Mathematics - 2020-06-08

Balarka Sen

12:00 AM

in T2(a, b) x T2(c, d) the diagonal class is like a x c - b x c + b x d + a x d or smth

anyway im sure theres a better way to understand this than bashing and computing, i'll think later

Mike Miller

life is about bashing and computing

Balarka Sen

i saw some very inspirational academia.se answer by one of the conrad brothers where they said for a lot of people results just come by piling computations one upon another and seeing some pattern

as i learn to compute more i find that to be true

oh maybe it was not conrad brothers it was ryan budney

compute the living daylight out of things, thats how u do math

3

i dont agree that all computations have to be Cotor_k[x,y](k, k) though

although that might be fun to compute

JESUS i need sleep

skillpatrol

@BalarkaSen in physics they take it one step further and tell you to shut up and compute

Thorgott

what's a computation? some special commutative diagram?

skillpatrol

12:58 AM

10

Q: What does "Mathematics of Computation" mean?

I visited this link: http://www.ams.org/journals/mcom/1950-04-030/S0025-5718-50-99474-9/ And I a bit confused by its title "Mathematics of Computation". I am not a native English speaker. Could anyone tell me what does this phrase really mean? What's the difference of: Mathematics Calculus ...

Cristopher

1:31 AM

Hi, any hint on how to solve this?

N. Maneesh

2:03 AM

chat.whatsapp.com/BsaYGaOTK7aI10aXQONUgx

Anyone can join for excercises in mathematics

Anyone can join for excercises in mathematics[Under graduate level]

Terran Swett

3:45 AM

Well, I've written up the question I've been wanting to ask for a couple of weeks. Here's the sandbox post: math.meta.stackexchange.com/revisions/5101/189

It ended up being extremely long, though.

In my browser, it appears to be about 6 pages long.

Terran Swett

4:02 AM

Later, I'll revise it and see if I can make it much, much shorter.

Alternatively, I'll make it much longer and turn it into a blog post. :D

KReiser

5:44 AM

Looking for a quick sanity check: the system of equations given by $x^4+y^4-x^3+y^2=0$, $4x^3-3x^2=0$, and $4y^2+2y=0$ over an algebraically closed field has solutions which aren't $(0,0)$ iff the characteristic is 2, 7 or 13. (Happy to share details if asked, but I'm looking for an independent check first, so I don't want to spoil you.)

mijucik

6:42 AM

hi ive been self studying more advanced courses in mathematics since i recently graduated high school, and i was wondering if it's common for individuals to be disheartened by the leap in difficulty of the exercises (computational stuff to proof based stuff)

like is it normal for ppl to be stuck on an exercise or something for long periods of time without feeling like they're making progress?

Knight

11:55 AM

Can somebody please explain me what happened in that module step? “Prove that $2^n + 3^n$ , $n\in \mathbb Z$, can never be a perfect square”

“Working with modulo 3, we have $2^n +3^n= (-1)^n $ since all squares mod 3 are equal to 0 or 1 so, we conclude that $n$ is even. We could have got the same result for working with modulo 4 ...”

@CowperKettle Hello!

Can someone please tell me what happened when the author just said “working with modulo 3, we have ...”

How he got $$2^n +3^n = (-1)^n$$ ?

Thorgott

it's a straightforward computation, what are you stuck at?

Knight

@Thorgott How we got ?$$2^n +3^n (-1)^n$$ we were not given that

Thorgott

yes, it's a computation

try computing $2^n+3n\mod 3$ yourself

Knight

Let me try:

Should I plug some value for $n$?

For $n=1$ our expression becomes $2 +3=5$ and $5~mod~3 = 2$

Thorgott

12:11 PM

sure, you can try computing explicit examples first if that helps you see the pattern

Knight

For n=1 it doesn’t give me -1

Did I do some mistake ?

Thorgott

no, there is no mistake. why?

Knight

How he got $$2^n + 3^n = (-1)^n$$

Because for his part, if $n=1$ we will get $2^n +3^n ~mod ~3 is ~-1$

Thorgott

yes, I'm asking you why that is the same as what you got

Knight

@Thorgott No, I got 2

8 mins ago, by Knight

For $n=1$ our expression becomes $2 +3=5$ and $5~mod~3 = 2$

Thorgott

12:20 PM

I know

Knight

But he got -1

Thorgott

I know

I'm asking you to figure out why that's the same

you need to think about what we are doing in the first place

Knight

@Thorgott Yes, that’s an important question. Here, we are trying to prove that $$2^n + 3^n$$ can never be a perfect square.

