Mathematics - 2020-04-08 (page 1 of 2)

12:00 AM

But yeah so continuing the problem, that's our setup, we want to extend by $0$ to get a sheaf on $X$. Define $\mathcal{G}(V) = \mathcal{F}(V)$ if $V\subset U$, and $0$ otherwise. Then sheafify. Okay well do we need to sheafify? Turns out yes, and the example I thought of was to take a tiny bump function within $(0,1)$, and then extend by $0$ way outside

Ted Shifrin

The usual proof if the deRham theorem is sheafical.

Why can't I edit? Grrr

Amin Idelhaj

Oh, that I didn't know. And that's weird, trying to edit the most recent message?

Ted Shifrin

If the group is discrete, what you're saying doesn't work?

Well, chat on iPad sucks.

Amin Idelhaj

Or not that it's necessarily not a sheaf, just that in general it isn't automatically a sheaf

So I just took that as a single counterexample of, take the sheaf of smooth functions etc

Ted Shifrin

Working with an open set is troublesome, anyhow, because things can blow up as you go to the boundary. So you get the sheaf of discontinuous sections or something.

Anyhow, if you look at Bott/Tu (e.g.) you see plenty of sheaf stuff in the non-complex setting.

Amin Idelhaj

12:08 AM

I see

I should check it out soon for sure

Ted Shifrin

Yeah, the presheaf of bounded holo fns (working on $\Bbb C$) is interesting, too.

Rithaniel

12:28 AM

Hmmm, what would be the general approach to show that if $\text{char}(K)\neq 2$ and $[F:K]=2$ then $F$ is Galois over $K$? (In particular, where does the $\text{char}(K)\neq 2$ come into play?)

Amin Idelhaj

Well, what's the most obvious example of a degree 2 field extension?

Rithaniel

$\mathbb{Q}(\sqrt{d})$ where $d$ is squarefree

Amin Idelhaj

That works, so what are the Galois automorphisms? Explicitly

Rithaniel

Can I say that the degree of the minimal polynomial must be 2? Not sure where to go with that, though

Well, the automorphisms would be the trivial automorphism and the automorphism that takes $\sqrt{d}$ to $-\sqrt{d}$ and vice versa.

Amin Idelhaj

Does something here sound like it could go awry in char 2?

Rithaniel

12:32 AM

Ah, right. $-1=1$

Good point

Amin Idelhaj

Play around and see what you can get

Thorgott

12:47 AM

Every degree 2 extension in characteristic=/=2 is generated by a square root (omplete the square)

Simple

3:12 AM

in Probability and Statistics, 37 mins ago, by Simple

Let $\{S_n\,:\,n\geq0\}$ be a simple random walk with $p$ be the probability of stepping up and $q=1-p$ be the probability of stepping down. Let $T_k=\min\{n\geq1\,;\,S_n=k\}$. If $p<1/2$, is $E(T_1\,|\,T_1<\infty)<\infty$?

Leaky Nun

5:29 AM

@Rithaniel $\chi(K) \nmid [F:K]$ so $F/K$ is separable; take a normal closure $N/K$, then $\operatorname{Gal}(N/F)$ is a subgroup of $\operatorname{Gal}(N/K)$ of index $2$, so it is normal, so $F/K$ is normal

deostroll

6:13 AM

Hello. Anyone know how to search the OEIS. I am looking for non-primes which do no have perfect squares...

Mathphile

@deostroll $n!+4$ is always a non-square

deostroll

I want the full sequence...

i have a few terms: 6, 8, 10, 12, 14, 18, 20, 21, 24, etc...

Leaky Nun

6:42 AM

@deostroll what do you mean by non-primes which do not have perfect squares?

ok I guess I know what you mean now

I don't see why you would find this in OEIS

Mathphile

@LeakyNun apparently $n!+4$ is never a perfect power

Do you know why?

Leaky Nun

no

Mathphile

I am not able to prove it either

user777

Hi I have a small question: in theoretical computer science, sometimes a problem is a hard to solve, but sometimes when we reduce the problem to metric space like in TSP, then it becomes hard but still we can approximate it. Now my question is: what is the space of graph? I mean in general a graph has vertices and edges but it did not say anything about space so what is the space of this graph.

So, since every graph can be considered as a groups/rings, then does this mean that graph is on complex space since ring C is the highest level of number and cannot go beyond that.

TSP is the traveling salseman problem

if my question is not clear, then let me know

Leaky Nun

8:30 AM

@AlessandroCodenotti halp why is $(a_1, \cdots) \mapsto (a_1/1, a_2/2, \cdots): \ell^2 \to \ell^2$ compact

Balarka Sen

8:54 AM

@LeakyNun What is your definition of compactness? Do you know that compact operators are exactly the ones which are limit of finite rank operators under the operator norm?

