Mathematics - 2021-02-20 (page 1 of 3)

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12:00 AM

analytic is the worst I ever encountered

@AlessandroCodenotti while browsing through the harvard exams I found a nice course

A problem in mathematics is classifying objects up to some notion of isomorphism. Famous examples include: the classification of compact orientable surfaces up to homeomorphism by their genus and classification of Bemoulli shifts up to isomorphism by their entropy. Descriptive set theory allows for a precise study of the complexity of various classification problems and the possible invariants which they admit.
Topics: Polish groups and their actions on Polish spaces, definable equivalence relations, classifications problems and invariants, and interactions between these topics and forcing.…

Alessandro Codenotti

@user2103480 uh that's amazing, do you have a link?

math.harvard.edu/graduate/?courseid=65

but no special link

the instructor is Assaf Shani

It's spring semester

So it started a few weeks ago?

You could ask him nicely to get a zoom link

Spring Term: January 25–May 15, 2021.

spring term starting in january??

Alessandro Codenotti

@user2103480 oof a month's worth of lectures is a lot of catching up to do

Yes, that's the US schedule for semesters. Quarters are different, but hardly any schools on quarters.

12:08 AM

@Thorgott How german to think they are the outliers

Basically everyone except germany starts around february

which, incidentally, messes up spring exchange semesters since people have to do exams in germany and homework in their exchange country

thanks for nothing, arbitrary german calendar

that's so weird though

january isn't winter

Exams at 38 degrees are another fine byproduct of the german schedule

At least nobody batted an eye when I came in sweatpants to the oral exam, as the prof himself was there in shorts

Alessandro Codenotti

Lücke teaching in shorts and flip flops

That June was insanely hot, that's exactly why in normal countries there are no lectures in the middle of the summer

This makes sense Alessandro

@user2103480 Assaf is an acquaintance of mine

Ran into him in a layover in Poland once

If you want me to get in touch with him I can

can't even take a bit of heat smh

12:15 AM

@TedShifrin This is my first semester school.

@AlessandroCodenotti man if he wore some psychedelic garn longsleeve he would've looked so woodstock

missed an opportunity there

@AlessandroCodenotti ooff I remember I once had to take a spot in the sun and I died. Not to mention the train ride there and back in tropical temperatures

Alessandro Codenotti

@MikeMiller Thanks but there's no need to, I'll send him an email tomorrow! It's time to sleep now though

An indian computer science student told me last year that the summer here is somehow worse than where he comes from in india

(in Bonn)

good night!

12:30 AM

@MikeMiller Granted, but google says under 15% on quarters.

I agree. Just a curiosity.

12:48 AM

Berkeley switched in 1980, I think, right after I finished. But the other UC schools stayed on quarters. All? Most?

1:40 AM

@Thorgott wat

I can't figure out whether that's a good or bad situation. Was this, like, a take home exam and the person automatically failed, so that they have to do an oral as a second try? Or was this just a "hah! I see you are preparing yourself" moment?

Hello World...

Hello @TedShifrin @Thorgott @copper.hat

I honestly don't know, but that makes it even better

I just love the immense swag of posting an answer on MSE saying "come find me in my office"

it feels like the virtual equivalent of T-posing

hello @123

1:56 AM

I think it was “I caught you cheating,” clearly.

Hi, @123

2:24 AM

@TedShifrin that does not explain what the consequence is, but I dont know US regulations

Is a second exam after cheating necessarily an oral oxam? Does one just fail?

Could also just be cheating for homework questions

@Thorgott math student moment #642250871

There are no universal regulations. Of course.

Most universities will fail you in the course or even put you on probation if you are convicted of cheating.

is that the universal property of regulations?

smacks Big

Our lecturer talked about the class reviews and one student said the workload was too high and he worked about 10 - 15 hours on sheets alone, + lectures, exercise sessions and revising lectures

@Ted can you tell me whether this question deleted by the OP? math.stackexchange.com/questions/4031528/…

2:30 AM

(which was realistic for that course)

snickers in the corner

Yes, @Thor

And the lecturer just answered "yeah it should be that way"

beautiful, so OP posted that question 17 hours ago, just literally reposted the same question 20 minutes ago, I point it out and vote to close and in response they deleted the original question

My students typically worked way more than that in every course I taught.

2:31 AM

The first law of teaching mathematics: Everybody takes only your course, and does not work on the side

Flag it, @Thor.

