Mathematics - 2020-04-28 (page 1 of 2)

Ted Shifrin

00:08

@r9m Oops. Sorry about that!

Akiva Weinberger

00:30

Today I learned of the "missing strip plane" - $\mathbb R^2$ with all points with $x$-coordinate in $(0,1]$ deleted. It satisfies almost all of the axioms of Euclidean geometry!

It doesn't satisfy the parallel postulate, for sure - what else does it not satisfy?

(We also throw away any vertical lines or circles that are entirely contained in the strip.)

Note that it does satisfy a completeness axiom: every line looks the same as $\mathbb R$, on its own.

(Another part of the model - the length of a line segment is defined to be the length of the portion outside of the strip. Which actually distorts the shapes of circles, I think)

topologicalorientablesurface

01:43

1

Q: Proper subsets of connected spaces and proper product

Let $A$ be a proper subset of $X$ and $B$ a proper subset of $Y$. If $X,Y$ are connected. Show that $X\times Y\backslash (A\times B)$ is connected. Lemma: Let $X$ be a space and $A_1,A_2...,A_n$ a finite sequence of connected subsets in $X$. If $A_j\cap A_{j+1}\neq \varnothing$ for each $j=1,2....

Balarka Sen

03:15

@TedShifrin Hi

Ted Shifrin

03:51

Hi, a @Balarka.

Balarka Sen

I am convinced I would be adrift without moving frames, @Ted

Ted Shifrin

LOL, you're just kissing up. :)

Balarka Sen

I just proved a Riemannian metric admits diagonalizing charts implies Weyl tensor vanishes

Ted Shifrin

Charts ... very different from moving frames, who are always diagonal :)

Balarka Sen

Sure used it to compute though

Ted Shifrin

03:55

Interesting.

Balarka Sen

It's in garbology, if you want to read walls of texts

I am currently going to try and compute sectional curvature of CP^2, I think. Moving frames again

Ted Shifrin

No walls this instant, but ....

Best way to do that is with Lie algebra stuff for symmetric spaces.

Balarka Sen

Yeah I don't know this stuff at all.

Ted Shifrin

But that's where you'll end up with unitary frames.

Balarka Sen

I was trying to come up with examples of conformally nonflat Riemannian manifolds; I can now show that conformally flat Einstein manifolds are constant sectional curvature and I "know" CP^2 is Einstein. It surely doesn't have constant sectional curvature; on complex 1-dim subspaces of the tangent space the sectional curvature is 1 clearly. The real planes will be a mess.

@TedShifrin I'll have to look at your notes carefully

I should do this calculation it seems extremely important to me

Ted Shifrin

03:59

1/4

Hermitian Einstein, not Riemannian.

Balarka Sen

Oh ok, that's because of the scaling. The complex 1-dim subspaces exponentiate to round spheres CP^1 in CP^2, which is why I guessed.

It's "homogeneous in the complex directions"

@TedShifrin Oh huh alright

Sorry, but Hermitian Einstein implies Riemannian Einstein, or am I confused? Just take the real part?

of Ric = kg

Ted Shifrin

I dunno. Actually, for me, the holo sectional curvature is $4$ and the real planes are $1$.

Balarka Sen

Oh

So it's quarter-pinched.

Wow.

Leaky Nun

05:11

@BalarkaSen 12 hours

Mike Miller

11:25

@TedShifrin Is there a relatively simple reason that $S^{4k+1}$ admits <= 1 linearly indep v.f.s?

Mike Miller

12:01

The obstruction theory approach should work. $\pi_{n-1} V_2(\Bbb R^n)$ is $\Bbb Z/2$ in dimensions 0,3 mod 4 and is $\Bbb Z \oplus \Bbb Z/2$ for dimensions n = 1,2 mod 4; the projection to $\pi_{n-1} V_1(\Bbb R^n)$ is an isomorphism on the $\Bbb Z$ factor. Assuming the existence of a section of a rank $n$ bundle $E \to S^n$, then, the obstruction to finding two lin indep sections lies in $\Bbb Z/2$ in all dimensions $n$ --- the part of the obstruction in the $\Bbb Z$ factor vanishes

But then we need to actually compute this guy

(Equivalently, take the complement of your trivial line summand --- this is a rank $n-1$ vector bundle; a global section of this bundle over $D^n$ gives rise to such a global section over the boundary sphere, which gives one an element of $\pi_{n-1} S^{n-2} = \Bbb Z/2$, so long as $n \geq 5$)

Seems possible but unpleasant

uhoh

13:28

0

A: Analytical expressions for acceleration due to zonal harmonics of a gravitational field?

tl;dr: I'm almost there. I can reproduce Wikipedia's result for $J_2$ and $J_3$ so far (only $J_2$ shown) except for a minus sign in both cases, which reminds me again of the unresolved issue mentioned in What is the sign of Earth's J2? Let's look at @mmeent's comment suggesting that the spher...

