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1:02 AM
@Ultradark HEY!
@ShineOnYouCrazyDiamond hi!
I'm learning some group cohomology, I have to prove that the first differential formula is such that $d^2 = 0$ (is a chain complex)
It's an 80 page notebook from Romyar Sharifi's lectures
Some people earlier in here tutored me on why it's $d^0(f)(g) = ga - a$ for the $i=0$ differential
It makes sense, it's the difference from the original to the acted upon result
$g \in G$ acting on abelian group $A$.
$f$ can be any function from $G^i \to A$, that's what's crazy about it!
In the $G^0 = A$ as an identification which is why you see $a$ in the first formula and not $f(*)$
@Ultradark do you know any group cohomology?
It might help with your diffential form / manifold studies
@Shine: I've been doing differential forms and manifolds for almost 50 years and I don't know any group cohomology. I highly doubt that.
1:19 AM
@TedShifrin Ah, I think I misinterpreted: it takes 6 or 13 steps to transform one expression into the other using the axioms.
I still count fewer than 6. But who cares.
1:33 AM
@TedShifrin I was mistaken
2:24 AM
Whelp, I wonder what's wrong with my answer here.
A: What is $0 \neq 1$ in the context of Algebraic Axioms?

Tanner SwettWell, the statement $0 \ne 1$ means that $0$ is not the same thing as $1$. In other words, it means that the additive identity ($0$) is not the same as the multiplicative identity ($1$). This statement is related to addition because it mentions the additive identity ($0$), and it's related to mu...

3 downvotes and no upvotes.
Maybe I interpreted the question too literally. I thought the asker was asking what the statement $0 \ne 1$ means. So I explained what it means.
After all, the asker wrote, "What does 0≠1 means?"
Meanwhile, the upvoted answer seems to be answering a totally different question: namely, what is the motivation behind including $0 \ne 1$ in the list of axioms?
Although, to be honest, I'm going to delete the answer.
Really, I have no idea what the asker was trying to ask.
I left a comment asking for clarification.
6 hours later…
8:27 AM
Morning all
9:02 AM
Hey @Alessandro
how's it going?
Pretty well thanks, I'm in Tuscany and I'm doing a lot of math! What about you?
I'm just back at work atm, been reading a lot of number theory again
as usual
You're reading about operator algebras right?
Technically I can call it algebra :P
iirc there are some books on operator algebras going free in my math department atm
could ninja them for you if you like
9:12 AM
That could be a good excuse to come to Heidelberg :P are they moving stuff in the library?
a bunch of the pure math group left the university
and they left lots of books
Ah makes sense
so I got a copy of Rudin's real and complex analysis
9:13 AM
I actually own a single physical math book
I've got too many
and have read too few
I only have Shoenfield's mathematical logic and I haven't even read it all
All the books I'd buy are way too expensive
I have lang's ANT
and Edward's "Genetic intro to ANT"
and Rudin and some other junk
Galois theory etc.
Genetic intro to ANT?
yeah it's like
basically it's the story of fermat's last theorem
so it develops Kummer's proof in terms of ideal numbers and such
and proves the equivalence of regularity and the divisibility of Bernoulli numbers and stuff
it's quite cool
9:20 AM
Sounds like it but what does it have to do with genetic?
well it's because it's how ANT was born
there are some cool little facts in there
the story of how Lame was laughed out of the paris academy and stuff
and how Kummer's notation all derives from a paper by Jacobi on reciprocity laws
Sounds very cool actually, I always considered the history of ideals to be very interesting, but I don't know anything else about ANT's history
It's very cool, but it's so hard to do anything without ideals
9:30 AM
Are Serre classes of abelian groups closed under extensions?
So if $\mathfrak{C}$ is a Serre class of abelian groups and $H \in \mathfrak{C}$ where $H$ is a subgroup of $G$, is $G$ also in $\mathfrak{C}$?
What is a Serre class?