@user2103480 thats not a lot tbh

oh right, I can flag instead of close voting, duh

@TedShifrin Way more in one course? The typical CP requirement is 3 of those courses which adds up to about 60 hours

in one semester

excluding having to work 10 - 20 hours for financial survival

Most courses at my university were not so demanding. Even most math classes. Truth.

But I get your point, of course.

2:35 AM

During the bachelor's, that was mostly the case for me as well. People say the first semesters are the hardest but that's a lie. For master's courses, that mostly got worse

Uni should be itself a full-time job, but times have changed a bit.

Especially since 5CP courses are now more common, which at my unis seem to have about 100 - 120 pages of lecture notes on average

@TedShifrin Yeah. But it's either full-time job and debt or full-time job + half-time job

I thought EU education was free. It isn’t here.

Depends on the country. US debt is crippling debt

In Germany, I'm just referring to the cost of living, especially at age 25+ when one has to pay insurance oneself

That's still waaaay better than in the US or UK

If I'm unlucky I have 10 - 15 000€ debt at the end of it, but nothing comparable to paying tuition etc

3:01 AM

yeah

people still here?

3:27 AM

@user2103480 Only because idiot UK brexited. Tromp love kills.

Okay, I am a bit confused at an asymptotic calculation. I am not very good with the notation, but in my notes it says something like $$(\frac{1}{z^2} + o(z^2))-\frac{1}{z^2}\left(\frac{-1 - \frac{1}{2}f'(v)z^3 + o(z^4)}{1 - 2z^2 f(v) + o(z^4)}\right)^2.$$

It then says this is simply $$=2f(v) + o(z)$$

this is all as $z\rightarrow0$?

Not sure if it is the right o/O/omega notation.

here I am just meaning o(z^4) is just z^4 times some smaller things.

@TedShifrin I think it was already expensive before. They never quite aligned with the rest of the EU

So z^4 or higher powers.

I think they are trying to show this is differentiable at z=0 or something.

I don't know how to work with the denominator in this. It's clear to me that the numerator is $1+o(z^3)$, but then I am not sure how to come out with $2f(v)$.

3:53 AM

You need to use $1/(1-u) = 1+u+u^2+ ...$

@TedShifrin Ah!

Thank you, this works immediately!

A trick I won't forget now.

(fast forward to a few years from now when Ted tells me again).

So to translate this in terms of showing it is holo at z=0, I thought we wanted to look at the coefficient of the order 1 part, not the order 0 part. That is, if we show $g(z) - g(0) = \psi z + o(z^2)$, then $\psi$ is the derivative and $g$ is holomorphic.

But what I showed here is effectively $g(z) - g(0) = \psi + o(z)$.

z^4 is in O(z^4) but not o(z^4). If f/g is asymptotically bounded f is in O(g). If f/g is asymptotically zero (limits to zero) f is in o(g).

@anakhro . o O ( years? )

@anakhro That would be true if $\psi=0$, then $g'(0)=0$ (if you use $o(z)$)

4:10 AM

@MikeMiller yes I am unsure of which of o or O it is. So it's supposed to be O that I am using here?

@robjohn Yes, that seems to make sense, since if you divide through by z, you get the usual expression.

@anakhro If you use $O(z)$, then we don't know if $g'(0)$ exists.

Is there a best notation for this that I am missing?

you wanna say $g(z)-g(0)=z\psi+o(z)$, that's equivalent to $g^{\prime}(0)=\psi$

@Thorgott Hmmm, okay, so then what rob says is probably what is being played off of.

@anakhro what Thorgott said

4:17 AM

Thanks everyone.

Also, on a totally different topic, can we honestly define the square integrable functions L^2 for Riemannian integrals?

In such a way that it is a Banach space.

Hmmm, and maybe without the use of equivalence classes/agreeing almost everywhere.

without that, you don't even have a metric space

@anakhro You can have null sets without Lebesgue integrals

Integrals over null sets must be $0$

@Thorgott Is there no way around it with a different kind of construction?

4:34 AM

I'm not exactly sure what you're trying to do, but I don't think so. My point is that the norm is given by integration and integration can't tell apart what happens on any set of measure 0. This is pretty much an ambiguity by design.

22

Q: Can $L^{2}$ be represented as a space of functions (not equivalence classes)?