Leaky Nun

i want to die, my riemann surfaces course is filled with functional analysis

Balarka Sen

ya lol its too much for me

Leaky Nun

@BalarkaSen image of bounded contains convergent subsequence

my text says I can choose a subsequence such that each component is convergent

and I'm like, huh?

Balarka Sen

yeah that should be workable

Leaky Nun

I see how to make finitely many components convergent

Balarka Sen

8:58 AM

yeah idk man im not going to try to do it this way altho it shouldnt be too hard

too annoying

can i draw affine vectors in geogebra

Ah I can

Nice

Leaky Nun

@BalarkaSen also my text claims that identity plus compact operator is fredholm of index 0; they spent 3 paragraphs proving the 3 conditions of fredholm; but they didn't prove index 0 ;_;

also why is a hilbert space isomorphic to infinite direct sum of itself?

(R^n reacc only)

Edward Evans

Tryna prove that a lattice $\Gamma \subseteq \Bbb R^n$ is complete iff $\Bbb R^n/\Gamma$ is compact. I got one direction (take the closure of a fundamental parallelepiped for $\Gamma$ and take the image of this guy under the canonical projection). The other direction seems harder (I think I just lack characterisations of compactness). I guess it would be easier to show that if $\Gamma$ is not complete then the quotient is not compact, but I can't draw any conclusions!

Leaky Nun

@EdwardEvans what is a lattice and when is it complete?

(I've seen 100 definitions of lattice and complete lattice)

Edward Evans

a lattice is a discrete subgroup of R^n and complete means it has full rank

or alternatively just $\Gamma = \Bbb Zv_1 + \cdots +\Bbb Zv_n$

rip

Leaky Nun

well if it doesn't have full rank then the quotient is just (R/Z)^rank x R^unrank right

Edward Evans

9:09 AM

unrank rofl

I guess so

and then the R/Z factors are compact but the R's are not

Leaky Nun

right

conclusion: always choose basis

Edward Evans

it just seems almost too easy

Balarka Sen

@LeakyNun That also becomes clearer if you use the fact about compactness I mentioned. If T is finite rank I + T has index zero because ker(I + T) = im T, and dim im (I + T) = dim ker (I + T)^* = im T^*, and those match

Leaky Nun

@EdwardEvans why must a discrete subgroup of R^n have a finite set of generators?

Balarka Sen

and then you just take a limit

the hard part as Leaky points out is classifying lattices in R^n

but also that's not too hard

you project to every coordinate and classify discrete subgroups of R instead

Leaky Nun

9:16 AM

ker(I+T) = im T?

Edward Evans

well, it's a theorem that a subgroup of an n-dimensional R-vector space is a lattice iff it is discrete. The original definition in Neukirch is just that it is a subgroup of the form Zv_1 + ... + Zv_m with m <= n

Balarka Sen

@Leaky Yikes, I am not paying attention

Let me fix that

Ok, dim im (I + T) also computes the dim of the eigenspace of -1 of T

Leaky Nun

isn't that dim ker I+T

Balarka Sen

Wait lol what was index again

dim ker - codim im right

codim im(I + T) = dim ker (I + T)^* = dim ker (I + T^*)

so that's dim of eigenspace of -1 of T^*

Everything's finite dimensional so these agree

Leaky Nun

why dim ker I+T* = dim ker I+T?

Alessandro Codenotti

9:32 AM

@LeakyNun Balarka's explanation of a limit of finite rank operators is the fastest way, I agree

Balarka Sen

dimension of eigenspace of lambda for an operator on a finite dimensional inner product space and it's adjoint are the same

facts from linear algebra yo

Alessandro Codenotti

fun fact: it was open for a long time whether "every compact operator is a limit of finite rank operators" holds in any Banach space, until Enflo constructed a counterexample in 1973 and won a goose from Mazur because of it

Balarka Sen

Oh

Alessandro Codenotti

There's also a photo of the prize being awarded en.wikipedia.org/wiki/Approximation_property

Balarka Sen

Mazur looks salty

Alessandro Codenotti

9:35 AM

True

Edward Evans

the fuh

lol Mazur is Barry and not Stanislaw in my mind

Barry is a good guy

o/

Baerry

he was a topologist before he went full nuts

Edward Evans

9:37 AM

nuts is an acronym for NUmber TheoriSt right

Balarka Sen

accurate

Khallil

Functional analysis is not very pogchamp

Alessandro Codenotti

@LeakyNun Because Hilbert spaces are classified by the cardinality of an orthonormal basis