I'm there except for a minus sign, I am beginning to wonder if Wikipedia is wrong

Alessandro Codenotti

14:26

I read that if $G$ is a compact group acting on a metric space $(X,d)$, then the metric can be averaged by integrating wrt the Haar measure, $d'(x,y) = \int_{G} d(g^{-1}\cdot x,g^{-1}\cdot y) dg$, to give a $G$-invariant metric $d'$. But is it clear that $d$ and $d'$ induce the same topology?

I'm mostly interested in the case where $G$ is a finite discrete group and we're just summing over the orbit. $d'(x,y)\geq d(x,y)$ is obvious

geocalc33

14:50

Anyone know how to work out the solution? $\varphi_1(B)=\exp\bigg( \frac{\log^2(C_1)}{\log(B)} \bigg)=B$

Alessandro Codenotti

Oh maybe this works. If $g\in G$ is fixed, then $d'(x,y)=d(gx,gy)$ induces the same topology as $d$, since $x\mapsto gx$ is an isometry, and sum of metrics inducing a topology induces the same topology

geocalc33

$B, C_1$ are matrices in the special linear group

is it as simple as I think it is?

just have to solve $\log^2(C_1)=\log^2(B)$

So $C_1=B$

Knight

16:17

Leaky where can I ask about some function of my mobile? I mean which SE site or chat room?

manooooh

16:52

Hello all! We know that if $A,B$ are sets then $|A\times B|=|A|\cdot|B|$. What happens if $|A|=\infty$ and $B=\emptyset$?

Can we use the mentioned property? Or it fails when one of them is infinite?

Thorgott

$|A\times B|=|A|\cdot|B|$ is the definition of the RHS

your last line is correct

manooooh

Okay, so the property is vaild if and only if both $A,B$ are finite, correct @Thorgott?

Thorgott

no, $|A|\cdot|B|=|A\times B|$ is always valid, because that is the definition of $|A|\cdot|B|$

if $A,B$ are finite, this multiplication agrees with the usual multiplication of natural numbers

manooooh

@Thorgott what happens when are infinite?

I can't do $|A\times B|=\infty\cdot0$ because $\infty\cdot0$ is not a number

Tobias Kildetoft

@manooooh $\infty$ is also not a cardinality

manooooh

17:02

@TobiasKildetoft in my textbook there is a proposition: True or false?: $|A|=\infty\to\forall B(|A\times B|=\infty)$

Thorgott

well, what do you think based on the example you brought up?

Tobias Kildetoft

@manooooh $|A| = \infty$ just means that $A$ is not finite

manooooh

Tobias Kildetoft

The statement about products of cardinalities only holds when you actually have those

manooooh

@TobiasKildetoft I have both: for $A$ is not finite, and for $B$ is zero

Tobias Kildetoft

17:04

(and even, then, the rule is somewhat counter intuiitive, though fairly simple)

Thorgott

yes, that is correct

can you figure out what happens if $B\neq\emptyset$

manooooh

@Thorgott @TobiasKildetoft the question I am asking if it is valid to do the following: $|A\times B|=|A|\cdot|B|=\infty\cdot0=0$

@Thorgott of course the product is not finite

Tobias Kildetoft

No, because the product there is meaningless. Exactly as you said

In general, if $A$ is infinite and $B$ is non-empty then $|A\times B| = max(|A|,|B|)$.

manooooh

@TobiasKildetoft ok. Thank you!!

@TobiasKildetoft interesting!

Thorgott

(assuming choice)

Tobias Kildetoft

17:08

right (as always)

topologicalorientablesurface

$[0,1]^{[0,1]}$ is not metrizable.

i suppose it is

then since its compact by tychonoff

it is seperable

and therefore second countable

now what?

Thorgott

it's compact, but not sequentially compact, which can't happen in metrizable spaces

topologicalorientablesurface

why is it not sequentially compact?

Balarka Sen

its not first countable, much less second countable

topologicalorientablesurface

alright, why not?

