12:00 AM
Should I call my former Analysis TA who's now an Assistant Prof. "Professor Last Name", when emailing them to touch base with something research-related?
We're on great terms, and he was a great TA to me. But I've only ever called him by his first name ... so it'd seem a little strange to call him Professor Last Name -- but I don't want to offend anyone either.

Nah, it's fine to call him as you always have. You can comment (humorously?) that you were tempted to call him Professor to celebrate his new status. :)

@TedShifrin Ok, cool -- thanks :)
@TedShifrin Can I ask you something?
I reread some of my writings -- journaling, sorta -- from about two years ago. I've taken quite a different direction since then for research. But I'm revisiting those ideas now, about doing abstract linear-algebraic research, such as functional analysis, operator theory, representation theory. Last night, I found an interview with a pretty famous prof. who remarked that Operator Theory was a dead field and that he urges his PhD students to enter into computational math / numerical analysis
@TedShifrin do you think it's an accurate thing to say that Operator Theory is kinda dead?

7 hours later…
7:14 AM
I wanna learn about fiber bundles. What's a good textbook for learning about that which would have simple examples?

2 hours later…
9:06 AM
@Joanna a lot of people study operator algebras in very different contexts

9:24 AM
Operator algebras are key in formalizing quantum mechanics I'm pretty sure

Hi everyone. One question. Is it important to pick good uni for undergrad maths degree for aspiring mathematician? I think of changing unis atm.

10:13 AM
@Astyx I don't know anything about the physics, but I've been told that the AQFT people also think a lot about operator algebras
But there's also people who apply operator algebras to group theory or rep theory by studying the group C*-algebras associated to a group, this has been generalized to semigroups, étale groupoids, etc.
Connes embedding conjecture was settled recently by quantum computing people
Doesn't look like a dead field

"étale groupoids"
I never wanted to know these two words can be combined that way

I'm forgetting everything. How do you prove that for a sufficiently nice $G$-representation on a vector space $V$, $k[V]^G$ is a finitely generated $k$-algebra?
Ah, alright. There's an operator $k[V] \to k[V]^G$ which mimics averaging that one can cook up which is surjective, and so $k[V]^G \cong k[V]/I$ for some ideal $I$
That isn't quite it

10:31 AM
Isn't that due to Hilbert

Yeah but it should follow directly from having that operator
need to use the basis theorem somewhere

(|G|,|k|)=1?

@Thorgott lol, they are the nice case, general groupoids are too hard apparently

Yeah k is algebraically closed characteristic 0

Then surely you can drop "sufficiently nice"

10:34 AM
I think Hilbert did this for subgroups of $GL_n(k)$; that operator does not exist in general

$\dim V < \infty$?

I'm blackboxing that of course

Sure, there are no infinite dimensional vector spaces

what does $k[V]$ mean and why isn't its dimension $|V|$

polynomial ring dude

10:41 AM
I have forgotten everything this is horrible

nvm

I'm confused if I have. The point is $k[V]/I$ is Noetherian? It's unclear to me where exactly I'm using Hilbert's basis theorem, I guess that is it.

so if $k = \Bbb C$ and $V = \Bbb C$ then $k[V] = \Bbb C[x]$?

I'm just using polynomial rings are Noetherian
Yes bro it's just the ring of regular functions over $V = \Bbb A^n_k$, jesus christ
Stop playing so much chess it's burning a hole in your brain
But nuts, am I correct? Is it this trivial?

oh I thought it's the group ring lol

10:51 AM
Hello people ! I have a vector space $\{V, +, *\}$ and I add to it an operation, $\circ$, such that $\circ$ is associative, distributes over $+$ and $(\alpha * x) \circ y = \alpha * (x \circ y) = x \circ (\alpha * y)$. Does someone knows the name of such a structure ?
My book is old school and calls this a "linear algebra"

then I think I read some proof like this: let $R$ be the subring generated by the coefficients of $\prod_{g \in G} (x - gt)$ for $t \in k[V]$, then $R$ is f.g. alg. and $k[V]^G$ is integral over $R$
something like this

