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12:45 AM
If $M$ is an $R$-module, does $End_{\Bbb{Z}}(M)$ just denote group homomorphisms of $M$?
Well, then what does it denote?
Oh, sorry. Just saw the group homomorfisms
Yeah, every group endomorphism is also $\mathbb{Z}$-linear
yeah, $\mathbb{Z}$-homomorphisms, which is the same as group homomorphisms (since $\mathbb{Z}$-modules are the same things as groups)
1:02 AM
Q: tangent plane and tangent space coincide

orientablesurfaceLet $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be a smooth function. Suppose $f^{-1}(0)$ is a regular level set of $\mathbb{R}^3$. If $X_p\in T_p\mathbb{R}^3$ such that $X_pf=0$ show that for any $h\in C^{\infty}(\mathbb{R}^3)$ such that $h|_{f^{-1}(0)}=0$ we have $X_ph=0$. I was able to prove the...

1:17 AM
Q: Complex Analysis and Bernoulli Numbers from $\frac{z}{2} \cot (\frac{z}{2})$

user193319 Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers. I understand why t...

1:40 AM
@TedShifrin so the exercise is correct?
@Lucas ... depends on the defn. I am not following the room.
Hiya @TedShifrin ..math.stackexchange.com/questions/3872942/… is this even true?
1 hour later…
3:05 AM
@orientablesurface I don't know what you mean by your final sentence/question. What is the "tangent plane about $p$"? ... But, yes, the result is correct.
@TedShifrin 'Evening, Ted - I have a question for you. Consider a measure space $(X, \mathcal{B}, \mu)$. Let's say I have a nonatomic measure (i.e., for each set $A$ with positive measure, there is a subset $B \subset A$ with $0 < \mu(B) < \mu(A)$) and I have some set $E$ with measure $\mu(E) < 1/3$, and another set with measure $\mu(F) > 2/3 - \mu(E)$. Is there any way I can construct a set using just $X$, $E$, and $F$ which is guaranteed to have measure between 1/3 and 2/3 inclusive?
The intuition behind this intermediate step is getting me more than anything
Oh, and assume $\mu(X) = 1$
At its crux, it's a probability problem
At least I think it is
If it is helpful, $\mu(E) = \sup\{\mu(B): B \in \mathcal{B}, \mu(B) < 1/3 \}$.
I thought about considering the set $E \cup (X \setminus (E \cup F))$. But in this case, I'm only able to demonstrate that it has measure between $\mu(E)$ inclusive and $2/3$. Close, but not enough.
3:22 AM
Clarinet: My days of proficiency with measure theory are 40+ years in the past. I like integration theory, but I'm useless for this. Take some geometry/topology. :)
Lol, some day, I will
3:39 AM
Some psychologists have done studies of math skills that range over 40 years.
iirc, they compared it to remembering how to speak a foreign language
6:03 AM
Lol yeah I'm sure there are some that have studied math for 40 years, they still chose the tertiary studies equivalent of the short bus to major in
Maybe communications in algebra is becoming less about miscommunication. They retracted the Division algebra article
I mean if QAnon was the best false flag they could come up with to take heat off the Epstein problem I don't know if psychological warfare is having a good year
1 hour later…
7:31 AM
@Thorgott lmao
lEarnIng oN mAnIfoLd
What is the proverbial description of a topologist who went from broke to millionaire after they switched to data science?
Ans: Earning on manifolds
7:59 AM
I would like to clear up some notation convention that wasn't explained well in school. Often I would see something like $y=f(x)=x+1$. So what this says is the set $y$ is mapped to the set $x$ by the function, specifically the function named f, that is x+1
Is this right?
Also I've seen $y(x)$...what's that supposed to mean?
8:12 AM
It means the variable y is the same as the function f applied to the variable x, which is equal to x+1
y(x) means y is a variable that depends on the variable x
so y(x) means variable y depends on variable x, BUT f(x) doesn't mean the function x depends on x...that wouldn't make sense, the the function is more of an artificial construct which doesn't depend on anything. It does take the argument x.
am I looking at it right?
TL;DR would it be incorrect to say given f(x), the function f depends on x?
