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12:04 AM
Hi @Ted!
Hi, a @Balarka!! You feeling better?
Yup. Dengue test didn't turn up anything.
Good to have that confirmed.
It is something to do with a chest infection, which I am recovering from definitely.
Owen's book is fantastic by the way. Love it.
Oh cool!
You can wax rhapsodically entropic.
12:06 AM
Hey @Ted and @Balarka
Hi @ÍgjøgnumMeg
hi @ÍgjøgnumMeg
Get well soon if you're ill! :P
Y'all know what is meant by "pseudo-isomorphism" of modules?
12:08 AM
@TedShifrin Apparently one of the equivalent formulations of the second law is that for a fixed state $S_0$ of a thermodynamic system, there are arbitrarily close states $S$ in the state-space such that one cannot reach $S$ from $S_0$ by a quasi-static adiabatic path (i.e., to get from $S_0$ to $S$ you need heat exchange happening)
That makes sense, yes.
Basically, if $M$ is the statespace, and $\delta Q$ is the heat 1-form, then there exists points arbitrarily close to $S_0$ which are not reachable from $S_0$ by paths tangential to $\ker \delta Q$
This actually implies $\delta Q$ is integrable
(If it's non-integrable somewhere, it's totally non-integrable on a chart around that, and then you're done, because Legendrian curves approximate every curve)
oh it's just if the induced morphisms on localisations at primes are isomorphisms
12:12 AM
So there exists functions $T$ and $S$ on the statespace such that $\delta Q = T dS$
We have just proved existence of temperature and entropy by magic
I haven't thought about this stuff in 47+ years :P
But it's cool how this is just a statement about forms, right? If $\alpha$ is a 1-form on $M$ such that for any point $p \in M$ there exists points arbitrarily close to $p$ which are not reachable by paths tangent to the distribution $\ker \alpha$, then $\alpha$ is integrable
@ÍgjøgnumMeg I have some algebraic number theory questions for you
I am not 100% sure about this.
I will not be able to answer them, but go ahead
well maybe
I'm worrying about bracket length (I've forgotten what this is called) ...
12:18 AM
Oh, maybe. The $k$ in $\alpha \wedge (d\alpha)^k$?
I thought about this when we did our sub-Riemannian geometry seminar, but now I forget.
I guess I have only seriously thought about the case when $M$ is a $3$-manifold. There integrability is equivalent to $\alpha \wedge d\alpha = 0$.
Right. In that case you're right.
Good point
@ÍgjøgnumMeg What setting is this in? I thought there was a standard way to show that if you get isomorphisms when you localize a module homomorphism at all primes (or just at all maximal ideals) you actually started with a legitimate isomorphism.
12:22 AM
@Karl in Iwasawa theory, i quoted it wrong, it's actually if the induced homomorphisms are isomorphisms for all primes of height at most 1
hi Karl
hey Ted
apparently this is equivalent to "pseudo-nullness" of the kernel and cokernel of a module homomorphism
where pseudo-null means localisations are 0 at all primes of height at most 1
12:28 AM
Ah, that makes sense. A module which is zero when localized at every prime has to be the zero module, yeah?
I'm still trying to comprehend what pseudo-null modules are like.
@BalarkaSen Yes, the 'standard way' I referenced above involves constructing exact sequences and localizing.
Got it. I see why you just need maximal ideals now; I can take the maximal ideal containing the annihilator ideal of any element and then localize. The kernel of $M \to S^{-1} M$ is all the elements of $M$ which are annihilated by $S$, which would say every element of $A$ annihilates $M$.
nonzero element, of course
12:33 AM
@Balarka what ANT question did you have?
Oh right hm let me see
(if I can't answer it maybe @Karl can)
it'll just be nice to see if I know what you're on about lol
I have something simple I think. Suppose $K$ is an algebraic number field and $\mathcal{O}_K$ be the ring of integers. I can prove every prime ideal is maximal in this following fashion: take $\mathfrak{p} \subset \mathcal{O}_K$ a prime, then $\mathcal{O}_K/\mathfrak{p}$ is a domain. Also, since $\mathcal{O}_K$ is a finitely generated $\Bbb Z$-module (under the natural incl $\Bbb Z \hookrightarrow \mathcal{O}_K$), $\mathcal{O}_K/\mathfrak{p}$ is naturally a finitely generated abelian group.