The example I'm studying is linear transformations from $X$ to $X$ and $\circ$ operation corresponds to composition

or rather $k[V]$ is integral over $R$ and $R \subseteq k[V]^G$

ah hm

@Astyx salut

10:54 AM
I had an answer that I have been writing for about 6 weeks now, and had saved it as a draft in the "ask a question" tab. Today I found that my draft had disappeared. Is there any way to retrieve old question drafts?

rip

Let $G$ be a compact group acting linearly on $\Bbb C^n$. Define $\Bbb C[x_1, \cdots, x_n] \to \Bbb C[x_1, \cdots, x_n]^G$, $f \mapsto \int_G g \cdot f d \mu$, where $\mu$ is the Haar measure. Is this well-defined?

@Thorgott This is exactly what i wanted, an associative algebra. Thanks a lot !

Seems to be, right?

10:58 AM
why do you need Lie

Thanks, I always have the Riemannian volume form in mind
But averaging over polynomials under a group acting linearly should still give me a polynomial, yeah?

I mean, isn't this just averaging coefficient-wise

yeah

So long as you have the right topology on the space of polynomials, so that polynomials parameterized by a compact group have bounded degree
@BalarkaSen k[V]/I is obviously finitely generated because k[V] is

yeah im just so confused, where the is the hard work going?
i seem to have constructed both the Reynolds operator and given the proof

11:03 AM
Is your averaging map really a ring map?

the action on polynomials being a ring action, yeah, right?

Take Z/2 acting on C[x,y] by swapping x and y. The averaging map sends x to (x+y)/2, and then sends x^2 to (x^2+y^2)/2. So...
Seems additional work is needed to construct a ring map

oh ok
wait no what
Ahh, it's not going to be a ring map. I see
For finite groups it's most definitely just averaging. The map $R \to R^G$ is going to only be an $R^G$-linear map.
That's why everything I said is nonsense

Yep seems correct to me
Remember the example of C[x,y] with negation action of Z/2 on x and y
The quotient is a cone, so requires three generators
So your map can't possibly always be a surjective ring map!

right...
OK, so all I have is an $R^G$-linear projection $\rho : R \to R^G$. Let's see how to reconstruct the proof of finite generation of invariants.

11:15 AM
@LeakyNun Salut

For any ideal $I \subset R^G$, $IR \cap R^G = I$, since if an element of $R^G$ is an $R$-linear combo of elements of $I$, applying $\rho$ to both sides gives that the same element is an $R^G$-linear combo of those elements of $I$ (since $\rho | R^G = \text{id}$ and $\rho$ is $R^G$-linear); this immediately gives that if $R$ is Noetherian so is $R^G$
If $I_1 \subset I_2 \subset \cdots$ is an increasing chain of ideals in $R^G$, $IR_k = IR_{k+1} = \cdots$ stabilizes eventually, then just intersect back with $R^G$
That's not enough for $R^G$ to be a finitely generated $k$-algebra if $R$ is as such
It's definitely not true but what's an example of a Noetherian $k$-algebra which is not a finitely generated $k$-algebra lol
@LeakyNun Do you know an example

@BalarkaSen is $\Bbb C(X)$ noetherian?

that's a field?

It's a field

we're a field

11:28 AM
$k[[x]]$, something like this
That's a Noetherian ring
Not f.g. $k$-algebra though, because powers of $x$ are an infinite collection of lin indep things

@BalarkaSen That doesn't seem like an argument
That shows it's not a f.d vector space

you're right

I guess your point is really that a finitely generated k-algebra is countable dimensional while the power series ring is uncountable dimensional
since it's the ring of functions $k^{\Bbb N}$ with convolution product
$|k^{\Bbb N}| \geq |2^{\Bbb N}| = \mathfrak c$

yeah

nice

11:38 AM
any finitely generated $k$-algebra has a countable basis as a $k$-vector space, that is the right argument

even a countably generated k-algebra i guess

Yeah you're right
I suppose the point is $k[x_1, \cdots, x_n]^G \subseteq k[x_1, \cdots, x_n]$ is graded
If $A \subseteq k[x_1, \cdots, x_n]$ is a graded subalgebra which is also a Noetherian ring, it should be a finitely generated $k$-algebra, look at the ideal $A_{\geq 1}$ which is finitely generated...

