Mathematics - 2019-07-18 (page 1 of 2)

12:06 AM

Would it be on-topic here to ask the number of possible combinations of inputs for a function that takes an input length of anywhere from 1 to 2^128 bits? I know the answer for a function that takes exactly 2^128 bits (it's 2^2^128), but I'm wondering if this basic question is acceptable here.

There's surely a better way to calculate it than 2^(2^1 + 2^2 + 2^3 + ... + 2^128).

Joe Stavitsky

12:19 AM

Can anyone confirm his claim that n leq 23 from a 90 degree bend radius? I cannot confirm his result from any of my calculations
https://defproc.co.uk/blog/lattice-hinge-design-minimum-bend-radius/

forest

I guess what I'm asking is whether there's a better way to say $2^{\sum_{i=1}^{128}2^i}$ .

Ted Shifrin

@forest: Of course there's the formula for the sum of a geometric series.

forest

@TedShifrin Not sure what that is, or if it'd be relevant here (my math needs work).

Ted Shifrin

$\sum\limits_{i=0}^n r^i = \dfrac{r^{n+1}-1}{r-1}$

forest

Yeah that's not what I need.

Ted Shifrin

12:26 AM

So, in your case, with $r=2$, you get $2^{129}-1$.

forest

oh

Ted Shifrin

Then $2^{2^{129}-1} = \frac12 2^{2^{129}}$, which is not what you typed originally.

forest

Would that answer how many possible inputs there are to a function that takes a 1 to 2^128 bit input?

Ted Shifrin

No, I think you need to put the sum sign somewhere else for that.

If you have a $2^k$-bit input, then you get $2^{2^k}$ possible inputs, right?

forest

Isn't $2^{\sum_{i=1}^{128}2^i}$ the correct formula then?

@TedShifrin Yes, but you can also have an input length anywhere from 1 to 2^k.

Ted Shifrin

12:29 AM

No, I think it's $\sum\limits_{i=1}^{128} 2^{2^i}$.

Which is WAY huger. Or is it?

forest

d'oh

Ted Shifrin

We're comparing $2^{a+b}$ with $2^a+2^b$. No, your original sum formula was way huger.

Aha, so now $2^{2^{128}}$ gives the right order of magnitude.

forest

Yeah that's the correct order of magnitude, but it's not exact.

Ted Shifrin

(Not the right answer exactly, of course.)

I don't know a closed-form expression for adding $2^i$th powers.

forest

So $\sum\limits_{i=1}^{128} 2^{2^i}$ is correct? For information, this is because I'm trying to correct an answer I have on another site where I incorrectly claimed that SHA-512, a hash function that takes an input message up to 2^128 bits, had an input message space of 2^2^128.

Ted Shifrin

12:33 AM

I don't know the computer jargon.

It seems to me this formula is wrong, though. You're way over-counting.

forest

It's just a (mathematical) function that takes an input that's typically considered to be arbitrarily sized, but is actually limited at 2^128. Why would that be over-counting?

Ted Shifrin

Isn't having $N$ bits the same as having $N-k$ bits with the last $k$ bits known to be $0$?

forest

Not for this function (it encodes the length internally).

Ted Shifrin

Because I think a $2$-bit function gets counted when you count $4$-bit functions.

Yeah, but encoding just the length is a tiny order of magnitude.

So you're saying that you can't absorb a $2$-bit function into a $4$-bit or $8$-bit function? If so, the summation formula I wrote down is correct.

forest

What do you mean?

Ted Shifrin

12:37 AM

I said overcounting because you can think of a length 2-sequence as a length 4 (or 8 or whatever) sequence where the last digits are all $0$s.

If you say that's wrong, then the summation formula is correct.

forest

The original one I posted or the corrected formula you gave me?

Ted Shifrin

The corrected one.

forest

$\sum\limits_{i=1}^{128} 2^{2^i}$ right?

Ted Shifrin

Yup.

forest

Then I guess there's no simpler way to say "There are X possible distinct inputs" without having $\sum\limits_{i=1}^{128} 2^{2^i}$ in place of X? I should calculate it...