A Serre class of abelian groups is a non-empty collection $\mathfrak{C}$ of abelian groups satisfying the property that if $0 \to A \to B \to C\to 0$ is a short exact sequence then $B \in \mathfrak{C}$ if and only if $A, C \in \mathfrak{C}$.
here they are closed under extensions by definition
Yeah, I donno if the definitions are equivalent though, I can show closure under taking subgroups and quotient groups though
I think I need to think about this some more
Note that the way you're using the term "group extension" is nonstandard, usually we say that $G$ is a group extension of $H$ (by $F$) if they fit an exact sequence $1\to F\to G\to H\to 1$
9:37 AM
Oh I didn't know that
I think in that case it is closed under extensions
Thanks @AlessandroCodenotti
@Perturbative definitely not
consider finite groups
Why are only Abelian groups considered?
I was thinking that amenable groups have this (and others) closure property, but then I realized that all abelian groups are amenable so it's not a very interesting example
10:10 AM
@AlessandroCodenotti The notion of Serre class is applied to homotopy groups and homology groups and most theorems about Serre classes fail when the space doesn't have abelian pi_1 acting trivially on all higher homotopy groups
I see, thanks
1 hour later…
11:25 AM
@TedShifrin thank you very much! As expected, your explanation is perfect. Thank you so so very much.
11:47 AM
@AlessandroCodenotti The key theorem is that Hurewicz theorem starts to hold modulo Serre classes. If $C$ is a Serre class, $X$ is said to be $n$-connected modulo $C$ if $\pi_i(X) \in C$ for all $0 \leq i \leq n$. Then $\pi_{n+1}(X) \cong H_{n+1}(X)$ modulo $C$, that is, the Hurewicz homomorphism $\pi_{n+1}(X) \to H_{n+1}(X)$ has kernel and cokernel in $C$
Higher homotopy and homology groups are all abelian, so that's the relevant context
12:29 PM
Let $X$ be a Banach space and $F(X)$ the set of finite rank operators over $X$. Is $F(X)$ spanned by rank 1 projections? I think this is false, but I know it holds in Hilbert spaces
you mean rank one maps, not projections right?
I mean projections
In the Banach space setting I just mean that $p^2=p$
If $u$ is a rank one map on a Hilbert space $H$ then it is the projection on $u(H)$ scaled by a scalar, right?
no (look at $|a\rangle \langle b|)$
but the thing you say is true for hilbert spaces
@s.harp I'm not sure what do you mean with $|a\rangle \langle b|$
@AlessandroCodenotti Let $a, b\in H$, then $|a\rangle\langle b| (v) = \langle b, v\rangle \cdot a$
its physics notation, where you would write $|a\rangle\langle b|v\rangle$
12:45 PM
Oh ok, Murphy uses $(a\otimes b)(v)=\langle v,b\rangle a$
Anyway I don't see what goes wrong with it
(in physics the first component is also the anti-linear one)
well $a\otimes b$ is rank one but not a projection onto its image
Isn't it a scaled projection?
if $a,b$ are orthogonal $(a\otimes b)^2 = 0$
Hmmm I'm confused now. In the proof that $F(H)$ is spanned by rank one projections Murphy writes every $u\in F(H)$ as a linear combination of $n=\dim u(H)$ rank one projections
you can use the polarisation formula $\sum_{k=0}^3 i^{-k}(a+i^kb)\otimes (a+i^kb) = 4a\otimes b$ (modulo errors of exponent and sign)
12:56 PM
I see where my mistake was. Murphy's proof assumes wlog that $u$ is positive, so in the general case I need to put together the linear combinations I got for $u^+$ and $u^-$
While $a\otimes b$ is not positive for orthogonal $a$ and $b$
that is true
@s.harp which thing are you referring to here?
Actually what's $\sigma(a\otimes b)$?
@AlessandroCodenotti to this thing
Oh ok, do you also know whether it holds for Banach spaces?
