Mathematics - 2025-02-23

Thorgott

01:37

@BalarkaSen the height function on the circle is not submersive at the interval boundary, unless I'm misunderstanding your example

@BalarkaSen I initially wanted to avoid going through the proof again to see this (the authors of the paper certainly seem to imply it's rather straightforward), but in any case this does sound convincing

Pizza

09:53

Hi

skullpatrol

10:05

yo

one potato two potato

10:52

Given a compact Riemann surface $S$, does a homotopy class of map $f:S\to\Bbb CP^1$ uniquely determine the complex line bundle on $S$ up to isomorphism?

psie

11:24

A bit out of context, but I'm reading the rank theorem in Rudin's book. We have $F$ and $H$ which are $C^1$ maps and then $A=F'(a)$ for some fixed $a$ in the domain of $F$. $P$ is a projection with range being the range of $A$, and $P$ has null space $Y_2$. He claims suddenly that $\psi(x)=F(H(x))-Ax$ is $C^1$. Is this simply an application of the chain rule and the sum law for differentiation?

Sine of the Time

@Pizza Hi!!

Pizza

11:55

@SineoftheTime hi! how are you? Anyway I passed the physics and algebra and geometry exams, now I'm doing electronic calculators

However, I had thought about also doing mathematical methods for engineering, although I was thinking of doing it in the evening, to give priority to this other exam.

Do you think I can take 2 exams together, but both are 9 CFU

I mean, I would like to do it because I think there are things that will also be useful for the subjects I'll be doing soon.

1010011010

I have made a rather complex system where things piece together to magically create the properties I want ( I have six points of a "discrete" CDF from which I make a "discrete" PDF). There is an error in the system and I want to post on the main page to discuss how to even troubleshoot it. Is this within the scope of the SE at this point? Hope it sort of makes sense why I'm doubting whether or not to post.

I ask this from a meta perspective; because I can post a massive wall of text which will probably lose most people so trying to sidestep the work that goes into it makes sense.

Sine of the Time

@Pizza Long time no see; I'm fine thanks :). Congrats for passing the exams

@Pizza yeah, I think you can do both

but methods is not easy

Pizza

12:24

Yes, I have to study the theory well

Sine of the Time

when is the exam?

Pizza

@SineoftheTime I'm definitely not ready by March, then I don't know the other dates :/

In March it would be the 17th

Sine of the Time

mmmh yeah it's almost impossible

Pizza

Yes, also because I'm studying another subject too

Sine of the Time

I see

Pizza

12:32

Ok, I can still start and see the new dates when they are

Sine of the Time

:)

Thorgott

14:21

@onepotatotwopotato yes, that's true

X4J

Let $\varphi: (R, +_R, \cdot_R) \rightarrow (S, +_S, \cdot_S)$ be rings (with identity) homomorphism.
Suppose $G \subset R$ s.t $(G, \cdot_R)$ forms a group.
Does it imply that $(\varphi(G), \cdot_S)$ also forms group?

Thorgott

@psie yes? what else would it be?

X4J

I know that if the identity of $G$ is $1_R$ then it is correct. Otherwise, does the structure is still preserved?

I've seen an example for which the ring $R$ has a subset that forms a ring in respect to a different $1$ but the same multiplication, hence I was wondering about the above question

Thorgott

does it matter what the identity in $G$ is?

one potato two potato

@Thorgott could you explain?

X4J

14:25

@Thorgott I'm asking if it is always satisfied, .i.e, for any such $G$

Thorgott

@onepotatotwopotato $S$ is $2$-dimensional and $\mathbb{CP}^1$ is the $3$-skeleton of $\mathbb{CP}^{\infty}$, which is a $BU(1)$

@X4J yes, and I'm suggesting it doesn't matter whether the identity of $G$ is $1_R$ or not

try proving that $\varphi(G)$ is a group, is there any step you are stuck on?

Ben Steffan

you might want to recall that a group homomorphism $f\colon G \to H$ is simply a map with $f(gg') = f(g) f(g')$, no further requirements

in particular no additional requirement to preserve any given identity element

maths student

A circle of radius 1 is surrounded by 4 circles of radius $ r $. What is the least common multiple of the values of $ r $ that satisfy the following conditions:

1. Each of the four outer circles is tangent to the center circle and to two of its neighbors.
2. Two diameters of the center circle lie on the lines $ y = x $ and $ y = x $ and are integer distances apart. can someone help me regarding this question ?

X4J

14:54

I had confusions about the identity element but now I clearly see that $1_{\varphi(G)} = \varphi(1_G)$

Thanks tho

So ring homomorphism is much more demanding

one potato two potato

15:33

@Thorgott You mean this? ncatlab.org/nlab/show/BU(n)#smallest_classifying_space

Ben Steffan

@onepotatotwopotato indeed

Thorgott

yes

psie

16:32

Sorry for this basic question, but just learning about rank of a matrix and the (constant) rank theorem. Consider a function $f: E\subset \mathbb R^n\to\mathbb R^n$ which is $C^1$ and $f'(a)$ is invertible for some $a\in E$. Then by the inverse function theorem, there are open sets $U,V\subset \mathbb R^n$ where $f$ is 1-1 and onto, so it has an inverse $g$ which is also $C^1$ and whose derivative is $g'(y)=[f'(g(y))]^{-1}$ for $y\in V$ . My question is; does $g'(y)$ have rank $n$?