8:32 AM
yeah basically
9:02 AM
If $T$ is a linear transformation from $\mathbb{Z}_p \times \mathbb{Z}_p$ where $p$ is a prime >3, and $T$ satisfies $T^3=I$, then can I say that the minimal polynomial of the linear transformation divides $x^3-1$?
If $T$ is a linear transformation from $\mathbb{Z}_p \times \mathbb{Z}_p$ to $\mathbb{Z}_p \times \mathbb{Z}_p$ where $p$ is a prime >3, and $T$ satisfies $T^3=I$, then can I say that the minimal polynomial of the linear transformation divides $x^3-1$?
@BuddhiniAngelika sure
9:42 AM
Thanks @LeakyNun how can I justify it?
what's the definition of minimal polynomial?
The minimal polynomial m(x) is the monica polynomial of smallest degree such that m(T)=0
@BuddhiniAngelika do you know that it divides any such polynomial?
Yes, I got it now.
Thanks. :) Also, a matrix representing the above T should always be a 2 X 2 matrix isn't it?
T(v)=T.v for any vector $v \in (\mathbb{Z}_p \times \mathbb{Z}_p)$, all v's are 2 X 1 vectors, so then shouldn't T be having dimension 2 X 2?
10:07 AM
Q: Complex Analysis and Bernoulli Numbers from $\frac{z}{2} \cot (\frac{z}{2})$

user193319 Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers. I understand why t...

To calculate the $n$-th Bernoulli numbers, wouldn't I just take the $n$ derivative of $\frac{z}{2} \cot (\frac{z}{2})$, evaluate it at $z=0$, and then divide by $n!$?
Hi! Two questions are in need of urgent moderator intervention.
Who do I reach out to?
Two users have attempted to ask questions from an ongoing examination at my institution, in order to plagiarize the solution and compromise the sanctity of the examination.
I know the OPs.
I've requested for deletion by flagging one of them
flagged both now
10:27 AM
@user193319 $\frac z2\cot(\frac z2)=1-\frac{B_2}{2!}z^2+\frac{B_4}{4!}z^4-...$
@user91500 Yes, I know that. And if $f(z) = \frac{z}{2} \cot (\frac{z}{2})$, that means $\frac{f^{(2)}(0)}{2!} = \frac{B_1}{2!}$, $\frac{f^{(3)}(0)}{3!} = \frac{B_2}{4!}$, etc...right?
which gives me a way of calculating the Bernoulli numbers, right?
Actually, I should've written $\frac{f^{(1)}(0)}{1!} = \frac{B_1}{2!}$, etc....That is, the index of the derivative order should be one less.
In other words, more generally $\frac{f^{(n)}(0)}{n!} = \frac{B_n}{(2n)!}$, right?
1 hour later…
11:46 AM
does doing drug makes you learn math better?
12:05 PM
I have been on weed for a weeks and I found i learn rudin better than ever
@Stupidquestioninc I smoked like 4g a day for a year and it decimated my memory and it's only just recovering
everything in moderation
also waddup
I finally understood what blowing up was
when you eat a super hot curry
umm i don't want to be caught in school for smoking
... don't do it at school then? looool
12:08 PM
I think smoking will only get you bad consequences in the long run
I'm no expert though
Wasn't there some story about TSA catching math people on an airport who were talking about blowups?
Dude the explanation that would proceed after that would be hilarious
"Have you tried blowing up the real plane ?"
12:10 PM
erdos on amphetamine
@Astyx This reminds of The Dictator helicopter scene lol
I haven't seen this movie yet. Is it good ?
I think amphetamine helps you do math too but haven't try it yet
Anything by Sacha Baron Cohen is gold. Though this one is relatively okay compared to his other work
i only liked ali g
12:13 PM
Dude did you see Who is America
no man
The best part of Borat is that it caused Sacha Baron Cohen to be declared persona non grata in Kazakhstan
Watch it dude
ill pass haha
his type of comedy is too retarded for my taste
but also thats exactly what makes it funny
Lolol, I am a massive sucker for Political Comedy.
