@GaurangTandon you're right, instead of summing over prime divisors, sum over square free divisors (since $\mu(d)$ doesn't contribute anything otherwise) and then count how many squarefree divisors of $n$ there are. That should probably give you something :P

@ÍgjøgnumMeg Alright, i'll check it out. Thanks :)

This is at least how one finds a closed form for $\sum_{d \mid n} \mu(d)$

if you weight the summands with $\varphi(d)$ something probably changes but idk

@ÍgjøgnumMeg Isn't it just [n=1] anyway (The dirichlet identity)?

which? That sum?

It's $0$ for $n \neq 1$ and $1$ otherwise

$\sum_{d \mid n} \mu(d)$ isn't it the dirchlet identity?

Idk what you mean by Dirichlet identity

do you mean the function that leaves a function $f$ invariant under convolution?

Yes

I believe that is actually the Kronecker delta

or, well, the function that assigns $n$ to $\delta_{1, n}$

Yah, that's right. The PDF didn't name the function to be Kronecker delta.

I see :P

@GaurangTandon have you done problem $5$ on that PDF?

I suspect you'll probably wanna use that result

@ÍgjøgnumMeg Nice! I think you're right. Yeah I had skipped problem 5 hehe :P

Let me know how it goes, i'm interested but have to do problem sets lol

Sure, I'll let you know by today or tomorrow! :)

Nice one

I need to read the proof of Barratt-Priddy-Quillen theorem eventually

Are all closed set in R^2 with usual topology compact ?

@BalarkaSen

Take any unbounded closed set

Unbounded closed set are not compact ? But i guess bounded closed set are compact

No, yes.

That's called the Heine-Borel theorem

ook

Thanks

@Balarka did you give any thought to the trace for general field elements?

@ÍgjøgnumMeg Yeah I think I got it

Nice :)

It's $[L : K(\alpha)]\sum \sigma(\alpha)$, right?

well it depends if $L$ is a separable extension or not but yeah

if $L$ is separable then you can ditch the factor at the front

Oh my $\sigma$ is running in $\text{Gal}(K(\alpha)/K)$ though

Yeah sure, but you can do better than that by looking at how many ways there are to extend a $K$-hom of $K(\alpha)$ to a $K$-hom of $L$

(if $L$ is separable this is just $[L:K(\alpha)]$)

Ah I see your point.

Wait, separable? You want normality, right?

and then $\operatorname{Hom}_K(L, \overline{K})$ decomposes into equivalence classes modulo $\sigma \sim \tau \iff \sigma|_{K(\alpha)} = \tau|_{K(\alpha)}$

No you want separability

by some algebra fact that I can't actually remember the proof of

No $\Bbb Q$-automorphism of $\Bbb Q(2^{1/2})$ extends to a $\Bbb Q$-automorphism of $\Bbb Q(2^{1/4})$

You get from the block matrix thang the characteristic polynomial $f(X) = \prod_{i = 1}^m(X - \sigma_i(\alpha))^d$ where $m = [K(\alpha):K]$ and $d := [L:K(\alpha)]$ and then using the equivalence relation I just wrote and separability you get $f(X) = \prod_{i=1}^m\prod_{\sigma \sim \sigma_i}(X - \sigma(\alpha)) = \prod_{\sigma \in \operatorname{Gal}(L/K)}(X - \sigma(\alpha))$

I don't follow; didn't I just write a separable extension $L/K(\alpha)/K$ where you cannot extend a $K$-aut of $K(\alpha)$ to a $K$-aut of $L$?

Hmm

oh you don't need them to be automorphisms, just embeddings into an algebraic closure of $K$

which coincide with $K$ automorphisms if your extension is Galois

OK, corrected formula is $\text{tr}_L(\alpha) = [L : K(\alpha)] \sum_{\sigma \in \text{Gal}(\overline{K}/K)} \sigma(\alpha)$

because in the proof you're actually factoring the minimal polynomial over $\overline{K}$

There's no embeddings in $\overline{K}$

Oh yeah you're right

OK, I'm happy. That was a careless typo.

in the thing I wrote above I should actually take the product over $\sigma \in \operatorname{Hom}_K(L, \overline{K})$ but this is the same as $\operatorname{Gal}(L/K)$ in the Galois case lol

nise

No your formula was right in the first place, it's just that $\operatorname{Gal}(K(\alpha)/K)$ isn't a thing if the extension isn't Galois

Whenever I write Gal I mean Aut, which always makes sense.

alright lol

I guess I am always abusing notation when I am writing sum over all Galois automorphisms

$\text{tr}_{K(\alpha)}(\alpha)$ is sum of the distinct Galois conjugates of $\alpha$, is what I really mean. This is a notational mess.

yeah it is

And then $\text{tr}_L(\alpha)$ is $[L : K(\alpha)] \text{tr}_{K(\alpha)}(\alpha)$.