@AlessandroCodenotti nope, no clue
1:03 PM
I see, maybe I'll ask on MSE later
@AlessandroCodenotti if $a,b$ are orthogonal its $\{0\}$, else it should be $\{0\}\cup \{\langle a,b\rangle\}$
If I didn't mess up my calculation $(a\otimes b)^\ast=b\otimes a$, so it is self adjoint for $a=\lambda b$, which is needed to be positive
yes, i believe that that is correct
Makes sense, since $a\otimes a$ is a projection iff $\|a\|=1$, so for $a=\lambda b$ we get a scaled projection
In my noob finite-dimensional brain: every matrix can be written as a nonsingular matrix times a projection. This is also known as a Hermite canonical form. Does this hold for finite-rank operators on Hilbert spaces? (Is that part and parcel of the question?)
If $H$ is a Hilbert space, $V, W \subset H$ closed subspaces, $T : V \to W$ an isomorphism between these subspaces, does there exist an extension $\widetilde{T} : H \to H$ to an isomorphism?
You just need an isomorphism of the orthocomplements
1:21 PM
@BalarkaSen yes, looks like it. If $A$ is finite rank then let $K^\perp$ be the complement of its kernel and $P$ the projection onto that space. Let $B: K^\perp \to \mathrm{im}(A)$ be given by restriction of $A$, so $B$ is invertible. Now choose a linear iso $C$ between $K$ and $\mathrm{im}(A)^\perp$ and look at $(C\oplus B)\cdot P$
Yeah my idea was, suppose $T : H \to H$ is finite rank. Now choose a section $G : \text{im}(T) \to H$ of $T$ (easy), and extend that using the above lemma to an isomorphism $\widetilde{G} : H \to H$.
Then $TGT = T$, i.e., $G$ is a generalized inverse of $T$
$T' = GT$ is your projection
It seems for Banach you just need Hahn-Banach or something to extend instead
in banach spaces extending operators from closed subpsaces to the entire thing has problems, hahn banach works if the target is $\Bbb C$ or $\Bbb R$
But if the closed subspace has finite rank, I have no issue, yes?
Because I can choose a basis and every coordinate projection is map to $\Bbb C$
that makes sense
@BalarkaSen not necessarily, since $V=H$ and $W$ a proper subspace is possible
1:37 PM
I think it's true if $V$ and $W$ have the same (possibly infinite) codimension though
Very annoying
@AlessandroCodenotti I'm thinking about the right shift on $\ell^2(\Bbb N)$ here
@AlessandroCodenotti if they have the same (Hilbert) co-dimension their complements are isomorphic $T_\perp: V^\perp\to W^\perp$, and you can define $T\oplus T_\perp$ to be a linear isomeorphism on $H$
1:43 PM
In particular no issues if $V$ and $W$ are finite dimensional subspaces
If we’re on the topic of functional analysis, here’s a challenge: Give a simpler proof of Gleason’s theorem
The proofs I’ve (tried to) read are pretty laborious
(See sections 2,3 for the essential points)
I suppose now my question would be if this is true for compact operators as well? If $T : H \to H$ is compact, can I write $T = AP$ where $A$ is an isomorphism and $P$ is a projection?
@Semiclassical proof by appealing to more powerful result: Every self-adjoint in $B(H)$ is a linear combination of 10 projections
@Semiclassical "A gap of that proof has been fixed in 2018 by V.Moretti and M.Oppio." this might be the first time I see a result by one of the professors from my bachelor in the wild
1:50 PM
this guy says 4 projections: arxiv.org/pdf/1608.04445.pdf but mispells the word orthogonal 3 times on the first page...
@s.harp what is this dark magic
I mean the title has a typo
Holy hell
The proof in the POVM context (see later in the nLab article) is much simpler
As it should be, since you’re assuming much more
1:56 PM
honestly that result (and the paper) is kind of weirding me out right now
its full of things that dont make any sense
@s.harp What's the source for this weaker result?
@Semiclassical P. A. Fillmore, Sums of operators with square zero, Acta Sci. Math. (Szeged) 28 (1967), 285–288.
or P. A. Fillmore, On sums of projections, J. Funct. Anal. 4 (1969) 146–152. or K. Matsumoto, Self-adjoint operators as a real span of 5 projections, Math. Japon. 29 (1984) 291–294.