Mike Miller

@BalarkaSen Unnecessary argument

Balarka Sen

Everything about my life is unnecessary so why not

You should talk to Leaky I am gonna duck out

Mike Miller

9:47 AM

If you know that Fredholm + compact = Fredholm and you know homotopy invariance, then i(F+K) = i(F) because F+tK are all Fredholm

Balarka Sen

O yeah I never was able to prove homotopy invariance

Someday I will try it again

Leaky Nun

hmm

Mike Miller

@BalarkaSen Homotopy invariance for surjective/injective Fredholm operators is clear; let's take injectivity for simplicity. Then we may think of the cokernel as a subspace $J$ of the codomain, so that the map $F \oplus I: V \oplus J \to W$ is an isomorphism, and hence of index zero. Now for small $t$ we also have that $F_t \oplus I$ is an isomorphism, because isomorphisms are open; and this $i(F_t) = i(F_t \oplus I) - i(I) = - \dim J = i(F)$, as desired.

Balarka Sen

10:04 AM

Clever

What do you do for the general case

Mike Miller

Now suppose $F$ is a general Fredholm operator --- again let's say negative index for simplicity. Pick $J_F \subset V$ as the kernel. Then using projection we have a map $\tilde F: V \to W \oplus J_F$ which is injective, and $i(\tilde F) = i(F) - \dim J_F$. Now for $F'$ near to $F$, we have that $\tilde F'$ is near to $\tilde F$ (using the same $J_F$ --- don't replace it with $J_{F'}$)

So $\tilde{F}'$ is also injective --- and by the first argument, because it is near to $\tilde F$, we have $i(\tilde F) = i(\tilde F')$. Putting this together, we get $$i(F) - \dim J_F = i(\tilde F) = i(\tilde F') = i(F') - \dim J_F,$$ so that as desired $i(F) = i(F')$.

Thus index is locally constant and hence constant on path components

Instead of analyzing dimensions of kernels and cokernels everything is reduced in the end to stability of isomorphisms

Leaky Nun

my text says the space of fredholm operators is weakly homotopy equivalent to Z x BU lol

oh and it used the Mazur swindle to show that U(H) is contractible

Mike Miller

the adjective weakly is unnecessary, but sure

the statement just means that maps to the space of fredholm operators give rise to vector bundles up to iso.

Secret

86

Q: Why is $\frac{987654321}{123456789} = 8.0000000729?!$

Many years ago, I noticed that $987654321/123456789 = 8.0000000729\ldots$. I sent it in to Martin Gardner at Scientific American and he published it in his column!!! My life has gone downhill since then:) My questions are: Why is this so? What happens beyond the "$729$"? What happens in bas...

Well looks like I learned a new way to investigate the cause of mathematical coincidences

by rewriting any funny number in base n

and see what pops out in the resulting algebraic expression

Still...

Mike Miller

10:21 AM

so suppose $X$ is compact, and $X \to \text{Fred}(H)$ a continuous map. You can find some large finite-dimensional subspace $J \subset \text{Fred}(H)$ so that $\text{Im}(F_x) + J = H$ for all $x$. Thus the map $\tilde F_x = F_x \oplus I: H \oplus J \to H$ is surjective for all $x$, and hence its kernel gives a vector bundle $V$. The cokernel is zero, but we added on this stupid extra term $J$, so that we call the "virtual index bundle" by the difference $\text{ker}(\tilde F_x) - J$

This is a virtual vector bundle, which is what $\Bbb Z \times BU$ classifies

Secret

$$ab = \sum_{k=0}^{\infty} \frac{a_k}{n^k} \sum_{\ell=0}^{\infty} \frac{b_\ell}{n^{\ell}} = \sum_{k=0}^{\infty}\sum_{\ell=0}^{\infty} \frac{a_kb_{\ell}}{n^{k+\ell}}$$ this said little about how multiplication propagates

unless I need to take a mod somewhere...

Mike Miller

the point of the above construction is that you'd really like to just say $I(F) = \text{ker}(F) - \text{coker}(F)$ is a virtual vector bundle because both ker and coker are vector bundles, but the rank of the "kernel bundle" can jump, as can the rank of the "cokernel bundle". So you instead just add a factor you'll formally subtract later so that the cokernel bundle is just zero, and ker(F) is locally constant rank.

Khallil

10:47 AM

This all sounds like the precursor to trying to jump into the Atiyah-Singer index formula (via heat asymptotic stuff)

Or at least that's the context in which I learned stuff like this

Mike Miller

This is just indices of Fredholm operators in general, of course knowing that is a pre-requisite to knowing how to calculate indices of differential operators ;)

If you're doing heat kernels you won't need to know so much of the topology, like that $\Bbb Z \times BU \simeq \text{Fred}(H)$

I didn't know you learned that stuff

Khallil

Yea that was the last bit of maths I did at university!