Let $X$ be the vector space of all Lebesgue-measurable functions $f:\left[a,b\right]\rightarrowℝ$ such that $\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx<\infty$ (Lebesgue integral). Then we can define an equivalence relation on $X$ as follows: $f \cong g$ if $f(x)=g(x)$ almost everywhere on $\...

fa.functional-analysis ca.classical-analysis-and-odes sp.spectral-theory

I was thinking sort of like this.

9 hours later…

Bohemian relativist

1:42 PM

is the difference between a vector space and and an affine space just that a vector space has a distinguished origin while an affine space doesn't?

2:11 PM

Proove/Disproove : There exist a subgroup of $A_4$ isomorphic to $S_3$

Clearly the subgroup $H \leq A_4$ has to be of order 6, but im not sure if one exists.

2:27 PM

i want to ask if i can write "integers between 0 and 100" as $(0, 100)\cap\mathbb{Z}$.

@MikeMiller Thanks, for your comment. So, I have the following: Let $F$ be a free group of rank at least two and $a,b\in F$ be two non-trivial elements such that $b\not\in \{a^n:n\in \Bbb Z\}$. So, $a^kba^l\not =b$ for all $k,l\in \Bbb Z$ unless $(k,l)=(0,0)$.

Alessandro Codenotti

@EdwardEvans are you around?

2:49 PM

@Bohemianrelativist I think so

@Eminem hint: look at the class equation of A4

Bohemian relativist

@LeakyNun thank you very much for your answer.

Alessandro Codenotti

Leaky do you want to think about an ANT exercise that's giving me an headache?

@LeakyNun Im not familiar with that term...

@Eminem it means, look at the sizes of the conjugacy classes

@AlessandroCodenotti perche non

hmmm...

first the statement is true, right?

3:06 PM

@Eminem second hint: look at the index of the subgroup

Alessandro Codenotti

I want to determine the ring of integers of $\Bbb Q(\alpha)$ where $\alpha$ is a root of $x^3-3x+1$. The exercise has a bunch of steps, so far I've shown that $(\alpha+1)$ is a prime ideal of norm $3$ and now I'm asked to show that $O_K=\Bbb Z[\alpha]+(\alpha+1)O_K$ and then to deduce from there that $O_K=\Bbb Z[\alpha]$

hmm

Alessandro Codenotti

So I know that $O_K/(\alpha+1)O_K=F_3$, but I'm not sure how that helps

doesn't that complete the first step then?

Alessandro Codenotti

Does it? What am I missing?

3:09 PM

perche lo che tu sai implica che $\mathcal O_K = \{0, 1, 2\} + (\alpha+1) \mathcal O_K$ credo?

Alessandro Codenotti

Ok that's fair

or did I misinterpret what you need to show

is the $\subseteq$ harder or the $\supseteq$?

yeah I think what I said is right

Alessandro Codenotti

No ok I agree

great

Alessandro Codenotti

I see now, thanks

3:11 PM

e tu sai come fare il segundo passo?

Alessandro Codenotti

Also stupid question: $K=\Bbb Q(\alpha)$ does not imply $\alpha\in O_K$?

non, perche $\Bbb Q(i) = \Bbb Q(\frac{1}{100}i)$

Alessandro Codenotti

ah that's a nice example

($K = \Bbb Q(i)$, $\alpha = \frac{1}{100}i$)

Alessandro Codenotti

@LeakyNun You multiply both sides by $(\alpha+1)$ and do some manipulation with already known identities

3:13 PM

volo sapere come lo faccio lol

p-adically you just make the coefficient smaller I guess, with what you said

Alessandro Codenotti

Hm no ok maybe it's not as straightforward as I was thinking

Well $|A_4|=|A_4:H||H| \implies |A_4:H|=2$

Im not sure how to continue..

not sure what Leaky is trying to get at, but here's an alternative hint: $S_3$ is generated by an element of order $3$ and an element of order $2$

@Eminem what do you know about subgroups of index 2?

Oh. H is a normal subgroup of $A_4$

But $S_3$ isnt, so they are not isomorphisms?

no

idk

lol

3:27 PM

@Thorgott von der Klassequation kan man dann sehen, dass es unmoglich ist, dass es eine Untergruppe gibt, die sechs Elemente hat

@AlessandroCodenotti ricordi la prova dello fatto che $\mathcal O_K = \Bbb Z[\zeta]$ where $K = \Bbb Q(\zeta)$ and $\zeta^p = 1$?

@User873110 You are just restating your original question at me? As I mentioned before the condition on a^n is superfluous.