yeah you can actually prove that any quotient of $\mathcal{O}_K$ is finite I think
12:45 AM
any ideal $I$ contains a non-zero rational integer, and then third isomorphism theorem
Because I can choose any $\alpha \in I$ and take the last coefficient of it's minimal polynomial
OK, so I was thinking of a more explicit way to do this but I can't seem to carry it out
and $\mathcal{O}_K/(a) \cong (\Bbb Z/(a))^n$
so that guy mod $I/a\mathcal{O}_K$ is finite
$\mathcal{O}_K$ is a rank $n$ free $\Bbb Z$-module
12:48 AM
Suppose $I \subset \mathcal{O}_K$ is an ideal and I look at all the $\text{Gal}(K/\Bbb Q)$-conjugates of $I$. Let's denote these by $I^g$ for each $g \in \text{Gal}(K/\Bbb Q)$. Shouldn't it be true that $\prod I^g$ is principally generated by some integer?
I know how to do this if $K$ is imaginary quadratic. But trying to do the same thing feels like a pain
by the norm of some element?
looks plausible lol
if you take the ideal generated by an inert prime it works...
This is enough for ideal division I think.
12:52 AM
And once you have $I \subset J$ implies $I = JK$, prime ideals are indeed forced to be maximal
For imaginary quadratics, the argument is something like, if $I \subset \mathcal{O}_K$ then $I$ has a lattice basis $\{\alpha, \beta\}$ in $\Bbb C$. So $I = (\alpha, \beta)$. $\overline{I} = (\overline{\alpha}, \overline{\beta})$, and $I\overline{I} = (\alpha\overline{\alpha}, \alpha\overline{\beta}, \overline{\beta}\alpha, \beta\overline{\beta})$.
Take the gcd of the integers $\alpha\overline{\alpha}, \beta\overline{\beta}, \alpha\overline{\beta} + \overline{\beta}\alpha$. Let's say that's $n$. My claim is $I\overline{I} = (n)$.
This is some trick, say $\gamma = \alpha\overline{\beta}$. We know the coefficients of $(x - \gamma/n)(x - \overline{\gamma}/n)$ are integers, so $\gamma/n$ and $\overline{\gamma}/n$ are algebraic integers.
Shouldn't a similar argument work in any case? All ideals in a Dedekind domain are either principal or bigenerated
This implies $I\overline{I} \subset (n)$ indeed. The reverse inequality is obvious.
@ÍgjøgnumMeg The Galois conjugates are vast in number though
It got messy when I tried it but maybe I was dumb
idk I'm rusty af so I wouldn't attempt it atm
I didn't know the result you mentioned, interesting
"vast in number" competes with my "wax rhapsodically entropic" :P
12:57 AM
Oh ok, the result about bigeneratedness is just because Dedekind domains are 1-dimensional, right?
The various meanings of dimension
Maybe not.
Maybe I have to factorize into prime ideals or something
something like that
it's because a principal ideal is a product of exactly two non-zero ideals
i'm just re-reading it
so take a non-zero $a \in I$ then $I \mid (a)$ so $(a) = IJ$ for some other non-zero $J$
also there is a non-zero $I^\prime$ s.t. $II^\prime = (b)$
and $I^\prime + J = \mathcal{O}_K$
1:05 AM
Clever stuff!
and then $(a) + (b) = I$
= $(a, b)$
because $(a,b) = (IJ, II^\prime)$
and that's just $I(I^\prime, J) = I$
bad notation
but yeah
This is pretty cool.
and the coprimeness just comes from the prime factorisationiness of Dedekind domain magic
1:08 AM
I'm glad I'm ignoring all this.
ignores that too
I attended some classes in an algebraic number theory course this semester. They proved that ring of integers of Q(zeta) is Z[zeta] for a root of unity zeta
I didn't get what happened, felt like random trick math
I at least understand, visually, the ring of integers of Q(sqrt(d)) for d < 0
1:10 AM
That's in Artin :P
yeah I mostly did stuff with quadratic fields in my diss lol
cuz they're "easy"
@TedShifrin I was reading Artin actually
It's very good
lol Keith Conrad writes "the inclusion $\Bbb Z[\zeta_p] \hookrightarrow \mathcal{O}_{\Bbb Q(\zeta_p)}$ is an isomorphism when tensoring with $\Bbb Z_p$ for all $p$
why tho
so all ya have to do is show $\Bbb Z_p[\zeta_p] \subset \Bbb Q_p(\zeta_p)$ is the ring of integers for all $p$, which is just computations in unramified/totally ramified extensions of local fields!"
