen to them (though you should if you like post rock), but god is an astronaut released a new single a few days ago and there will be a new album in February @Balarka
Hey everybody
I had a differential equations exam yesterday
I wonder if i solved the questions correct
Could you please check my math, please?
1. Questions: Find the lowest rank differential equation that has the following solution; $y=a\cos(\ln{x})+b\sin(\ln{x})$, a and b are arbitrary fixeds
@AlessandroCodenotti OK I will listen
Speaking of post rock also pg.lost and collapse under the empire released new albums this year
2. Question: Find the solution of $ydx+(2x+3y)dy=0$
3. Question: Find solution of $(2+2x^2y^{1/2))ydx+(x^2y^{1/2}+2)xdy=0$ using $u=x^2y^{1/2}$ variable replacement.
4th and last question: Find the solution of $y'+y^2-(1+2e^x)y+e^{2x}=0$ by seeking a special solution in form of $y=e^{mx}$.
Thanks If you can check my solutions.
2:03 PM
there are very few people who are going to bother looking at such a wall of calculations like that in this chat
2:15 PM
Yeah, it is a trend that a good few of us come around to just hang out. The more commitment a problem looks likely to require, the less likely people are to engage with it. That being said, never say "never."
It's more about the ratio of expected energy to spend compared to interestingness of the problem
and walls of ODE calculations are among the least interesting things I've encountered in my time at university
@user2103480 more interesting than whatever shit you're up to
laughs arithmetically
to quote mohamed ababou
not as you think,
2:24 PM
oh, an infidel
numbers have an end
i am unaware of this meme
@user2103480 If you believed in those words you wouldn't have taken models of hott
2:28 PM
@EdwardEvans you're already 2 years behind in ababou lore
elliptic curves with complex multiplication have an end-OMORPHISM RING LARGER THAN $\Bbb Z$
there are better things to concentrate on
oh god
it's like wanting to start watching one piece now
the size 30 font makes it look genuine
to explain the various phenomena that
the human understand from its significance the
ability of the creator whom there is nothing like
of course
2:29 PM
the greater mystery than mohamed ababou is still nafissa atlagh
who is she
what does she know
and why was she drafted in to translate when she clearly cannot do so
Ababou was obviously duped into believing this person was a translator and gave money to someone who just dumped the entire text into google
I'm not sure whether mohamed ababou has the money to do that
but lets not delve deeper into that
The words of the prophet are hard to understand, but the fault lies in your mind, not in his mouth nor in his translator
and let's not hate on nafissa atlagh
i'm supposed to be doing psets but i wanna know why the numbers have an end
2:32 PM
without nafissa, we would not have access to our prophet's writings
Ask Wildberger
i have a controversial opinion
@AlessandroCodenotti or, as ababou would say
@AlessandroCodenotti i'm gonna get really into this meme and quit my degree and become a disciple
edwards personality collapsed
@EdwardEvans literally 2 or 3 years too late
2:33 PM
it's never too late to join a cult
ababou is great but his conclusions are boring
we've all done that and came back
he's overhyped
@BalarkaSen he is if you hype him today. his meme density from 2 years ago was crazy though
this is like
the absolute complement of that paper that describes addition as a cocycle
2:34 PM
he's nothing compared to Thierno M Sow
or carrying or whatever i can't remember
Wildberger is under hyped
@user2103480 Are going to write down a spde to model his meme denseness over time or something?
@BalarkaSen you know nothing
I've been in a chat where we had ababou with several crackpots and let them argue
ababou won
2:36 PM
Speaking of weird papers (as NSFW as a researchgate link can be)
Lmfao what
What a title
I have the pdf wait, I'll send it to you on discord
oh lmfao i can see it
lmao that's a classic
a peer of mine during my bachelor tried to get my dissertation supervisor to critique his water-tight proof of the Collatz conjecture and when he refused he just submitted his paper to some journal with the profs name on it
he also tried to get me to read his proof and the first line was like.. some incorrect modular arithmetic
2:39 PM
"1 publication, 0 citations, 1.133.744 reads" the author's profile lmao
sneaky 1337 in there
Read the acknowledgements
Okay, so the upper triangular matrices can be written as a direct sum of their columns, and these columns will form left submodules/ideals. How does one show they are indecomposable at modules over the upper triangular matrices (everything is over a field $k$, btw).
3:04 PM
Let $T_n(k)$ be the algebra of upper triangular matrices over the field $k$, and define $$S_j = \{A \in T_n(k)$ \mid \text{ every column of A, except possibly the j-th column, is 0 } \}$$. Then it's clear that $T_n(k) = S_1 \oplus ... \oplus S_n$. I am trying to show that the $S_j$ are indecomposable ($T_n(k)$-modules?). Is it possible that they are actually simple modules?