Ted Shifrin

12:39 AM

I don't know a formula for that, as I said.

forest

oh

If there are $\sum\limits_{i=1}^{128} 2^{2^i}$ possible lengths in bits, then that would be the exponent for 2. Unless I misunderstood.

Ted Shifrin

No, you have to count the functions for each length and then add up.

That is your original mistake.

forest

For length 1 it's 2^1, for length 2 it's 2^2, for length 2^128 it's 2^2^128.

Then I just sum those, right?

Ted Shifrin

That was the formula I wrote down.

forest

But then that is the formula that I'm looking for (the number of combinations).

Ted Shifrin

12:42 AM

I think you're wrong, but I am not continuing this.

I am packing for a move.

forest

Ok. Thanks for taking your time and putting up with a novice like me. :P

Ted Shifrin

Good luck.

forest

thanks

forest

1:15 AM

It was a lot simpler than I expected:

1

A: What is SHA-512's input space, taking into account variable message size?

There is one zero-length message, two 1 bit long messages, four 2 bit long messages, eight 3 bit long messages, and so on. Since these messages are distinct, you can simply use the formula: $$1 + 2 + 4 + 8 + \dots + 2^{n-1} = 2^n-1$$ Note that the maximum message length is the same as the maxim...

2^(2^128 - 1) - 1

Joshua Barron

1:53 AM

Trying to hash this out in some code I'm writing that models the probability of die rolls in a pool of die rolls (like "roll 6d6") - how do I model an conditional effect like "remove a die and reroll any amount of remaining dice"? Do I have to split the entire pool by conditional probability, like P(using the effect) * P(die outcomes minus the die removed by reroll) + P(not using the effect) * P(current outcomes)? Sorry if this makes no sense, my probability skills are way out of date.

Ryan Unger

5:45 AM

@BalarkaSen kurims.kyoto-u.ac.jp/~motizuki/…

ÍgjøgnumMeg

7:11 AM

Morning all

Alessandro Codenotti

7:46 AM

Morning @ÍgjøgnumMeg

ÍgjøgnumMeg

Hey @Alessandro, you in Italy yet? :P

Alessandro Codenotti

Yep, I landed around midnight

ÍgjøgnumMeg

Cool, have fun :) How long you there for?

Alessandro Codenotti

Until the winter semester begins I guess

ÍgjøgnumMeg

Fair

Leaky Nun

8:06 AM

Dschinghis Khan - Moskau 1979

►

Leaky Nun

9:56 AM

@BalarkaSen spectral sequence computation of K(Z,2) is dank

Mathphile

11:00 AM

The solution of $e^x=x$ is expressed as a complex number of the form $a+bi$. Are $a, b$ transcendental?

Tobias Kildetoft

@Mathphile Are you saying it has a unique solution?

Mathphile

@TobiasKildetoft yes

Tobias Kildetoft

Interesting. Is that obvious?

Mathphile

according to WA it is $-W(-1)$

although i do not know how we get that result

also i wonder if there is a closed form for $a$ and $b$

@Semiclassical any ideas?

Leaky Nun

11:21 AM

no it has infinitely many solutions

WA says it is $-W_n(-1)$

I'm thinking of using Rouche's theorem to prove it

anyway

@Mathphile if a and b are both algebraic, then $z=a+bi$ is also algebraic, so $e^z$ is transcendental (Lindemann theorem), contradiction

so at least one of them must be transcendental

Mathphile

@LeakyNun okay

@LeakyNun would $-W(-1)$ be one of the infinite solutions?

Leaky Nun

sure

Mathphile

btw what does $n$ mean in $-W_n(-1)$?

Semiclassical

The WA solution is a bit misleading: the Lambert-W function has an infinite number of branches, so you need to say which one you’re using. It’s analogous to saying that $y=x^2$ implies that x is the square root of y: Which square root?

Mathphile

@Semiclassical i see

If we take logarithm on both sides of the first equation:$$\ln e^x = \ln x \implies x=\ln x \implies e^x = \ln x$$

also WA shows only two solutions to $\ln x=e^x$

does this mean that there are only two solutions to $e^x=x$?

Semiclassical

11:39 AM

There are two real-valued solutions

Mathphile

@Semiclassical what do you mean by real-valued?