Why don't you do the example of the cone to see where your generators come from
this seems good though

11:56 AM
@MikeMiller I'm confused about your cone example. $k[x, y]^{\Bbb Z_2} = k[x + y, xy]$
That's just Newton's theorem
It's 2-generated
You probably meant $(x, y) \mapsto (-y, x)$ or something
Rotation by 90 degrees

I gave two examples, the first was the swap map
The second was $(x,y) \mapsto (-x,-y)$

Ah OK

Ring of invariants should be minimally generated by $x^2, y^2, xy$

Yeah that's correct

So where did the ideal these generate come from

12:00 PM
It's the ideal of all positive graded invariant elements of $k[x, y]$, but that's not saying much

You mean $\geq 2$-graded?

Yeah because there's no invariant element in grade 1

I missed the word "invariant" above

This is some induction fact, if $f$ is a non-constant invariant element, $f = x^2 g + xy h + y^2 k$ because $(x^2, xy, y^2)$ is the ideal of all positive graded invariant elements, and $g, h, k$ have degree strictly less than that of $f$.
Maybe it should be possible to read off the quantitative Hilbert's invariant theorem from here, let me think.

Is the argument in general supposed to be: "Take the ideal generated by invariant polynomials; it's finitely generated; in lowest grading it agrees with the invariant polynomials; induct upwards to see that it agrees with invariant polynomials in all gradings"
?

12:05 PM
Yeah that should be it

The induction idea was clever
Nice

Well, OK, the $g, h, k$ need not be invariant elements. But apply Reynold's operator to get $f = x^2 \rho(g) + xy \rho(h) + y^2 \rho(k)$
So WLOG they are invariant elements
in strictly lesser degree

Aha

seems solid

This seems like you're telling me that the generating set for $k[V]^G$ is going to be homogeneous
I'm feeling skeptical
For instance, if you get some $V$ and $V'$ for which you get homogeneous generators in degrees 2 and 4, then $V \oplus V'$ should have non-homogeneous generating set i think

12:07 PM
Yeah that's what I think this should relate to the quantitative version, Noether proved that the generating set is of size at most $\binom{n + d}{d}$ where $d$ is degree of the largest homogeneous element or something

Aha

@MikeMiller Hmm OK I need to think more

I think my idea might work with your induction approach though
Like choosing an arbitrary (possibly non-invariant) generating set and proving that it must in fact be generated by invariants

Fucking hell how did I get here, all I wanted to see was that if $G$ is a compact Lie group acting linearly on $\Bbb R^n$ and $\zeta_1, \cdots, \zeta_n$ are generators of $\Bbb R[x_1, \cdots, x_n]^G$ then $(\zeta_1, \cdots, \zeta_n) : C^\infty_0(\Bbb R^n) \to C^\infty_0(\Bbb R^n)^G$ is surjective
Just Malgrange things yaknow

Oh does this all work well over R
I guess the argument probably doesn't use alg closedness

12:21 PM
yeah doesnt seem like it

12:33 PM
@BalarkaSen I was sure this was obvious until it wasn't
This is awful
You should write $\zeta_k$ for the last term probably

You're right
It should be an easy Malgrange argument but I have been putting it off
$G$-equivariant germs unfold to germs on a fundamental system of $G$-invariant germs
Beautiful
Pure geometry

2 hours later…
2:37 PM
Hey @BalarkaSen ! If we have an integer $a$ and when we divide it $b$, we have the following relation $$a = bq +r$$ Where $q$ is the quotient and $r$ is the remainder. But how can we prove that $r$ will always lie between $[0,9]$

that ain't it

Means $r$ can be any integer ?

well, it does depend on $b$
but there's no reason whatsoever for it to be constrained in the interval from $0$ to $9$
in general, $r$ will be an integer between $0$ and $b-1$ (both included) and any of those cases can be realized

And if $a \lt b$ can we talk of negative remainders ?