'Pseudodifferential Operators and the Atiyah-Singer Index Formula' was the name of my dissertation at the end of it all

It was actually really fun despite the lack of supervision for most of the time I had to do the project

Also yea the topology stuff above is super new to me!

Mike Miller

Impressive that you were able to do it without much supervision --- pretty difficult material! What have you moved on to since then?

Khallil

It eventually reduced to calculating stuff on a Riemann surface after I did some of the general stuff

I have done very little since then!

I took a break for a year to try and weigh up my options and realised I didn't want to study anymore so I've been learning how to program for the last month :)

skullpatrol

iirc robjohn did his thesis on Pseudodifferential Operators

Khallil

11:00 AM

Needless to say I don't remember very much of stuff like that ↑

@skullpatrol Long time no see!

skullpatrol

Hi pal @Khallil

Khallil

Wow I haven't robjohn in ages

I tried to tag @skullpatrol and realised 'skillpatrol' was another option

Have you been roaming these parts with more than one account? xD

skullpatrol

yup :D

i picked up the habit from anon

Mike Miller

@Khallil Good for you!! I have good friends who decided they were at the right spot to stop studying math, and made the same move, and they're very happy now.

I hope it's rewarding for you too

Khallil

Thank you! Programming doesn't come too naturally for me but it's good fun

I've actually been doing some math related problems on Project Euler which has been interesting

Never been a massive fan of number theory but it seems quite important for programming (at least on an elementary level)

Mike Miller

11:10 AM

Really, outside of project Euler? How odd

Khallil

Yea it's quite strange! It seems like optimisation (for the practice problems I'm doing) usually boils down to using number relations to reduce the number of checks/conditionals

Mike Miller

I don't think Project Euler is meant to be representative of skills necessary for programming in general, just that one needs to get better at programming (and specific programming skills) to do the problems as they get harder, but I might have misinterpreted you

Khallil

Ahhh I meant to write 'important for programming similar problems to Project Euler'

Mike Miller

Haha, fair enough

Perturbative

Apart from complex geometry, have there been applications of algebraic geometry that's yielded results for manifolds?

Mike Miller

11:21 AM

It's hard for me to interpret that question, since I'm not sure what exactly qualifies as results for manifolds

Perturbative

Okay yeah true that question is a bit vague, I guess like just one aspect I'm thinking about was if there been applications of alg geom to construct new invariants of manifolds

But I think I do have an answer for that, moduli spaces of manifolds seem to be a growing area of research currently

Mike Miller

sure

moduli space isn't a term exclusive to algebraic geometry, it just means "space of objects"

the space of all squares in Euclidean space up to isometry could be called a moduli space; it is homeomorphic to $(0,\infty)$, the homeomorphism given by sidelength

Perturbative

Oh I see, I just assumed moduli spaces were from algebraic geometry

Thanks for linking that paper btw

Mike Miller

Here is a much older nice paper

you could look at walter neumann's work, who has studied 3-manifolds arising from singularities of complex polynomials and their associated invariants

or you could start by looking into the Casson invariant, which is defined by a (perturbed count of) intersection points in a certain algebraic variety, the representation variety

Perturbative

11:38 AM

Thanks for those suggestions!

Though I do need a bit more background before I could tackle some of those things I guess

Mike Miller

probably not for the milnor paper

anyway, the point is just that there are things to do, keep going with your studies

Khallil

This kinda stuff has really piqued my interest lol

Michael Spivak

Michael David Spivak (born May 25, 1940) is an American mathematician specializing in differential geometry, an expositor of mathematics, and the founder of Publish-or-Perish Press. Spivak is the author of the five-volume A Comprehensive Introduction to Differential Geometry. In 1964 Spivak received a Ph.D. from Princeton University under the supervision of John Milnor. In 1985 Spivak received the Leroy P. Steele Prize. Spivak was born in Queens, New York.Spivak has lectured on elementary physics. Spivak's most recent book, Physics for Mathematicians: Mechanics I, which contains the material...

Can somebody explain this picture?

Astyx

Sanity check : are the irreducible representations of $SO_3$ in bijections with the irreducible representations of $SU_2$ of odd dimension ?

Thorgott

11:54 AM

Ted can explain that, I think

skullpatrol

> I think it's a picture of him doing a stretching exercise, which some Wikipedia vandal has given a funny caption.

from tweeter

Astyx

I think he just came back from Australia and had a hard time readjusting to not being upside down

skullpatrol

he does look flexible

geocalc33

What does "math is a reflection of the universe mean"?