Semiclassical

solutions which are real

There’s complex solutions as well

Mathphile

well aren't both of those solutions complex numbers?

$x ≈ 0.3181315052047640944061139753 \pm 1.337235701430689415856543101 i...$

i don't see any real solutions

also i don't think there plots can ever intersect

Semiclassical

Yeah, I was being careless

Mathphile

@Semiclassical so are there only two complex-solutions to $x=e^x$?

Leaky Nun

12:34 PM

@Semiclassical how do you prove that there are infinitely many solutions?

(saying that they are $-W_n(-1)$ is cyclic)

Tobias Kildetoft

@LeakyNun Well, if you can show that those are solutions, then that gives you a proof

Leaky Nun

I thought they're just defined as the $n$-th solutions

Tobias Kildetoft

So if $x$ is a solution and $a$ is some number such that $ax = x^a$ and $e^a = a$ then $ax$ is also a solution. Right?

Probably not a great way to proceed

uhoh

12:53 PM

5

Q: Poincaré map of orbital trajectories

A Poincaré map, by definition, is supposed to be a reduction of a phase diagram on a codimensional plane cutting normal to the original trajectory. In this form, one should plot points from the phase space (xdot vs. x), maybe in this case it makes sense to plot a-dot versus a (semi-major axis). B...

Mathphile

if there are infinitely many solutions then why does wolframalpha only show two?

uhoh

I've just added a +200 bounty to the Poincaré map question. It needs a practical (rather than pure mathematical) answer.

Mathphile

@TobiasKildetoft i am really confused. Are there two solutions or infinte?

could anyone explain why would there be infinite solutions to $e^x=\ln x$?

Tobias Kildetoft

@Mathphile When did you change the equation?

Mathphile

@TobiasKildetoft i didn't

Tobias Kildetoft

1:04 PM

@Mathphile You started with $e^x = x$

Mathphile

as i showed above there are equivalent

1 hour ago, by Mathphile

If we take logarithm on both sides of the first equation:$$\ln e^x = \ln x \implies x=\ln x \implies e^x = \ln x$$

Tobias Kildetoft

ahh, right, sure

but why complicate it further?

Mathphile

@TobiasKildetoft because wolframalpha only gives an exact solution to $e^x=\ln x$

Joe Stavitsky

Can anyone confirm his claim that n \geq 23 when Theta = 90deg and t=3? defproc.co.uk/blog/lattice-hinge-design-minimum-bend-radius

Mathphile

@TobiasKildetoft i.e. $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

Tobias Kildetoft

1:07 PM

@Mathphile How is that an exact solution?

And didn't it give a solution involving the Lambert omega before?

Mathphile

@TobiasKildetoft not exactly exact

@TobiasKildetoft nope

wolframalpha.com/input/?i=e%5Ex%3Dlnx

Tobias Kildetoft

Didn't it say that $-W_n(-1)$ were solutions?

Mathphile

you can check it out above

@TobiasKildetoft that was when i typed in $e^x=x$ not $\ln x=e^x$

Tobias Kildetoft

Right, so how did this in any way improve things?

You went from a nice complete solution to an approximate one

Mathphile

what i am trying to say is that i think there may be only two solutions and not infinite

and the only two solutions are : $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

Tobias Kildetoft

1:11 PM

@Mathphile Why would you conclude that from this?

You have even taken a logarithm, which in itself basically ensures you get infinitely many values

Mathphile

@TobiasKildetoft because according to wolframalpha, there are the only two solutions that exist for $e^x=\ln x$

Tobias Kildetoft

@Mathphile Does WA claim it to be a complete solution? Because it does only give it as numerical

Mathphile

thus there are only two solutions to $x=e^x$

Tobias Kildetoft

@Mathphile Getting from the first to the last involved taking a logarithm, which is something you have not accounted for yet

Mathphile

@TobiasKildetoft could you give me an example of another numerical example other than this?

Tobias Kildetoft

1:16 PM

@Mathphile Not sure what you mean. I am just saying you have not shown the equations to actually be equivalent.

Mathphile

@TobiasKildetoft i mean that could you prove that $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$ are not the only two solutions?

Assume u satisfies $e^u=u$
then $\ln (e^u)=\ln u$
which means $u=\ln u$
which we get $e^u=\ln u$

i don't understand how is this not true?