1:13 AM
Huh that's not how we did it
Seems clearer
idk anything about local fields so
Me neither, just that the yoga "do it for each prime locally" makes sense
It's a good slogan
there's some crazy group that measures failure of local-global principles
which is the kernel of some galois cohomological magic
which means f*ck all to me
holy shit that sounds hype
see this is the problem with algebraic number theory
you go into it thinking you'll learn mega magical machinery with cool sounding names to it like "adeles", "Hasse local global", "Tate's thesis"
$$\bigcap_v \ker \left( H^1(\operatorname{Gal}(\bar{K}/K, A) \to H^1(\operatorname{Gal}(\bar{K}_v/K_v, A_v)\right)$$
1:17 AM
but all the first couple chapters of Marcus does is olympiad tricks with steroids
it seems all to just be "do stuff locally and stick it all together" tbh
can't wait to do more in Heidelberg :)
also $A$ is an abelian variety so
#elliptic curves
argh one of the tempo increases in Hungarian rhapsody momentarily coincided with the flashing of the cursor in this text box and for a minute I was enjoying life
u should teach some of this schtick to me
Mathein is a much better person to talk to than me
ask me in about 2 years
man Iwasawa theory is my current holy grail, it looks so cool
1:37 AM
too complicated in here
dont worry we're just about to start talking about perfectoid spaces
1:53 AM
alright so obv $(a) \subseteq \prod \sigma_i I$
where the $\sigma_i \in \operatorname{Gal}(K/\Bbb Q)$
What's $a$
just some rational integer
this is the question you were talking about a while ago
Ok, but you should choose an appropriate $a$ if you want the other inclusion, right?
Pick a non-zero rational integer $a \in I$, then $a \in \sigma_i I$ for all $i$ so it is contained in the intersection of all those guys, which is contained in the product
yeah exactly
so you want an $a$ that divides something like $\sum \prod_i \sigma_i \ell_i$
where $\sigma_i\ell_i \in \sigma_i I$
Oof m8
Maybe something like gcd over all symmetric polynomials on the conjugates
1:57 AM
you can just do this for primes
well everything simplifies when your extension is Galois because all inertial degrees and ramification indices are the same so just show that the product over conjugates of a prime is principally generated by a rational integer
and it works because the Galois group acts transitively on the primes
and norms are multiplicative
and the Galois action is multiplicative
and the norm of the prime is the guy you want as suspected
Hm I see
I wanted to use this to prove factorization of ideals though, so kind of kills my point. But at least it proves it
idk about non Galois
oh I see
there's a cool argument for factorisation of ideals based on localisation in my diss
which I lifted from Milne so idk why I'm still plugging that
but yeah, another "patched together locally" argument
2:13 AM
basically every ideal contains a product of powers of primes $\mathfrak{b} = \mathfrak{p}_1^{r_1} \cdots \mathfrak{p}_m^{r_m}$ so by some lemma/CRT you have $\mathcal{O}_K/\mathfrak{b} \cong \prod \mathcal{O}_K/\mathfrak{p}_i^{r_i} \cong \prod \mathcal{O}_{K, \mathfrak{p}_i}/\mathfrak{p}_i\mathcal{O}_{K, \mathfrak{p}_i}$
and then the same thing replacing $\mathcal{O}_K$ by $\mathfrak{a}$ (where $\mathfrak{a}$ is the ideal containing $\mathfrak{b}$)
Oh that's clever
but the nice thing is that the $\mathcal{O}_{K,\mathfrak{p}_i}$ are discrete valuation rings so you can immediately write $\mathfrak{a}$ as a power of the unique prime ideal
and then use the correspondence between ideals of $\mathcal{O}_K/\mathfrak{b}$ and ideals of $\mathcal{O}_K$ containing $\mathfrak{b}$ to compare the powers of primes dividing $\mathfrak{a}$ and $\mathfrak{b}$
or smth to that effect
I liked that one the most, the first proof I saw was just some ugly inclusion computations in an intro book

Does my question here make sense?
2:23 AM
Anyone I'm gonna go to sleep lol, I get overexcited when someone has a question I vaguely know something baout
G'night! And thanks, this did help a lot
No worries, night night, get well soon
Thanks man
2:39 AM
If I have subspace of $\mathbb{R}^3$ defined as $W=\mathbb{R}^3$ and I take a compliment from $W$ isn't the compliment in $\mathbb{R}^4$?
why would it be
if you take a compliment from W you should just blush
Well in order something to be orthogonal to this space it needs to be on 4th dimension?
complement is the correct term I believe
I don't understand what your line about the 4th dimension means.