Masterphile

12:13 PM

just that you can find math possibly everywhere if you search for it, and also apply it

geocalc33

so math is everywhere. Is that why they call it the field of mathematics?

skullpatrol

"field" means "subject area" in that sentence

Masterphile

"field" is to denote the particular domain of study, i am not sure why exactly that word is "used", but work is done in fields (and was more in ages before, i mean like agriculture and similar), so that could be a clue

that´s more a question about genesis and evolution of particular languages

skullpatrol

@geocalc33 here

geocalc33

thanks

skullpatrol

12:25 PM

np, pal

geocalc33

Sai ak un borstan, pero marque zelarico moi tambose paco, nosga saime a toi to yenne

Sai ak un soire, moise nocto pranagarte duelo crisee normale, eso tris poino ben

okay that's enough out of me.

that's the language i'm making lol^^

Khallil

12:41 PM

Is it some kinda cypher of the Latin alphabet?

Leaky Nun

no it seems to be some sort of mix of Romance languages

VJ123

@Thorgott When you say "anything that is improperly Riemann-integrable is also HK-integrable", does that mean there are no restrictions at all? So the function could be non-negative (I've seen this as a condition both in Tonelli's theorem and the corollary of Fubini's theorem for improper integrals)?

Alessandro Codenotti

1:00 PM

@Khallil Ted has explained it before in chat, stretching exercise despite the wiki caption

Jan 16 at 0:18, by Ted Shifrin

It was about limberness (he did all sorts of marshall arts stuff) ... not about shoe-sniffing.

Masterphile

what is usually considered to be a neighborhood of a point in metric spaces? some open set containing that point?

Khallil

I've seen neighbourhood denote both an open set containing the point and just a set containing the point.

I've also seen the term open neighbourhood to mean an open set containing a point.

I think it just depends on how your lecturer/text defines the convention to begin with!

@AlessandroCodenotti I guess only the highest up can practice marshal arts xD

Masterphile

this is in the context of neighborhoods homeomorphic to $\mathbb R^n$ so i think they need to be just open sets containing that some point

Alessandro Codenotti

@Masterphile Either an open set containing the point or a set containing an open set containing the point, depending on definitions (but it makes no real difference)

Khallil

ahhh I forgot the open set in the set that contains the point

Thanks, @AlessandroCodenotti!

Masterphile

1:14 PM

am i observing something is imprecise in the following definition of the manifold:

a manifold is a metric space $M$ with the following property:

if $x \in M$ then there is some neighborhood $U$ of $x$ and some integer $n \geq 0$ such that $U$ is homeomorphic to $\mathbb R^n$

i think this could be better phrased

Alessandro Codenotti

Why? It's quite clear

Masterphile

but this allows the option that $U=\{x\}$, just choose $n=0$ for every point

Leaky Nun

do you know what a neighbourhood is?

Masterphile

yes, i almost always take it to mean an open set U containing some point

Alessandro Codenotti

$\{x\}$ is not open in most metric spaces though

Thorgott

1:20 PM

If $\{x\}$ is an open subset of $M$, then sure, this is true

Masterphile

yes, but then $n \geq 0$ can be changed into $n>0$

Thorgott

@VJ123 well, I'm not sure which restrictions you mean. Surely such a function can be non-negative

@Masterphile since you usually want the same $n$ for each point $x\in M$, this is just noting that discrete spaces are $0$-manifolds

Leaky Nun

@Masterphile the reason we allow $n=0$ is because a discrete metric space is also a manifold

Masterphile

how is discrete metric space defined? what it means discrete in that context?

Thorgott

a metric space such that every subset is open

Alessandro Codenotti

1:23 PM

A metric space in whose topology is discrete, meaning that $\{x\}$ is open for all points of the space

So you can take $d(x,y)=1$ for all $x\neq 1$ as metric for example

Masterphile

why is the choice of metric not specified in the definition? does that means that if $U$ is homeomorphic with $\mathbb R^n$ with some metric $d$ then it is homeomorphic with some other metric $w \neq d$?

or that´s irrelevant for homeomorphisms?

Alessandro Codenotti

In the definition you start with a metric space, so a metric has been given

Masterphile

yes, but in this context all metrics are, sort of, equivalent in some precise sense, so i can think in terms of "Pythagorean distance", or, "the usual metric" as it is sometimes called

Khallil

(The definition of a manifold is something more general than a metric space whose open neighbourhoods are homeomorphic to a copy of $\mathbb{R}^{n}$)

VJ123

@Thorgott for example, on the following link: math.libretexts.org/Courses/Montana_State_University/…

If you scroll down to equation 21, it says underneath that "It is very important to note that we required that the function be nonnegative on D for the theorem to work" (dealing with improper integrals).

This was the kind of restriction I was talking about, though I'm not sure why this is
necessary.

