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00:00 - 22:0022:00 - 00:00

00:01
I guess some people like to put false results as exam problems.
Oh, I missed that bit of context. That's not nice...
I asked the OP ages ago where he’d found this.
One of my very smart ex-colleagues notoriously wrote real analysis problems he “thought” he knew how to do. It was because of him that we changed the system — not that the change always worked. We made 3-oerson committees write the exams — 2 experts, one non. The non-expert was supposed to hold the feet to the fire. Most didn’t bother.
When I was the non-expert (real or algebra), I worked the exams and raised hell if problems were bad. But most people didn’t care enough … or trusted the experts.
By definition, isn't a set unordered? What would you call an ordered set? Is it assumed a set is unordered until it's qualified with "ordered"? For example in a card game what would you call a hand that must be played in ascending order (e.g. a run like 3,4,5 but not 5,3,4)?
@northerner you call ordered set, ordered set
and that's not even a joke
note that enumerating a set and ordering a set are different things
all you can ask of sets is, does a particular element belong to the particular set
you can't ask, is this particular element before that particular element in this set
00:15
@Jakobian Nuh uh!
You can also ask if it belongs to another set!
uh... okay
sku
sku
00:26
Hi All, I don't understand why there are multiple upper bounds leading to the concept of the least upper bound. Can someone give me an example where there are so many upper bounds and hence the concept of a LUB matters? Thanks
In $\Bbb R$, the set $(0,1)$ has zillions of (uncountably many) upper bounds.
You can give me a specification of them all.
sku
sku
What is an example of an upper bound for the set (0,1). You mean 2, 3,4 are all upper bounds?
Yes, and plenty of other numbers.
So is $\sqrt 2$.
all numbers greater or equal than $1$
sku
sku
But isnt it obvious that 1 is the LUB. So why are even considering these other numbers....
I am missing the intuition I think.
00:31
How do you pick that one if you haven’t first talked about all upper bounds?
In some worlds, there is no LUB for a bounded set.
you don't want to consider those other numbers, you want the least one
thats why you take least upper bound in the first place
Maybe. If you can.
Sku, in the world of only rational numbers, take the set of all numbers $x$ with $x^2<2$. Is it bounded above?
sku
sku
yes. In this example of $x^2 < 2$, if I say $t$ is the LUB, I can always find another rational between t and sqrt(2). This logic correct?
@sku We live in the world of rational numbers. What is this $\sqrt{2}$ you speak of?
I hear tell of such things in the world of real numbers, but 'tis a strange land, I have not ventured there.
sku
sku
good point... let me think
00:40
But that didn't really answer Ted's question, which was "Is the set of rational numbers $x$ with $x^2 < 2$ bounded above?"
Sku skipped ahead :)
maybe Sku didn't see the definition of least upper bound for ordered sets
but just for $\mathbb{R}$
Yes, I believe that is the assumption that we are working under. The generalization to generic ordered sets doesn't seem appropriate or helpful, here.
Best to stick to concrete examples, such as $\mathbb{R}$ and $\mathbb{Q}$ until the idea is more clear. :D
sku
sku
Ok. This is where Rudin uses t = x - (x^2-2)/(x+2) which implies t^2 = 2 + 2(x^-2)/(x+2)^2 and we can get a contradiction. If x^2 - 2 < 0, we have t > x and t < 2. Hence contradiction.
So the answer is that the set of Q such that $x^2 < 2$ has no lub.
00:46
Indeed.
Indeed.
Indeed.
But we need to talk about upper bounds before we can broach the subject of LUB.
sku
sku
Rudin in theorem 1.11: Suppose S is an ordered set with the least-upper-bound property,
B c S, B is not empty, and B is bounded below. Let L be the set of all lower
bounds of B. Then .......
00:48
There seems to be something wrong with your keyboard...
yeah its like if there was a lake and you want to feel the edge, you need to put your hand into the water first and then slide towards you till you grab the edge
so with LUB you need to go into the upper bounds and then you slide down until you approach the least one
*assuming its more stoney lake and not sandy
Great. Now my feet and hands got cut up.
no, like the big flat rocks
your feet are fine
Can I wear shoes? I don't like going barefoot...
I think beaches like that are in Scandinavia somewhere?
sku
sku
00:53
thank you. Let me read some more and get back.
Take care.
01:43
Wow, I've got an amazing post for you all ༼ つ ◕_◕ ༽つ
0
Q: Let me get this straight: For any odd prime number $N$, the sequence $( (X + i)^2 + 1 \pmod N, i=1...n)$ eventually will converge to $1 \pmod N$?

Daniel DonnellyTake the function: def f(x0:ZZ_N) -> tuple: x = x0 while True: y = x**2 + 1 if isprime(y): return x x += 1 That's what got the wheels turning in my 🧠. Let our computer memory integers typed ZZ_N be modeled finitely ie as usual on a modern machine t...

Kind of arrogant to call your own post "amazing", no?
Not if it’s pure sarcasm.
Ah... tone is not easy to read on the internet... :/
Well I got one upvote ╰( ̄ω ̄o)
I’m not saying he meant it as such.
 