I didnt read any of those papers tho
btw, when I say that the proofs I've seen are 'pretty laborious'
I mostly have in mind that an entire appendix of one book on quantum foundations is apparently devoted to a simplified proof of such
the audience for that is not mathematicians, so that's maybe a bit exagerrated
but it still seems like the proof of classical Gleason is hard by quantum foundations standards
2:03 PM
the nlab article has two formulations taht seem to say different thigns
@s.harp ?
one says "a state on B(H) is uniquely determined by its value on projections" and the other says "every finitely additive measure on Proj(B(H)) can be extended to a state on B(H)"
well, note that the theorem in section 3 has both a forward and backwards direction
"If dim($\mathcal{H}$)≠2 then each finitely additive measure on $\mathcal{P}$ can be uniquely extended to a state on $\mathcal{B}(\mathcal{H})$. Conversely the restriction of every state to $\mathcal{P}$ is a finitely additive measure on $\mathcal{P}$."
the intro say the second part of what you wrote + restriction is injective
but the first part is missing and the second part incomplete
for comparison, the version on Wikipedia is:
"Theorem. Suppose $H$ is a separable Hilbert space. A measure on $H$ is a function $f$ that assigns a nonnegative real number to each closed subspace of H in such a way that, if $\textstyle \{A_{i}\}$ is a countable collection of mutually orthogonal subspaces of $H$, and the closed linear span of this collection is $B$, then $\textstyle f(B)=\sum _{i}f(A_{i})$.
If the Hilbert space H has dimension at least three, then every measure f can be written in the form $\textstyle f(A)=\mathrm{Tr} (WP_{A})$, where $W$ is a positive semidefinite trace class operator and $P_{A}$ is the orthogonal proj
2:14 PM
ie every state corresponds to a trace with a density operator?
well, for me a state = a density operator. so I'd say it more as "existence of measure => existence of state"
for me a state is a positive norm one linear map from the algebra to $\Bbb C$
that is an expectation value functional
which norm?
$\|\omega\|=\sup_{A\in \mathcal A, \|A\|≤1}|\omega(A)|$
As I look at this, I find myself wanting a constructive version of this
i.e., given info about the state (in your sense of the word), construct what the density operator would have to be
2:32 PM
@Semiclassical Let $\{e_n\}$ be a Hilbert basis and $P_{n,m}=|e_n\rangle\langle e_m|$. Then $W=\sum_{n,m} W_{n,m} P_{n,m}$ (convering weakly) and $P_{n,m}WP_{n,n}= W_{m,n}P_{n,n}$ so $Tr(WP_{n,m})=Tr(WP_{n,n}P_{n,m}) = Tr(P_{n,m}WP_{n,n}) = W_{m,n}$, thats how you can get the desnity operator
2:48 PM
makes sense
Any time you can bring two systems together in such a way that the final state of one particle depends on the input state of the other, you can make an entangled state by making that input state a quantum superposition.
Entanglement is created in a _______ manner?
a "local" manner
Is there an operator that continuously deforms a pattern into a non pattern?
$\mathscr D[x]$
perturbation theory?
3:30 PM
@J.Doe Hey J. Doe
'Sup @Ultradark
can i tropicalize some curves for you
I charge 10 per curve
How many curves can you get
3:40 PM
I can knock out 100 curves every hour
Then that's a steep price, ma man.
i said 10
not 10 dollars
I charge small rocks
I deal in Zimbabwean dollars.
I suppose there's no fair price for anything of which there's an infinite amount.
It's called martingale economics or something like that.
3:44 PM
I am going to degenerate some functions
Often to understand a shape
we are going to embed it somewhere
@J.Doe You might ask: What are the simplest equations that will embed a shape into space? By “simplest,” mathematicians want to know two things: How many variables does it need? And what’s the highest power of those variables?
Go on...