Tobias Kildetoft

@Mathphile Probably they are the only solutions to the new equation. But you are taking a logarithm to get from one to the other

Mathphile

@TobiasKildetoft yes but as long as i am taking the logarithm on both sides it should be fine

Tobias Kildetoft

no, it will not, since these are complex numbers

Anyway, even once you pick a branch, you can not just go back again, so you have a correct implication, but no reason to expect to get all solutions (which would then be obtained from other branches)

Mathphile

@TobiasKildetoft didn't know that

$x=\ ln x$ also seems to have the same solutions

@TobiasKildetoft will doing something like $e^(x)=e^{(\ln x})$ also change the solutions due to complex numbers

Semiclassical

1:33 PM

Btw, if I type “solve x==e^x” into wolfram alpha, I get “-W_n(-1) for integer n”

Which is to say: infinitely many solutions, indexed by n

Mathphile

@Semiclassical okay but are there only two solutions for $e^x=\ln x$?

Semiclassical

Dunno. WA only reports two solutions to “solve x==ln x” tho

Both in terms of the W function

Mathphile

@Semiclassical but if click on approximate forms you would get $x ≈ 0.3181315052047640944061139753... \pm 1.337235701430689415856543101 i...$

Tobias Kildetoft

@Semiclassical Presumably one solution per branch (it just picks one before solving)

Semiclassical

My guess is that it depends on which branch of the log function you apply to both sides

@TobiasKildetoft something like that, yeah

Mathphile

1:40 PM

so we can prove that there are infinite solutions?

Semiclassical

The solutions it reports are $e^{-W(-1)}}$ and $e^{-W_{-1}(-1)}$

Mathphile

@Semiclassical yes

Semiclassical

For the rest I’d guess one must consider x +2pi i n = ln(x) for integer n

Any solution to one of those will be a solution to x==e^x

Mathphile

@Semiclassical okay

Semiclassical

There is an interesting question there: which branches of the log function yield which solutions of x==e^x?

Mathphile

1:47 PM

@Semiclassical are there an infinite number of branches which yield solutions?

Semiclassical

Yep

Looking at the numbers for the x==e^x solution, the rule of thumb seems to be “which multiple of 2pi is the imaginary part closest to”

The 0.318 +/- 1.33i solutions are closest to 0, so they get grouped together upon taking logs

Mathphile

@Semiclassical you could ask this on MSE if you want

Semiclassical

The next pair of solutions are 2.06 +/- 7.56i, and the imaginary parts are closest to +/- 2pi respectively

So they’ll get grouped differently

So it’s not quite “one per branch of the log” but the only exception are the first two which land in the same branch @TobiasKildetoft

Mathphile

2:05 PM

leakynun proved that if $a+bi$ is the solution for $x=e^x$ then at least one of $a, b$ must be transcendental

Can we prove that both $a,b$ are transcendental?

ÍgjøgnumMeg

@Mathein Mein Zulassungsbescheid ist gekommen :D

Tobias Kildetoft

@ÍgjøgnumMeg My German is a bit rusty. Something about a message related to lassos?

(or does it mean letter of approval?)

Semiclassical

The wiki page for the W-function (en.m.wikipedia.org/wiki/Lambert_W_function) gives an argument in the section “Special values” that W(x) is transcendental for any nonzero algebraic x

Via Lindemann-Weierstrass

Tobias Kildetoft

@Semiclassical But that still only accounts for one coordinate

Semiclassical

?

Tobias Kildetoft

2:13 PM

@Semiclassical it only implies that at least one of $a$ or $b$ is trancendental, not that they both are

Mathphile

yes

ÍgjøgnumMeg

@Tobias letter of acceptance :)

Semiclassical

ah

Tobias Kildetoft

@ÍgjøgnumMeg Ahh, so my guess was not that far off

ÍgjøgnumMeg

I had a big problem with some bureaucracy that meant I received a rejection letter late last month

lol

Tobias Kildetoft

2:16 PM

@ÍgjøgnumMeg Actually, would zulassung not be closer to "admission" than "acceptance"? Or is my German truly rusty?