Thorgott

and what were you trying to calculate

Knight

And then we took the help of modulo arithmetic to do the prove. So, when we write $$2^n + 3^n ~~~mod~3 ~~is ~~x$$ we mean that when $2^n+3^n$ is divided by 3 we get the remainder as $x$

Thorgott

12:27 PM

that's not how modular arithmetic works

Knight

If you have a little time, please explain where my fundamental went wrong.

Thorgott

working mod 3 means that we are considering integers up to multiples of 3

two numbers are congruent mod 3 iff their difference is divisible by 3

of course, if you perform Euclidean division by 3 on a number, it is congruent to its remainder, but it can be congruent to a lot of other numbers too

for example if you divide 4 by 3, you get a remainder of 1, so 4 is congruent to 1 mod 3, but you also have, for example, that 4 is congruent to 7 mod 3

Knight

Yes. I fully understand this notation $$a \equiv b ~~~~~(mod 3)$$

Thorgott

then you should know why your calculation does not contradict the authors'

Knight

But I’m unable to understand statements like these $$a ~~mod ~3 ~is~b$$

Thorgott

12:32 PM

that means the same thing

excepts it's worse phrasing that I wouldn't recommend using

because it sounds like such a $b$ would be unique, even though it's not

Balarka Sen

"a mod 3 is b" is just a way to write "a = b (mod 3)" is just a way to write "3 divide a - b"

Knight

So, $5~mod~3 ~is~2$ means $5\equiv 2~~~~mod 3$

?

Balarka Sen

Sure

Knight

I still don’t understand what $$\text{working with modulo 3} \\ we’ve~~~~~ 2^n+3^n =(-1)^n$$

Means ?

Thorgott

it means $2^n+3^n\equiv(-1)^n\mod3$

Balarka Sen

12:35 PM

Means 3 divides (2^n + 3^n) - (-1)^n, by definition.

Knight

And how he got that

Balarka Sen

By arithmetic

3 always divides 2^n - (-1)^n, and 3 always divides 3^n. Add these up

Knight

Oh okay! But how did get that? Was it a guess ?

Balarka Sen

No it's not a guess. You can prove it.

Knight

Okay, how?

Edward Evans

12:37 PM

waddup

Knight

Morgen

Thorgott

calculate it

Balarka Sen

@Knight You really should be able to do it on your own

There's a factorization formula for a^n - b^n you know, use that.

Knight

$a^n +b^n$ involves $i$ in its expansion

Balarka Sen

I said a^n - b^n

Knight

12:40 PM

But we have $2^n +3^n$

Balarka Sen

Go back and read what I wrote

Knight

4 mins ago, by Balarka Sen

3 always divides 2^n - (-1)^n, and 3 always divides 3^n. Add these up

Thorgott

@Edward if $\rho$ is a rep of $G$ and $\tau$ is a rep of $H\le G$, why is $\rho\otimes\mathrm{Ind}_H^G(\tau)=\mathrm{Ind}_H^G(\mathrm{Res}_H^G(\rho)\otimes\tau)$

Knight

Yeah it seems natural, but how to go for guessing it

Thorgott

equality should mean sth like naturally isomorphic as bifunctors, but w/e

@Knight like you guess most formulas, compute small examples and guess a pattern

Balarka Sen

12:44 PM

I can't understand what you mean by guessing. You can prove the formula, using $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, which I have hinted at how to prove above.

Knight

@Thorgott Yes.

Balarka Sen

Why would you guess something you can rather prove?

Seems silly

Knight

Okay. So, we got $$2^n + 3^n \equiv (-1)^n~~~~~~~(mod ~3)$$

Thorgott

I mean, sometimes you want to see where a formula is coming from to gain intuition

but in this case, the proof itself gives the intuition, so yeah

Knight

Then he wrote “since all squares mod 3 are 0 or 1 therefore we conclude that $n$ is even”

How he know that all squares mod 3 are 0 or 1?

Thorgott

12:48 PM

have you carried out the previous computation

Balarka Sen

Just first prove this man.

All your questions can be answered using that

Knight

Okay, $$a^n -b^n == (a – b)(a^n– 1 + a^n – 2b + a^n -3b2 + ··· + ab^n – 2 + b^n-1)$$

Then? What should I do?

Thorgott

that's not how the factorization goes

Balarka Sen

Your LaTeX is unreadable, and you didn't write a proof yet - you just wrote some symbols. Please write a proper proof.

Knight

Let’s start afresh

Tell me exactly how should I begin? Do I need to expand $a^n- b^n$?

Thorgott

12:55 PM

think about what you're given, think about what you want to conclude, think about how you can get from one to the other with a factorization

Balarka Sen

I won't give anymore hints beyond what I have said. Take your time, write a precise proof, and we can continue.

You're trying to prove the following claim: $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$.

Knight

I know that.

I did it just in the morning

Thorgott

then you should have no issues replicating it

Balarka Sen

Write a proof for me.

