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00:15
I have a professor that mentioned something about a connection between the de Rham cohomology and dynamical systems. Does anyone know generally about their relation?
@KarlKronenfeld Do you have another idea?
@PedroTamaroff There's a chance the degrees of the purported factors of $x^p-x-1$ are equal (contradiction). I am trying that out.
@KarlKronenfeld Explain, explain! =D
who discovered the riemann functional equation?
I am still insisting on my idea of differentiation.
@user4140 Probably not Riemann.
00:19
riemann?
it was Riemann
in 1859
Cool.
"An equivalent relationship had been conjectured by Euler over a hundred years earlier, in 1749..."
Oh Leonhard!
@FernandoMartin Agh, those use a lot of Galoish things.
Let's wait then.
What's the problem? Seems like it's the natural setting for the problem
The first one looks elementary, but I haven't read it
@PedroTamaroff Jyrki's answer is elementary indeed.
00:26
@KarlKronenfeld It says "$r$ is one of the roots of $\Bbb F_p$" but $X^p-X-1$ has no roots at all. Does he mean we should adjoin one or something?
I think he doesn't know that it has no roots
and he's looking at what would happen if it had one
@PedroTamaroff Yep, you can work in the algebraic closure of $\mathbb F_p$, where the roots exist.
@FernandoMartin Showmatch restarts today. SIGH.
YES
best tv show ever
can't wait.
@KarlKronenfeld Right.
00:32
for which values does the functional equation hold?
riemann functional equation that is
@user4140 Read the article.
it doesn't say
The Riemann zeta function or Euler–Riemann zeta function, ζ(s), is a function of a complex variable s that analytically continues the sum of the infinite series \sum_{n=1}^\infty\frac{1}{n^s}, which converges when the real part of s is greater than 1. More general representations of ζ(s) for all s are given below. The Riemann zeta function plays a pivotal role in analytic number theory and has applications in physics, probability theory, and applied statistics. This function, as a function of a real argument, was introduced and studied by Leonhard Euler in the first half of the eight...
@user4140 Every complex number but $s=1$.
so then shouldn't 4 be a 0?
@user4140 Ah?
00:34
$sin(2\pi)=0$
Not sure if trolling.
no, i'm not
I'm confused
Then explain yourself clearly, please.
f*ing debugging this *** damn prog (that I wrote).
$\zeta(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s)$ so $\zeta(4)=2^4\pi^{4-1}\sin(\frac{\pi 4}{2})\Gamma(1-4)\zeta(1-4)$
00:37
impossible
$sin(\frac{4\pi}{2})=sin(2\pi)=0$ or am I really messing up here?
@user4140 But $\Gamma(-3)=\infty$.
Careful.
Oh, I see
@Mike
Back to black.
@PedroTamaroff nice photo, what cellphone do you have?
00:50
@user4140 XPERIA ZL.
who here's done exterior algebra stuff?
how the heck would the riemann hypothesis help us find primes?
@AlexanderGruber like for ten minutes one day
@user4140 why would you think that it would?
@user4140 The zeta function is intimately related to the distribution of primes. @seaturtles might be able to explain something.
00:56
@AlexanderGruber I watch tv
@KarlKronenfeld well, i'm trying to prove that exterior product commutes in a certain way with direct sums
that there's a functorial isomorphism $$\wedge^n\left(V\oplus W\right)\cong \bigoplus_{r+s=n}\wedge^rV\otimes_k\wedge^sW$$
anyhow i'm thinking maybe this can be done with the universal property... "the $r^\text{th}$ exterior product of $M$ is defined as the module $N_0$ having a universal alternating $r$-fold multilinear map $f_0:M^r\rightarrow N_0$ that satisfies the property that every alternating $r$-fold multilinear map $f:M^r\rightarrow N$ factorizes as $f=hf_0$ for a unique $A$-linear map $h:N_0\rightarrow N$"
i'm thinking it may be easier to use that than actually construct an isomorphism
what do you think, @Karl? have i gone too deep into the rabbit-hole to be helped by other mortals?
@AlexanderGruber I agree.
@AlexanderGruber That's how I'd do it, for sure.
01:04
You'd have to use the UP of the tensor product too.
and direct sum :P
@AlexanderGruber In a simpler situation, I had to use three UPs there.
