Mathematics - 2019-05-17 (page 1 of 2)

@BalarkaSen so a rectifiable varifold is a pair $(S,\theta)$ where $S$ is some rectifiable set and $\theta$ is a function on $S$ that tells you about multiplicities

Balarka Sen

Gotchu

What do you do with these

Ryan Unger

12:12 AM

so you can associate a measure to this situation

via $$\mu(U)=\int_{S\cap U}\theta(x)\, dH^k(x)$$

This is a Radon measure, so you get a compactness theorem for these varifolds

One thing I didn't mention is we usually take $\theta$ to be integer-valued.

Now there's a theorem which says that if a varifold is "close" to a plane, then it's in fact a graph

And you get an explicit estimate on the $C^{1,\alpha}$ norm

Balarka Sen

That's interesting

Ryan Unger

It turns out that "mass minimizing" varifolds satisfy this -- and hence will be smooth almost everywhere

And that's in a nutshell the "elliptic theory" for mass minimizers

Balarka Sen

so they are "limits of submanifolds" in some sense?

Ryan Unger

No, actually, that's a conjecture

They're "amost submanifolds"

Just almost everywhere from what I just said. It's a hard theorem of Federer that the codimension of the singular set is $7$ (for hypersurface currents)

Well I haven't said what a current is but it's a varifold with an orientation basically

Balarka Sen

That's a lot of codimension

Ryan Unger

12:17 AM

Right, so (area-minimizing) minimal hypersurfaces in manifolds up to dimension 7 are smooth

Érico Melo Silva

balarka should read simon's notes

Ryan Unger

There's a conjecture that this is true in all dimensions generically

Balarka Sen

Oh because those are solutions to some weird PDEs which have solutions with only varifold regularity?

Ryan Unger

There's an explicit cone in R^8 which is stable minimal and has a singularity

Balarka Sen

@ÉricoMeloSilva only if you read SGAI

Érico Melo Silva

12:19 AM

iunno what that is

Balarka Sen

grothendieck

Érico Melo Silva

oh u mean SGA I

Balarka Sen

sry

yes

Ryan Unger

I thought it was EGA

Érico Melo Silva

that's also a thing

Balarka Sen

12:20 AM

so this is what y'all fuckers do because minimal surfaces don't have enough regularity eh

madmen

Ryan Unger

Well the madman paper is Schoen--Yau 17

Érico Melo Silva

@BalarkaSen simon's notes are probably like way easier but ok :(

Ryan Unger

They prove a theorem by descending through singularities

@ÉricoMeloSilva do you know how they prove regularity in the min-max construction

that's one of my goals for the summer

Érico Melo Silva

no idea

i need 2 learn min-max stuff my man

Ryan Unger

I have the Willmore conjecture survey on my desk

Been putting it off for algebra

I might be sick

@BalarkaSen anyway the best GMT theorem is that convex functions have second derivatives a.e.

Balarka Sen

12:25 AM

what kind of singularities are they? just a bad measure 0 set outside of which the varifold is smooth or do the singular set have varifold structure themselves

is it a "stratification" in any sense of the word

Ryan Unger

there is, that's a very recent result of Aaron Naber and others

together with some classical work of Leon Simon

I don't know specifics sorry

Balarka Sen

sounds interesting

Ryan Unger

Lohkamp has also shown that if you "blow up" along the singular set you get a Gromov hyperbolic space

Balarka Sen

O_O

Ryan Unger

he studies singulartities as the Gromov boundary of this space

no one knows if what he's doing is right

Balarka Sen

12:27 AM

Okay I might want to read more about these in the near future but I'm afraid I have 0 background

@ÉricoMeloSilva what's the title of the notes?

Ryan Unger

web.stanford.edu/class/math285/ts-gmt.pdf

^

Érico Melo Silva

ye those

Balarka Sen

tnx

Ryan Unger

@ÉricoMeloSilva did you upload your vaccination records

Érico Melo Silva

not yet

Ryan Unger

12:30 AM

It was a journey for me because I seem to have disjoint records

they seem to be very strict about it

presumably if you went to an undergrad in the US it should be fine

Érico Melo Silva

i think that's the only thing i havent done yet and im too lazy to figure it out

Ryan Unger

You sent them your picture?