Edward Evans

@Thorgott I'm not sure but perhaps identify your rep with a $K[G]$-module $V$ and the induced rep as $K[G] \otimes_{K[H]} M$ (with $M$ a $K[H]$-module) and then do some tensor product magic

Knight

12:57 PM

$$a \equiv b~~~~~(mod ~m) \\
a \equiv b ~~~~~~(mod~m) $$ if moduli are same we can multiply corresponding sides without affecting the congruence

Balarka Sen

I don't know modular arithmetic. You have to tell me why that's true.

Knight

Proof: Let's say we have $$a\equiv b ~~~~~~~(mod~m) \\ c\equiv d ~~~~~~~~(mod~m) \\ i.e. ~~~~~~~~~\frac{a-b}{m} = t_1 \\ \frac{c-d}{m} =t_2$$

Thorgott

I'm trying to avoid the module pov, modules make me cry

Edward Evans

oh :( I have avoided working with the homomorphism pov hahaha

Balarka Sen

@Knight I don't understand. What is $t_1$ and $t_2$?

Thorgott

1:00 PM

my pov is the direct sum pov lol

probably the least elegant

Edward Evans

wait what's that?

Knight

$$c\frac{a-b}{m} = ct_1 $$ $$\frac{ac-bc}{m} =ct_1~~~~~~(1)$$ $$ b\frac{c-d}{m} = bt_2 $$
$$\frac{bc-bd}{m} =bt_2~~~~~~~~~~~~~(2)$$

@BalarkaSen They are positive integers

Balarka Sen

So $1$ is not congruent to $4$ mod $3$, because $1 - 4 = -3$, which upon dividing by $3$ is $-1$, a negative integer?

Knight

Okay we can allow it to be negative also

Astyx

But 4 is congruent to 1

Thorgott

1:03 PM

if $\tau\colon H\rightarrow GL(W)$ is a rep of $H\le G$, you have $\mathrm{Ind}_H^G(\tau)\colon G\rightarrow GL(V)$ with $V=\bigoplus_{\sigma\in G/H}W$ and the action is defined as follows:
fix a system of representants $g_{\sigma}$ and write $v=\sum_{\sigma\in G/H}g_{\sigma}w_{\sigma}$ for $v\in V$, then define the action via $\rho(g)(\sum_{\sigma\in G/H}g_{\sigma}w_{\sigma})=\sum_{\sigma\in G/H}g_{\sigma^{\prime}}\tau(h_{g,\sigma})(w_{\sigma})$, where $gg_{\sigma}=g_{\sigma^{\prime}}h_{g,\sigma}$ (and $g_{\sigma^{\prime}}$ is the unique representant and $h_{g,\sigma}$ is the unique elemen…

Edward Evans

@Astyx this is pedagogy lol

Balarka Sen

@Knight So you should mention that. $t_1, t_2 \in \Bbb Z$.

Just pretend I don't know anything and you're explaining me every step

Edward Evans

@Thorgott ohhhhh, yes that is a vile pov, which I also had to trudge through :(

Astyx

I know

Balarka Sen

@Knight I agree. continue

Knight

1:04 PM

Adding equation (1) and (2) we have $$ \frac{ac -bd}{m} = ct_1 +bt_2$$ that is $m$ divides $ac-bd$ that is to say $$ ac \equiv bd ~~~~(mod ~m)$$

Thorgott

in the previous setup, the thing I'm trying to show comes down to $\rho(g)(v)\otimes\tau(h_{g,\sigma})(w_{\sigma})=\rho(h_{g,\sigma})(v)\otimes\tau(h_{g,\sigma})(w_{\sigma})$

Balarka Sen

Why does it follow that $m$ divides $ac - bd$?

Thorgott

and I have no clue why that ought to hold

Knight

So, in words "if moduli are same we can multiply corresponding sides of two congruences without affecting the congruence sign"

Edward Evans

Jantzen-Schwermer define an injective $K[H]$-module homomorphism $\gamma_M : M \to \operatorname{ind}_H^G M$ with $\gamma_M(x)(g) = g^{-1}x$ if $g \in H$ and $0$ otherwise, and then show that $g\gamma_M(M)$ is just the set of $f \in \operatorname{ind}_H^G M$ such that $f$ vanishes on $G \setminus gH$, and then identify $\operatorname{ind}_H^G M$ with $\bigoplus_{i=1}^{[G:H]}g_i\gamma_M(M)$ lolo

Knight

1:07 PM

@BalarkaSen Because $c, ~t_1 , ~b, t_2 \in \mathbb{Z}$ therefore $ct_1 \in \mathbb{Z}$ and $bt_2 \in \mathbb{Z}$ and hence $ct_1 + bt_2 \in \mathbb{Z}$

Balarka Sen

Yes, OK

Nice.