Thorgott

1:34 PM

That just seems like a weaker version of Tonelli and I genuinely don't understand what about those integrals is "improper"

@Masterphile both the notion of a discrete space and of a continuous function are topological notions, so it doesn't really matter which metric the space carries, only what topology this metric induces

Masterphile

yes, that seems understandable

Alessandro Codenotti

@Khallil Is it? I'm not convinced

If you define a manifold to be a locally euclidean hausdorff second countable space, then all manifolds are metrizable, but Masterphile's definition also considers as manifolds uncountable discrete spaces, so it seems to be the more general one

Thorgott

and that one can be made even more general if one replaces metric space with topological space as in that case, not all manifolds are metrizable, no?

Alessandro Codenotti

Depends on what you ask, specifically in whether you ask either paracompact or second countable or only locally euclidean

Thorgott

locally euclidean topological space would be the most general definition

and then you can take stuff like the long line

Edward Evans

1:44 PM

@Thorgott Hey, weißt du ob der Begriff "F invariant" das Gleiche bedeutet wie "commutes with F" auf Englisch? Bzw wird es normal so benutzt auf Deutsch? :)

Khallil

Yea I think my definition always omits second countability so I always allowed for manifolds whose topology isn't metrisable @AlessandroCodenotti

Edward Evans

Oha zwei mal bedeutet geschrieben lol

Khallil

Yea precisely what @Thorgott said. Thanks!

Alessandro Codenotti

@Thorgott I think everyone requires Hausdorff, but apart from that I agree

Otherwise you get the line with two origins and that kind of weird stuff

Masterphile

quoting:

"[If you know anything about topological spaces, you can replace “metric
space” by “topological space” in our definition; this new definition allows some
pathological creatures which are not metrizable and which fail to have other
properties one might carelessly assume must be possessed by spaces which are
locally so nice."

Thorgott

1:46 PM

@Edward In welchem Kontext?

Alessandro Codenotti

@EdwardEvans Is F a group acting on something?

Khallil

Was your original question just that you wanted to relax the dimension from $n \geqslant 0$ to $n > 0$?, @Masterphile?

Edward Evans

Ja also vom Kontext her ist es irgendwie offensichtlich: Ich krieg Tr(Fz) = FTr(z) lol, wollt nur fragen ob invariant normal so benutzt wird lol

Thorgott

one must be mad to consider non-Hausdorff spaces to begin with

Alessandro Codenotti

@Thorgott D:

Edward Evans

1:48 PM

@AlessandroCodenotti F is just an involution on a C-vector space

Doing Minkowski theorx hehe

Thorgott

ich würde unter F-invariant eher Tr(Fz)=Tr(z) verstehen

Edward Evans

Okey ich auch, aber irgendwie benutzt der Neukirch invariant für kommutiert mit

Masterphile

@Khallil i am just considering some variations of the definition, i started to "do" differential geometry with only some basic knowledge of calculus and linear algebra, so i could "run" into some difficulties with understanding of the material

Edward Evans

Zb <.,.> ist ein hermitesches Skalarproduct das unter F auch invariant ist, und in Neukirch steht <Fx,Fy>=F<x,y>

Produkt*

Lol ah well

Alessandro Codenotti

F-invariant to me sounds like applying F or not makes no difference, not that it commutes with F

Edward Evans

1:53 PM

Yeah exactly

W e i r d

Thorgott

if $F$ is an involution on the vector space, what even is $F\langle x,y\rangle$?

Edward Evans

Loooolmhe then goes on to describe points that are invariant under F as points that are unaffected by the action of F

Khallil

Oh sweet. I really enjoyed diffgeo when I studied it a while back. I hope you enjoy it too :) @Masterphile

Masterphile

@Khallil haha, i am not a student, could be mine age is strictly larger than yours , i have much free time and in it i much do math :)

Alessandro Codenotti

@Thorgott Good question

VJ123

1:58 PM

@Thorgott Hmmm. Even for Tonelli's theorem, the condition is that the function must be non-negative and measurable (en.wikipedia.org/wiki/Fubini%27s_theorem). Why can the function not be negative?

Thorgott

maybe he fixed an embedding of the base field into the space

Alessandro Codenotti

This is starting to sound like you have $G=\Bbb Z/(2)$ acting on the vector space and the base field and you want to talk about $G$-equivariant things

Edward Evans

Hang on I'll give you more context: we're looking at the space $K_\Bbb C = \prod_\tau \Bbb C$ where $\tau$ are embeddings of a number field $K$ into $\Bbb C$. There's a hermitian scalar product on this C-vector space called $\langle \cdot, \cdot \rangle$. Then we define $F : z_\tau \to \overline{z}_{\overline{\tau}}$.