I think so
If the $S_j$ were simple, then $T_n(k)$ were semisimple which is not the case
That makes sense...because later in the problem I am asked to determine the Jordan-Holder series for $S_j$.
If I could show any two ideals/submodules are included in each other, would that work? What's the best way to approach this problem?
3:19 PM
Oh ok $\bigoplus_{i=1}^k k E_{i,j}$ with $k\le j$ is a proper submodule of $S_j$
My bad
@user193319 right, the submodules are determined by leaving out entries from bottom rows
Essentially, the action of the algebra can move terms up but not down
So the easiest argument for these being indecomposable is that they have simple (in fact 1-dimensional) socles
one way to show that the $T_n(k)$ are indecomposable is to compute $\mathrm{soc}(T_n(k))$. If an Artinian module has simple socle, then it is indecomposable
oh, sniped
What's a socles?
the socle of a module is the sum of all simple submodules
In this case the socle is fairly easy to identify explicitly
3:22 PM
basically if you can show that $P_j$ has a unique simple submodule, then it follows that it is indecomposable. That's because if we had $P_j=A \oplus B$ with $A$ and $B$ non-zero, both $A$ and $B$ have a simple submodule, so there's more than one simple submodule
@LukasHeger Well, we are not showing the the algebra is indecomposable here, since it is not. Just the pieces already found
You can also (more elementarily) show that all submodules are of the form I gave above, and it's clear that the intersection of these submodules is never 0
oops I confused the notation
I meant $P_j$ where I said $T_n(k)$ above
So, for $n=3$, we have $$S_3 = \{\begin{pmatrix} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & c \\ \end{pmatrix} \mid a,b,c \in k \}$$. Would one submodule of this be $$\{\begin{pmatrix} 0 & 0 & a \\ 0 & 0 & b \\ 0 & 0 & 0 \\ \end{pmatrix} \mid a,b \in k \}$$?
3:27 PM
Ah, okay.
$\Bbb e$
Let $\xi$ denote the symmetric group
Consider the wreath product of $\xi$ with itself
Try writing this on a blackboard
$\xi \wr \xi$
3:42 PM
my numerics prof would write that as $\wr\wr\wr$
squiggles squiggly squiggles
- wreath product of xi with itself
i think this is officially better than Xi over Xi bar
fight me
Do people write $\xi$ as the symmetric group?
3:47 PM
no lol
Let $C$ be a smooth manifold and $c \in \Bbb N$, then we have $C^c(C) \supset C^c_c(C) \subset C_c(C)$. Now write that sloppily on a blackboard so that $($, $\subset$ and $C$ look the same
I recently denoted a sequence of spheres as $(S_n)_{n\in\mathbb{N}}$
@LukasHeger Ok this is epic
@LukasHeger I stared it because everyone was doing it, I don't know what it is...
3:56 PM
How do I show that any vector field on $S^n$ can be extended to a vector field on $\Bbb{R}^{n+1} \setminus \{0\}$? My intuition tells me that I should extend the vector field on $S^n$ radially throughout $\Bbb{R}^{n+1} \setminus \{0\}$, but how do I write this down?
compose a deformation retraction $\mathbb{R}^{n+1}\setminus\{0\}\rightarrow S^n$ with the vector field $S^n\rightarrow TS^n$ and then with the embedding $TS^n\rightarrow T\mathbb{R}^{n+1}\setminus\{0\}$ induced by the inclusion $S^n\rightarrow\mathbb{R}^{n+1}\setminus\{0\}$?
$X(x) = X(x/\|x\|)$...?
essentially that
@Thorgott lol
thats the opposite of writing
except you can't call them both $X$
but you're staying in line with the previous theme, so I respect it
3:59 PM
$\mathbf {X \otimes X \otimes X}$
@BalarkaSen Hmm that's it right? I was doubtful, the question was supposed to be a "long question"
Well upto identifications
A vector field on $S^n$ is naturally a map $S^n \to \Bbb R^{n+1}$
A vector field on any open subset of $\Bbb R^{n+1}$ is also naturally a map to $\Bbb R^{n+1}$
You just define the latter from the former by polar coordinates, constant on radial coordinates
Right, that feels better
a vector field on any submanifold of R^n is a map into R^n by identification
4:05 PM
vector bundles are projective basically
The $S_j$ form the Jordan-Holder series, because each $S_j/S_{j-1}$ is $1$ dimensional?
hmm, I guess you can put it that way
4:18 PM
If I have a set of $n$ points in $\mathbb{R}^2$ what is the maximum number of segments of unit length between two points of the set?
hmm it has an oeis
Oh nice it was propsed by Erdos
5:13 PM
Hmm...not sure that makes sense anymore. $S_j/S_{j-1}$ is a cyclic module...but cyclic doesn't always imply simple...
5:30 PM
what types of questions can we ask about 4-manifolds?