Thorgott

starts sweating

Knight

yeah

Astyx

(I've always been taught to avoid writing divisions when doing arithmetic)

Balarka Sen

So indeed, that is a good way to do it. $a \equiv b \pmod{m}$, $c \equiv d \pmod{m}$ implies $ac \equiv bd \pmod{m}$. Multiplication preserves congruences. In particular, by inductively applying this, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$.

Knight

1:08 PM

So, if we're given $$a\equiv b ~~~~~~~(mod~m)$$ then by continuous use of that property we can arrive at $$a^n \equiv b^n ~~~~~~~~~~(mod~n)$$

Edward Evans

@Thorgott unfortunately that point of view makes for disgusting computations and my Seminarvortrag ran over by 15 minutes because I spent a lot of time going through the computations so things made sense hahaha, then the prof asked "wHy DiDn't U jUsT dEfInE iT vIa $\otimes$"

and I'm like "MATE I WANTED TO BUT YOU SAID WE HAD TO GO BY THE BOOK"

Thorgott

LMAO

I'm glad I won't have to do this stuff in my talk

Balarka Sen

Here's an alternative one-line proof: $a^n - b^n = (a - b)(a^{n-1} + a^{n-2} b + \cdots + a b^{n-2} + b^{n-1})$. Thus, $a - b$ divides $a^n - b^n$. Thus, if $a \equiv b \pmod{m}$, $m$ divides $a - b$ thus $m$ in turn divides $a^n - b^n$. Hence, $a^n \equiv b^n \pmod{m}$.

Edward Evans

The talk I wanted to do was on Galois cohomology but somebody else got it and I was landed with half a talk on induced reps lol

Knight

Yes, thats also a property "if $d$ divides $m$ then we have $a \equiv b ~~~~(mod ~d)$

Balarka Sen

1:11 PM

Yes. Alright, so time to figure out your squares question.

Knight

I did it 8 hours ago

@BalarkaSen YES my captain

Thorgott

Galois cohomology sounds cool

Balarka Sen

You want to prove $x \in \Bbb Z$ implies $x^2 \equiv 0, 1 \pmod{3}$.

Thorgott

cause it has Galois in it

Balarka Sen

What do you know about $x \pmod{3}$?

Edward Evans

1:11 PM

@Thorgott I'm more excited for my talk on number theory though, where I'll be proving local Kronecker-Weber and Hasse-Arf (wobei I have no idea what Hasse-Arf is even about)

Knight

We can have just three possibilities $$x \equiv 0 \\
x \equiv 1 \\
x \equiv 2$$

(all are mod 3)

Balarka Sen

So use the result we just proved.

Thorgott

Kronecker-Weber is exciting

though I don't yet know what the "local" means

Balarka Sen

over $\Bbb Q_p$ I assume

Edward Evans

just that your extensions are extensions of $\Bbb Q_p$ instead of $\Bbb Q$

Knight

1:14 PM

So, for $x^2$ we can have only two possibilities
$$
x^2 \equiv 0 \\
x^2 \equiv 2 $$

Balarka Sen

@Knight No, you did something wrong

Knight

What and where?

Thorgott

yeah, my comfortability with $\mathbb{Q}_p$ is infinitesimal

Balarka Sen

How did you get 2?

Edward Evans

@Thorgott same tho

Knight

1:16 PM

We have $$x \equiv 1 \\
x \equiv 2 $$ if we multiply them side by side $$x^2 \equiv 2$$

Balarka Sen

Huh??? $x$ cannot be both congruent to $1$ and $2$ modulo 3 at the same time

That is nonsense

Knight

Yes, I agree with that

4 mins ago, by Knight

We can have just three possibilities $$x \equiv 0 \\
x \equiv 1 \\
x \equiv 2$$

What we should do after ^ ?

Balarka Sen

Think.

Knight

I got it

$$ x^2 \equiv 0 \\
x^2 \equiv 1 \\
x^2 \equiv 4 $$

Thorgott

I don't even really know why anyone cares about $\mathbb{Q}_p$

scary number theorists n stuff

Knight

1:18 PM

We have to multiply same things

Balarka Sen

What is 4 congruent to modulo 3?

Knight

1

Balarka Sen

Amongst many other things, yes.

Edward Evans

@Thorgott because Hasse principle, or some other reason that I'm simply quoting because I don't really know (y e t)

Balarka Sen

So the only possibilities for x^2 modulo 3 are?

