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00:00 - 20:0020:00 - 00:00

12:00 AM
@Jacksoja I forget algebra all the time
 
idk i think most of them dont have tests and stuff
 
Right maybe those courses run in the spring
 
i think no one gets grades iirc from the open house
 
FA was offered this spring
 
12:00 AM
u two should switch to arithmetic geometry
 
@Balarka idk I've never read that
 
pls frend
 
dont insult me like this
 
@BalarkaSen explain that course desc and I will
 
12:02 AM
@ÍgjøgnumMeg I'm wondering because then the fundamental theorem of arithmetic at the level of ideals will be a primary decomposition, I guess
@RyanUnger "adelic polarization" o my
 
@Ryan do u know anyone else going besides me lol
 
I met a guy Will (?) at UT Austin
He wasn't sure
I didn't go to the open house so no
 
Will he go then?
 
@TedShifrin okay right! :)
 
boo
 
12:04 AM
@RyanUnger okay good am not alone on this
 
oh i think i met him somewhere but i didnt talk to him
 
each time , i need to take last years book to re read haha
 
He's a pure PDE guy
 
@ÉricoMeloSilva You'll prove the Collatz conjecture by doing geometric measure theory on anabelian schemes one day
 
Balarka what the hell is a diagonal Green's current
because when I google that I do get GMT
 
12:04 AM
i die
 
what is a current anyway (serious q)
 
but that course is about arithmetic geometry
 
sounds like thinking of Green's functions as currents
 
Like a submanifold but messed up
 
Is there a math map ? something like what courses to take in what order? or is it not something very important ? @TedShifrin
 
12:05 AM
You should first think about integral currents
 
I would love such map for algebra and analysis
 
Which are integral varifolds with an orientation attached
 
@Jacksoja Yeah it starts like scheme theory -> ...
 
currents are linear functionals on differential forms, so they're either (generalized) submanifolds or locally $L^1$ differential forms.
 
they're the misunderstood bad boy analogue of submanifolds
 
12:06 AM
Ah I see
 
@BalarkaSen you joke too much so i cant tell when you are being serious
is that a fact Ted?
 
that's what advisers are supposed to be for, @Jacksoja, although I found that most of them don't do a great job
 
Integral varifolds are just rectifiable sets with some "density" to be thought of as the surface overlapping on itself
 
Is what a fact?
 
@TedShifrin so true ! since day 1 i feel like am taking topics out of order
what balarka said
i should do scheme theory first
I have no idea what that is
 
12:08 AM
That is a joke
 
I knew it
I so dislike that guy balarka
 
@BalarkaSen Think about an immersed surface that's not too bad, like some curve with some self-intersections
 
@RyanUnger Mhm
 
you can associate to that some rectifiable set and hence it's Hausdorff measure
 
@Jacksoja hurt
 
12:09 AM
but this doesn't tell you about the multiple points
 
@Ryan our committees will presumably end up looking v similar lol
 
we can tag team it
 
word up
 
@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
 
Gotchu
What do you do with these
 
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
 
That's interesting
 
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
 
so they are "limits of submanifolds" in some sense?
 
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
 
That's a lot of codimension
 
12:17 AM
Right, so (area-minimizing) minimal hypersurfaces in manifolds up to dimension 7 are smooth
 
balarka should read simon's notes
 
There's a conjecture that this is true in all dimensions generically
 
Oh because those are solutions to some weird PDEs which have solutions with only varifold regularity?
 
There's an explicit cone in R^8 which is stable minimal and has a singularity
 
@ÉricoMeloSilva only if you read SGAI
 
12:19 AM
iunno what that is
 
grothendieck
 
oh u mean SGA I
 
sry
yes
 
I thought it was EGA
 
that's also a thing
 
12:20 AM
so this is what y'all fuckers do because minimal surfaces don't have enough regularity eh
madmen
 
Well the madman paper is Schoen--Yau 17
 
@BalarkaSen simon's notes are probably like way easier but ok :(
 
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
 
no idea
i need 2 learn min-max stuff my man
 
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.
 