There is a set of vectors $V \subset \Bbb R^3$ so that for any $v \in V$ and $w \in W$, you have $\langle v, w \rangle = 0$.
This $V$ is a vector subspace.
Your job is to find what it is. There is no fourth or fifth or millionth dimension in this definition at all
hmm yes
I mean the problem is that what the intuition behind this is
2:51 AM
Behind the orthogonal complement construction in general? Or this particular case?
If it's the latter then maybe finding the solution will help provide you with intuition
If you have 2 dimensional plane and you want something to be orthogonal to it lets say a line. Doesn't the line need to be in $\mathbb{R}^3$ in order to this be possible
If you think of normal vector for plane
cross product for example
orthogonal complement is set of $$ W = \{ \vec{v} \in V | \langle \vec{v}, \vec{w} \rangle = 0, \quad \forall \vec{w} \in W \} $$
Yes. The 2-dimensional subspace has orthogonal complement a (3-2) = 1 dimensional subspace, aka, a line.
The orthogonal complement by definition always lives in $V$
And what you meant to write there is $W^\perp = \cdots$
yes I don't know the syntax for this symbol
2:58 AM
All I am saying is that whatever symbol you use to denote orthogonal complement you should not call the orthogonal complement of W by the name W...
If I have $W = \text{span}((1,-2,0))$, where $W$ is subspace of $\mathbb{R}^3$ then, "describe how $W^{\perp}$ looks like?"
I can quite easily tell that W resembles a line
but the complement is exactly what in this?
I think you should compute the answer to your question.
It is doing mathematics that gives us intuition for mathematics.
You can start by trying to do it for $W = \Bbb R^3$, and see if you see a pattern in the dimension of $W$ vs the dimension of the complement.
yes the 4th dimension thing was trying to figure out pattern just based on this "geometric intuition" but it went into the woods.
well the description of this problem stated that "this problem is suppose to be done without any computation"
Then it's expecting that you know something from the chapter that you're not using or that you already know the pattern from previous problems. But that doesn't matter to me. You just gotta do the math. You have the formula for the complement, you just need to write down what it says about the vectors.
3:21 AM
Still I somehow don't know how to percieve this.
7 hours later…
10:27 AM
Can anyone explain the marked passage pasteboard.co/Itf8Amg.jpg
2 hours later…
12:37 PM
1:15 PM
Hey @ÍgjøgnumMeg
1:33 PM
waddup @Balarka
$\{(x,\frac1x)\in (0,1]\times\Bbb R: x\in(0,1]\}$ is closed in $[0,1]\times\Bbb R$, right?
I mean, this is closed in $[0,1]\times\Bbb R$, right?
@Silent what would you say?
@Misha.P i'd say yes, because its complement seems open, because, even if x is taken closer to 0, we can find tinier balls in complement of that graph with $[0,1]\times\Bbb R$ strip.
@Silent yes, exactly
1:49 PM
Hi @Balarka @ÍgjøgnumMeg
Hey @Alessandro
Hi @Alessandro
How are you?
Better than before. Still recovering.
2:08 PM
Hello. Good to see the above three folks again!
That's good. What are you doing mathematically?
bunch of random shit idk lol
2:24 PM
Ah probability
Hardly, but I did learn the definition of a Poisson point process
I don't think I want to know what that is
Haha ok
I've been learning about operator algebras lately
2:37 PM
Nah it's quite interesting
Hi guys
I have a little problem with a simple trigonometric problem
anyone can help me?
@s.harp are you here by any chance? I have an extremely stupid question
2:53 PM
@AlessandroCodenotti What book are you using?
Murphy to get the background for Brown-Ozawa
Never heard of Murphy. It means this is above my pay grade.
It's called C*-algebras and operator theory
Aha! I like Conway's 'A course in operator theory' which follows on from his 'A course in functional analysis'.
I like Conway's FA, but I picked Murphy's book because that's what Brown and Ozawa suggest as a prerequisite to their book
3:20 PM
i am here @AlessandroCodenotti, but i might give you an exteremly stupid answer :)
If I have subspace in $\mathbb{R}^3$ which is defined as $W^{\perp} = \begin{bmatrix} 2a \\ a \\ c \end{bmatrix}, \text{ when } a,c \in \mathbb{R}$, what kind of subspace this is?