12:15 PM
why watch political comedy when i can watch US politics live
@Sayan have you heard of "The Thick of It"
Okay on the list
btw morales' party got elected in bolivia again
it's quite funny
12:16 PM
@BalarkaSen Facts. Trump just said that people shouldn't vote for Biden because Biden would listen to scientists.
after they got couped by CIA last year
memes golden
"just" meaning "one or two days ago"
The commons just voted through a bill that essentially establishes a UK Stasi
hopefully the Lords shit on it
@BalarkaSen Maybe it's another Pinochet waiting to erupt
12:42 PM
So, I'm told that $\mathcal{H}$ is a Hilbert space, but then asked to show that $\text{Ran}(T)$, where $T:\mathcal{H}\to\mathcal{H}$ is a bounded operator, is a subspace of $\mathcal{H}$ that isn't always closed. Isn't this contradictory because $\mathcal{H}$ is a Hilbert space?
Why is that contradictory?
Because any subspace of a Hilbert space is closed, right?
What's Ran ?
Why is that true?
The range of $T$
Because a Hilbert space is complete. I thought one of the main descriptors of a Hilbert space was "all subspaces of a Hilbert space are closed"
12:44 PM
@Rithaniel What's your favourite example of an (infinite dimensional) Hilbert space?
You're telling me a lot of things about Hilbert spaces but you still haven't given me an argument that all subspaces are closed
I may have misplaced my memory, because I immediately thought of $l^2(\mathbb{N})$, but then $l_0(\mathbb{N})$ is a non-closed subspace
What is the main "attractive" property of Hilbert spaces, then, @MikeMiller
The inner product
You can have an inner product space that isn't Hilbert, though. Like $C[0,1]$
Sure, but then you lose completeness, so you might have sequences that want to converge but don't.
I assume you're using the inner product from $L^2[0,1]$.
Every separable (inf-dim Hilbert space) is isomorphic to $\ell^2(\Bbb N)$, aka has a complete countable basis. You can use the basis (and the inner product) in arguments.
Many nice properties are shared between Hilbert and Banach spaces but the existence of a basis makes Hilbert spaces stand out.
12:49 PM
Sure but in an Hilbert space the inner product and the topology of the space are related in a good way
Okay, so completeness/existence of a basis are the the things you get from a space being Hilbert
(Completeness is part of the definition)
Yeah, so I should say "what is meant when you say a space is Hilbert"
I seem to recall there was something of importance about closed subspaces of a Hilbert space
I'm sure there's more intuitive good properties of closed subspaces of a Hilbert space, but the thing that comes to mind right now is that they are always complemented
Sanity check : do continuous functions on a closed interval with $L^1$-norm form a Hilbert space ?
12:57 PM
Wait, is it that Hilbert spaces themselves are closed?
But a non-Hilbert infinite dimensional vector space isn't necessarily closed, as with $C[a,b]$
Whatever the topology on $X$ is, $X$ is closed, so no
Not with $L^1$ I think, maybe parallelogram law doesn't hold. You have to check though to be sure.
Every finite dimensional subspace of a Hilbert space is closed, but this is true in any vector space, right?
@Astyx Maybe take $x$ and $1-x$ on $[0,1]$
I just need to show the set of continuous functions is a closed subset of $L^1([0,1])$, which is true because continuous functions on [0,1] are uniformly continuous (?)
1:14 PM
@Astyx No dude, approximate the indicator of [1/2, 1] by piecewise linear functions
Oh yeah
I'm dumb
But isn't it also enough to say that the norm is not coming from an inner product?
Just asking
I think it does come from an inner product because it is induced by the norm on $L^1([0,1])$ which is a Banach space
@Astyx Why so? A banach space norm need not come from an inner product
Ignore me, I'm saying nonsense
1:28 PM
(Currently trying to come up with a proof that $\text{Ran}(T)$ might not be necessarily closed)
(I am thinking about constructing an example bounded operator $T$ in $l^2(\mathbb{N})$ for which $\text{Ran}(T)=l_0(\mathbb{N})$, but I have a sneaking suspicion that there is an easier way)
1:49 PM
Measure theory question for you all:

Consider a measure space $(X, \mathcal{B}, \mu)$. Let's say I have a nonatomic measure (i.e., for each set $A$ with positive measure, there is a subset $B \subset A$ with $0 < \mu(B) < \mu(A)$) and I have some set $E$ with measure $\mu(E) < 1/3$, and another set with measure $\mu(F) > 2/3 - \mu(E)$. Is there any way I can construct a set using just $X$, $E$, and $F$ which is guaranteed to have measure between 1/3 and 2/3 inclusive, assuming $\mu(X) = 1$?