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
 
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
 
sounds interesting
 
Lohkamp has also shown that if you "blow up" along the singular set you get a Gromov hyperbolic space
 
O_O
 
he studies singulartities as the Gromov boundary of this space
no one knows if what he's doing is right
 
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?
 
ye those
 
tnx
 
@ÉricoMeloSilva did you upload your vaccination records
 
not yet
 
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
 
i think that's the only thing i havent done yet and im too lazy to figure it out
 
You sent them your picture?
 
im also still in school so im busy and just never have the energy
ya
 
12:44 AM
@BalarkaSen why is Nakayama's lemma important
 
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
 
Proposition 2.8 in AM?
 
maybe i dont have AM with me right now but this is a corollary of whatever form of Nakayama's lemma they prove
 
yes
 
$m$ is the Jacobson radical of $A$ after all
 
12:47 AM
right
 
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
 
Somewhere in Vakil there's a 'geometric Nakayama lemma' which makes things more geometric
 
@BalarkaSen ok I buy it
I'll worry about geometry later haha
 
It's an oft used tool, specifically when the ring is local.
 
well im destroyed, cannot write any further tonight
 
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 $m = n$.
 
@BalarkaSen umm but that $\cong$ just means iso right
not homeo in any sense
 
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
 
lol its not often you hear
zariski topology and homeomorphic in the same sentence
 
same for R^n so why not
 
uh
 
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
 
I think you can just argue by Krull dimension, it's defined topologically so it's preserved under homeomorphisms
 
How is it defined topologically? Isn't it defined by the highest tower of subvarieties?
 
balarka stop im still on modules (definitely review)
 
@BalarkaSen longest chain of irreducible closed subsets
 
@loch Oh I see, right.
Interesting
 
1:39 AM
well there goes your claim
even I understood that one
 
i claim youre a nincompoop
see if you can break that
 
deletes account
 
how many years of suspension did you have to go thru
 
uh how old are you
 
i lost count at some point
im 57
 
1:44 AM
last time we spoke you were 13
damn, 44 years
 
airhorn
r u still an anarcho conservative or did you switch ideologies
 
What's anarcho conservative
 
idk man
milo yiannopolous style
 
haha
is he still ostracized?
 
i think so
did u watch zizek vs peterson lmao
 
1:49 AM
I haven't watched a Peterson video in over a year
 
damn ur missing out
 
I heard he's gone crazy
 
its too funny
 
the Brouwer thing is an exercise in ch 2
 
yep
 
2:06 AM
@BalarkaSen do you have a favorite part?
 
2:13:31
 
@BalarkaSen have you read the book?
 
no lol
 
it seems marketed towards incels
 
yeah thats pretty obvious
he's the father of basement dwellers
 
 
3 hours later…
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.
 
@BalarkaSen prove the equivalent statement about rings by quotienting by a max ideal and reducing it to a statement about vector spaces?
 
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.
 
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?
 
Topological spaces not schemes!
Wrong category
 
Ahh, ok
Then I guess Krull dimension is the right way to go as loch said!
 
It's quite interesting in that light that Krull dimension works
Never thought of it as a purely topological invariant
 
It makes sense though
 
5:49 AM
Yeah
 
So that the dimension of $\mathrm{Spec}R$ as a scheme (or top space) agrees with its Krull dimension as a ring
 
Not dimension as a topological space in any dimension theoretic sense, is it?
 
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
 
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
 
Should be the length of the longest chain of irreducible closed subsets
 
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
 
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
 
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\\
 
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 $P$
 
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 $B=P^tAP$
 
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?

Sam WilliamsSo 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...

 
Thank@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

djangologyGiven 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...

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?
 
@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
 
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 $x_n^i$?
 
$x_n = \sum x_n^i u_i$
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.
 
8:13 AM
ok
 
user131753
How do you know that $\lim x_n^i$ exists @LeakyNun?
 
because $\lim x_n$ exists
projections are continuous
 
user131753
Yes by projections, right?
 
right
 
user131753
@LeakyNun Oops, I see you commented earlier.
 
user131753
8:16 AM
Ok. Thanks @LeakyNun.
 
9:02 AM
The answer above is d, right?
I took $A$ rational numbers and $B$ irrational numbers
oh! that won.t work!
 
@Silent the answer is B
 
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.
 
what if there is no $b \in B$
 
oh!
So, if $B$ empty then $C$ empty as well?
 
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
 
9:17 AM
Its a multiple select question, so its ok
 
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
 
I gone head and messed this up, does anyone see a better simplication here: math.stackexchange.com/a/3229363/60900
 
10:22 AM
Mornin'-ish alal
all*
 
 
1 hour later…
11:33 AM
Good evening :)
 
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
 
@Silent but $\Bbb Z$ is not a multiplicative group
 
oh
So, that statement is false. Thank you, @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
 
yeah
 
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)?
 
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. ;)
 
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?
 