Ugh, I find the unmentioned switch from cohomology with values in abelian groups to values in rings confusing

3:43 PM

@user2103480 how do you do cohomology with values in rings

Does the künneth theorem hold regardless of any conditions if we have values in an abelian group?

it isn't even an abelian category

modules, rings, idk I dont algebra

cup product requires you to work with values in a ring to make sense

Ah yeah yes we actually work with rings

3:49 PM

There are two ways to do cohomology with values in a ring $R$. If $C_{\ast}(X)$ denotes the chain complex on $X$, you can look at the cochain complex $\operatorname{Hom}_{\mathbb{Z}}(C_{\ast}(X),R)$ and the homology thereof is the cohomology of $X$ with values in $R$ (this is the approach you might as well take for arbitrary groups $G$ instead of $R$).
You can also take the chain complex with coefficients in $R$, i.e. $C_{\ast}(X;R)\cong C_{\ast}(X)\otimes_{\mathbb{Z}}R$, which is naturally an $R$-module and then take the homology of the cochain complex $\operatorname{Hom}_R(C_{\ast}(X;R),R…

Ay that's smart

The latter perspective is occasionally useful, because it entails that it makes sense to evaluate a cochain with values in $R$ on a chain with values in $R$. And with the usual boundary formula, this implies it makes sense to evaluate cohomology classes with values in $R$ on homology classes with values in $R$.

To be explicit for a change, this means that if I take a cochain $\varphi$ and think of it as an element $\operatorname{Hom}_{\mathbb{Z}}(C_n(X),R)$, I can still evaluate it on a chain in $C_n(X;R)$, say $\alpha=\sum_ir_i\sigma_i$ where the $r_i\in R$ and $\sigma_i$ are simplices, via $\varphi(\alpha)=\sum_ir_i\varphi(\sigma_i)$. Note that this uses multiplication and won't make sense for groups.

Ah smart. But what exactly wont it make sense for groups? The commutative diagram? Since $r_i \sigma_i$ hides a tensor product

But $\varphi(\alpha)=\sum_ir_i\varphi(\sigma_i)$ make sense regardless no?

$r_i\varphi(\sigma_i)$ is multiplication in $R$

ahh adding a group element would just result in $\varphi(\sigma) + a$

1. do an algebra course in semester 3
2. wait 7 semesters
3. do algebraic topology after forgetting everything
4. ???
5. profit

4:01 PM

@LeakyNun With coefficients in rings, your chain complexes are R-modules

In homology one often parses all coefficients as being Abelian groups because one doesn't tend to care about the coproduct structure

But if you want to write down a cohomology ring you'd better be able to multiply your coefficients

The cohomology ring naturally incorporates the possibility to multiply cohomology classes with ring elements right

through 0th cohomology

1. do commutative algebra the previous semester
2. solve algebraic topology homework by drawing diagrams and spamming "nAtuRaLitY"
3. ???
4. profit

@user2103480 Uh sure what you're saying is it's an R-algebra so contains a canonical copy of R in it which acts as scalar multiplication

And what you mean there is that $\varphi_r \in H^0$ corresponds to the function which eats a point and spits out r

not every R-algebra contains a canonical copy of R

but this one does, of course

R-algebra means unital to me

4:05 PM

ok. What happens in the case when there are several connected components? Then $H_0$ isn't just $R$ right

@MikeMiller absolute unit

I carefully phrases it for you so that I didn't use connectedness

How does this affect the cup product

The unit element is the functional which takes any point and spits out 1. It's the constant function 1 on X.

Ok so this gives us the copy of R in any case

This doesn't make sense bro the cup product doesn't care whether something is connected or not there's no connectedness in the formula

4:06 PM

thanks

@MikeMiller ...but... I care...

I don't understand why you're obsessed with this copy of R this isn't how we are thinking of the R-action

@MikeMiller Z/2Z is a unital Z-algebra

We are just multiplying our functional by an element of r

Oh gotcha sure I meant it receives a homomorphism

Thanks

I'm not obsessed, the answer "we don't care about connectedness" is enough. I just didnt know if we care

I haven't actually thought about this, but the cohomology ring of a space surely splits as the coproduct of the cohomology rings of its connected components

no wait, product, not coproduct

4:10 PM

@Thorgott prolly doesnt matter in finite cases

so just gotta figure out what the product in the category of graded commutative R-algebras is, gotta think for a sec

@user2103480 I'm trying to say that if this is not obvious yet you should stare more at the definition of the R-algebra structure :)

@Thorgott coproduct is tensor product I think

Oops

Fine you're right direct product

@MikeMiller R-algebra? you're operating one level of generality above our exposition

4:16 PM

Yeah, it's just taking direct product in each degree and multiplication is continued in the obvious way by making the disjoint components multiply to 0

We just said it's a graded ring I think

"It should be obvious that if you have functions which eat chains and spit out elements of R, you can scale them by elements of R."