The intuition behind this intermediate step is getting me more than anything.
With respect to the natural action of $M_n(k)$ on $k^n$, where $k$ is a field, I am asked to "describe" $Hom_{M_n(k)}(k^n,k^n)$.
What exactly do you think is meant by describe? Completely characterize?
Since the natural action of left multiplication by matrices includes scalar multiplication, it's not hard to see that any $f \in Hom_{M_n(k)}(k^n,k^n)$ is also a linear map.
@Rithaniel Maybe this works, take $T : l^{\infty} \to l^{\infty}$, $T(\{x_n \}) = \{\frac{x_n}{n} \}$. Then consider $x_n = (1 , \sqrt{2}, \sqrt{3}, \cdots , \sqrt{n}, 0,0, \cdots )$ then, $T(x_n) = T((1 , \sqrt{2}, \sqrt{3}, \cdots , \sqrt{n}, 0,0, \cdots )) = (1, 1/\sqrt{2}, 1/\sqrt{3}, \cdots, 1/\sqrt{n}, 0, \cdots, 0, \cdots)$. Then $\{T(x_n) \}_n$ converges to $x = \{1/\sqrt{n} \}_{n}$. But there is no element in $l^{\infty}$ which maps to $x$.
Is there anything else that can be said about $Hom_{M_n(k)}(k^n,k^n)$?
Hence the $\operatorname{Ran}(T)$ is not closed.
Hmmm, okay, I think that works. All I need to do then is to show that $T$ is bounded
2:00 PM
@user193319 For every $y$, there's an $M\in M_n(k)$ such that $y=Me_1$ (where $e_1=(1,0,\dots,0)$)
@Astyx Yes, I used that to show that fact to show $k^n$ is a simple $M_n(k)$-module.
But are you saying that fact can also be used to describe $Hom_{M_n(k)}(k^n,k^n)$?
Oh, wait...a module homomorphism between two simple modules is either trivial or an isomorphism, right?
This means that any $f\in Hom_{M_n(k)}(k^n, k^n)$ is completely determined by $f(e_1)$
Which it does right, if $||x||_{\infty} \leq 1$, then for all i $sup(|x_i|) \leq 1$, but then $|x_i|/i \leq |x_i|$, so you have that it is bounded
Oh, so $Hom_{M_n(k)}(k^n,k^n)$ is cyclic?
in some sense?
@Astyx Does this mean that $f \mapsto f(e_1)$ is an isomorphism from $Hom_{M_n(k)}(k^n,k^n)$ to $k^n$?
Module isomorphism?
We can view $Hom_{M_n(k)}(k^n,k^n)$ as an $M_n(k)$-module, right?
What structure do you have on the Hom?
2:09 PM
Nevermind...$Hom_{M_n(k)}(k^n,k^n)$ has a ring structure, not a module structure.
So, every $f \in Hom_{M_n(k)}(k^n,k^n)$ is completely determined by $f(e_1)$...is there anything further I can say?
Well, you were right about some of the above things
it has both a ring and a module structure
hmm...so is $Hom_{M_n(k)}(k^n,k^n)$ isomorphic to $k^n$ in any sense?
Yes, in the sense you would most commonly expect given the notation
As modules?
Thorgott seems to suggest that $Hom_{M_n(k)}(k^n,k^n)$ has a module structure.
2:13 PM
I need new haters, old ones became my fans
Well, a k-module structure
Oh, so isomorphic as vector spaces.
Ah, okay. Thank you! Let me see if I can write this up.
I don't believe this
Maybe I'm missing something, but the only elements of $\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ should be homotheties, no?
2:21 PM
should be what?
scaling transformations
multiples of the identity
Cause an $f\in\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ commutes with scalars from $M_n(k)$, so is an element of $Z(M_n(k))\cong k$
Ahh, right
So what is preventing us from just sending $e_1$ to $e_2$ and extending this to a map on the entire space by the above mentioned uniqueness?
I don't understand...Are you guys saying that $f$ just acts by scalar multiplication?