If $L/K$ is Galois, then I can rewrite this as $\sum_{\sigma \in \text{Gal}(L/K)} \sigma(\alpha)$.

when I did this exercise I had something like $\sum_{\operatorname{Hom}_K(L,\overline{K})/\sim}\sum_{\sigma \in \bar{\tau}}\sigma(\alpha)$

Because every Galois conjugate of $\alpha$ appears $[L : K(\alpha)]$ many times.

rofl

yeah exactly

@ÍgjøgnumMeg Yikes

for some reason I operatornamed $K$ instead overlining it

lawl

hah

But yeah the whole thing follows from extending a basis of $K(\alpha)$ to that of $L$ by choosing a basis of $L$ over $K(\alpha)$ and doing the usual product of basis stuff, in which basis the matrix of $\times \alpha$ has companion matrices of the minimal polynomial of $\alpha$ along the diagonal.

exactly

Cool stuff

Alright how dis: prime ideals of $R_1 \times R_2$ are $R_1 \times \mathfrak{p}_2$ and $\mathfrak{p}_1 \times R_2$ cuz... $R_1/\mathfrak{p}_1 \times R_2/\mathfrak{p}_2$ ain't an integral domain unless one of the factors is $0$

you should throw away the ideals which aren't product of two ideals first, yeah?

All the ideals of the product are products of ideals anyway aren't they?

Ah yeah you are working with unital rings yes :P

I shan't live in a world where rings aren't unital

You can also argue directly, by saying if the ideal doesn't contain either factor completely, then choose $(a, 0)$ and $(0, b)$ outside of the ideal.

Their product is zero, contained in the ideal, so it's never prime

Nice

My tutor prefers argumenting in the way I have, he often marks down if one argues with elements instead of making structural arguments hahaha

Weird

yeah

How very algebraist of him

I think he's trying to force people to think structurally though, which is fair. A lot of people try computing quotients by literally listing elements and not just using an isomorphism theorem

which is alien to me

Eh, both approach have their merits, I think

Yeah sure, but in these easy cases it's just silly

I was honestly never that good at general nonsense arguments.

I have to get a feel for the stuff I have at hand

that's fair

@ÍgjøgnumMeg Solve the next sheet by super high level topos theory or infinity category arguments

Hahahaha

We had a problem to show that $y^2+5=x^3$, one could argue that the existence of a solution would give rise to an elliptic curve to which one could assign no modular form, which is a contradiction to Wiles‘

theorem

(Probably idek)

Sorry, that that guy has no solutions in integrrs*

I‘m on my phone lel

Why can't you assign a modular form if it has an integral solution

idk I don’t understand the modularity theorem, it’s just a dumb joke based on Fermat‘s last theorem

Lol I see

(Because that was essentially Wiles‘ argument)

I really really want someone to turn up with like a disgusting 5 line elementary proof of FLT

it would be hilarious

I have an ugly question that you probably know the answer to @Balarka

You're scaring me

Go ahead

So I have a Riemannian manifold $(M,g)$ and I can define the integral of an $f\in C^\infty_c(M)$ since $M$ comes with a volume form for free

But $M$ also comes with a metric for free, so it has an $n$-dim Hausdorff measure ($n$ is the dimension of $M$) and I can integrate with this one instead

Do the two notions agree?

No reason they'd agree if the metric chosen on $M$ has nothing to do with the Riemannian metric on $g$!

The metric on $M$ is the one coming from $g$ I meant

Ah, the arclength metric.

I think it's called Riemannian distance or something like that

Yes, I think it's true.

I would expect it to hold

I don't know a proof immediately

Brb let me check Federer

LOL

I was reading something a few days ago where they gave a result without proof and referenced Federer for a proof, yeah thanks I'll trust you on that one

Can we prove the Riemannian volume agrees with the Hausdorff measure on geodesic balls?

I dunno

Me neither

Maybe @TedS knows

(I think I pinged the wrong Ted by mistake as well)

@AlessandroCodenotti he might be the wrong Ted but he is a very good Ted

I think of "Ted" as more than just a name, it is a title :D

Technically my name is also Ted

rofl

@ÍgjøgnumMeg Technically Ted's name isn't Ted!

deep

He's actually called Theodore

Indeed lol

Ted is also a nickname for Edward

So are you called Edward or?

Yes hahaha

Isn't Ed a nickname for Edward? English nicknames are weird

Yeah Ed is my usual nickname

but Ted is one used by older generations

Makes sense

Hey

How would you prove that given a colection fo sets

of same cardinality, that you need n-1 bijections to show that they are of same cardinality ?

@ÍgjøgnumMeg @AlessandroCodenotti I used induction

the second part of the question asks for what is a follow up question that can come from this

my answer is , what is the minimal number of bijections needed

I modeled the seifert surface of the Hopf link and used a custom shader (and some touch ups in GIMP) to turn it into pixel art: i.imgur.com/hnOuDod.png

Damn, nice.

Danke. I wanted to do the borromean rings instead of the hopf link, but that one was a bit too difficult to do off-the-cuff. I'll need a plan for it.

winterbash2019.stackexchange.com

Huh, you can go back and self-answer a question you asked?