Knight

1:21 PM

Yes, it congruent to $7$

Leaky Nun

@Thorgott disagrees

Knight

@BalarkaSen $$x^2 \equiv 0 \\x^2 \equiv 1$$

Balarka Sen

So done

Thorgott

huh, interesting

Edward Evans

I guess it's the same principle that is used all over number theory (prove things for primes and then stick them together with tape) but I'm sure it's a lot more subtle rofl

Knight

1:23 PM

Okay, so you mean for $2^n + 3^n$ to be a perfect square we have to have $$ 2^n + 3^n \equiv 0 \\
{\Large OR} \\
2^n +3^n \equiv 1 $$ ha ?

Balarka Sen

Modulo 3, yes.

Thorgott

that makes sense, I guess

still scary

Edward Evans

agreed

Knight

@BalarkaSen So, what's after that? I want to learn from you and @Thorgott not from that author who assumes me to know everything (actually its not his fault, its mine only)

Edward Evans

I'm gonna go make some rice with höllenfeuer level chillis and maybe die brb

Balarka Sen

1:26 PM

@Knight What are you reading

Thorgott

lol hf

Knight

qbyte.org/puzzles/p155s.html

Please guide me further about proving that $2^n +3^n$ can never be a perfect square.

Balarka Sen

Read a proper book. Try "An introduction to the theory of numbers" by Niven-Zuckerman-Montgomery

Knight

I'm reading Niven's

It's just a question that I took up (for some purpose)

Thorgott

Niven-Zuckerman-Montgomery got some evil exercises

Knight

1:35 PM

Can we move on? (I mean on the question not from this platform)

geocalc33

2:42 PM

Hi🗿

Balarka Sen

4:08 PM

Bi-invariant forms on Lie groups are closed, right? Let $\omega$ be a bi-invariant $k$-form on $G$, then for $k+1$ left-invariant vector fields, $d\omega(X_0, \cdots, X_k) = \sum (-1)^i X_i \omega(\cdots, \hat{X_i}, \cdots) + \sum (-1)^{i+j} \omega([X_i, X_j], \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots)$

The stuff in the first sum vanishes by invariant

The stuff in the second sum are constant, but they should be zero for some reason... hm

Ah, that just comes from $\text{Ad}$-invariance of $\omega$ on $\mathfrak{g}$?

Balarka Sen

4:24 PM

Yeah so let's see $\text{Ad}(g)^* \omega = \omega$ as a form on $\mathfrak{g}$, differentiating this at the identity in the direction $g = \exp(tX)$ gives $\omega([X, X_0], X_1, \cdots, X_k) + \omega(X_0, [X, X_1], \cdots, X_k) + \cdots + \omega(X_0, X_1, \cdots, [X, X_k]) = 0$ by $\text{ad}$-invariance.

Plugging in $X = X_0, \cdots, X_k$ and then adding them up with various signs should give me the identity

Hm, not really

Let's do it for 2-forms. Need to see $-\omega([X, Y], Z) + \omega([X, Z], Y) - \omega([Y, Z], X) = 0$

Knight

@BalarkaSen Can you give me some resources where I will find some examples where modular arithemtic is used to prove that no integer solutions exist for an Eqaution.

Balarka Sen

$-\omega([X, Y], Z) = \omega(Y, [X, Z])$ and that cancels with stuff

@Knight dunno its probably in niven zuckermann montgomery

Knight

@BalarkaSen Okay I will check, by the way thanks for today.

Balarka Sen

Damn, how do I do this manipulation it should be something obvious

Meh that can't be good

Ted Shifrin

4:50 PM

@BalarkaSen I think I told you the more general result once. On a homogeneous space, the cohomology of the complex of invariant forms gives the usual cohomology. On a symmetric space (e.g., a group with bi-invariant metric), the complex of invariant forms is already the cohomology.

Balarka Sen

Yes, I know the result. Trying to reproduce it.

Getting stuck in a stupid computation

Ted Shifrin

I usually do it with facts about the structure constants.

In the case of the symmetric space setting (which might be easier) you get a split of the Lie algebra and you know stuff about Lie bracket of those factors.

Balarka Sen

On a Lie group with a bi-invariant metric you can average a form to be bi-invariant, and that's a chain homotopy equivalence. But the chain complex has zero differentials (by fact I am getting stuck on lmao)

Ad-invariance of, say, a 2-form is exactly the statement that $\omega([X, Y], Z) + \omega(X, [Y, Z]) = 0$, right?

On the Lie algebra

Ted Shifrin

Probably, but I'd need to check. Running off right now.

Balarka Sen

No worries

Oh wow maybe my formula for the exterior derivative in terms of the Lie bracket is wrong lol

Nah can't be

Knight

6:00 PM

Can somebody help me with simple modular arithmetic ?

I have this equation $$15m^2 -7k^2 =1$$

and someone said:

you have 0 ≡ 15m^2 = 7k^2+1 ≡ k^2+1 (mod 3), but that is impossible because you can check all 3 possibilities for k mod 3.

and I couldn't understand what does that mean.

user193319

Is an arbitrary direct product of amenable groups amenable?