"what types of questions can we ask about 4-manifolds?" is an example
How about what types of questions can we ask about the manifold: $\zeta^{3,1}$ with metric, $g=\frac{dxdy}{xy}+\frac{dudv}{v-uv}.$?
@geocalc33 is that even a metric? I believe that that allows two different points whose distance is zero.
maybe I don't understand the topology of $\zeta^{3,1}$
@robjohn this is allowed in the semi-riemmanian case. I should've stated that
Ah, I was just going by the classical definition of metric space.
5:50 PM
if i take the $\frac{d}{dx} \int_{a}^{b} n \,dx $ what would i get?
$n*(a-b)$ or $n*(x(a)-x(b))$ or $n(a)-n(b)$?
@MadSpaces the scope of $x$ is inside the integral, it does not make sense to differentiate outside the integral by $x$.
@robjohn i dont know what you mean. i am dumb physicists and i dont understand math.
You are asking $\frac{\mathrm{d}}{\mathrm{d}x}n(a-b)$
arguably you get 0
bad notation nonetheless
what if x is way function such that x(a) is a path and x(b) is path, would that change anything
thus they have no concrete values)
5:53 PM
@MadSpaces the $x$ in the integral does not have an existence outside the integral
I cannot make sense of that
@MadSpaces Then how do you integrate with respect to $x$?
I dont know my dude.
Thanks for the help anyway :D
@MadSpaces I am trying to figure out what you are asking. Aside from the $x$ inside the integral and outside the integral conflicting, is $n$ a constant, or a function of $x$?
I do not know. apparently the notation provided by my prof is plane crap and hides the true mathematical discrebtion of the matter... i guess i cant do anything about it.. dumb physicts right?
5:57 PM
From the answers you are considering, it appears that $a$ and $b$ might be functions of the $x$ that exists outside the integral.
Yup.. i just checked wikipedia page.. and theres ton of mathematics that wasnt shown. this integral is planely embarassing to write
Nvm my stupid questions, kinda not my fault :D
So, if that is the case, then $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^bn(t)\,\mathrm{d}t=\frac{\mathrm{d}b}{\mathrm{d}x}\cdot n(b(x))-\frac{\mathrm{d}a}{\mathrm{d}x}\cdot n(a(x))$$
I see.
1 hour later…
7:16 PM
Can somebody explain to me the concept of codistribution in differential geometry, or link me to a good source where it is explained clearly? I am a real beginner in differential geometry. Thanks in advance.
Hello everyone. I have a very simple question about the simple proof that is the uniqueness of a limit of a real function $f\colon X\to \mathbb{R}$, with $X\subseteq\mathbb{R}$.

I have proved it before, but this book I'm utilizing gives us the following step with an additional information concerning $a\in X'$.

Ok, we have $\lim_{x\to a}f(x) = L_1$ and $\lim_{x\to a}f(x) = L_2$, and then for every $\varepsilon > 0 $ there exists $\delta_1,\delta_2 > 0$ such that $x\in X, 0 < \left|x-a\right| <\delta_1 \implies \left|f(x) - L_1 \right| < \varepsilon$ and $x\in X, 0 < \left|x-a\right| <\delt
@J.D. I have been a differential geometer for 40+ years, and this is the first time I've heard that. This is a distribution as a subbundle of the cotangent bundle rather than as a subbundle of the tangent bundle?
@Attractor The key is to choose $\epsilon$ small enough so that you cannot have both inequalities.
What is $X'$, anyhow?
@TedShifrin Ohh ok! This will discard the $x$ that may be within $\delta_1$ of $a$ but not within $\delta_2$ of $a$. $X'$ is the derived set
The set of all accumulation points of $X$
But the key thing is to choose $\epsilon<|L_1-L_2|$. All the other stuff is silly. You just use the min as usual and get a contradiction.
Hi Ted
7:25 PM
Salut @Astyx
Oh ok! I just never saw this proof using the fact that $a\in X'$ to get a $\overline{x}$ such that (...). So there's actually nothing new. Thanks
Yeah, I suppose the author wants to choose a particular $x$ and calls it $\bar x$, but it's silly.
Do you know about geometrically integral/reduced/irreduicble schemes?
My days of scheming are long gone. I know what reduced and irreducible mean, and integral is about integral ring extensions, but I don't remember much.
I'm looking for geometrical intuition behind the notion of "geometrically integral" etc
7:29 PM
Yeah, I cannot help on that. I wrote precisely one paper where scheme structure was actually relevant.
You understand reduced and irreducible, though?
Not geometrically
Irreducible is easy. Doesn't look like the union of two components (e.g., $xy=0$ is reducible).
Reduced means the only nilpotent is 0 and irreducible that there is a dense point
7:33 PM
Reduced means that intuitively that the implicit function theorem applies: you're looking at $x=0$ rather than $x^2=0$, which is the same underlying set but has nilpotents in the coordinate ring.