Semiclassical

it does imply that a+bi and a-bi are both transcendental, but yeah—no guarantee that both coordinates are

ÍgjøgnumMeg

@Tobias well Zulassung is admission yeah

Zulassungsbescheid is just "notification of admission"

but letter of acceptance is more standard in Englisch

Semiclassical

Eg, e+i and e-i are both transcendental

But 1 certainly isn’t

Tobias Kildetoft

right, the bescheid means "message", right?

Ganja

Hello, people

Mathphile

2:17 PM

@Semiclassical it does guarentee that at least one of the coordinates is transcendental

ÍgjøgnumMeg

Bescheid geben means "notify"

Semiclassical

Sure

And if I had to guess, it’d be that both coordinates are transcendental for the x==e^x solutions. But I have little hope to prove that

Mathphile

@Semiclassical yes

and that's what i want to prove

Tobias Kildetoft

@ÍgjøgnumMeg Anyway, congrats. Where did you get admitted to?

Mathphile

although i think it may fall in the same category of difficulty as proving $\pi+e$ as transcendental

ÍgjøgnumMeg

2:22 PM

@Tobias Thanks :) Heidelberg

Tobias Kildetoft

cool

ÍgjøgnumMeg

Yup! So my scholarship is now officially takes effect

since it was tied to admission to Heidelberg

Tobias Kildetoft

So this is for a master's, or for PhD?

ÍgjøgnumMeg

For a Masters

Tobias Kildetoft

neat

ÍgjøgnumMeg

2:24 PM

2 year masters, hopefully I can get a PhD position somewhere nearby, or in Heidelberg afterwards

Otmar Venjakob is a big name in Iwasawa Theory so that would be neat

Mathphile

@Semiclassical i wonder if an example exists where one of $a,b$ is algebraic

@Semiclassical also if $e^{-W(-1)}=a+bi$, then are there some closed forms for $a, b$?

Tobias Kildetoft

@ÍgjøgnumMeg I could not seem to find a list of their math faculty anywhere. Who else is there?

ÍgjøgnumMeg

Gebhard Boeckle (Mathein's supervisor)

mathi.uni-heidelberg.de/fg-sga/index-en.html

@Tobias mathinf.uni-heidelberg.de/personen.html this is the list of faculty I believe

Tobias Kildetoft

I see. No names I recognize, so presumably not anyone in my (previous) field.

ÍgjøgnumMeg

2:39 PM

I guess so

Tobias Kildetoft

@ÍgjøgnumMeg Do you know who your advisor will be, or will you not decide that until you start the thesis in a year or so?

Semiclassical

@Mathphile one has to be a bit careful with the statement here. There are certainly values of x for which x==W(x) e^W(x) has nice solutions

ÍgjøgnumMeg

@Tobias I don't know how it works yet, presumably one picks later in the year

Tobias Kildetoft

ok, that makes sense

Semiclassical

For instance, if x==2e^2 then W(2e^2)=2 is algebraic

Tobias Kildetoft

2:46 PM

(was the same way it worked when I did my masters in Denmark)

Semiclassical

But 2e^2 isn’t

ÍgjøgnumMeg

Ah fair, I'm hoping to link my master thesis to my undergrad

so it would be nice to have Venjakob

but obv one can't necessarily guarantee that lol

Semiclassical

By contrast, -1 is certainly algebraic and so W(-1) is transcendental for any branch

The question is whether any of the real/imaginary parts happen to be nice

Tobias Kildetoft

@ÍgjøgnumMeg There is also something to be said for branching out a bit from your undergrad, in order to gain a broader base.

Semiclassical

My guess would be no but I can’t back that up

ÍgjøgnumMeg

2:48 PM

@Tobias well my undergrad was mostly applied mathematics, I mean from my undergrad dissertation

Tobias Kildetoft

Still

ÍgjøgnumMeg

from which all the material was self-taught lol

ÍgjøgnumMeg

3:03 PM

I'll see what I like lol

@Tobias I haven't had enough exposure to other areas of mathematics because my undergrad was so inadequate, so maybe I'll find something else interesting :)

Balarka Sen

3:14 PM

@LeakyNun Try K(Z, 3)

Leaky Nun

@BalarkaSen doesnt the same approach work

since ΩK(Z,3) = K(Z,2)?