Doesn't exactly cover all of $\mathbb{R}^3$
I'm looking at a result that says that if $\|a|\\leq t$ and $a\geq 0$ then $\|a-t\|\leq t$, where $a$ is a selfadjoint element of a unital C*-algebra and $t\in\Bbb R$
Hi, can someone help me to understand why ${f(x, y, z)} ={ x^2 + y^2 − z^2}$ is a smooth map between manifolds of dimension 3 and 1. How we show that the map $f from ${R}^3$ to ${R}^1$ is a smooth map?
I solved my doubt while typing it out
@AlessandroCodenotti make use of $\|a\|=\sup_{\lambda\in\sigma(a)}|\lambda|$ (which holds for normal $a$)
@AlessandroCodenotti oh :o)
3:28 PM
@s.harp Uhm I don't see how that helps me
It looks like I can get a bound on $\|a-t\|$ from below by looking at the spectral radius of $a-t$
@AlessandroCodenotti since you solved it already i dont need to give any hints: As $a≥0$ you have $\sigma(a)\subseteq [0,\|a\|]$, then $\sigma(a-t)\subseteq [-t,\|a\|-t]$, which consists entirely of negative numbers and is bounded in absolute value by $|-t|$.
Oh it's plane
Murphy has a different proof which I quite like: look at the commutative unital C*-subalgebra generated by $1$ and $a$, using the Gelfand representation this is $C(\sigma(a))$ and then the result is obvious for real valued functions
If I have $w=\text{span}((1,-2,0))$ then orthogonal complement is on plane $w^{\perp} = \begin{bmatrix} 2a \\ a \\ c \end{bmatrix}, \quad a,c \in \mathbb{R}$ right?
3:34 PM
@Tuki yes (you can visualise it here for example wolframalpha.com/input/… )
yes thanks for the link
@AlessandroCodenotti yes that is elegant, but I think that these two proofs are the same proof!
Hmm maybe you're right
or not quite, the statements about the spectrum can be achieved with less machinery than the gelfand transform
but the gelfand transform is more elegant
4:07 PM
You guys are all great mathematicians
@Alessandro You know Wedderburn's theorem on matrix rings by any chance
@Balarka think Mathein has that in his blog
Oh, nice
What doesn't he have in his blog
idk man
@BalarkaSen nope
4:20 PM
@Alessandro If $D$ is a division ring, $M_n(D)$ is a simple ring, i.e., has no nontrivial 2-sided ideals. This is because if there is a nonzero ideal $I \subset M_n(D)$, it must contain a nonzero matrix $A \in I$. You can multiply by the "elementary basis matrices" $E I E'$ on the left and right to get some nonzero matrix with a single entry at the $(i, j)$-th term.
Then you can just make that an elementary basis matrix as well, with nothing anywhere except a $1$ at the $(i, j)$-th term, because you're working over a division ring
Use row and column operations to get all the elementary basis matrices with a 1 at the various diagonal positions, and finally get the identity matrix in your ideal $I$. This says $I$ is the full ring.
Wedderburn's theorem is, I think, the exact converse: say $A$ is a finite dimensional $D$-algebra for some division ring $D$ which is simple. Then $A = M_n(D)$.
Makes sense
Is that Wedderburn's theorem?
I think so
you can also find wedderburns theorem in Lang, corollaries 16. 3.5 and 16. 3.6 (here for example docs.google.com/file/d/0B0XWNg1VDJucMVUtNzN3Z0xFR3M/edit , page 649 )
i remember this specifically because a prof remarked about the theorem in a lecture, i thought hte statement was quite nice and then i used it some time later but had trouble finding a useful version of it
4:31 PM
I get confused by varying hypothesis. One thing I know is if $A$ is simple left-Artinian then it's a matrix ring
Because grab a minimal left ideal $I \subset A$. Then $IA$ is a 2-sided ideal, forcing $IA = A$. Choose $a_1, \cdots, a_n \in A$ and $r_1, \cdots, r_n \in I$ such that $\sum r_i a_i = 1$
Look at the map $I^{\bigoplus n} \to A$ by thinking of $(r_1, \cdots, r_n)$ as a basis of $A$ over $I$. This is going to be an isomorphism, I believe.
Uh, maybe I want $n$ to be minimal, to get injectivity
Anyway now you have $A \cong I^{\oplus n}$ as left-$A$-modules, so $A = \text{End}_A(A) = \text{End}_A(I^{\oplus n}) = M_n(\text{End}_A(I))$.