For example, the simplest equation that describes the circle requires two variables raised to the second power. Mathematicians say that its “rank” (the number of variables needed) is 2, and its “degree” (the largest exponent needed) is also 2.
so understanding the nature of the equations required to embed a shape in space also provides mathematicians with a deeper understanding of the shape itself — similar to how you could learn something about the shape of a physical object by trying to fit it inside various boxes
@Ultradark I (sort of) know what isometric embedding means. What other types of embeddings are there?
topological embeddings
conformal embeddings
linear embeddings
4:13 PM
Geometry/topology is too weird en.wikipedia.org/wiki/Non-orientable_wormhole
4:47 PM
@s.harp I did ask on MSE in the end if you're interested math.stackexchange.com/questions/3328029/…
2 hours later…
6:35 PM
@J.Doe: You can't say the Möbius strip is "too weird." You should have been playing with it for many years.
The Möbius strip is something you can even create in real life. That basically disqualifies it from being "weird" in the context of math
Also, it is useful, which practically disqualifies it from being math
@TobiasKildetoft its useful? (in part this is curiosity about what you understand the word "useful" to mean)
@s.harp It is used for certain types of drive belts
The Klein bottle is a better example in that you can’t create a physical model of it
@TedShifrin It's the physics bit that I find too weird. Maybe it sounds too weird to me because I don't know any advanced physics.
6:45 PM
@TobiasKildetoft cool, google gives this related tthing (skeptics.stackexchange.com/questions/12260/…)
ok i screwed up the url
i wonder what 'physical' applications one could appeal to with the Klein bottle
i mean, there's a lot of 'topological' stuff in physics nowadays (e.g. topological insulators) so I'm sure there's -some- way it shows up
@s.harp Yeah, that was what I had in mind
but it's necessarily rather more abstract than applications of the Mobius strip would be
One of the main points is that while you don't really get longer life out of the material, you get to have an "active" side with better area
And if the belt is placed at a location where taking it out and flipping it is impractical, then this can be a benefit
6:55 PM
I've actually worked at a place that had huge machine with mobius belt, to think of it.
@Semiclassical I agree. That sure is a paper.
What does it mean to say that 'general relativity doesn't make any requirements about the topology of spacetime'?
@Semiclassical You could even present it as "a paper by Rapoport"!
@Semiclassical There's that thing about the world being a Klein bottle
whatever that means
They found out using topological data analysis(TM)
7:03 PM
@AlessandroCodenotti And I could quite easily say "I'm not Rapoport" if anyone asks
There should be an accessible article by Carlsson somewhere where he talks about this
Hi @Ted
@J.Doe Indeed, some automobile fan belts are (thickened) Möbius bands.
Hi, a @Balarka!
hi, demonic Alessandro ... So is there some reason you're thinking about Cayley-Hamilton?
@TedShifrin That's an application I always liked
7:10 PM
I don't know if you remember the Car Talk puzzler at the beginning of chapter 8 of my book, @Balarka.
(@J.Doe Basically it has the advantage that since it has only "one side", both - well, one - sides equally disintegrate due to friction, increasing sustainability)
Oh, a @Balarka, care to answer this?
@TedShifrin Yeah that's where I must have learnt it
@TedShifrin I'm thinking about operator algebras
and ... ? @Alessandro
7:14 PM
I saw the result I told you about the spectrum and realized you can get Cayley-Hamilton (for $M_n(\Bbb C)$) from it
Oh, right.
Makes sense @BalarkaSen.
@TedShifrin Hm, is it integrable? I can't tell by first glance. It's integrable on the unit $n$-torus in $\Bbb R^{2n}$ for sure.
I'd have to calculate
The best approach of calculating seems to be by doing it implicitly; we know $TS^{2n-1}$ is given by $\sum x_i dx_i = 0$ in $\Bbb R^{2n}$.
Let's do $n = 2$. $\alpha = xdy - ydx + zdw - wdz$, then $d\alpha/2 = dx \wedge dy + dz\wedge dw$.
Uh $\alpha \wedge d\alpha/2 = x dy \wedge dz \wedge dw - y dx \wedge dz \wedge dw + z dw \wedge dx \wedge dy - w dz \wedge dx \wedge dy$
Seems right
We have $xdx + ydy + zdz + wdw = 0$.