2:29 PM
uniqueness doesn't give existence
aha, turns out I was wrong about something, tho
$\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ is not a $M_n(k)$-module
apparently non-commutative rings are that ugly
So elements of $Hom_{M_n(k)}(k^n,k^n)$ are clearly linear operators and therefore have matrix representations. So are you saying that a matrix representation commutes with everything in $M_n(k)$?
You generally need a Hopf-algebra to get a proper module structure on Hom spaces
but we have $\operatorname{Hom}_{M_n(k)}(k^n,k^n)\cong k$ as rings
2:33 PM
Wait...so I'm confused now...
it will probably help if you ignore everything I have said so far
(that holds in general, not just for this conversation)
hmm...okay...so back to square one then...What can we describe $Hom_{M_n(k)}(k^n,k^n)$ as?
@user193319 If $F$ is the matrix of $f$, then for every $M, x$, we have $f(Mx) = FMx = Mf(x) = MFx$, leading to $FM=MF$ so F commutes with every matrix
Okay, so $f$ lies in the center of all linear operators, which means that $f$ is a scalar mutliple of the identity.
2:42 PM
Can you do this if you don't impose $f(x+y) = f(x)+f(y)$ ?
Oh but that's induced by the $M_n(k)$ action
2:58 PM
@user193319 The series I've written is different. Check your question once again. Also note that $B_{2n+1}=0$
@Astyx yes, I believe you don't need additivity
3:33 PM
@Astyx A normed space which is completely metrizable is in fact complete to begin with
I don't remember the proof but you can google it
It follows that the $L^1$ topology on $C[0,1]$ is not even homeomorphic to the standard topology
That makes sense, cheers
4:10 PM
Is there an easy/standard example showing that Diff doesn't have equalizers? I think I have one, but it's not particularly pretty.
Hey @Tobias, do you have any idea about smooth representations?
if you're here lol
Not much
Now, algebraic ones I know about
tbh I don't think you need to know much to tell me I'm an idiot with the question I'm about to ask
Let's find out, shall we?
Suppose I have a rep $(\pi, V)$ of a (locally profinite but who cares) group G and define $V^\infty = \bigcup_{K\ \text{compact open subgroup}} V^K$ (the spaces fixed by the action of $K$). Is it obvious that $V^\infty$ is $G$-stable? Since every $v \in V^\infty$ is fixed by some compact open $K$, can't I just write $G = \bigsqcup gK$, so every $g^\prime \in G$ is $g^\prime = gk$ and then $\pi(g^{-1}g^\prime)$ fixes $v$
and then just.. do that for all the K's
(This $V^\infty$ is probably going to be the maximal smooth subrep of $V$)
4:23 PM
But your $g^{-1}g'$ is in $K$
So all you concluded was that the elements of any such $K$ fix that subspace
(which is certainly true)
(but not what I wanted)
yeah thanks lol
On the other hand, you don't want it to be fixed, you just want it to be stabilized
oh I read "G-stable" as "fixed by G"
4:25 PM
So you need to show that if $v$ is fixed by all elements of some open compact subgroup, then so is $gv$ for any $g$ (but not necessarily the same open compact subgroup)
that sounds easier
and/or possible
I assume the point here is that compact open subgroups will only have trivial smooth reps?
I'm not sure, I think the point is just to construct a smooth rep from a not necessarily smooth rep.
I'll regret this, but what's a smooth rep
Just a representation such that every $v$ is fixed by a compact open subgroup
4:31 PM
why's that called smooth
idfk I think it's equivalent to a continuity condition
like $G \times V \to V : (g, v) \mapsto \pi(g)v$ is continuous
where $\pi, G, V$ are the obvious notations
there's also "admissibility" which is that the spaces fixed by compact open subgroups are finite-dimensional, but I don't think I'd have guessed that by the word admissible
it's all just made up to confuse me
I mean, people call everything admissible when it's nice
but I don't see the connectino between this and classical smoothness
shrugs locally profinitely
like, uh, if $G$ is Lie and $G\rightarrow\operatorname{Aut}(V)$ is smooth, then is that also smooth in your sense?