And that onto implies epic in $\bf Ring$.
@PedroTamaroff Whaddya want
@Mike Me? Let me think.
I will rephrase.
Why'd ya ping me
01:10
I hardcover press printed version of Lang's "Algebra."
@Mike Oh, I was thinking about how to show $X^p-X-1$ is irreducible over ${\rm GF}(p)$.
@PedroTamaroff if we lived a couple centuries ago they'd burn us for drawing stuff like this.
boo/
@AlexanderGruber True. But first they'd burn chemists for four colour oscillators and such things.
@Ted: Oh yeah, I have looked at the Eisenbud, Harris book on schemes that you suggested a couple of weeks back. Their treatment of sheaves is fantastic; I have high expectation for the rest of the book. Thank you very much for the suggestion!
@KarlKronenfeld YAY KARL!
@KarlKronenfeld I think Lang has a typo here.
01:18
@PedroTamaroff unsurprising :D
@KarlKronenfeld I am talking about the very last sentence.
But you can read the preparation. It's $\S 4$.
@PedroTamaroff The very last sentence.. of the proof?
@KarlKronenfeld I think I might be mistaken. He says the leading coefficients of blah blah generate $\mathfrak a_d$.
I think he means $\mathfrak a_r$. Or he wants to change $r$ to $d$s.
The idea of the proof is clear though.
$\mathfrak a_d$ is correct.
Then he wants to write $f_{d1},\ldots,f_{dn_d}$? @KarlKronenfeld
Oh, sorry.
I forgot the ideals had settled.
My bad. =)
Well certainly $\mathfrak a_d=\mathfrak a_r$!
I was going after an extra step. =P
@FernandoMartin (?)
01:26
sorry, wrong window
@PedroTamaroff Never heard of factorial notation for ideals :) Though I admit you typed ! outside of the $.
@FernandoMartin You got that one cheap.
Big pimple on dong, how to remove painlessly?
Oh no, wrong window!
oh my god this is so messy
poor timing @AlexanderGruber
3
LOL
I AM DYING RIGHT NOW
01:27
... oh no :(
In mathematics, the explicit formulae for L-functions are relations between sums over the complex number zeroes of an L-function and sums over prime powers, introduced by for the Riemann zeta function. Such explicit formulae have been applied also to questions on bounding the discriminant of an algebraic number field, and the conductor of a number field. Riemann's explicit formula In his 1859 paper On the Number of Primes Less Than a Given Magnitude Riemann found an explicit formula for the normalized prime-counting function π0(x) which is related to the prime-counting function π...
what is $\mu$ here?
@user4140 Möbius $\mu$.
the Möbius $\mu$ function?
thanks
01:33
GAH I DONT UNDERSTAND ANYTHING
i hate math. i'm signing up to dig holes for a living.
@PedroTamaroff you can tell she's never done commutative algebra
@AlexanderGruber HAHAHAHAHA
that is exactly how i feel
@AlexanderGruber There there.
Seriously, just take some more time.
@AlexanderGruber This is a tune I like.
01:43
maybe i should try constructing the map
this universal properties thing isn't working out.
jesus christ
@AlexanderGruber Oh jeebeezuss.
maybe i'll just submit that with "the isomorphism is obvious from the definitions" written under it :)
ahahaha
@AlexanderGruber Hehhe, is this some assessment?
01:49
@PedroTamaroff it's my last commutative algebra assignment.
i would not be subjecting myself to this willingly, believe me.
@AlexanderGruber $M$ is getting ganged up by arrows there.
@AlexanderGruber I believe you!
@AlexanderGruber This might sounds silly, but did you try inducting on $n$?
And using some kind of distributive law on wedges and tensors and sums.
@PedroTamaroff i haven't. i'll try that. but, the problem is i have a short memory. there's only so much that can stay in here at once.
Let's see if this helps, @AlexanderGruber.
Suppose we take $n=2$.
What does an element of $\bigwedge^2(V\oplus W)$ look like?