Balarka Sen

Looped K(Z, 2) is K(Z, 1) = S^1, which has much simpler homology groups

Compute it, you'll see

Leaky Nun

oh I can foresee the problem

but let me revise K(Z,2) first

so we take the fibration ΩK(Z,2) -> PK(Z,2) -> K(Z,2)

Balarka Sen

@RyanUnger Lol @ Mochizuki

Leaky Nun

somehow PK(Z,2) is contractible

and ΩK(Z,2) is S^1

Balarka Sen

3:16 PM

You should be able to prove PX is contractible for any X.

Leaky Nun

just contract through the paths right

Balarka Sen

Ya

"Slurping the spaghetti"

Leaky Nun

lol I like that

so there's a spectral sequence E with $E_{p,q}^2 = H_p(K(Z,2); H_q(S^1))$ that converges to $H_\ast(PK(Z,2))$

Balarka Sen

Mhm

Leaky Nun

since $PK$ is contractible, we have $E_{p,q}^\infty = 0$ except $E_{0,0}^\infty = \Bbb Z$

and beyond $E^3$, the arrows either point to 0 or start from 0, so $E^3 = E^4 = E^5 = \cdots$, so $E^3 = E^\infty$

Ryan Unger

3:21 PM

@BalarkaSen did you look at it

IUTT is just the arithmetic version of computing the Gaussian integral

Balarka Sen

I saw some Gaussian integral stuff

Ryan Unger

turns out Scholze and others just don't understand it because they never learned calculus properly

smdh

Leaky Nun

so $E^3_{p,q} = 0$ except $E^3_{0,0}=\Bbb Z$

Balarka Sen

Yup

Leaky Nun

this means that $E^2_{p,1} = E^2_{p+2,0}$

Mike Miller

3:26 PM

What's the news

Leaky Nun

so $H_{p+2}(K(Z,2)) = H_p(K(Z,2))$

Balarka Sen

Leaky is learning spectral sequences.

Leaky Nun

so it is $\Bbb Z$ when $p$ is even and $0$ when $p$ is odd

and in the case of K(Z,3) we don't have finite support on K(Z,2)... this is not good

Balarka Sen

Ya

Leaky Nun

but on even pages the differentials all point to 0 / away from 0

so $E^{2r} = E^{2r+1}$?

how do we know its limit even exists

Balarka Sen

3:28 PM

Ya but that doesn't tell you much.

I don't actually know if K(Z, 2) -> PK(Z, 3) -> K(Z, 3) spectral sequence even degenerates at a finite page. I doubt.

@MikeMiller surely knows

Ya it doesn't.

You're going to need the ring structure for this

Leaky Nun

but this is homology not cohomology

page 2: $\begin{array}{c} H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ 0&0&0 \\ H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ 0&0&0 \\ H_0(K(Z,2)) & H_1(K(Z,2)) & H_2(K(Z,2)) \\ \cdots\end{array}$

Balarka Sen

Right, that's a problem, I guess. For K(Q, 3) with rational coefficients, homology and cohomology are the same. The Ext term doesn't exist.

Leaky Nun

how do you know the rank is finite...?

Balarka Sen

K(Q, 3) is a rational space. It's integral homology is a Q-vector space

Leaky Nun

but the spectral sequence must converge to the homology right, according to the theorem

so it must degenerate...?

Balarka Sen

3:34 PM

No, why?

Degeneration is like the sequence eventually becomes constant

It can converge without being degenerate

Leaky Nun

oh ok I see

Balarka Sen

3:45 PM

$H^n(K(\Bbb Z, 2), \Bbb Z) = [K(\Bbb Z, 2), K(\Bbb Z, n)]$ is in bijection with natural transformations $H^2 \Rightarrow H^n$ by Yoneda lemma. If $n = 2k$ this has to be generated by $\alpha \mapsto \alpha^k$, but if $n$ is odd there's no such thing.

Ryan Unger

using yoneda lemma unironically?

Balarka Sen

It's so useful m8

Ryan Unger

@BalarkaSen if you want to read a quick paper today, check out "the unknottedness of minimal embeddings" by Lawson.