Edward Evans

It means check all 3 possibilities for $k \bmod 3$

Balarka Sen

Was messing up some sign and that totally confused me. Sigh. $\omega([X, Y], Z) + \omega(X, [Z, Y]) = 0$ is correct $\text{ad}$-invariance.

user193319

My thought was to try to view $S_{\Bbb{N}}$, the full symmetric group on $\Bbb{N}$ which is not amenable, as the direct product of amenable groups.

Knight

We have $$k \equiv 0 \\
k \equiv 1 \\
k\equiv 2$$

But what happened to 7 and why just k not $k^2$?

Edward Evans

6:09 PM

$7 \equiv 1 \bmod 3$

Didn't you have this conversation earlier?

Knight

No it was quite different

What happened after $$7k^2 +1 \equiv 0 ~~~~~~~~~~~(mod ~3)$$

Edward Evans

$7 \equiv 1 \bmod 3$

Knight

Okay

Balarka Sen

@user193319 Nah it's not true. For every word $w \in F_2$ in the free group on two generators, come up with a 2-generated finite group $G_w$ where $w$ is nontrivial. Then $F_2$ embeds in $\prod_{w \in F_2} G_w$ (factorwise quotient homomorphism $F_2 \to G_w$), so the product is not amenable. But each factor is a finite group, so amenable.

I guess you're using $F_2$ is residually finite. Its OK, you can prove it by hand

I mean, you need some covering space theory. Find a finite-sheeted regular cover of the wedge of two circles such that the word $w$ is not image of a loop upstairs

Ted Shifrin

6:26 PM

@Balarka: If I have a symmetric space $G/H$, this means that I get a splitting $\mathfrak g = \mathfrak h \oplus \mathfrak m$ with $[\mathfrak h,\mathfrak m]\subset\mathfrak m$ and $[\mathfrak m,\mathfrak m]\subset\mathfrak h$. Now I want to choose a basis of the left- (or, usually for me, right-) invariant forms with $\omega^\alpha\in\mathfrak m^*$ and $\omega^\mu\in\mathfrak h^*$.

Then the bracket conditions tell me that $c^\gamma_{\alpha\beta} = c^{\nu}_{\alpha\mu} = c^\alpha_{\mu\nu} = 0$. Thus, we have $$d\omega^\alpha - \sum c^\alpha_{\beta\mu}\omega^\beta\wedge\omega^\mu \equiv 0\pmod{\omega^\mu}.$$

Any invariant form $\phi$ on $G/H$ must be a polynomial in $\omega^\alpha$ with constant coefficients, and so $d\phi\equiv 0\pmod{\omega^\mu}$. Since $\phi$, and hence $d\phi$, is horizontal for $G\to G/H$, we must have $d\phi = 0$. :)

Balarka Sen

That's black magic.

Very nice

I will look at this computation more carefully after dinner, thanks!!

Ted Shifrin

(I'm using the usual structure equations $d\omega^i = \frac12 \sum c^i_{jk}\omega^j\wedge \omega^k$. Modulo sign convention on left-right invariance stuff.)

Balarka Sen

Yeah.

Ted Shifrin

Happy dinner!!

Knight

6:45 PM

@TedShifrin Hello

Ted Shifrin

hi Knight

Knight

Would you please explain me this?

45 mins ago, by Knight

you have 0 ≡ 15m^2 = 7k^2+1 ≡ k^2+1 (mod 3), but that is impossible because you can check all 3 possibilities for k mod 3.

Ted Shifrin

What are the possible values for $k\mod 3$?

Knight

0,1 and 2

Ted Shifrin

Do any of those satisfy $k^2+1\equiv 0\pmod 3$?

Knight

6:50 PM

I didn’t get the question

Ted Shifrin

What is $0^2+1\pmod 3$?

-1

Huh?

Sorry it’s 1

OK, now do it with the other two values of $k$.

Knight

6:53 PM

We will get 2 and 2

Ted Shifrin

Done.

Knight

But what we actually did? Why we replaced $k$ with 0,1 and 2?

Let’s begin from here:

Ted Shifrin

Because the mod 3 universe has those three elements.

You need to learn the basics.

Knight

$$k \equiv 0 \mod 3 \\ k\equiv 1 \mod 3 \\k\equiv 2 \mod 3$$

We have the expression $k^2+1$. Now, how did we proceed?

Ted Shifrin

You answer that for yourself.