The notation is weird

Alessandro Codenotti

Where's Mathei when we need him

Edward Evans

ye

Alessandro Codenotti

2:01 PM

This is getting too algebraic for me and I need to study for my exam, I'm out for a while

See you later

Edward Evans

No worries, cya

Khallil

Best of luck with your exam, Alessandro!

Alessandro Codenotti

Thanks

Edward Evans

Now Neukirch says that $\langle \cdot, \cdot \rangle$ is $F$-invariant and writes "so $\langle Fx, Fy\rangle = F\langle x, y\rangle$, which works out, and also that the trace I described is F-invariant, which also works out, and then on the next line he talks about F-invariant points of $K_\Bbb C$, which are those such that $\overline{z}_\tau = z_{\overline{\tau}}$ lol

Alessandro Codenotti

It's the last one I need to finish my Masters and it's going to be on skype which doesn't sound great

You could check what the English translator of Neukirch's book decided to use

Edward Evans

2:04 PM

yeah good idea lol

Thorgott

@VJ123 Well, this has something to do with a kind of fundamental asymmetry we often see in measure theory. If $f$ is a non-negative, measurable function, the integral $\int f$ always makes sense, but it may be infinite. To deal with not necessarily non-negative measurable functions, we split them up into a positive part, $f^+$, and a negative part, $f^-$, for which the integrals are defined as previously and then define $\int f=\int f^+-\int f^-$.
For this to work out nicely, we require both integrals on the RHS to be finite (we don't want something like $\infty-\infty$ to stand there) and …

3

Khallil

2:21 PM

I can vouch for this ^

Seriously well-written, @Thorgott!

Thorgott

thanks a lot :)

Masterphile

the books starts with a theorem without a proof of it! :)

it seems the theorem is needed for a nice start

Khallil

Which book are you reading through, @Masterphile?

Masterphile

2:37 PM

one professor sent me his book, i am reading Spivak´s book

he said he will endorse me if needed

on arxiv

now i only need to prove something new! :P

or something old with new methods

anyway, it´s a challenge

Khallil

3:39 PM

That sounds pretty cool!

VJ123

@Thorgott Wow that's a really nice explanation. Really helped me understand where all of the stuff about Lebesgue integrability comes from. Thank you very much.

Masterphile

@Khallil yeah, i am already trying to prove something, but am distracted by NT stuff i also do

VJ123

@Thorgott So with Fubini's theorem and improper integrals - say you have a function that becomes unbounded at some point in the region, then you don't have to worry about the function being non-negative as long as the integral of the absolute value of the function is finite?

Masterphile

mostly elementary NT

Astyx

4:29 PM

How do I show that the adjoint representation of $SU_2$ has $SO_3$ as its image ? I've shown that the image is included into $SO_3$ but I don't see why it has to be all of $SO_3$

Thorgott

@VJ123 yeah

VJ123

@Thorgott And just a final clarification, for a function to be Lebesgue integrable, does it also have to be measurable? Is that also part of the definition?

Khallil

Yea it's part of the definition!

VJ123

@Khallil Thanks :)

Is there any easy to way define what it means to be measurable?

(I don't have much knowledge of measure theory at all)

Khallil

Have you constructed a measure before?

Oh right, then it's probably best to consult an introductory text

That way you'll be able to follow the whole development of integration, starting with sigma algebras, measures, measurable functions etc.

VJ123

4:38 PM

Okay cool. And I read somewhere that practically any function that can be described is measurable. How accurate of a statement is that?

Leaky Nun

what is the possessive form of "Tom and I"?

Khallil

Tom's and my?

@VJ123 honestly it's probably best for you to learn about the preliminary stuff like sigma algebras, measures and measurable spaces before you move onto measurable functions

That way you can build your intuition up properly!

VJ123

@Khallil All right cheers. I appreciate your help.

Thorgott

if I have a non-measurable set, I can easily describe a non-measurable function

but the point really is that essentially all functions you will naturally encounter will be measurable

Khallil

e.g. a Vitali set ^

Thorgott

4:46 PM

(in the context of euclidean spaces)

the folklore is that if you construct a function without thinking about the axiom of choice, it will be measurable

Érico Melo Silva

@AlessandroCodenotti you sent this ten thousand years ago but the paper is called "hyperbolic groups" lol

Alessandro Codenotti

I know, Balarka told me in the meantime

But thanks

Leaky Nun

‘Amazing’ Math Bridge Extended Beyond Fermat’s Last Theorem, Quanta Magazine

Khallil

That image is amazing, @LeakyNun

Érico Melo Silva

gotcha

Astyx

5:03 PM

If I have a morphism between two Lie groups $\phi:G\to G'$, such that its derivative is a Lie algebra isomorphism, how can I show that the image of the morphism is the connected component of the neutral element in G' ?