I might just do that

sure, it's called continual improvement :P

My only question on the Mathematics stackexchange is a very poorly worded one about rational approximations of points on a sphere

It was a while back

go for it, pal

Gotta go up and help proctor an exam in an hour. I'll definitely start looking into it, though

New weird modular forms fact of the day: the $n$th Fourier coefficient of the discriminant modular form is congruent modulo $691$ to the sum of the $11$th powers of the divisors of $n$ for all $n \in \Bbb N$...

@ÍgjøgnumMeg Hi, so I solved all three more parts of prob. 13 using the identity you showed me today. And now I am stuck on two more parts: $\sum _{d|n} \mu^2(d)\phi^2(d)$ and $\sum _{d|n} \frac {\mu(d)}{\phi(d)}$

Any ideas on how to go about them?

uhhm, not sure but you could write $\mu^2(d)\varphi^2(d) = \mu(d)(\mu(d)\varphi^2(d))$ and then use the same identity from before (since product of multiplicative functions is multiplicative)

and I guess the same idea for the other one? $\varphi(d) \neq 0$ for any $d$ so

I see. Can we say that $\frac 1 {\phi(d)}$ is multiplicative? Oh, of course, we can

Sure

Nice reduction. Thanks. That's the only problem I'll solve then. I don't have the energy to solve the others :(

Well try it out, it might be kinda ugly lol

hehe exactly

@Rithaniel Neat

Danke

Next I need to try something a bit more difficult

the Klein bottle?

Treefoil

I was thinking the seifert surface of the borromean rings, or maybe a plot of roots of polynomials in the complex plane with height corresponding to the number of polynomials with that particular root

Trefoil might be cool, too

Wow

I thought i was good at math for my age but theres a lot of stuff left to learn i guess

There always is

Very true

@Rithaniel How did you make that?

Some 3D modeling program?

Yeah, Blender

Was about to explain the process before you specified what I used. :P

Yeah, I'm curious about the details

Well, I made two circles, offset them, rotated one by 15 degree and the other by -15, made a bunch of quads with the vertices of the two circles, used the solidify modifier to turn the object 3d, separated the inner/outer edges from the rest of the geometry, solidified it and smoothed it, and then put together some materials to color the two pieces

Then, I fooled around with lighting and camera position to get a good image, rendered it, and touched it up in GIMP to make it look sharp

is $R^n$ Connected ? for usual topology .provided n is natural number

@Mike Well, what do you think?

Does anyone know where I can find a proof that |z^k| ≤ |z|^k for any complex z?

They’re equal. Use polar coordinates.

hi pal

@Mike Note that R^n is path connected for all n

@RyanUnger Thanks!!

Are there are multiple Seifert surfaces for any particular knot?

@Rithaniel yes, because you can add a handle and the genus will increase

Alright, that works for me.

This is knot theory, right?

apparently

tfw you get pings explaining why R^n is connected

Yeah, among all the fields I want to study, but haven't had a chance to yet, knot theory and differential topology are towards the top of the list

Also, I totally thought that was you, Mike

@MikeMiller rip

@MikeMiller just in case you forgot

So sorry for that @MikeMiller

@SayanChattopadhyay Haha don't worry about it, it's funny to me

There's no way to avoid it

he has a cyber mind

:-)

now, that would make a nice hat

Seifert surface for borromean rings. Feels a little flat

hello

Is there any relation between $\binom{n}{k}$ and $\binom{n^2}{k}$ ?

Well, they're both natural numbers

Are you maybe wanting to write one in terms of the other or in terms of $n$ and $k$?

like some identity, like pascals relates $\binom{n}{k}$ and $\binom{n}{k-1}$

Yeah, I don't know if you'll find one very easily for a squared term

Yeah, fooling around with numbers a little bit, I don't see any immediate patterns

yes I also couldnt find any pattern, seems like there isnt any direct relation

Say we have a space $X$

why do we care about a space $Y$ which is isometric to $X$

@Rithaniel: Can you get the orientable surface (of minimal genus) whose boundary is the boundary of the Möbius band?

Hmmm, maybe. I assume this involves me figuring out what that looks like. But isn't the boundary of the Mobius band just the unknot?

No, it's actually a trefoil.

Wait. I'll give you the pic I put in my book.

Hmmm, that is surprising

Wait: math.stackexchange.com/questions/1084388/…

One more question, is there a known closed form for lacunary series for finite n: $\sum_{j = 1}^{n}z^{a^{j}}$

$z, a, n$ are known

No, @jeea, not so far as I know.

Ah, the image on the left is the unknot. The upper right loop is entirely above the line passing below it, so you can straighten it out

I guess you're right.

Still, the Seifert construction gives you something somewhat interesting.

Also, I forgot the Seifert surface had to be orientable. My Borromean ring attempt is 100% unorientable

Aha.

Also, I'll attempt a trefoil

I met the guy who wrote the Knot Book

Hi @TedS! I pinged you earlier but maybe I pinged the other Ted instead

InsTed* so many Teds

Hi, demonic @Alessandro.

No, you pinged me, in fact.

But it was the middle of the night.

I'll have to re-attempt the borromean rings some time

This is my all-time favorite picture of a knot