Blah @ dense point ...
Doesn't a reducible one have that too?
You just want to say you cannot decompose into pieces.
Are step functions dense in $L^{\infty}[0,1]$?
Right yes
Don't think so
Step functions with finitely many pieces to them, I presume?
Hmm, why don't you think so?
Heya @Lukas
Hi @Ted
7:36 PM
Oh I guess I was a bit confused.
@Astyx As far as I recall, dense point implies irreducible, and the converse holds for affine schemes
Oh should be true actually
Oh, I see why dense point fails if two components.
Yeah, @user, I think it's true. Remember that two functions are the same if they differ on a set of measure 0.
Howdy, @MikeM.
If there is a dense point you never have $X= F_1\cup F_2$ because then $x\in\emptyset = F^C_1 \cap F^C_2$, no?
But can't Linfty functions be quite wild?
7:38 PM
I think they aren't dense
I mean the only condition is that they are bounded
And measurable.
something like xsin(1/x)
on [0,1]
take a characteristic function of infinitely many disjoint intervals
Oh, no problem with that one.
7:39 PM
how are you gonna approximate that with step functions in the infty-norm
@Astyx Right, or put another way, one of those sets will contain the dense point and thus be the entire set since it is closed
Yeah, that's the kind of thing I was thinking about, @Thorgott, but I took the characteristic function of the rationals.
So take a Cantor set of positive measure. That should do it.
I am fairly certain irreducible does not in general imply that there is a dense point. It might hold fro schemes in general, and I at least recall showing that it holds for affine schemes
it should be true in L^p, cause then step functions are dense in simple functions
It holds for scheme in general I think
7:40 PM
But this is stuff I have not actually really thought about since taking a class on intersection theory at UGA
That's still far more recent than for me, @Tobias :D
schemes are sober, for sober spaces every irreducible closed subset has a unique generic point
<--- prefers drunk schemes
I wonder what the English term is there ... I've never heard of this.
Right, because irreducible => every open is dense and irreducible, so you can consider an affine cover
7:43 PM
For affine this is simple enough (take the prime ideal $0$)
Howdy, @copper.
Hi @TedShifrin!
I'm procrastinating from work.
Highly unusual.
But coming back to reduced, why is it that the implicit function theorem applies?
7:46 PM
I was just saying that for intuition. But it's true in that setting.
Implicit function theorem says you have a zero set of something like $z-f(x,y)$, not $(z-f(x,y))^2$.
@TedShifrin sober space seems to be standard terminology
I've never ever ever heard of it, @Lukas.
Maybe it's too new-fangled for me.
yeah, I've heard of sober spaces before
@TedShifrin I'm not sure I understand, what's the difference between those two?
Same point set, different scheme structure. The second one is not reduced.
7:51 PM
In fact, there's a soberification functor, if you apply that to the maximal spectrum of an affine algebra A over an algebraically closed field (which is "drunk" in general), you get Spec(A) back.
So, the point is that the implicit function theorem may fail to prove something is a manifold even though it is ... just because you're giving the set a non-reduced scheme structure.
LOL, @Lukas. I was never taught that.
I mean, that statement just boils down to say that points in Spec(A) are irreducible closed subsets of the maximal spectrum
I already mentioned this recently in chat, but my favourite paper shows that the Scott topology on a poset is not necessarily Sober and it is called "Scott is not always sober". The first letter in each paragraph spell out the sentence "I'm not being personal Dana"
LOL ... good for those of us who actually knew Dana Scott :P
sober spaces are interesting because they're a class of spaces which you can reconstruct from the categories of sheaves on them
7:55 PM
@TedShifrin Do you know him? He is a big name in logic and related areas
Yes, I met him when I was very young.
For non-mathematical reasons, in fact.
Sober spaces are moderately interesting
8:39 PM
Hi, all.
Hi there
Hi everyone
anybody around that can point me to a mathematically meaningful definition of boltzmann's discretization of phase space in statistical mechanics
is there such a thing
out of my space, sry
Hi @BalarkaSen
8:47 PM
@user2103480 can you be more specific? i have had a course in statmech but never heard of the terminology "discretization of phase space", i can only guess what you mean
anyone here know much about partial differential equations?
If this sigma-algebra F is really generated by a finite partition, if we work in a finite volume
sounds fancier in French
8:54 PM
yeah i guessed thats what you meant
i dunno what kind of math formulation you want tho
I wonder if French call these "sets statistiques"
then all the real-valued measurable (borel) functions are constant on the "microstates"
you just declare $h^{3N}$ volume cubes are your elements of the prob space
yeah or you can quotient the measure space or whatever, doesnt matter
its some physically meaningful hypothesis
i can tell you an example in math tho
id have to recall the example first
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