Érico Melo Silva

im also still in school so im busy and just never have the energy

ya

Ryan Unger

12:44 AM

@BalarkaSen why is Nakayama's lemma important

Balarka Sen

it's sort of an inverse function theorem

if $A$ is a local ring with max ideal $m$ and $M$ is an $A$-module

then Nakayama's lemma says if you have a set in $M$ which pushes forward to a basis for $M/mM$ as a $A/m$-vector space, then the set in $M$ is a generating set

Ryan Unger

Proposition 2.8 in AM?

Balarka Sen

maybe i dont have AM with me right now but this is a corollary of whatever form of Nakayama's lemma they prove

Ryan Unger

yes

Balarka Sen

$m$ is the Jacobson radical of $A$ after all

Ryan Unger

12:47 AM

right

Balarka Sen

if you think of $M$ as a local patch over $A$, whatever that means, this says that infinitisimally if a set of elements for that local patch span the tangent space $M/mM$ to that local patch then it's local coordinates for that patch

loch

Somewhere in Vakil there's a 'geometric Nakayama lemma' which makes things more geometric

Ryan Unger

@BalarkaSen ok I buy it

I'll worry about geometry later haha

Karl Kronenfeld

It's an oft used tool, specifically when the ring is local.

s.harp

well im destroyed, cannot write any further tonight

Balarka Sen

1:22 AM

@Ryan u can for example prove the Brouwer invariance of domain for rings using Nakayama, i.e., $A^m \cong A^n$ iff $m = n$.

Ryan Unger

@BalarkaSen umm but that $\cong$ just means iso right

not homeo in any sense

Balarka Sen

ofc

isom as $A$-modules

@loch That makes me wonder, it's not easy to prove that the affine spaces $\Bbb A^m_k$ and $\Bbb A^n_k$, with Zariski topology, are not homeomorphic, right? You'd have to end up showing $\Bbb A^m_k \setminus 0$ and $\Bbb A^n_k \setminus 0$ are not homeomorphic, which would require some cohomological tools I think

loch

lol its not often you hear

zariski topology and homeomorphic in the same sentence

Ryan Unger

same for R^n so why not

loch

uh

Balarka Sen

1:35 AM

except the cohomology theories are much harder here!

@RyanUnger Well the correct version of the question here is that $\Bbb A^m_k$ and $\Bbb A^n_k$ are not biregularly isomorphic which is way easier

Their coordinate rings have field of fractions of different transcendence degree

(so not even birationally isomorphic)

It's the mixing of categories that's forcing you to apply big tools

loch

I think you can just argue by Krull dimension, it's defined topologically so it's preserved under homeomorphisms

Balarka Sen

How is it defined topologically? Isn't it defined by the highest tower of subvarieties?

Ryan Unger

balarka stop im still on modules (definitely review)

loch

@BalarkaSen longest chain of irreducible closed subsets

Balarka Sen

@loch Oh I see, right.

Interesting

Ryan Unger

1:39 AM

well there goes your claim

even I understood that one

Balarka Sen

i claim youre a nincompoop

see if you can break that

Ryan Unger

deletes account

Balarka Sen

how many years of suspension did you have to go thru

Ryan Unger

uh how old are you

Balarka Sen

i lost count at some point

im 57

Ryan Unger

1:44 AM

last time we spoke you were 13

damn, 44 years

Balarka Sen

airhorn

r u still an anarcho conservative or did you switch ideologies

Ryan Unger

What's anarcho conservative

Balarka Sen

idk man

milo yiannopolous style

Ryan Unger

haha

is he still ostracized?

Balarka Sen

i think so

did u watch zizek vs peterson lmao

Ryan Unger

1:49 AM

I haven't watched a Peterson video in over a year

Balarka Sen

damn ur missing out

Ryan Unger

I heard he's gone crazy

Balarka Sen

its too funny

Ryan Unger

the Brouwer thing is an exercise in ch 2

Balarka Sen

yep

Ryan Unger

2:06 AM

@BalarkaSen do you have a favorite part?

Balarka Sen

2:13:31

Ryan Unger

@BalarkaSen have you read the book?

Balarka Sen

no lol

Ryan Unger

it seems marketed towards incels

Balarka Sen

yeah thats pretty obvious

he's the father of basement dwellers

user76284

5:03 AM

Anyone have any ideas about how this problem could be tackled?
https://mathoverflow.net/questions/321355/darkness-in-the-lamplighter-group

The best I've been able to do so far is come up with some lower bounds.