An introduction to Lie groups and the geometry of homogeneous spaces
 
1:41 PM
what does terms like $x*\int f(x)dx$ mean? Can I pull the x inside the integral?
 
@user170039 (page 79, I forgot to say sorry )
 
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.
 
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.
 
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?
 
What's the name of the theory whereby you study manifolds according to the smooth fields that can be defined on them? I forgot.
 
2:28 PM
I first tried working with $(123)$ of $S_3$, but $S_3$ has two elements of order 3
 
You're familiar with Cauchy's Theorem, right, Silent?
 
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 $L$?
 
Also, if an element $x$ has order n then so does $x^{-1}$, in fact $gxg^-1$ as well
 
Think about $\langle n\rangle\leq G$
 
@Rithaniel that there exists element with order of each prime divisor in $G$? yes.
 
2:33 PM
@Silent Correct, but if there's a unique element of order $n$, then...
 
Then Lagrange's theorem?
 
No.
 
@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
 
:)
Well, $Z(G)$ has order more than $1$ because it contains both $1$ and $x$.
 
Hmmm, I thought it was going to be D.
 
2:37 PM
@Rithaniel me too!!
 
You can't guarantee that there are more than 2 elements, because $x$ might have order $2$.
 
I see.
@BalarkaSen you are talking about center, right?
 
Yes.
 
Wait why D? Seems too restrictive given the loose condition.
 
Not to be repetitive but: Can G\F denote a quotient group? I’m reading a (Physics) paper that seems to be doing so
 
2:39 PM
Depends on left or right action but yeah
 
Mmkay, just wanted to check
 
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
 
@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
 
Well, my thinking was that if there is a unique element of order $n>1$ then $n=2$, though perhaps that isn't true?
 
Wow! Balarka's example seems to clarify all our doubts!
 
2:42 PM
Hmmm, well fair enough.
 
@user76284 you got it, right?
 
@KarlKronenfeld Thanks.
 
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?
 
Well, that's true. If $x$ is an element of order $n > 1$ then $x^{-1}$ is as well.
 
@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).
 
2:47 PM
The only way to get uniqueness is $x = x^{-1}$, which forces $n = 2$.
 
(See I was about to start going a roundabout way. Just looking at the inverse is much more efficient)
 
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 $L$?
 
@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.
 
yes:)
 
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 :)
 
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.
 
@Rithaniel You may be right.
 
What would be antonym for "conservative estimate"? Except "optimistic estimate".
 
3:16 PM
liberal estimate :3
 
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
 
lmao
rip dude
 
4:38 PM
It's very nearly the worst when things are "geometrically obvious", but no obvious rigorous proof seems to present itself.
 
I think depending on scenario it can say that the thing may or may not be trivial, even if geometrically obvious
 
Definitely.
 
anyone familiar with linux?
 
@quallenjäger I am sure there are linux chatrooms, but what is your issue?
 
@BalarkaSen do you like probability theory and mathematical stats ?
 
4:52 PM
@GabrielRomon I'm still learning but yeah I do
 
@quallenjäger there is a room here: chat.stackexchange.com/rooms/26/dev-chat
 
what is the name of the file with alias
.\bash_after?
or whats like again?
 
5:04 PM
Greetings, @GabrielRomon. Long time! Howdy, @anakhro, a @Balarka, @quallenjäger.
 
Hi @Ted!
 
Howdy, demonic @Alessandro.
 
Hey Ted
 
Hey Ted, yeah I've not been here in a while
 
5:08 PM
@Ted
Has been a while
 
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?
 
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
 
@AlessandroCodenotti thank you!
 
Isa
I have a measure theory question about a borel measurable function
 
5:39 PM
Just ask it
 
 
2 hours later…
7:44 PM
Is there a name for the group with elements $\frac{1}{n}$ and inverse elements $n$?
 
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
 
@Ultradark Are both 2 and 1/3, say, supposed to be elements of that set?
 
@Semiclassical 2 would be an inverse element and 1/2 would be an element
 
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.
 
except 0
 
7:55 PM
E={1/n} and I={n}
 
@skillpatrol If zero doesn't have an inverse in your group, it's not a group
@Ultradark If you take the elements of a group, and write down all their inverses, you'll just get the same set of elements back again.
@Ultradark Again, no: A group is not a list of elements and inverse elements.
There is only one set associated with a group, and that's the set of group elements. The inverses of these elements are not distinct from this set.
The set of inverses in a group is the same set as the group elements themselves.
 
1/2 times 2 equals the identity element: 1
 
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