$R$-algebra just means it's a ring with an $R$-module structure in such a way that ring multiplication and scalar multiplication are compatible

how would it work if it weren't commutative hmm

I was about to nitpick your wording there @BS

4:17 PM

yeah, I made myself uncomfortable

I think it doesn't matter

probably bc of the underlying abelian group helping you

somehow

just like product of rings is the same in the commutative and the not-necessarily-commutative category

the category of definitely noncommutative rings

cause the inclusion CRing->Ring has a left adjoint

coproducts are what gets uglier

same way as it is for Ab vs Grps

without checking, it's probably the same pattern in graded commutative R-algebras vs graded R-algebras

4:21 PM

yeah, probably. things go between tensor product $\leftrightarrow$ free product

monoidaltransform

if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and increasing monotonically, does it mean that for each $y\in Y$ $f^{-1}(y)$ is connected?

@MikeMiller yea sure but one can say that in a more useful way. I mean, you could have said "calculate for finitely many path components $\mathrm{Hom}(H_0(X),R) = \mathrm{Hom}(\Bbb Z^n, R)$ for some commutative ring and then look what the unit in the cohomology ring corresponds to, and relate this to the generators of the cohomology groups of the path components"

so say you want to make an infinite algebraic extension over $\Bbb Q$, $K$, but you want $Aut(K/ \Bbb Q)$ to be finite. Can you do that? I can't seem to find how. I think if you toss in all the roots of unity you get $p$-adic integers, which is infinite. I kind of thought all of these would be infinite. Is there a nontrivial way to get a finite guy from this?

Which is more in the spirit of what my question was about

@monoidaltransform it's a singleton

Alessandro Codenotti

4:33 PM

If $L/K$ is Galois and $\sigma\in\mathrm{Gal}(L/K)$ fixes $O_L$ pointwise, must it be the identity?

But again, all the versions of künneth I find are either with some weird module conditions, or just for $\Bbb Z$ coefficients

@AlessandroCodenotti yes, $O_L$ generates L as a K-vector space

the point is that cohomology with coefficients in R naturally are R-modules, not just groups

Alessandro Codenotti

@Astyx is this clear? I think I'm missing something

It's an order

Alessandro Codenotti

4:35 PM

What do you mean?

monoidaltransform

not strictly monotonic @Thorgott

Order (ring theory)

In mathematics, an order in the sense of ring theory is a subring O {\displaystyle {\mathcal {O}}} of a ring A {\displaystyle A} , such that A {\displaystyle A} is a finite-dimensional algebra over the field Q {\displaystyle \mathbb {Q} } of rational numbers O {\displaystyle {\mathcal...

@monoidaltransform it's an interval

then it's at worst an interval

@Thorgott [a,a] and (a,a) count as intervals

fite me

4:38 PM

I don't disagree

@BigSocks Is that possible?

I feel it isn't because you can construct a subgroup of arbitrarily large size

(by looking at Galois subextensions)

@Thorgott man I just want to know whether this sequence is exact and splits $$0 \rightarrow \bigoplus_{p+q=n} H^{p}(X;G) \otimes H^{q}(Y;G) \rightarrow H^{n}(X \times Y;G) \rightarrow \bigoplus_{p+q=n+1} \operatorname{Tor}\left(H^{p}(X;G), H^{q}(Y;G)\right) \rightarrow 0$$

monoidaltransform

if for every $y\in Y$, $f^{-1}(y)$ is connected for $f:\mathbb{R}\rightarrow \mathbb{R}$

continuous

then $f$ is either increasing or decreasing

right?