01:54
either zero or all entries distinct.
and by all entries i mean, pairs $(v_1,w_1)\ne (v_2,w_2)$, so the $v$'s can be the same and the $w$'s can too but not both at once
so we're trying to show that it's $\wedge^2 V\oplus \left(\wedge^1 V \otimes_k \wedge^1 W \right) \oplus \wedge^2 W$
@AlexanderGruber Sorry, I thought $\wedge^k(V)$ was something different. The definition is taking the quotient of $V^{\otimes k}$ by a certain span of things that induces the alternating relations?
@PedroTamaroff well it's entirely possible i have no idea what i'm talking about, the professor lectured about it for less than $5$ minutes and the book only has the definition by universal property
@AlexanderGruber holy shit
That's just for the tensor product.
@KarlKronenfeld yep, that's how the entire year has been.
@KarlKronenfeld well i've got the morphism to $M$ indexed by $r$ and $s$ so it's the cone for the direct sum of them anyhow
I think you want to use maps like $$\wedge^r V\to\wedge^r(V\oplus W)$$
And then piece them together
02:02
@KarlKronenfeld well, that's technically in there, with compositions.
i think it actually kind of makes sense? the vertical line is all unique morphisms
from the various properties
but how to prove it's equivalent to the LHS diagram's chain of unique morphisms is something i have no idea about
@AlexanderGruber $(v_1,w_1)\wedge (v_2,w_2)\to (v_1\wedge v_2,v_1\otimes w_2+v_2\otimes w_1,w_1\wedge w_2)$?
@PedroTamaroff what i'm reading is this "the quotient of the $r$-fold tensor product $M\otimes \cdots \otimes M$ by the submodule generated by elements of the form $x_1\otimes \cdots \otimes x \otimes \cdots \otimes x \otimes \cdots \otimes x_r$"
What about explicitly going from $\wedge^r V\times\wedge^s W$ to the LHS, instead of doing UP crap for that. @AlexanderGruber
02:06
@AlexanderGruber Yes.
@KarlKronenfeld yeah i'm thinking just making the map is maybe the way it's gotta be, those diagrams make me want to die.
That's essentially what I said? =D
@PedroTamaroff so i guess i'm mixing it up with the definition of an alternating map
@AlexanderGruber Yes, I did that too.
Then I realized my mistake, so I asked.
but this is supposed to be "the simplest alternating map" right? the $f_0:V^r\rightarrow \wedge^r V$ map
02:08
Yes, I was seeing each $x\wedge y$ as maps.
I studied things like this from Spivak.
Briefly.
i've never worked with it before concretely, i just learned the definition on Friday.
let me try to digest what you wrote
I need help understanding this article:en.wikipedia.org/wiki/…
@PedroTamaroff i don't understand where the middle part comes from
@AlexanderGruber It's really a hunch. I "multiplied" that thing out. You shouldn't be listening to me. =D
The cat is back :)
0
Q: Show that $\log(x)$ is a Bounded Mean Oscillation (BMO)

Jessy CatAs an extension of our class notes, we were asked to show that the function $w =\log(x)$ is a Bounded Mean Oscillation (BMO). First off, I believe our professor made a mistake, and really wanted us to show that $w = \log|w|$ was a BMO, just to keep everything real. That being said, by the John-...

02:15
okay wait a minute here
@AlexanderGruber Is there even a canonical way to send the middle term of that direct sum decomposition into $\wedge^2(V\oplus W)$?
If not, then that will pose problems for your UP approach.
If $M$ is a free $A$-module of rank $n$, with basis $e_1,\ldots ,e_n$, then $\wedge^r M$ is $0$ for $r>n$, and for $r\leq n$ it's the free $A$-module of rank ${n \choose r}$ with basis $\{e_{i_1}\wedge \cdots \wedge e_{i_r} | 1 \leq i_1 < \cdots < i_r \leq n \}$
that's lookin like somethin that could be $\oplus_{r+s=n}$'d.
does this mean I have to substract all the elements on the right?
no, the sign is determined by $\mu(n)$.
02:19
also, are the ones in the right all the natural numbers or the prime?
all natural numbers
so then how do I know which ones are added or substracted?
calculate $\mu(n)$.
(This is all from reading the $\sum$ expression)
Okay, now that the typos are all fixed...
@JessyCat I don't do that kind of analysis, sorry. =D
02:21
Neither do I apparently
And the snarks are out of the tank :(
oh, ok, thanks karl
@user4140 well what are you trying to do?