@Thorgott Lie groups have very few compact open subgroups
4:37 PM
I think a lot of examples of locally profinite groups are p-adic Lie groups
"p-adic Lie group"
don't ask me I just say what the book says man
@Thorgott What else could you have possibly meant when you said Lie group
p-adic Lie group in the sense that the groups have the structure of a p-adic analytic manifold obviously
p-adic quantum mechanics is a collection of related research efforts in quantum physics that replace real numbers with p-adic numbers. Historically, this research was inspired by the discovery that the Veneziano amplitude of the open bosonic string, which is calculated using an integral over the real numbers, can be generalized to the p-adic numbers. This observation initiated the study of p-adic string theory. Another approach considers particles in a p-adic potential well, with the goal of finding solutions with smoothly varying complex-valued wave functions. Alternatively, one can consid...
who even considers groups that aren't given by the classifying etale topos of a p-adic scheme
4:54 PM
A primary ideal Q of a commutative ring Q is an ideal such that if $xy\in Q$, then $x\in Q$ or $y^n\in Q$ for some $n>0$.
Why is the definition of a primary ideal not symmetric in x and y ?
it is , is it not
since the ring is commutative
I always get the urge to play Skyrim when I hear the music, and as soon as I start playing Skyrim I realise that I find it kinda boring and what I really wanted was to literally be there
and then I cry in the foetal position
Oh ok I get it
I can't play skyrim cause it gives me motion sickness
5:12 PM
Primary ideals are amazing
Is there an easy proof of invariance of domain for smooth functions?
My favorite kind of commutative algebra
What is your version of invariance of domain
smooth injection U->R^n with U open in R^n is open map
Ya it's definitely easier than topological invariance of domain
Hm, maybe not.
Not sure, sorry.
how about an easy reason for why there is no smooth injection $M\rightarrow N$ if $\dim M>\dim N$?
5:17 PM
Find a regular point by Sard's theorem
The preimage is a $\dim M - \dim N$ dimensional manifold
In particular more than a singleton.
What's the precise statement you're asking about? Is it

Let U be an open subset of R^n given induced smooth manifold structure, and f:U-->R^n be a smooth injective map, then with induced manifold structure f gives an isomorphism of smooth manifolds between U and f(U)?
@Balarka this sounds incredibly silly, but how do we preclude that $M$ doesn't inject as a measure zero subset
@NeverEnoughTime yes
What? I don't understand. Sard's theorem is the statement that there is always a regular value.
Why are you thinking measure zero etc at all.
take the zero map R->R, the differential is everywhere zero, so no point is regular
of course there are regular values, but they have empty preimages, so are useless
Oh fine. Yeah I don't know you can argue somehow
5:25 PM
Sard's says that if f:R^n->R^m is smooth enough, the image of the set of critical points has measure zero. For your argument, we want a regular value with non-empty preimage, so the existence of a regular point. To get that from Sard, we need that the image has positive measure.
Yeah I don't care haha
You can argue
Sard's works to show that there is no surjection N->M
Find some measure theory theorem which says some preimage has positive measure lol
This is a waste of my time
does anybody know how to strike through an integral in latex?
doesn't work
There's $\fint$?
Oh that uses a package
5:35 PM
package.. yeah
Apparently the esint package
but what package?
It's not a straight strikethrough though
I want a flat line
wtf why is that so difficult
Did you try googling it lol, I find lots of answers doing that in the first results
5:37 PM
I did
A: Average integral symbol

Heiko OberdiekThe following example defines macro \mint{<symbol>} (\limits|\nolimits|\displaylimits)* _{...} ^{...} The first argument is the symbol that is put in smaller math style in the middle of the integral symbol. Then a limits specification follows, any number and order. The last one is used. Then ...

You saw that?
I was hoping for something easier..
whatever. that'll have to do
If something easier existed, why would it have 11 upvotes lol
Someone would write a comment "Wow, this is retarded, you can do this in 1 line by blah"
You can make a terrible version of it by hand
as in, theres gotta be a package out there
5:39 PM
you suck at this
Is that not what you want?
lol.. what do you think?
the answer on SE worked fine
5:40 PM
now you screwed it up again
keep trying. I'll check back in in an hour
Not sure if you're being a dick on purpose
you were playing with the integral weren't you..? you even almost had it at some point
6:10 PM
If $R$ is a commutative ring and $M$ and $R$-module, does the following define an $R$-module structure on $Hom_{R}(M,M)$? $(r \cdot f)(x) := f(rx) = rf(x)$?