Alessandro Codenotti

@BalarkaSen prove the equivalent statement about rings by quotienting by a max ideal and reducing it to a statement about vector spaces?

Balarka Sen

5:45 AM

@Alessandro How do you pass to a question about rings? Note that I asked if they are homeomorphic or not.

There is no induced map at the level of coordinate rings.

Alessandro Codenotti

Affine schemes and commutative rings with unity are equivalent categories, you have an homeo in one iff you have an iso in the other, no?

Wait what's $\Bbb A^n$ for you here?

Balarka Sen

Topological spaces not schemes!

Wrong category

Alessandro Codenotti

Ahh, ok

Then I guess Krull dimension is the right way to go as loch said!

Balarka Sen

It's quite interesting in that light that Krull dimension works

Never thought of it as a purely topological invariant

Alessandro Codenotti

It makes sense though

Balarka Sen

5:49 AM

Yeah

Alessandro Codenotti

So that the dimension of $\mathrm{Spec}R$ as a scheme (or top space) agrees with its Krull dimension as a ring

Balarka Sen

Not dimension as a topological space in any dimension theoretic sense, is it?

Alessandro Codenotti

I remember there's some subtletlies in dimension for schemes, like dimension and codimension of subschemes not adding up to the dimension of the ambient one and stuff like that in bad cases

@BalarkaSen good question, I don't know

The inductive dimensions don't seem to work well at all because open sets are huge

Balarka Sen

Right.

What's the standard definition of dimension of a scheme? On an affine chart it's just Spec R, where it's the maximal height of a chain of prime ideals in R

Alessandro Codenotti

Should be the length of the longest chain of irreducible closed subsets

Balarka Sen

5:57 AM

I am sure there's some global definition. It should be true eg that every scheme has a dense open subscheme consisting of points at which the stalk is normal

Then dimension of the scheme = dimension of that smooth subscheme, which is just dimension of a Zariski tangent space at a point

If this is not true algebraic geometry should not exist

Alessandro Codenotti

Lmao, I don't know though

I guess with one or two adjectives before scheme (compact?) it agrees with the dimension of the irreducible component with biggest dimension

N. Maneesh

6:21 AM

$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0\\
\end{bmatrix}$ if we do $R_3\to 1/2 R_3$ and $R_4 \to 1/2 R_4$ we get \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & \\
0 & 0 & 0 & 1/2 \\
0 & 0 & 1/2 & 0\\
\end{bmatrix}. Can we say that there is a matrix $P$: $P^T \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & \\
0 & 0 & 0 & 1/2 \\
0 & 0 & 1/2 & 0\\
\end{bmatrix}P$ =$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0\\

Secret

6:44 AM

Recall that each row operation corresponds to an elementary matrix. You might discover that by multiplying these two elementary matrices together, you get $P$

N. Maneesh

7:12 AM

Yes. Let $A$ be the matrix. Let $E_1,E_2,E_3$ are the elementary matrices. Let B is obrained by sequence of elemntary operation(say row operations only). $B=E_{i_1}E_{i_2}...E_{i_k}A, i_k=1,2,3,\forall k$. Let $P=E_{i_1}E_{i_2}...E_{i_k}$. then still I am not able to write $B=P^tAP$

Secret

7:50 AM

There are going to be matrices such that $B$ is similar to $A$, but you might need something more to ensure that $P$ is a permutation matrix

3

Q: Do elementary row operations give a similar matrix transformation?

So we define two matrices $A,B$ to be similar if there exists an invertible square matrix $P$ such that $AP=PB$. I was wondering if $A,B$ are related via elementary row operations (say, they are connected via some permutation rows for example) then are the necessarily similar? Obviously swapping...

N. Maneesh

Thank@Secret

Secret

A permutation-similar matrix is a much stricter requirement and not all matrices can do that

The matrix A and B you given above seemed to preserve the trace, so it might be possible you can get something permutation-similar, but I kinda doubt it but I don't know how you can prove that

3

Q: How to tell if two matrices are equal up to a permutation

Given two real rectangular matrices A, B how can I tell if they are equal up to a permutation of their rows/column without trying all possible permutations? (This is closely related to the question I asked yesterday but which seems too specific/complicated to receive a response. Maybe I'll be lu...

linear-algebra combinatorics matrices permutations numerical-linear-algebra

In fact it is a hard problem in general

user131753

Does anyone know about any sequential approach of showing that a $n$-simplex is closed?