Closed minimal surfaces in $S^3$ are isotopic to the standard genus $g$ surfaces.

Leaky Nun

page 2: $\begin{array}{c} Z&Z&Z \\ 0&0&0 \\ Z&Z&Z \\ 0&0&0 \\ Z&Z&Z \\ \cdots\end{array}$

page 3 the same

how to compute the differentials? @BalarkaSen

Balarka Sen

You need to do cohomology spectral sequence.

Ryan Unger

3:53 PM

@BalarkaSen a related idea is that in any Ricci positive manifold, compact minimal surfaces must intersect

Balarka Sen

@RyanUnger That sounds very nice. I will look at it soon, but not today.

Ryan Unger

a cool application is that a minimal surface in $S^3$ has to intersect every equator

Balarka Sen

@LeakyNun Hold on a second. It's H_*(Base, H_*(Fiber)). What did you write there?

Leaky Nun

yeah I know lol

I didn't bother to correct it

Balarka Sen

try to be careful if you want my friendly advice

i usually make tons of errors

and it takes me ages to figure out what went wrong

spectral sequences are the worst tool of mankind

Ryan Unger

3:59 PM

jstor.org/stable/1970471?seq=1#page_scan_tab_contents

this is also a quick read

and uses similar ideas

it's all about the morse theory for normal geodesics

Balarka Sen

i gotta scurry. i'll check these out later for sure

thanks

Ryan Unger

@BalarkaSen im preparing you to read my work in preparation

Leaky Nun

ok so there's a spectral sequence converging to Z on (0,0) and 0 otherwise, whose page 2 is: $\begin{array}{c} H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) & \cdots \\ 0&0&0 \\ H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) \\ 0&0&0 \\ H_0(K(Z,3)) & H_1(K(Z,3)) & H_2(K(Z,3)) \\ \vdots\end{array}$

or cohomology idk

I haven't learnt about the cohomology one

@BalarkaSen I still don't know how you compute the differentials

anyway from this we already can deduce that $H_0(K(Z,3)) = \Bbb Z$, $H_1(K(Z,3)) = 0$, $H_2(K(Z,3)) = \Bbb Z$

Akiva Weinberger

5:02 PM

Mike Miller

@LeakyNun This is a small part of what's called the Hurewicz mod C theorem. In fact any simply connected CW complex $X$ with finitely many cells in each dimension necessarily has all homotopy groups finitely generated. (Clearly this extends to finite fundamental group.) It is not true in general: $\pi_2(S^1 \vee S^2) = \pi_2(\vee_{\Bbb Z} S^2) = \Bbb Z^\infty$.

@LeakyNun In the approach you're using, you compute them by knowing what the $E^\infty$ page is, and so get through formal arguments that the differentials have to do something or another

Using that to compute a lot more differentials is why Balarka is saying to use cohomology; you have a ring structure and the differentials satisfy a Leibniz rule

Perturbative

5:46 PM

@LeakyNun Are you looking at the Leray-Serre spectral sequence?

Semiclassical

Spectral sequences seem like a mathematical equivalent of Cthulhu, in the sense that if you don't know enough algebraic topology then you can go on in life blissfully ignorant that such things exist.

(I count myself in the blissfully-ignorant crowd.)

Leaky Nun

@Perturbative I think so

loch

@Semiclassical I feel like this applies to most subjects in math

Rithaniel

Another example would be near-rings

Semiclassical

6:14 PM

@loch yeah

heck, this line out of Hermite is very nearly Lovecraftian: “I turn away with fear and horror from the lamentable plague of continuous functions which do not have derivatives.”

Rithaniel

You know, Lovecraft was commented as having a "constitution too weak for math."

He was sent to some secondary school to specialize in the subject and ended up leaving because he just couldn't handle it, if I recall correctly.

Which sheds light on his writing about "impossible, four dimensional geometries that drive people mad to look upon them."

Balarka Sen

6:51 PM

Hah

Ryan Unger

imagine lovecraft reading Freedman's Poincare paper

Balarka Sen

I should read some Lovecraft. I know very little about him. He seems to have a lot of influence on modern horror.

@LeakyNun Do you want me to walk you through cohomology spectral sequences using this example?