Thorgott

@Balarka So here's an attempt at being a bit more precise. With a PoU, we can construct vector fields $X_1,\dotsc,X_n$ on $W$ s.t. a) they push forward to $\partial/\partial x^i,\,i=1,\dotsc,n$ on $X=\mathbb{R}^n$ (this is w.l.o.g. by using charts) and b) they push forward to $(\partial/\partial x^i,0),\,i=1,\dotsc,n$ on $X\times(a_0-\varepsilon,a_1+\varepsilon)$ at points over, say, $(a_1-\varepsilon/2,a_0+\varepsilon)$ and $(a_0-\varepsilon,a_0+\varepsilon/2)$ (have to take /2 to allow a bit of wiggle room to glue with the PoU).

@psie is it an invertible matrix?

psie

@Thorgott yes, $g'(y)$ is invertible so I guess its rank is $n$. So a $C^1$ function doesn't automatically have full rank derivative?

Thorgott

correct to the first part. to the second part: of course not

psie

ah ok

Balarka Sen

18:40

@Thorgott Take a circle $S^1 \subset \Bbb R^2$ centered at $0$ and of radius $1/2$ and consider $f : S^1 \to [-1, 1]$ as the height function. Then $f$ is submersive on $f^{-1}[\pm 1-\varepsilon, \pm 1 + \varepsilon]$, as this is the empty set. Any map from the empty set is vacuously submersive(?)

@Thorgott Looks right to me.

Thorgott

18:58

@BalarkaSen yeah ok, that checks out

@BalarkaSen thanks

i might have to type up (a more complicated version of) this argument, but I really dont want to lol

psie

19:23

Rudin has this nice result about linear maps from $\mathbb R^n\to\mathbb R^m$; the operator norm is finite and they are uniformly continuous functions. Now, in the rank theorem, I'm facing a linear map that maps from a subspace of $\mathbb R^m$ into $\mathbb R^n$. I'm confused how to modify the proof of the operator norm being finite and show they are (uniformly) continuous.

My initial block is; how do we define the operator norm for a linear map that maybe does not include the unit ball in its domain?

To be more precise, it concerns $S$ defined in the proof above.

Maybe the domain of $S$, i.e. $Y_1$, contains the unit ball after all? Hmm.

I just don't understand why $S$ is continuous with the tools in the book given so far.

Thorgott

20:45

@psie any finite-dimensional normed vector space is isometric to some $\mathbb{R}^n$ with the Euclidean norm

Xander Henderson

21:03

@Thorgott $\mathbb{Q}$?

Soumik Mukherjee

21:21

Over R i guess

psie

21:34

@Thorgott ok. That is kind of what Rudin does in the proof, right? It seems like $S$ is constructed as a bijection between $Y_1$ and $\mathbb R^r$, except that Rudin seems to write $\mathbb R^n$ instead of $\mathbb R^r$.

actually, perhaps not

the bijection would have to be $$S(c_1y_1+\ldots+c_ry_r)=c_1e_1+\ldots+c_re_r,$$me thinks.

psie

21:48

I still don't quite see how this isometry allows us to say that $S$ with domain $Y_1$, a subspace of $\mathbb R^m$ in this case, is continuous because linear maps in $L(\mathbb R^m,\mathbb R^n)$ are.

Joe

Let $A$ be a commutative ring. In the definition of the structure sheaf on $X=\operatorname{Spec} A$, many textbooks begin by defining it on principal open sets: $\mathcal O_X(D(f))=A_f$ for all $f\in A$. Does this definition not bother anyone other than me? If $f,g\in A$ and $D(f)=D(g)$, then $A_f$ is not necessarily equal to $A_g$ (though they are isomorphic).

Ben Steffan

22:11

@Joe there is no such thing as equality :)

a canonical isomorphism is as good as an equality for most purposes

Joe

I agree that if $X$ and $Y$ are isomorphic, then anything meaningful one can say about $X$ will also true be of $Y$. And if a canonical isomorphism $X\to Y$ has been specified, then I am usually happy to conflate $X$ and $Y$ in speech and writing. But to me, it just seems formally incorrect to not have an unambiguous value of $\mathcal O_X$ at each principal open set.

Ben Steffan

Sure, but you're not punching this into lean

whether it's formally correct is not too relevant, as long as you know that it can be made formally correct if need be

you can protest this, but this is a common AG thing

Joe

Fair enough. At least I do know another definition of the structure sheaf that doesn't have this problem

Ben Steffan

I recommend opening up EGA to see how much Grothendieck does this

in some parts he really seems to be pushing for the record of how many things you can canonically identify and then pretend they're equal per page :))

Thorgott

22:41

@psie continuity does not change if you replace one space with a homeomorphic one

@Joe they are canonically isomorphic in this case

so the answer is "just pick one and it doesn't matter which"

Xander Henderson

23:29

@Thorgott I wish they were cannonically isomorphic. As in, they are the same, even after being fired from a cannon.

Ben Steffan

canonically isomorphic just means the catholic church has sanctioned one particular isomorphism

psie

@XanderHenderson "with Canon you can!"...or do you have Nikon? :) Or neither perhaps?

Xander Henderson

@psie I used to have a really nice Minolta, but the mount for the lenses eventually wore out, and film had become so expensive, and so hard to get developed, that I sold it.

I had other things going on in my life that make photography something that I just didn't have all that much time for any more (grad school and such).

Ben Steffan

@XanderHenderson part of me was expecting that sentence to end in "[...] that I ate it."

psie

I see :(

Xander Henderson

23:34

@BenSteffan No, I only eat Black Lotuses.

Black Loti?

Black Lotusen?

Blacks Lotus?

In other news, my sister has a three legged cat. She named the cat Nikon.

Because Tripod would have been too obvious.

Ben Steffan

@XanderHenderson apparently "lotus" comes from the latin adjective "lotus" so this is correct :)

but this itself comes from greek "lotos" which makes for the cooler plural lotoi (imho)

Xander Henderson

It is Latin? Well, then. Blacks Lotus it is!

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$Mathematics$ Mathematics