Balarka Sen

17:16

Why don't you try to prove that? It's a direct set theory argument

Take the 0 sequence. There's too many open neighborhoods around that

The point is separable does not imply second countable. That's only true in metrizable setups

Well, first countable and separable implies second countable (EDIT: Nah, the lower limit topology)

Alessandro Codenotti

18:18

@BalarkaSen Do I spot a Balarka doing set theory?!

Balarka Sen

lol

Right now I am doing geometry

That is to say Taylor expansions

Alessandro Codenotti

I'm also doing some ugly computations tonight so I can't really say anything

geocalc33

What do you call a space where each point is a matrix

and as you vary through the space the matrices smoothly vary

Balarka Sen

A submanifold of $M_{m \times n}(\Bbb R)$

geocalc33

ah okay

Alessandro Codenotti

19:22

And my ugly computations are not even working out as I was hoping

Balarka Sen

Same

Alessandro Codenotti

:/

Do you want to think about some GGTish stuff with me?

Balarka Sen

Not right now, but I'd be glad to ponder after I wrap up computations on my end

Alessandro Codenotti

Fair

Balarka Sen

No guarantees I can help though

Alessandro Codenotti

19:27

No problem

Balarka Sen

Hey @ÉricoMeloSilva!

I sent you something on hangouts but I guess you didn't notice

Ted Shifrin

howdy, demonic @Alessandro and a @Balarka

Alessandro Codenotti

Hi @Ted

Balarka Sen

19:45

Hi @Ted!

Érico Melo Silva

yo

Ted Shifrin

Oh, yo, @Eric.

Érico Melo Silva

@BalarkaSen I did not notice yeah, I haven't logged onto my gmail in weeks

Balarka Sen

No worries. It was just some Riemannian geometry question. Feel free to look when you get time

Érico Melo Silva

gotcha

Ted Shifrin

19:47

Balarka knows better than to ask me Riemannian stuff. :)

Balarka Sen

Heh I will ask you soon when I do moving frames computations

Érico Melo Silva

have you checked if the first part of your question is correct since then @Balarka

visualizing it is hard and so I'd have to compute

Balarka Sen

yeah I haven't, I might try to do the computation

Érico Melo Silva

the geodesics that dont emanate along the meridians are weird

so im not sure

Balarka Sen

Yeah I think they might spiral like the (p, q)-torus knots, but this is all guess. I'll have to recall Clairaut's relation to do computations

Érico Melo Silva

19:55

yeah, I might do this too because it's a cute visual question

Joe Shmo

@TedShifrin I see here that we defined $H_{*}(X, A) \cong H_{*}(X / A)$ for $* > 0$, but we skipped the definition as far as I can see in dimension $0$. So how do you do it in dimension $0$?

Érico Melo Silva

also with regards to doing a Morse theory for d_p I've never seen anyone try such a thing but I think it's not hard to prove the homotopy type lemma for annuli that stay away from the cut locus for these guys but idk if you can get diffeomorphism the way you do in standard morse theory, the derivative of d is so bad @Balarka

mind you i havent written anything down

Ted Shifrin

@JoeShmo: Balarka knows more algebraic topology than I do, but I don't get "defined." The relative homology $H_*(X,A)$ is defined from its chain complex, as is $H_*(X/A)$. Are you saying this is a proposition?

Thorgott

Does $IJ=I\cap J$ for ideals $I,J$ in a commutative ring $R$ imply that $I+J=R$ if they're both nonzero?

Balarka Sen

@ÉricoMeloSilva yeah I don't know how to write this thing down

@JoeShmo That's a weird definition. Like Ted said, relative homology is defined using the relative homology chain complex, and it is not always true that $H_*(X, A)$ is isomorphic to $H_*(X/A)$ using that definition. You need $A$ to be have a neighborhood in $X$ which retracts back to $A$ for this to hold

Érico Melo Silva

20:09

you need to figure out some definition of "critical point" and "index" that works here since all those things happen for d_p where it isn't smooth

Balarka Sen

I'll try to work the torus example out to gain more insight

I think it's true that cut locus of $p$ is homotopy equivalent to $M \setminus p$, by joining points on the cut locus to $p$ by minimal non-extendible geodesics, but it seems annoying to prove this is continuous

Ted Shifrin

@Thorgott: I haven't thought about this in ages, but what if we take $I = (x)$ and $J=(y)$ in $k[x,y]$?

Balarka Sen

So maybe one only gains insight about the top cell.