So is it true that finite dimensional algebras over division rings are Artinian? Feels right to me
Yes, clearly
5:31 PM
Hi folks, I want to know what is the internationally adopted notation for representing "the factors of 12 is 1, 2, 3, 4, 6, 12"?
@MoneyOrientedProgrammer Notation for which part of that?
I need something like $F(12)={1,2,3,4,6,12}$.
@MoneyOrientedProgrammer nope, just write it however seems natural for your setting
It's a set so that works
You have to make a note about definition of $F$ though
I don't think I have seen any widely adopted notation for the set of divisors
Also in LaTeX you use \{ for set
5:33 PM
OK. Thank you @ShineOnYouCrazyDiamond and @TobiasKildetoft
I would use $\text{factors}(12) = \{1,2, \dots \}$
hover over with mouse to get LaTeX code
I have no mathjax enabled
or $\text{divisors}$ or $\text{prime_factors}$ if you just want the primes
Ask someone how to get that link
The book I taught from used div(n) for the set of positive divisors
I'ts a JavaScript code that you can bookmark then you click the bookmark in the future
$\text{div}(n)$ works too
5:35 PM
but there is no notation that would not need to be explicitly introduced
But remember to use $\text{ on "div"
@ShineOnYouCrazyDiamond Actually, it is better to make it an operator in regards to spacing
$div(n)$ is not right
@TobiasKildetoft if you know how to do that
so \operatorname inline
Or you can just put a [space][space] after \text{div}: \text{div} \ (n)
looks exactlyl the same to me
@ShineOnYouCrazyDiamond I mean spacing in various contexts, such as the space between the div and the ( for example
@TobiasKildetoft the operatorname is the first one above, it doesn't do anything here
does same thing as \text
5:38 PM
I don't recall which it is that \text ends up with slightly odd spacing
It is better to create operator in the preamble.
new operator
$\text{div} \ n$
$\textbf{divisors} \ (n)$
\textbf{...} \ (n)
@MoneyOrientedProgrammer Sure, i you will be using it multiple times
Thank you all, I have to go to bed. It is already 12:40 AM ( I am in GMT+7).
5:41 PM
K, night have nice dreams, hopefully :)
My dreams have been weird lately, but not terrible
If differential $d^i(f)(g_0, \dots, g_i) = g_0 \cdot f(g_1, \dots, g_i) + \sum\limits_{j=1}^i (-1)^j f(g_0, ..., g_{j-2}, g_{j-1}g_j, g_{j+1}, \dots, g_i) + (-1)^{i+1} f(g_0, \dots, g_{i-1})$, then $d^0(f)(g_0) = g_0 - f(g_0)$ correct?
In group cohomology, where $d^i$ is a differential from $C_^i(G; A) = \{ \text{ functions } f : G^i \to A \}$ and $d^i : C^i(G;A) \to C^{i+1}(G, A)$?
Not sure what's wrong with the LaTeX there
Just wondering if I have the formula for $d^0$ correct given that I have the general formula
The weird notation in the middle term just combines the argument $j-1$ with $j$
(the latex error is from " _^ ")
i dont know group cohomology, but it may also be that your formula is especially for the case $i>0$ and there may be another formulation of it that can take care of the case $i=0$ as well. But the way I read the formula I would expect $-f(g_0)$
Oh sense $f() = 0$ since it maps to an abelian group $A$?
the question is how to interpret "f with no argument", whether you want the multiplicative unit or the additive one
5:53 PM
Has to be $0$ I guess then
Thank you!
Isn't it weird that this cohomology development is with arbitrary functions and not necc. homs?
I wonder what the book will do next, this is exciting!
ok if I begin reading the definition of group cohomology on wikipedia a term like $g_0- f(g_0)$ should be $g_0(0) - f(g_0)$ which is $-f(g_0)$, as $G$ should act by group automorphisms
$d^i(f)(g) = g_0 f(g[\hat{g_0}]) + \sum\limits{j=1}^i (-1)^j f(g[\widehat{g_{j-1}}, g_{j-1g_j},\widehat{g_j}]) + (-1)^{i+1}f(g[\widehat{g_i}])$
$0$-cochains are just elements of $A$
$d^0(a)(g) = g \cdot a - a$, that's all
How do you get $g\cdot a$ in there?
Like we said, $f() = 0$
so it's $g \cdot f() - a$
00:00 - 18:0018:00 - 00:00

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