Yeah I don't think $\alpha \wedge d\alpha$ is zero on $S^3$
7:32 PM
@Alessandro I think I found a proof for your question
Consider $e_2, e_3, e_4 \in T_{e_1} S^3$. $\alpha \wedge d\alpha$ evaluates to $2$ on this triple, if I am not wrong. These vectors lie on the yzw-plane in $\Bbb R^4$, so out of all those terms above only $xdy \wedge dz \wedge dw$ will eat it and give something nonzero (the rest will make the parallelpiped degenerate under other coordinate plane projections). $x(e_1) = 1$, $dy \wedge dz \wedge dw(e_2, e_3, e_4) = 1$.
So I should be right, unless I am wrong, which is also likely
8:00 PM
Hello. I asked this on math stack exchange, but it hasn't received a satisfactory answer yet. Consider the inequality lim inf x+ lim inf <= lim inf (x+y)<=lim sup(x+y)<=lim sup x + lim sup y. In each of the three inequality signs, we can replace them by either a equal sign or a strict inequality sign. That gives eight possibilities. Can each of those eight possibilities occur, and if so can anyone give me eight pairs of sequences where those possibilities occur.
The possibilities are ===, ==< , =<=, =<<, <==, <=<, <<=, <<<
this is something you would profit from a lot more if you thought about it yourself, some of the cases are very easy
@s.harp It seems to work to me, but I'll check the argument carefully tomorrow morning! Thanks
8:15 PM
@AlessandroCodenotti at the moment there is a problem, in the edit I had modified the projection $P$ (which was originally one with image $\mathcal C + \mathcal I$) to have image $\mathcal C$, now $p_i\circ P$ no longer is a projection, I'm thinking about if that can be fixed
@BalarkaSen Hysterical. See my posted solution. :)
@s.harp ah I see
I'm too tired to think about those details now, I'll postpone this problem to tomorrow
@TedShifrin Cool, our conclusions agree!
I figured there should be a nice symmetry argument.
First I was going to try to write things down in terms of the complex structure, but it's not worth the bother.
You know I don't like to evaluate forms on vectors unless it's an absolute necessity :D
I like the $SO(4)$-invariance!
Very nice argument
8:23 PM
Well, you should double-check me, but I think I'm right, unless I'm wrong :P
I seem to remember that quote from someone auspicious.
I'm confident in my results until such a time as I have reason to not be confident in them.
Not quite so catchy as a Balarka's :P
@TedShifrin re: that cylinder question on main, I get tired of people expecting mindreading
8:33 PM
I think it's someone's homework, but that someone doesn't understand the question. Duh.
The sad fact is that I'm not nearly so patient as I was when I started on this site 6 or 7 years ago.
i have some patience for it, when I'm looking for something to distract myself with
but uh
that's rather conditional
I still try not to be rude (most of the time) :)
it's healthy to walk away from it
8:52 PM
@AlessandroCodenotti ok I think I cleaned it up to be both more readable and more correct now
9:20 PM
What does "sin^3(x)" mean? Is it another way of writing "sin(x^3)"?
typically, it's another way of writing (sin x)^3
as in, the cube of sin(x)
it's not the best notation but it's common
Ahh okay, thank you! :)
9:40 PM
I need a help to show that the $X:={(x,y,z)∈ $R^3$∣x^3+y^3+z^3−3xyz=1}$ is an embedded manifold which is 2-dimensional in R3?
While I can't help you much there, it is interesting to note that $x^3+y^3+z^3-2xyz=1$ does not give a manifold. (it's got singular points)
I began to show that the rank of the matrix of #df# has rank 2, I try to use the regular value theorem. but I do not succeed!
Is there a way to mathematically write the following. X = a list of 2D locations in matrix A where the value is greater than y.
I'm converting code to equation
and not quite sure how to do this here?
usually you'd do set builder notation
{(i,j) : A_{i,j}>y}
ahh i see.
Got it, thank you!
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