Érico Melo Silva

@BalarkaSen this is true and you're right that continuity is nontrivial

Thorgott

@TedShifrin then that's a counterexample and I don't like that it is

thanks

user714630

20:17

heh

Balarka Sen

$IJ = I \cap J$ holds iff $\text{Tor}^1_R(R/I, R/J) = 0$, because if you tensor $0 \to I \to R \to R/I \to 0$ by $R/J$ you get $I/I \cap J \to R/J \to R/(I + J) \to 0$ and if $I \cap J = IJ$ that left thing is in fact the kernel

If $I$ and $J$ are coprime the Tor term vanishes but this is usually much weaker

Thorgott

maybe algebra was a mistake after all

Érico Melo Silva

correct take

Alessandro Codenotti

@Thorgott Going back to analysis?

user714630

Typo: First term is $I/IJ$ @BalarkaSen

Balarka Sen

20:24

Thanks! Sorry

geocalc33

what do you call a space in which each point is a category and the categories smoothly vary from point to point in the space?

Balarka Sen

asking the right questions

@user714630 Does this manifest as some flatness criterion for the varieties $V(I), V(J)$ in $\text{Spec}(R)$?

I am not quite sure what this means yet

I remember doing something similar once

fomin

if 𝐸𝐻=4×𝑂𝐼

user714630

Should be weaker than that. I thought I came across an AG connection before though.

fomin

𝐴𝐼=4×𝑂𝐻

how do you know from that that O is 1 or 2?

geocalc33

20:27

is that a chemistry question?

fomin

no a math question

A, I, O and H are variables

Anonymous

Hi. Does anyone here know under what condition the order of a group $G$ can be written as product of orders of its factors (in its composition series)?

Thorgott

the differential computations on the next analysis sheet are certainly starting to look like greener pastures than this algebra sheet

fomin

it's meant to be simple but I don't see it

Thorgott

@SanchayanDutta always?

fomin

20:31

anyone see what is going on?

Thorgott

@fomin what if $I=H=0$

fomin

@Thorgott well exactly.. but see puzzling.stackexchange.com/a/97625/68951

Balarka Sen

@user714630 I expect it means some sort of transversality of $V(I)$ and $V(J)$

Thorgott

the way the question is phrased, I would assume that "Eh", "Oi", "Ai" and "Oh" are each a single variable

which makes my skin crawl

Balarka Sen

$\ht$

Hrm height doesn't work

Anonymous

20:37

@Thorgott Could you point me to a proof?

Thorgott

Doesn't this just follow from Lagrange? You don't even need a composition series, this works for any subnormal series $1=G_0\triangleleft G_1\triangleleft...\triangleleft G_n=G$. Then $|G|=\frac{|G_n|}{|G_0|}=\prod_{i=0}^{n-1}\frac{|G_{i+1}|}{|G_i|}=\prod_{i=0}^{n-1}|G_{i+1}/G_i|$.

Anonymous

@Thorgott Ah, nice!

Balarka Sen

@Thorgott Yeah haha this is horrible

quickmaffs

geocalc33

what objects can the log operate on?

Anonymous

@Thorgott Though, is subnormality required there? Wouldn't it hold for any arbitrary subgroup sequence too?

Balarka Sen

20:47

what does factor group mean if the sequence isnt subnormal

Thorgott

well, the equation still remains true, but $G_{i+1}/G_i$ doesn't carry a canonical group structure anymore then

Anonymous

Heh, right. Thanks

Balarka Sen

Time to do more geometry

aka more Taylor series

loch

it's weaker than transversality, probably easiest to see this is e.g. take I = y-x^2 and J = y
(but transversality does imply this i think)

Balarka Sen

Yeah I was thinking of this example

loch

20:55

but if Tor = 0, it means that (say when the intersection is zero), you can define the intersection multiplicity at a point by the length (dimension as a vector space) of the corresponding ring (localized at the point)

doesn't work if Tor non-zero though, and Serre tells us that we need to correct the number with a bunch of Tor terms

geocalc33

Let's say you have $\log(X)$ where $X$ is some mathematical object. I know that you can take the logarithm of some numbers, some matrices, some groups, some rings, etc.
Where does it stop? If $X$ is some category, can you take the logarithm of a category?

Balarka Sen

Something horrendous like I = xy and J = x - y then IJ = (x^2, xy, y^2) and I cap J is not this

so I guess it means three lines intersecting at a point is excluded by this

@loch Huh i see

Ted Shifrin

@Thorgott You talking about differential forms?