Leaky Nun

@user170039 define an $n$-simplex

maybe barycentric coordinates would help

user131753

8:08 AM

@LeakyNun "Suppose the $k + 1$ points $u_0,\ldots, u_k∈\mathbb{R}^n$ where $n\ge k$ are affinely independent, which means $u_1−u_0,\ldots,u_k−u_0$ are linearly independent. Then, the simplex determined by them is the set of points,

user131753

$$\displaystyle C=\left\{\theta _{0}u_{0}+\dots +\theta _{k}u_{k}~{\bigg |}~\sum _{i=0}^{k}\theta _{i}=1{\mbox{ and }}\theta _{i}\geq 0{\mbox{ for all }}i\right\}$$" - Wikipedia

Leaky Nun

ok great

then it's quite clear isn't it :P

suppose $x_n \in C$ is a sequence with $\lim x_n = x$

then $\sum x^i = \sum \lim x_n^i = \lim \sum x_n^i = \lim 1 = 1$

and $x^i = \lim x_n^i \ge 0$

user131753

What is $x_n^i$?

Leaky Nun

$x_n = \sum x_n^i u_i$

the superscript is just another index and does not denote exponentiation

user131753

Yea. I was not sure what that would mean because in the definition it was written as $\theta_k$ and you didn't say this earlier.

Leaky Nun

8:13 AM

ok

user131753

How do you know that $\lim x_n^i$ exists @LeakyNun?

Leaky Nun

because $\lim x_n$ exists

projections are continuous

user131753

Yes by projections, right?

Leaky Nun

right

user131753

@LeakyNun Oops, I see you commented earlier.

user131753

8:16 AM

Ok. Thanks @LeakyNun.

Silent

9:02 AM

The answer above is d, right?

I took $A$ rational numbers and $B$ irrational numbers

oh! that won.t work!

Leaky Nun

@Silent the answer is B

Silent

Can d also be an answer? taking $A=\Bbb R$ and $B=\Bbb Q$

@LeakyNun How? suppose $A$ unbounded then $A+b=\{a+b:a\in A\}$ is unbounded for some $b\in B$, I think.

Leaky Nun

what if there is no $b \in B$

Silent

oh!

So, if $B$ empty then $C$ empty as well?

Leaky Nun

well I suppose your example shows that D is wrong too

@Silent yes

therefore the only thing that is wrong is the question itself lol

Silent

9:17 AM

Its a multiple select question, so its ok

Abdullah UYU

9:43 AM

Hi, in this euler.ac-versailles.fr/baseeuler/lexique/notion.jsp?id=96 page, I'm interested in proving that $(a,b,c)$ is a normal vector of the plane $ax+by+cz+d=0$. But, in the definition of normal vector, there is the statement of a line being perpendicular to a plane. I don't know what it means and and I couldn't find it in the very page.

And it gets circular when I try to write a definition

Nick

I gone head and messed this up, does anyone see a better simplication here: math.stackexchange.com/a/3229363/60900

ÍgjøgnumMeg

10:22 AM

Mornin'-ish alal

all*

Nick

11:33 AM

Good evening :)

Silent

I have this T/F problem: Let $R$ be a ring with unity the the set of all non zero divisors in $R$ form a group under multiplication.

I don't think I understand the question. is this correct interpretation of non zero divisors: e.g, look at $\Bbb Z$, it has no zero divisor then every element is non zero divisor

except 0

ÍgjøgnumMeg

@Silent but $\Bbb Z$ is not a multiplicative group

Silent

oh

So, that statement is false. Thank you, @ÍgjøgnumMeg

ÍgjøgnumMeg

@Silent in fact the units of a ring form a group, and a unit has to be a non-zero divisor (but the converse is false)

e.g. in $\Bbb Z$ you have $\Bbb Z^\times = \lbrace \pm 1\rbrace$ which is a group, because these guys have multiplicative inverses

Silent

yeah

N. Maneesh

12:44 PM

For option (a), I took $\frac{1}{z}$ in $\Omega=\{z\in \mathbb C:0<|z|<1\}$

For option c)$f(z)=e^{-z}$ in $\{z\in \mathbb C:x<0,y\in \mathbb R\}$

Isn't $b$ equivalent statement to (a)?

Karl Kronenfeld

1:26 PM

@N.Maneesh "only if" is synonymous with $\imlies$.

So, you get, (c) is equivalent to (a) and (b).