@Knight on the number line, multiples of $b$ form dots that are $b$ distance away from each other (for social distancing measures let's require $b$ > 1.5 metre), so wherever you are on the number line you can walk leftwards for less than $b$ steps to find a multiple of $b$
and less than $b$ means at most $b-1$

2:47 PM
I know it is unconventional, but if we write $$2 = 5\times 1 -3$$ can we say $-3$ is the remainder ?

In optimization technique, if we use integer indicator variables in the objective function multiplied with some constant value which is sampled from a non-linear function, would the objective function remain linear?
Since, there are no product of variables involved

@LeakyNun Couldn’t understand the last line. Please explain
“Wherever you’re on the number line, you can walk leftwards ....”

$a$ is a point on the number line
you're on $a$
you walk leftwards to find a multiple of $b$
since they are spaced $b$ units apart from each other, it should take you less than $b$ steps
i.e. at most $b-1$ steps

Got you Leakey ! Thank you so much

when will you spell my name correctly...

2:57 PM
I’m sorry :( I just inserted an extra “e”

1 hour later…
3:59 PM
Can we talk of negative remainders? For example if $a\lt b$ then, can we write $$a = bq - r ~~~~~~~~~~~~~r\in [0, b-1]$$ given that everything is an integer.

4:13 PM
@EdwardEvans I have learned how to write Gauss in German, it is $Gau\beta$

@Knight AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

4:39 PM
lol

5:05 PM
ß $\ne$ β

thiẞẞ

5:28 PM
I think he wrote Gauſs himself.

I think I’m too late for the joke

5:59 PM
@Knight you wrote a beta

@LeakyNun I thought they look same

6:17 PM
A banquet has to take place tomorow in the morning. There are 1000 bottles of wine, and one only is poisoned. You have rats to test the bottles, but if a rat ingest poison now, it will die just a few hours before the banquet. What is the smallest number of rats that have to be used in order to be sure to be able to determine which bottle is poisoned?

Seems like unjust treatment of rats.
Not to mention unjust treatment of wine ... opening all the bottles too soon.

No actually we are trying to use the smallest number of rats :)

So your point is that if you wait for the result of the test now, there is no time to do a further test?

@TedShifrin :') haha but totally fair. Wouldn't pass an IRB

Yes that's the point

6:30 PM
I'm not a good puzzle person, but I don't see how to be nice to rats.

I don't know if this puzzle will interest many people, but maybe I should wait and give a hint later if it is necessary. Otherwise, the number is really small :)

So you're also telling me that an individual rat has essentially infinite wine capacity?

will a poisoned rat necessarily just die a few hours before the banquet or is there a fixed time period that it takes for a poisoned rat to die
in the former case, one rat suffices (granted it has infinite wine capacity and can take 999 sips of wine in a couple hours)

I think the point is that there's a huge time lag before the rat dies, @Thorgott.

the poison is very efficient on rats, a drop is enough :)

6:46 PM
Hint: You could have 1024 bottles without increasing the number of rats needed

yes the rat will die if it takes the poison, that's the only way to know if he has drunk from the poisoned bottle

But we're wondering how quickly the rat will die, @Nûr.

ok the point was that you have to make the rats drink all in one moment and then you'll just see which rat die without another experiment

Ah, @Alessandro's hint is good. We encode the wines in binary and use one rat for each "decimal" place. Rat $k$ drinks from all the bottles with a $1$ in the $k$th slot.

yes :)

6:51 PM
Very cute. So eleven rats.

I thought 10, who's off by one?

Did you count $2^0$?
Oh, I don't need $2^{10}$.
I'm off.
Your hint was wrong. You meant 1023 bottles.
Well, no, I guess you can do 1024.

yes there is no n°0 bottle

But I still need $2^0$ to get odd number bottles.

yes ok

6:54 PM
I hope the rats don't get confused with one another.
I guess they're mathematical rats.

do you have an idea how to prove it is the smallest number actually?