Knight

6:57 PM

Really?

$$ 1 \equiv 1 \mod 3 \\ k^2 \equiv 0 \mod 3$$

Adding them will give us $$k^2 +1 \equiv 1 \mod 3$$

Similarly, $$ k^2 \equiv 1 \mod 3 \\ k^2+1 \equiv 2 \mod 3$$

And finally $$k^2 \equiv 4 \mod 3 \\ k^2 \equiv 1 \\ k^2 +1 \equiv 2 \mod 3$$

Is this what we really did?

Ted Shifrin

Yes.

If you want to be lazy, you notice that $k\equiv 2\pmod 3$ is the same as $k\equiv -1\pmod 3$, and $(-1)^2 = 1^2$ (whether mod 3 or not).

Knight

$k^2 \equiv 1 \mod 3$

Thank you so much Theodore sir.

Sha Vuklia

7:18 PM

Does anyone know here what they mean by "there exists a non-zero linear form vanishing on the image of $\tilde\rho$"?

I was trying something with the fact that if we have a strict subspace $W$ of $\mathbb C^n$, then we can find a basis $\{a_1,\dots,a_k\}$ of $W$. If we extend this to a full basis $\{a_1,\dots,a_n\}$, then on $W$ the coefficients for $a_{k+i}$ have to be zero, so we would get $n-k$ relations on the standard basis vectos of $\mathbb C^n$

I was thinking that those are the relations they're thinking of. but I can't exactly see how that translates to relations of the matrix coefficients of $\rho_i$

btw, $\tilde\rho$ is this canonical map which sends $g\in G$ to $\rho_g$ (where $\rho$ is a given representation), and then is extendedly linearly on $\mathbb C[G]$

Sha Vuklia

8:14 PM

ah nvm, I see that my question was asked before on stack, and it has been answered very clearly

Sha Vuklia

8:35 PM

(my problem was that I didn't realise that by linear form they meant linear functional, but apparently I could have googled that and found the answer immediately)

Ted Shifrin

8:47 PM

Funny, isn't it, that you recognize "bilinear form" immediately but not "linear form" :P

AbstractAlgebraLearner

10:19 PM

0

Q: Given certain $\Bbb{Z}$-matrices can we efficiently minimize the sum of entries under possible operation by an certain type of column operation?

This is about a possible exact algorithm of the smallest grammar problem over a singleton alphabet $\Sigma = \{a\}$. Presumeably all current literature is about inexact solutions or approximations to smallest grammars $g$ of an input string $s$. Suppose we're given an $n\times n$ matrix $A$ of...

linear-algebra computer-science formal-languages context-free-grammar discrete-optimization

AbstractAlgebraLearner

10:32 PM

If anyone is interested I'd be happy to explain the method()

Added to the 2x2 example

There's a slight error where the matrix entries for $a$ in variable $A_i$ have to be given along row $1$, but that shouldn't change the rest of the post (still valid)

Fixed error in post

Akiva Weinberger

11:01 PM

Earth is said to be in the Goldilocks zone 'cause that's where the bears are

Balarka Sen

Clever

user76284

How should I denote the set of natural-coefficient polynomial functions? $\mathbb{N}[x]$ is the set of natural-coefficient polynomials, but they're not equivalent. I think I need to use the evaluation map.

Balarka Sen

They're the same thing for natural coefficients. If $f, g$ are two polynomials with natural coefficients and $f(x) = g(x)$ for all $x \in \Bbb N$, $f - g$ is a polynomial which has every natural number as a zero. But a finite degree polynomial has infinitely many zeroes iff it's the zero polynomial

So in fact $f = g$ as polynomials.

user76284

Yeah, I just want to leave open the case that I'll switch from natural coefficients.

I guess I should just say "Let $S$ be the set of natural-coefficient polynomial functions..." if there's no standard notation.

Thorgott

why would you want natural instead of integer coefficients

user76284

11:11 PM

@Thorgott I need the set to be well-ordered under a specific kind of order.

Balarka Sen

i have not seen a standard notation.

The polynomial functions over $\Bbb Z_p$ in your sense is exactly the ring $\Bbb Z_p[x]/(x^p - x)$

user76284

Just to give a little bit of context:

For all $f, g : \mathbb{N} \rightarrow \mathbb{N}$ let
$$ f \leq g \leftrightarrow \exists x {\in} \mathbb{N} \, \forall y {\in} \mathbb{N} \, (x \leq y \rightarrow f(y) \leq g(y)) $$
Let $S \subseteq \mathbb{N} \rightarrow \mathbb{N}$ be the set of natural-coefficient polynomial functions. Then the ordered semiring $(S, \leq, +, \cdot)$ is isomorphic to $(\omega^\omega, \leq, +_H, \cdot_H)$ where $+_H$ and $\cdot_H$ are the Hessenberg sum and product.