Anyway, your counterexample works for part (b) as well. Though, the question never says $f$ is nonconstant. ;)

christmas_light

Hey chat, I'm studying homogeneous spaces from the book of Arvaniteyergos, but I have a doubt
Consider $M\cong G/H$ a homogeneous reductive ($\mathfrak{g}=\mathfrak{h}+\mathfrak{m}$) space, and $\pi_{\star,e}:\mathfrak{g}\rightarrow T_o(G/H)$ (where $\mathfrak{m}\cong T_o(G/H)$)
Now, given $X\in \mathfrak{g}$, $\pi_{\star}(X)=X^{\star}_o=\pi_{\star}(X_{\mathfrak{m}})$ (the $\mathfrak{m}$ component), and the author says immediately that $X^{\star}$ is a Killing vector field
Can anyone help me understanding why?

user131753

@christmas_light What is the name of the book?

christmas_light

An introduction to Lie groups and the geometry of homogeneous spaces

quallenjäger

1:41 PM

what does terms like $x*\int f(x)dx$ mean? Can I pull the x inside the integral?

christmas_light

@user170039 (page 79, I forgot to say sorry )

quallenjäger

it is just a bad notation isnt it? is it equivalent to $x*\int f(y)dy$?

user131753

@christmas_light Actually I asked because the name of the author didn't return anything when I searched for it in google.

christmas_light

I misspelt the name, it's pretty common for me ahahah

user131753

2:11 PM

Is it true that a finite group is always isomorphic to a direct product of simple groups? If not then can someone give me an example.

user131753

Ok. Nevermind. I figured it out.

Semiclassical

Am I right in thinking that factor groups sometimes are notated as G\F rather than F/G, depending on whether G acts on the left vs the right?

user76284

What's the name of the theory whereby you study manifolds according to the smooth fields that can be defined on them? I forgot.

Silent

2:28 PM

I first tried working with $(123)$ of $S_3$, but $S_3$ has two elements of order 3

Rithaniel

You're familiar with Cauchy's Theorem, right, Silent?

abenthy

If $K$ is a field contained in a ring $R$, and $A \in M_2(K)$ is a matrix which has a one-dimensional eigenspace $L$ to some eigenvalue $\lambda \in K$. Must then the eigenspace of $A$ in $R^2$ be the $R$-span of $L$?

Silent

Also, if an element $x$ has order n then so does $x^{-1}$, in fact $gxg^-1$ as well

Rithaniel

Think about $\langle n\rangle\leq G$

Silent

@Rithaniel that there exists element with order of each prime divisor in $G$? yes.

Balarka Sen

2:33 PM

@Silent Correct, but if there's a unique element of order $n$, then...

user76284

Then Lagrange's theorem?

Balarka Sen

No.

Silent

@BalarkaSen Oh wow!! then $gxg^{-1}=x$ hence $x\in Z(G)$, and since each power of $x$ are also in $Z(G)$ and order of $x$ greater than 1, $Z(G)$ has order larger than 1. So, option b. Thank you

Balarka Sen

:)

Well, $Z(G)$ has order more than $1$ because it contains both $1$ and $x$.

Rithaniel

Hmmm, I thought it was going to be D.

Silent

2:37 PM

@Rithaniel me too!!

Balarka Sen

You can't guarantee that there are more than 2 elements, because $x$ might have order $2$.

Silent

I see.

@BalarkaSen you are talking about center, right?

Balarka Sen

Yes.

user76284

Wait why D? Seems too restrictive given the loose condition.

Semiclassical

Not to be repetitive but: Can G\F denote a quotient group? I’m reading a (Physics) paper that seems to be doing so

Balarka Sen

2:39 PM

Depends on left or right action but yeah

Semiclassical

Mmkay, just wanted to check

Earth Cracks

is G acting on F?

G\F/H would denote the double orbit space of left G-action on F and right H-action on F

Balarka Sen

@Silent It's always good to have examples. Take $Q_8$, which has $-1$, the unique element of order $2$.

$Z(Q_8) = \{1, -1\}$, exactly order 2

Rithaniel

Well, my thinking was that if there is a unique element of order $n>1$ then $n=2$, though perhaps that isn't true?

Silent

Wow! Balarka's example seems to clarify all our doubts!

Rithaniel

2:42 PM

Hmmm, well fair enough.

Silent

@user76284 you got it, right?