Induction, maybe?

it's starting to look like a rat race in here

Or well-ordering principle.

induction and I think then we can suppose we have infinite time so we would split in two parts of 500 bottles, use a rat that drinks all bottles from the fist part and reiterate

7:00 PM
No, obviously with $2$ bottles it takes only one rat. But we cannot do $3$ with one rat (proof?). Now well-ordering?

When I said reiterate I mean use another rat if the other die so there is log_2(n) rats we need
I suppose there is infinite time now because what we do with binaries seems to be a trick to do this process in one time
the integer just greater than that of course

I don' t follow.
Hi, @Stan.

If you had infinite time to determine the bottle, you will use one rat that will drink the 500 first bottles. If it dies you know the poisoned one is in that group, if not it is in the other one. Then you use another rat to drink in the first 250 bottles of the group of 500 bottles in which you know there is the poisoned one ... etc
We also have to use 11 rats (because is the integer just greater than log_2(1000). I think binaries is just a trick to make all that in one time. So we don't have to suppose there is a limitation of time anymore. So if there is n bottles, you split in two groups with approximately the same number of bottles and you use one rat to eliminate one of the group, it seems that you can't do better with one rat. And then induction.

My algebra book arrived

@Nûr: Still, you haven't proved that there's not a supremely clever algorithm that's better. You've just shown a second algorithm is no better (granting extra time).

7:15 PM
yes it is true, i hoped it was clear this one can't be improved :/

Not clear. I think you have to show that a superior approach would lead to a superior approach to 3 bottles.

7:33 PM
it doen't seems easy for me :/ I will try to think about that. Thank you. Btw it is 10 rats I think (2^0, 2^1, ... 2^9) so 10 digits hence 10 rats

Yes, Alessandro and I agreed on 10 earlier :P

ok :D

8:12 PM
I have a question about terminology (or maybe about the name of a subfield)...
So to preface, there are sequences of calculations which I will, for lack of a better term, label as 'arithmetic'.
Think of a succession of assignments, like in a programming language function, that for a few inputs goes through some directed acyclic graph of computation, and outputs a single number.
But suppose you only have partial knowledge of some of the inputs, let's say a range. And you want to see what the range of the answer is, dependent on the operations This is traditionally called 'interval arithmetic' (because instead of operating on point values you're operating on a range, like maybe an error range.
This might come up in numerical computations for seeing what the error in the output is depending on the error in the input.
So the question is... what is the arithmetic called when the input isn't just some point with error range, but an actual probabilistic distribution?
I'm not saying automatically/symbolically compute the theoretical lmiting distribution of a computation on distributions (eg sum of two normals has mean = sum of means and variance = sum of squares of variances..
I'm saying input is some arbitrary distribution (but histogram known).
Is this what is termed 'probabilistic programming'?
That's it. That's the question.

8:50 PM
@Mitch that's called error analysis and is tought to physics students in a lab
not the formal math but useage of the resulting formulas
it has probability stuff in there
I'm liking Serge Lang's Algebra. It has homology theory as well as modular functions etc, aka advanced math stuff which is what I wanted.
I'm going to do every exercise in the book :D
It is my bible now :D
Fixed some bugs in BananaCats. Still more to debug. It's never-ending... :D

9:05 PM
@EnjoysMath Nice. Thanks.

@Mitch you're welcome
There's some math involved in deriving the error formulas.
Partial derivative chain rule application etc.

It seemed obvious to me that there should be such an area of computation, but I just didn't know of a name for it or had come across such a thing explicitly written about as an area of study.
@EnjoysMath Well, Im also just interested in -computing- the sample distribution from instances (of course figuring out a symbolic calculus for given theoretical distributions is very interesting too).

That's neat. I don't understand any of the math involved, I dropped out of that physics course. :D
I'm studying abstract algebra now myself

2 hours later…
10:45 PM
just tried reading a text and immediately dropped it when they started "applying functions from the right"
why on earth would anyone do that

Algebraists used to do that in the 50's ...
So much for you, Mr. Algebra.

ouch