Sounds reasonable

@Astyx that's along the lines of what I was thinking

2

Q: Finite automorphism group of infinite extension of field of positive characteristic

Here is the question I started with. Give an example of an algebraic extension $L$ of a field $k$ of positive characteristic such that $\left|\text{Aut}_kL\right|<\infty$ but $[L:k]$ is not. Since the fundamental theorem of Galois theory gives a one-one correspondence between extension dimensio...

abstract-algebra galois-theory

but this guy talks about the infinite extension $\Bbb Q [2^{1/3}, 2^{1/9}, ...]$ having trivial automorphism group so idk

4:43 PM

ok I guess you can break my argument by removing all Galois subextensions

Yeah his construction makes sense

how is it trivial tho

$2^{1/3^n}$ is the only root of $x^{3^n}-2$

@user2103480 You definitely want to take $G$ to be a PID $R$

(all the other roots are complex, and the extension is real)

I think you need a finiteness hypothesis in any case, but I'm not familiar with this level of generality

4:47 PM

yes we need a finiteness hypothesis

the homology groups need to be finitely generated

This doesn't work for powers of 2, because then you could take opposites

@Thorgott ..... dunno what that means

@Astyx but the next power of 2 restricts the current one

That's true

you want coefficients in a ring for cup product to make sense

you want the ring to be a PID for the proof to work

Bohemian relativist

4:51 PM

Does Minkowski space have a manifold structure? I am much confused by this post.

@Thorgott this does not involve any cup products

this is just about pure group structure

then explain your first map to me

@Bohemianrelativist it's a Lorentzian manifold

@Thorgott You are right, there is ring structure hidden in the map

Is the homomorphism $\phi:GL_2(\mathbb Z_7) \rightarrow U_7$ such that $\phi(A)=det(A)$ a well defined homomorphism?

We multiply cochains on simplices

4:59 PM

what's $U_7$

$U_7=\{x \in \mathbb Z_7:(x,7)=1$

ok, what are you worried about?

So for example if we let $A=\begin{bmatrix}4 & 0\\4 & 4\end{bmatrix}$ than $det(A)=16 \notin \mathbb Z_7$

Perhaps we need to do $16 \mod 7$?

yes

everything in sight is a residue class

What do you mean?

5:02 PM

the entries of the matrix aren't $4$, they're $4\mod7$

you're looking at a matrix with entries in $\mathbb{Z}_7$

the way you add and multiply these elements when calculating the determinant is the addition and multiplication in $\mathbb{Z}_7$

and the result is an element of $\mathbb{Z}_7$

everything is $\mod7$

@Astyx same reason as why $\Bbb Q [ 2^{1/3}]$ has trivial automorphism group over $\Bbb Q$ I guess

Oh ok

yes

cool ok thanks. I was just staring at it for a while

@Thorgott ok thanks a lot that helped, so I can use this for $\Bbb Z$, any field and polynomials in a field

5:05 PM

yeah

so is there any chance you could get something in between? probably if you did something sneaky like put in all the $2^{1/3n}$, but then also you toss in finitely many $2^{1/2n}$

Z,Q,Z_p will be the most important examples

by which I mean probably the only ones you ever need to consider

What do you mean by "in between" ?

yeah probably

Bohemian relativist

@Thorgott what do you mean? You mean Minkowski space has a manifold structure which is Lorentzian? But in that thread, Walterscoote said Minkowski space is just an affine space, which may or may not be a topological space, but Andreas Cap said an affine space is just a special case of a smooth manifold and any affine space can be canonically made into a smooth manifold by using the identification with the modelling vector space obtained by choosing a point in the affine space as a global chart.

5:06 PM

good to know it works for those

@Astyx finite automorphism group that is not trivial

I think you can add j (cube root of unity)

also, if $\Bbb Q [2^{1/3n}]$ is our extension, what would be the primitive element of the extension? $1$ since it has trivial automorphism group? probably not I think it's just $2^{1/3n}$

@Bohemianrelativist Tell me your definition of Minkowski space and your precise question.

@Astyx those too yeah

$n$ is fixed

5:09 PM

Let $G$ act on the probability measure space $(X,\mu)$ via probability measure preserving automorphisms. Let $\sigma : G \to U(L^2(X,\mu))$ be the unitary representation $(\sigma_g f)(w) = f(g^{-1}w)$ (analogue of the left regular representation). I am trying to show that each $\sigma_g$ is in fact a unitary operator. Linearity of $\sigma_g$ is obvious.

For unitarity, let $f \in L^2(X,\mu)$. Then $$\int_{X} |(\sigma_g f)(w)|^2 d \mu (w) = \int_{dom (g)} |f(g^{-1}w)|^2 d \mu(w) = \int_{dom (g)} |f(x)|^2 d \mu( gx).$$ Is it true that $d \mu (gx) = d \mu (x)$? If so, how does one show this?