Thorgott

yeah

Ted Shifrin

Nice. One of my favorite topics :)

Of course, there's still (multilinear) algebra in them :D

Alessandro Codenotti

21:15

I've read some weird things on "formed spaces" and forms on sets yesterday evening

Balarka Sen

YIKES

Alessandro Codenotti

here

Anonymous

Could someone point me to a proof of the fact that a group can have only finitely many different subnormal series?

Alessandro Codenotti

The terminology is very weird, but I'm translating it to things I'm more familiar with in my mind and it's fine. It's a cool framework to deal with both large scale and fine scale topology

Thorgott

I can proudly announce that after multiple hours of feeling stupid, I have finally solved an algebra exercise... by Cauchy induction

this is the first time I have ever used Cauchy induction for anything

Alessandro Codenotti

21:18

What is it even

Thorgott

you prove n=>2n and n=>n-1

Balarka Sen

the AM-GM trick

I have used double induction once

beat that

Alessandro Codenotti

Oh, I've heard that called forward-backward induction in competition math things

Thorgott

double induction? does that mean induction over NxN?

Balarka Sen

yeah

Alessandro Codenotti

21:20

There's also the classic thing where you do a proof by induction and the base case needs to be done by induction on some other variable

Balarka Sen

just do induction on ordinals am i right

Alessandro Codenotti

Indeed

Why stop at $\omega$

Ted Shifrin

Hmm, I've done double induction before, but this makes no sense to me.

Thorgott

at least double induction feels like a natural thing to do

but who on earth actually thinks about doing induction forward-backward

Alessandro Codenotti

Proofs by transfinite induction are extremely common in set theory

Balarka Sen

21:21

yeah

Thorgott

I learned transfinite induction recently, it's really cool

Alessandro Codenotti

Proofs by induction on well founded relations in general are fairly common, with $\in$-induction being the most common case

Balarka Sen

@TedShifrin I had to induct over (depth, dimension) of strata

it was horrible

Astyx

How does transfinite induction work ?

Alessandro Codenotti

Like induction but longer

Balarka Sen

21:24

instead of just successors you have to deal with limit ordinals as well

thats the only extra step

Ted Shifrin

@Astyx: You're never done. :D

Balarka Sen

think of inducting over $\{1/n : n \in \Bbb N\} \cup \{0\}$

Thorgott

I remember when we proved that the ring of symmetric polynomials is the polynomial ring in the elementary symmetric polynomials, we first inducted over the number of variables and then over the total degree

Astyx

That seems fun

Balarka Sen

Oh yeah that horrible @Thorgott

I hate that proof

Alessandro Codenotti

21:25

You want to prove that some property $P$ holds for all ordinals. You prove that $P(0)$ holds, that $P(\alpha)\implies P(\alpha+1)$ and that $\forall \alpha<\beta P(\alpha)\implies P(\beta)$

Ted Shifrin

You can sometimes turn those into single induction proofs by embedding the other quantifier into the inducting sentence.

Alessandro Codenotti

(the second case is really already covered in the third, but I like to write them separately to distinguish successor and limit)

Jingeon An

Hi, anyone who's doing Doctorate program in Europe?

Alessandro Codenotti

Hm I don't think any of the regular users here is a PhD student in Europe

Thorgott

yeah, the proof was really convoluted

Balarka Sen

21:27

Best way to prove the fundamental theorem of elementary symmetric polynomials is obviously to show that $\Bbb Q(x_1, \cdots, x_n)^{S_n} = \Bbb Q(e_1, \cdots, e_n)$ which is immediate from Galois theory

Alessandro Codenotti

With some luck and some optimism I might be one in 5 months or so

Thorgott

I think it's the only time I've seen my algebra prof fail to freestyle a proof in lecture

there also was some additional statement on the degree that had to be included in the induction hypothesis just to make the inductive step work, but which was irrelevant to the conclusion

@BalarkaSen yeah, but how do you get to the rings from there?

Jingeon An

@AlessandroCodenotti Nice! I am starting master in EPFL this year, and I want to proceed PhD if possible. Do you think the selection process is harsh for the PhD program in Europe?

Thorgott

inbefore commutative algebra magic

Alessandro Codenotti

I don't really know, I just submitted my first application yesterday

Jingeon An

21:30

Good luck for you! Have any friends who's applying as well? I just want to have idea how much grades I need to go PhD program.

Alessandro Codenotti

One thing that I would really suggest though is to try and get to know potential supervisors during your masters personally, it's much easier to get a supervisor if he or she already knows you

Jingeon An

Yeah definately will do that.