$R$ being commutative being integral here
Cool. Thanks!
@Balarka oh, it's actually just a consequence of the rank theorem
6:19 PM
Salut @Astyx. Hi, @Lucas. Did you ever settle the definition in your problem?
Salut Ted
@JoeShmo: When I've done "by hand" LaTeX fudges (e.g., making a transversality sign), I often end up using rlap and llap.
hey @Ted. yeah, $\mathrm{Div}(f) := \{p \in \mathcal{P}(\mathbb{F}): \exists q\ (f = pq)\}$
Hmm, so that includes non-proper divisors, so your complaint was right.
6:21 PM
I was being dumb. The exercise does not imply that every space is cyclic.
Oh, I guess I didn't pay enough attention.
What does the vertical bar mean in a lot of maths that I see?

It typically looks like this: $f(x)\vert_0^\infty$
Actually, if you have a polynomial $p = f_1 ^{\alpha_1}\cdots f_k^{\alpha_k}$ and an operator primary decomposition $V_{p(T)} = \bigoplus\limits_{j = 1}^k V_{f_j^{\alpha_j}(T)}$, it's not hard to show that, given $v_j$ with $m_{v_j} = f_j^{\alpha_j}$ (also easy to show that exists), $\sum\limits_{j=1}^k v_1$ has min poly $p$
@Sal: $f(x)\big|_a^b$ is shorthand for $f(b)-f(a)$.
6:26 PM
@Lucas: Oh, $\sum v_j$ ?
@Sal you may also see things like $f\big{|}_A$ which means the restriction of the function $f$ to the domain $A$.
In this context, it's what Ted said
That was pretty clearly a calculus context, @Lucas :)
@TedShifrin yeah, since the minpolys of $v_j$ are each two coprime
I was just checking, Lucas. You had a typo.
oh! lol
this linear algebra class is not fun at all
every theorem looks so damn arbitrary
5000 preliminary lemmas which do not look anyhow related. that's probably what happens in research, though. you try everything
can you confirm, @Ted?
6:30 PM
Well, I'm not a fan of that teaching style.
Confirm what?
those apparently unrelated and arbitrary, not obvious lemmas do show up while in math research?
our professor said that it's a math thing. in physics, things look less arbitrary since reports have all the process to get the results. in math, people do a clean up and the dirty work disappears
No, in research, you do things as you stumble upon them. You don't stack up a pile of preliminary lemmas. That's a style of book-writing and teaching which, as I said, I do not like.
Often in research, the lemmas don't really become lemmas until you write up the final thing and realize that some part of the proof is more clear as a lemma on its own
Sometimes in research, people use the "method of wishful thinking" and assume a result (as reasonable) to see if it gets them an interesting result. If it does, then after they come back to try to justify it.
Wews, @TedShifrin, I saw you were in the chat and I just wanted to say hi.
6:35 PM
Hey there, @Sha. Nice to see you!
And sometimes, that wishful thinking becomes a conjecture and several papers are written assuming it. And then someone goes and disproves the conjecture :)
Also, olas @Astyx
Oh wow, a wild Sha has appeard
How are you ?
@Tobias, yes, there are instances of that :P
6:36 PM
(glad I was not the one to write those papers)
I'm doing alright! Am reading in Atiyah & MacDonald atm
alright, thanks @TedShifrin I ended up doing it by hand..
How are you's
Wow, @Sha, a far throw from all your physics days.
Ye, I sort of almost quit physics, tbh.
At least this year I'm only taking math courses.
6:37 PM
@Thorgott Sorry for interrupting but how exactly? The rank theorem needs the jacobian to have constant rank in a neighborhood, how do you guarantee that for a smooth function $f: M \to N$, dim(M)> dim(N) ?
every smooth map has constant rank somewhere
cause the set of points where it has maximal rank is open
Two points to Thor.
@Sha: I remember the young and innocent Sha. :)
Oh yes nice nice.
@TedShifrin xD
@Astyx: Is this like trying to land on a Möbius strip?
6:58 PM
Man I get distracted by the chat too much, I had set to do some AG, I ended up doing some functional analysis and smooth manifolds.

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