N. Maneesh

@KarlKronenfeld Thanks.

Rithaniel

Though, I'm still suspecting that if there is a unique element with a particular order greater than 1, then it's order is 2. Is there a counterexample?

Balarka Sen

Well, that's true. If $x$ is an element of order $n > 1$ then $x^{-1}$ is as well.

Silent

@N.Maneesh Where was that question from? You seem to be very good in complex analysis, I tried so many books to learn complex analysis, but could not solve a single problem, (except those plug n chug kind of 'calculate this integral' type problems).

Balarka Sen

2:47 PM

The only way to get uniqueness is $x = x^{-1}$, which forces $n = 2$.

Rithaniel

(See I was about to start going a roundabout way. Just looking at the inverse is much more efficient)

abenthy

Okay, i rephrased it now, if anyone has an idea, please let me know :) Linear algebra question: If $A \in M_2(K)$ is a matrix over a field $K$ which is contained in a ring $R$ and $L$ is a one-dimensional eigenspace of $A$ to an eigenvalue $\lambda \in K$. Is then the eigenspace of $A$ in $R^2$ to $\lambda$ contained in the $R$-span of $L$?

N. Maneesh

@Silent Bro I am still at the sea shore of mathematics. I try problems. Trying problems may improve problem solving skills.

You can feel that. More you work out. more your feeling is. Analogous to gym.

Silent

yes:)

N. Maneesh

But some people may be god gifted. less than 1% of the population. Surely, i am not in that 1%. So I have to work hard :)

Rithaniel

2:58 PM

I'm of the opinion that those 1% still work hard, it's just that they've always thought of "hard work" as just normal work.

N. Maneesh

@Rithaniel You may be right.

user161005

What would be antonym for "conservative estimate"? Except "optimistic estimate".

Balarka Sen

3:16 PM

liberal estimate :3

Semiclassical

3:49 PM

Generous/charitable estimate might work depending on context

Though that’s not so far from optimistic

Just saw the phrase “Gromov boundary” in something I’m scanning through

Am now filled with dread

Balarka Sen

lmao

rip dude

anakhro

4:38 PM

It's very nearly the worst when things are "geometrically obvious", but no obvious rigorous proof seems to present itself.

Balarka Sen

I think depending on scenario it can say that the thing may or may not be trivial, even if geometrically obvious

anakhro

Definitely.

quallenjäger

anyone familiar with linux?

anakhro

@quallenjäger I am sure there are linux chatrooms, but what is your issue?

Gabriel Romon

@BalarkaSen do you like probability theory and mathematical stats ?

Balarka Sen

4:52 PM

@GabrielRomon I'm still learning but yeah I do

anakhro

@quallenjäger there is a room here: chat.stackexchange.com/rooms/26/dev-chat

quallenjäger

what is the name of the file with alias

.\bash_after?

or whats like again?

Ted Shifrin

5:04 PM

Greetings, @GabrielRomon. Long time! Howdy, @anakhro, a @Balarka, @quallenjäger.

Balarka Sen

Hi @Ted!

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Howdy, demonic @Alessandro.

quallenjäger

Hey Ted

Gabriel Romon

Hey Ted, yeah I've not been here in a while

quallenjäger

5:08 PM

@Ted

Has been a while

isaac9A

5:23 PM

If a function's image is equal to the codomain, we say the function is onto, is there a similar term for when a function's domain is equal to the preimage?

Alessandro Codenotti

The domain is assumed to be the whole set usually so domain and preimage are equal by definition

You can use total function vs partial function if you want to stress it might not be defined on the whole domain

isaac9A

@AlessandroCodenotti thank you!

Isa

I have a measure theory question about a borel measurable function

Alessandro Codenotti

5:39 PM

Just ask it

Ultradark

7:44 PM

Is there a name for the group with elements $\frac{1}{n}$ and inverse elements $n$?

christmas_light

https://math.stackexchange.com/questions/3229524/killing-vector-fields-in-reductive-homogeneous-spaces

If someone wants to help me understanding some basic stuff about killing vector fields on reductive homogeneous spaces :) sorry for the spam

Semiclassical

@Ultradark Are both 2 and 1/3, say, supposed to be elements of that set?

Ultradark

@Semiclassical 2 would be an inverse element and 1/2 would be an element

Semiclassical

Groups don't work like that. There's not a list of elements and inverse elements: Each element has an inverse, which is some other element.