Balarka Sen

@Thorgott if $f \in \Bbb Q[x_1, \cdots, x_n]^{S_n}$ by above we have $f \in \Bbb Q(e_1, \cdots, e_n)$. But $f$ is a regular function on $\Bbb C^n$, so that forces the denominator of $f$ as a rational function on $e_1, \cdots, e_n$ to not vanish anywhere, which forces it to be constant polynomial by fundamental theorem of algebra, if you wish

Alessandro Codenotti

As good as possible, but it's not like there's a clear cut point, grades are just one part of the application

I guess it also depends on how well known the institution you study at is in the area you are applying to positions for

Balarka Sen

its a "basechange and use rational locally regular = globally regular" argument in AG

Thorgott

21:32

what's a regular function?

Balarka Sen

polynomial function, sorry

Astyx

From what I've heard it's not too hard to find a PhD at the EPFL if you're in the EPFL

Your professors probably will be able to help you in that regard

Jingeon An

@Astyx That's a good news, but where did you hear about it?

Astyx

I myself am going to start a master's next year

Jingeon An

Haha, nice, math department?

Astyx

21:34

Yup

Thorgott

hmm, that sounds not wrong to me

Ted Shifrin

thinks to self: Everyone is growing up ... well, most everyone.

Balarka Sen

It's not as simple as I wrote, actually

Astyx

EPFL didn't want me though

Thorgott

but I remember there was a reason for why he put in so much extra effort into proving it for rings

oh wait, this is only over Q

Balarka Sen

21:35

Yeah it's a little annoying. I will tell you a correct argument

Jingeon An

Oh damn. But I guess you got a different acceptance?

Thorgott

but any algebraic closure should do a similar job

alright

Astyx

Not yet but I'm confident

Jingeon An

Nice.

Ted Shifrin

I'm assuming that between our president and the lovely virus that foreign students won't be coming to the US for a while.

Jingeon An

21:37

I don't think that will effect much for the top schools.

Astyx

Yeah I agree

I was hesitating for a long time between EPFL and the one I'm going to

I've heard they're all pretty equivalent

Jingeon An

I don't even know how to judge the level of the math departments by schools since I am from different major.

Is it about faculty level?

Astyx

I don't either to be honest

Does it really matter in the end ?

Jingeon An

I don't think so.

Astyx

I think the only thing it changes is how easy it is to find a PhD in some places

Ted Shifrin

21:41

In the US the caliber of the school and the reputation/experience of the PhD adviser both play a rôle.

Jingeon An

Does money matters in math department? Did not know that.

Balarka Sen

@Thorgott Right but this is not a polynomial of single variable story, you can't apply FTA. You have to notice if $f = P/Q$ where $f, P, Q$ are polynomial of $n$ variables over $\Bbb C$ then zero set of $Q$ is contained in the zero set of $P$, i.e, $V(Q) \subset V(P)$. That forces $P$ to be contained in the radical of $Q$, so $P = R Q^n$ for some polynomial $R$, by Nullstellensatz

This works over any field, you just take algebraic closure and basechange to that setup.

Thorgott

Ok, that explains why it's more difficult

Balarka Sen

Ya

You have to be an AG chad to notice this that's all

Thorgott

I thought FTA would work by just evaluating all variables but one wherever, but I guess this doesn't guarantee you get something non-constant

Balarka Sen

21:46

Yeah that sounds fidgety to me, I am not sure that works.

Nullstellensatz is a good theorem u should learn the proof sometime

I can give you a sheaf interpretation of this later if you want

the n Point Of View

Thorgott

I think my algebra power level is too low for that

Balarka Sen

Lol I promise it's actually geometry just put in an algebraic wrapper

thats what algebraic geometers do

they take simple statements and make them impossible to understand

thats their secret

Jingeon An

I want to focus on analysis for my studies and I am weak in algebra. Do you think it is mendatory to learn Galois?

Balarka Sen

geometry is the art of making good mathematics out of bad pictures and algebra is the art of making bad mathematics out of good pictures

thats literally it

Thorgott

is there anything you don't call "actually geometry"

"yes, higher topos theory"

Balarka Sen

21:55

Lmfao

Ted Shifrin

@JingeonAn Unless you're planning to do applied mathematics, yes.

Jingeon An

What do you mean by applied math? I want to study in PDE, do you call it applied math?

Transcript for

$Mathematics$ Mathematics