$2^{1/3^n}$, yeah

mmm ok, thank you loads

@user2103480 I don't follow what your point is and I don't think I could have said it in a more useful way

@user193319 I don't understand the second equality

what's $\operatorname{dom}(g)$

@Thorgott The domain of the automorphism $g$. $g$ isn't defined on all of $X$ but a conull (full measure) subset.

5:18 PM

ok, then what you're asking for is just a different way of saying that g is measure-preserving

measure-preserving automorphism means the same thing as that the pushforward measures equals the original measure

Oh...yeah, you're right...I am being a knucklehead.

@MikeMiller "the unit in the cohomology ring is the sum of the generators of the 0th cohomologies of the path components" would have been a non-BS reasonable answer

not quite

What's a generator in the 0th cohomology of the path components.

there's more than one generator on each component

5:27 PM

You haven't specified anything yet. The moment you do you say what I just said.

the fact that the unit is distinguished is exceptional

If I have a Laurent expansion of f around zero, is looking for the Laurent expansion around another point obtained by substituting $z:=z-a$ in for $z$? I feel like it would actually have to deal with derivatives to do this.

The power series expansion of $f(z)=z$ at $z=0$ is just $z$, but at $z=a$ it's not $z-a$, but $(z-a)+a$

maybe I'm misunderstanding

Okay, let $f$ be the Weierstrass $\wp$ function. Then $\wp(z) = \frac{1}{z^2} + g_1z^2 + \dotsc$. I want an expression for $\wp(z - a)$ now, and I feel like it should introduce $\frac{1}{(z-a)^2}$ into it.

Do you want a Laurent expansion at z=a or at z=0 ?

5:37 PM

z=a, I already have it at 0.

In a book, all I am given is that near $a$, it "has an expansion $\wp(z-a) = \frac{1}{(z-a)^2} + \dotsc$", and they don't give me any terms. I can't figure out what they are actually doing to get that expansion.

If it's just plugging in $z:=z-a$ into the previous series, then I am curious about later steps in their procedure (they get higher derivatives of $\wp$ introduced).

What is the Power set of
$\{\phi,1\}$?
$\{\phi,\{\phi\},\{1\},\{\phi,1\}\}$ or $\{\phi,\{1\},\{\phi,1\}\}$?

That might be because they are trying to replace $\wp(z)$ with this series, so they are using the Taylor series maybe.

@Wolgwang What do you think?

What you're describing is the Laurent expansion at a of a different function, f(z-a). Of course that ends up being the same (up to substitution) as the Laurent expansion at 0 of f(z).

@MikeMiller so it's as easy as substituting it in, no issues arise?

back on my question- if you extend $\Bbb Q$ by all the $p$th roots of unity, you get the $p$-adic integers, but if you extend by all roots of unity do you get the Prufer group since it is (topologically) isomorphic to the product of all the $p$-adic integers?

5:48 PM

@anakhro Well I think the later one, but my teacher had taught the former.

@MikeMiller ugh, unit of the ring R, respectively. was good that you both pointed that out, but you know what I mean

@Wolgwang so the difference is that you don't think $\{\emptyset\}$ is in the power set. Why is that?

But what is your isomorphism H^0(X; R) ~ R for a path connected space X?

You're telling me I'm being pedantic, but you know I hate pedantry. I'm saying that the moment you spell out what you're saying, you get what I'm saying. I'm writing down the more explicit statement!

If you want "the unit supported on a path component" that's the functional which spits out 1 for points on that component and 0 elsewhere.

More explicit than an abstract isomorphism

More clear, too, since then you can understand how it behaves in the cup product!

@anakhro Because we don't consider $\{\phi\}$ in the power set of $\{\phi \}$...

@anakhro No, the opposite, what you're doing gives you the right answer to the wrong question. Forget wp. Understand what Thorgott is saying.

5:52 PM

@MikeMiller yes I agree

@anakhro Think about composition functions. What is $f(z-a)$?

If you take f(z) = z, then yes, "substituting" we get g(z) = f(z-a) = z-a. That's the Taylor expansion of a different function of z at z=a.

But that's not what you want...

If you could just "substitute" like this to find taylor expansions of f(z) at every different point it would tell you that all functions are constant

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