Leaky Nun

@Astyx consider $G = G' = C_2$?

Astyx

What's $C_2$ ?

Khallil

Cyclic group of order 2?

Leaky Nun

yeah

Astyx

I only wanted it to contain the neutral component

Leaky Nun

5:18 PM

@Astyx I guess it's because the component of the identity is locally homeomorphic to the Lie algebra via exponential

I don't know

Astyx

Yes that's what I'm thinking as well

I can't convince myself though

Khallil

I did some digging and found this

It's been so long since I studied Lie stuff that I genuinely don't remember the basics haha

Stan Shunpike

5:33 PM

My professor wrote "Define $\mathcal{L}$ to be the set of vectors that can be obtained as linear combinations of the columns of $X \in \mathbb{R}^{n\times p}$

does that just mean the column space?

Khallil

5:54 PM

That's literally the definition

:)

The 'set of vectors that can be obtained by linear combinations' is also called the span of those vectors

(Replace columns with rows to get an equivalent definition for the row space!)

Stan Shunpike

@Khallil super! i thought so, but i had no idea why he was calling it L

@TedShifrin hey Ted!

Ted Shifrin

6:22 PM

Hi Stan

L for linear span, I s’ppose.

@Astyx: presumably an open/closed argument.

Mike Miller

@Astyx Yes, because the map has $df_e$ an isomorphism (would be sufficient for it to be a surjection), by the inverse (implicit) function theorem you find that the image is open. Then you just need to know that a subgroup of a Lie group that contains a neighborhood of the identity is also closed (equivalently, contains the identity component).

(Of course, because it's a subgroup and multiplication is continuous, containing an open neighborhood of the identity is equivalent to being open.)

And the rest just follows because, if $g_n \to g$, then consider $g_n g^{-1}$; for large $n$ this is contained in your subgroup $H$ because it contains a neighborhood of the identity; say $g_n g^{-1} = h_n$. Then in particular we have that $g = h_n^{-1} g_n$ for all sufficiently large $n$, and in particular $g$ lies in your subgroup, so it's closed.

Stan Shunpike

@TedShifrin his notation just generally sucks so i'm not sure its intentional

Lucas Henrique

hey chat

Ted Shifrin

Hi Lucas

Stan Shunpike

@TedShifrin what have u been up to lately?

Ted Shifrin

6:35 PM

Nothing fun ... trying not to get sick.

Lucas Henrique

Let $\mathbb R^{\mathbb R}$ be the set of all real-valued functions. What's the probability that a random element of this set is continuous?

Ted Shifrin

0

Lucas Henrique

(I'm not sure if my question is well-defined)

Masterphile

0?

Lucas Henrique

@TedShifrin that's what I guessed, but I'm not sure of how I'd prove that

Ted Shifrin

6:37 PM

Of course, you didn't tell us a topology or measure ....

Lucas Henrique

Measure theory looks cool

Masterphile

for every continuous one f choose an infinite number of those that are the same as f but the difference is only that those are discontinuous at one point, and you´re done

Alessandro Codenotti

Just by cardinality reason it should be 0 for any reasonable probability measure

Ted Shifrin

Still 0, I bet, if discontinuous just somewhere.

Masterphile

yeah, continuity is rather rare :D

but if you asked for a probability that it is continuous at at least one point or more, then...hm?

Ted Shifrin

6:49 PM

Yeah, I typed the wrong thing, most likely. Not paying attention.

Lucas Henrique

What do people mean when talking about "differential groups"? What's the differential part of this group?

I'm sorry about the sporadic, random questions. I guess my low effort questions are welcome for people who are also wasting time :p

Thorgott

hmm, what does the product-$\sigma$-algebra on $\mathbb{R}^{\mathbb{R}}$ look like

Astyx

Thank you @TedShifrin and @MikeMiller

mick

Hi all

0

Q: Fixed point question for branches

Let $A(z)$ be a real-entire function. $$ A(z) = a_0 + a_1 z + a_2 z^2 + ... $$ Also for real $x$ we have $$ A’(x) > 0 $$ Let $f$ be The functional inverse : $ f(A(z)) = A(f(z)) = z $. More specific $f$ is the main branch inverse. An example could be $A(z) = exp(z) $ and $f(z) = ln(z) $ ( no...

abstract-algebra complex-analysis riemann-surfaces fixed-points

CaptainAmerica16

7:32 PM

@TedShifrin Are there old practice problems for your Linear Algebra course online? Lately, I've been thinking about purchasing the textbook.

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$Mathematics$ Mathematics