I'm trying to explain the Riemann hypothesis to a high school student,
I don't get it myself, but such is my burden
how do I see the following sum?
from my previous image?
Well, if you know anybody who does that kind of analysis, tell them to visit the cat: math.stackexchange.com/questions/773645/…
okay let $r+s=n$ and $V$ have basis $e_1,\ldots , e_r$ and $W$ have basis $f_1,\ldots, f_s$, $V\oplus W$. then $V\oplus W$ has basis $e_1,\ldots,e_r,f_1,\ldots,f_s\triangleq g_1,\ldots, g_{n}$. then $\wedge^n\left(V\oplus W\right)$ has basis $\{g_{i_1}\wedge \cdots \wedge g_{i_n} : 1 \leq i_1 < \ldots < i_n \leq n\}$.
no
02:27
@AlexanderGruber you need to wedge those...?
$g_{i_1}\wedge\cdots\wedge g_{i_n}$.
@Mike @AlexanderGruber A friend is trying to access ams.org/books/memo/0451
@user4140 Pretty sure you're summing over roots of $\xi$.
@KarlKronenfeld @seaturtles
okay so then the $g$s with indices less than $r$ correspond to the $V$ and the others correspond to the $W$.
02:30
yes @AlexanderGruber
@KarlKronenfeld why?
filing these in the appropriate boxes gives us $\wedge^r V\otimes \wedge^s W$
and since we're just letting $r$ and $s$ be arbitrary that somehow leads to a direct sum
best proof ever, totally rigorous
@user4140 why not? :P I can't help you there, look carefully at whatever text comes before that equation, it didn't come from nowhere.
I found something similar in Edwards, which is why I think it's summing over the roots.
@KarlKronenfeld Yes, the sum is definitely over roots.
what is edwards?
02:34
ech i have no idea how to make that right. this has to be the method though.
It is right, I think @AlexanderGruber
I just have no idea what is happening in that step
@user4140 A quick glance at my book on the zeta function (Edwards, as well), seems to reference $\rho$ almost only in connection of the roots of $\xi$.
Oh. Nevermind--I see that's been answered...
@AlexanderGruber You have to be able to rearrange the terms of an arbitrary wedge in $\wedge^n(V\oplus W)$ into the form you have there, but that's it.
@KarlKronenfeld my hands are waving harder than a hurricane day beach here
i need to refit this into something that is a proof and not me saying things that just sound good
02:38
@AlexanderGruber That sounds like something I've said before while doing homework... :-P
I'm gonna use van-molgoldts formula
@anorton it's shocking to me how many professional mathematicians seem to have skipped that part while doing theirs.
the part that is still bothering me about this is why we can have varying $r$'s and $s$'s in $\wedge^n (V\oplus W)$ instead of fixed.
hey wait a minute.
this is stupid. if $r+s=n$ then we just have $$\wedge^n \left(V\oplus W\right)=\wedge^r V\otimes_k \wedge^s W$$
@AlexanderGruber No, I don't think so.
i think so. i think so.
because, my construction above is exactly that. and what if we have say $p+q=n$ where $p=r+1$ and $q=s-1$?
then $\wedge^p V=0$, so $\wedge^p V\otimes \wedge^q W=0$
does that make sense?
so the general case, the $\wedge^n\ldots = \bigoplus_{r+s=n}\ldots$, that is covering the cases when $n < r+s$.
why is that zero for $p$?
You're assuming the things are fin dim?
02:51
@PedroTamaroff yes i am
@PedroTamaroff i think it's by definition
@AlexanderGruber yes, if $p$ surpasses $\dim V$ yes.
But not elsewhen.
@PedroTamaroff right, so what i mean is, if $\dim V+\dim W=n$, then every term in the $r+s=n$ sum except for $r=\dim V, s=\dim W$ has to die
because one of $r$ or $s$ is going to have to be greater than the dimension of the corresponding $V$ or $W$
@AlexanderGruber yes, so it'd be interesting to see what happens for exponents at most $n$.
@PedroTamaroff i'm thinking that's where the direct sum comes in, corresponds to taking $r\leq\dim V$ basis elements of $V$ and $s\leq dim W$ basis elements of $W$

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