I'm trying to see what "Hessenberg exponentiation" $\uparrow_H$ for ordinals would correspond to (order-theoretically), if it corresponds to ordinary pointwise exponentiation $\uparrow$.

user76284

11:36 PM

I guess in short: Let $S$ be the set of polynomial towers with natural coefficients. Let $F : S \rightarrow \varepsilon_0$ satisfy
\begin{align}
f \leq g &\leftrightarrow F(f) \leq F(g) \\
F(0) &= 0 \\
F(1) &= 1 \\
F(f + g) &= F(f) +_H F(g) \\
F(f \cdot g) &= F(f) \cdot_H F(g) \\
F(\mathrm{id}) &= \omega
\end{align}
$F$ is an ordered semiring homomorphism. Assume $F$ is also bijective, i.e. an isomorphism. Does such an $F$ exist? Is it unique?
If so, does there exist a $\uparrow_H$ such that $F(f \uparrow g) = F(f) \uparrow_H F(g)$? Is it unique?

AbstractAlgebraLearner

@user76284 what a great question

@user76284 what is $\varepsilon_0$ some kind of topos theory symbol?

user76284

$\varepsilon_0$ here refers to en.wikipedia.org/wiki/Epsilon_numbers_(mathematics). Sorry if that wasn't clear.

$\sup \{\omega, \omega^\omega, \omega^{\omega^\omega}, \dots\}$

AbstractAlgebraLearner

We think alike I think

So what is this for computational complexity or what topic?

user76284

set theory/order theory/ordinals/semirings I suppose

AbstractAlgebraLearner

But what is an application, therefore we automatically have two models to choose from

when working on your question

user76284

11:44 PM

I want to know what's the natural extension of the Hessenberg sum and product to exponentiation, the next hyperoperation.

AbstractAlgebraLearner

@user76284 are your polynomials multi or single variate?

user76284

And whether it has an order-theoretic definition, like the Hessenberg sum and product.

AbstractAlgebraLearner

Makes sense now, let me reread

Are your variables single or multivariate polys?

I see they're a subset of $\Bbb{Z}[X]$ you should say that to be more clear

Or you could say the semiring $\Bbb{N}[X]$.

user76284

Natural-coefficient polynomial functions $\mathbb{N} \rightarrow \mathbb{N}$, i.e. the evaluations in $\mathbb{N}$ of the natural-coefficient polynomials $\mathbb{N}[x]$.

AbstractAlgebraLearner

I see now

user76284

11:48 PM

But I then extended it to polynomial towers.

e.g. $1+2x+x^{3x+1}+x^{1+3x^2+x^{6+x}}$, etc.

AbstractAlgebraLearner

I think it would not be bijective

I think not injective, but surjective is easy

user76284

I think it should be

Ordinal arithmetic

In the mathematical field of set theory, ordinal arithmetic describes the three usual operations on ordinal numbers: addition, multiplication, and exponentiation. Each can be defined in essentially two different ways: either by constructing an explicit well-ordered set which represents the operation or by using transfinite recursion. Cantor normal form provides a standardized way of writing ordinals. In addition to these usual ordinal operations, there are also the "natural" arithmetic of ordinals and the nimber operations. == Addition == The union of two disjoint well-ordered sets S and T can...

I should probably rewrite all of this in terms of the Cantor normal form.

I guess what I should say is, once I have an isomorphism $F$ mapping each ordinal below $\varepsilon_0$ to its corresponding polynomial tower, what is the function $\uparrow_H$ on ordinals such that $\alpha \uparrow_H \beta = F^{-1}(F(\alpha) \uparrow F(\beta))$.

AbstractAlgebraLearner

@user76284

bottom of en.wikipedia.org/wiki/…

says you're correct

What is the problem you're running into?

I would say you extend to $-$ numbers first

user76284

Nothing in particular, just brainstorming/working out the question more precisely.

AbstractAlgebraLearner

Note that usually you learn: $\Bbb{N} \subset \Bbb{Z} \subset \Bbb{Q} \subset \dots$ in that order

What could a $-$ ordinal represent?

user76284

11:56 PM

I've already done a similar thing here math.stackexchange.com/questions/3284679/…

AbstractAlgebraLearner

Then you work with $\Bbb{Z}[X]$ a more familiar territory

user76284

But I didn't extend exponentiation.

Well first I need to define exponentiation on ordinals.

But not the usual exponentiation.

AbstractAlgebraLearner

Yes you should tell us how you've defined it, and that will not only say that but also help with understanding Mult since it will likely be in terms of mult.

I see, they say it's impossible to add in $-$'s

user76284

Basically I took the field of fractions of the Grothendieck group of the semiring of ordinals under Hessenberg arithmetic.

i.e. the usual sequence of constructions $\mathbb{N} \rightarrow \mathbb{Z} \rightarrow \mathbb{Q}$.

But starting with ordinals instead of just $\mathbb{N}$.

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