3 hours later…
04:44
Consider irreps of $sl(2; \mathbb{C})$ labeled by positive integer $m$.
I am confused about how to prove that given two irreps $\pi_m$ and $\pi_{m'}$ such that $m = m'$ are isomorphic
04:55
Isn’t that backwards?
05:22
@Jakobian I think your explanations are perfect. Perfect. Everything, down to the last minute details.
05:40
Hi chat!
How can we solve for solution to the differential equation $\frac{dx}{dt} = \lambda (x + A + \gamma \cos(t))$ with initial condition $(x_{0},t_{0})$
where $A, \gamma$ are constants. Seems like a non-autonomous system
05:53
@SillyGoose Well try to imagine maps between them? What might a kernel of such a map be?
@TedShifrin Not so much disagreeing with this, but what would you call $M_{r,s}(k)\times M_{s,t}(k)\to M_{r,t}(k)$? I thought of this family of maps for $r,s,t\in \Bbb N$ as being the 'matrix operations'
I don't know if it forms a structure on its own (although I feel like one could try to set up some bigraded algebra type object, and take as convention that multiplying 'illegal pairs' gives zero, or something - or maybe one includes all the finite matrix sets here into some infinite matrix set or something, I dunno)
I guess technically one is meant to think about these as being actions
@tigre Of course matrix multiplication is called an operation, but in your setting it is not an operation on a set.
06:08
Ah yeah I just saw on wiki https://en.wikipedia.org/wiki/Operation_(mathematics)

They introduce language that distinguishes between these types of things. Your message was about what they call 'internal operations', and they leave n-ary operation as $\omega:X_1\times X_2\times \dots\times X_n\to Y$ without assuming $X_i=X_j$ in any case
So I suppose 'operation on a set = internal operation' makes sense
 
1 hour later…
07:19
@TedShifrin I totally agree with this
 
3 hours later…
10:43
@SoumikMukherjee Oh! Thank you
 
1 hour later…
12:01
One of the big problems with smoothness seems to be that $x\mapsto \|x\|$ is not differentiable at $0$. But this can be avoided if we compose it with an analytic function of the form $f(x) = \sum a_n x^{2n}$ since $x\mapsto\|x\|^2$ is smooth everwhere. This seems to be one way to handle this, the other being separated from $0$
12:58
is the process of making cosets of group identical to the process we use in the proof of Lagrang's theore?
the only difference is that the subgroup is a normal subgroup now?
and the latter fact means the cosets form a group?
in both proofs, we do gH, using $g$ that are not in the previously constructed lists
@RyderRude calm down. One thing at a time
The proof of Lagrange theorem (that I'm thinking of) shows that $|H|\cdot |G/H| = |G|$ for any group $G$
Here $G/H = \{xH : x\in G\}$ is defined for any subgroup $H$
This set $G/H$ is interesting regardless of if $H$ is normal or not
Writing $[x] = xH$, as its the equivalence class of equivalence relation $x\sim y \iff y^{-1}x\in H$
We see that the operation $[x]\cdot [y] = [xy]$ is well-defined iff $H$ is a normal subgroup, in which case $G/H$ is a group
13:24
thanks. this is what i meant
@RussiaMustRemovePutin In my opinion? Yes.
14:25
@XanderHenderson I think the whole question should be deleted or at least closed
It doesn't contribute anything to the site
@Jakobian I don't know that I agree.
The question is unclear and makes no sense. Its just ramblings of someone deeply confused
It is a pretty common misconception / problem.
There is a lot of confusion between what is a number, and what is a representation of a number. It is an extremely ordinary and common thing for students to get stuck on.
Sure. If that was what the question is about I'd agree
I mean, it seems pretty clear to me that this is exactly what it is about...
14:29
The question is about "why can't I divide 100% into exactly three parts"
So clearly its not
If all you read is the title, you might get that impression, I suppose.
If you read into it too hard you might get an impression that the question is anyhow deeper than it is, I suppose
It is not a very well phrased question, and I suspect that if it were asked today, it would be closed and deleted, but it is from 2014, and we are generally a bit kinder on older questions. And, again, I think that the asker is clearly struggling to articulate a problem which is very common.
@Jakobian You seem to be extraordinarily dismissive of this problem, which pretty clearly demonstrates (to people who have been teaching for a while) a very common mistake which students make.
I don't think its a problem of author not knowing how to articulate it. Pretty sure the problem was before that
I don't think author thought much about how to articulate it, or thought about the problem
Its the negligence thats the issue, I'm pretty sure. The answers are generous in their interpretation
Just like how I think you were generous to interpret it as a problem of representation of number vs what is a number
Look, I've told you that this looks very much like the kind of question that students who are struggling with the representation of numbers vs the numbers themselves actually ask. My experience with exactly this kind of question, in in-person interactions with students, makes it very clear what the underlying problem is. You have basically responded with "Nuh uh! You're wrong!"
Which... okay, fine. I guess we'll have to agree to disagree.
14:42
@XanderHenderson You're missing the point
The point is that you're overinterpreting what the question is about
Its not an agree to disagree type of thing
Again, all you are saying is "Nuh uh! You're wrong!" This is not productive. I'm leaving, now.
No
The meaning of the question is not included in the question. You're giving it a meaning because of what you think is the meaning
If anyone here acts like "I'm right, you're wrong", its you and only you
The position of "I've interacted with students in person so I must be right" is just ignorant
15:34
For what it's worth, I think Xander is right.
If you're not willing to elaborate on it, that's completely irrelevant
That too is incorrect. People can voice their opinion in a binary like this, and leave it at that, especially when you are very confrontational, and I don't have the energy for that
15:49
That's null
Irrelevant to the current discussion
You can voice your opinion. Will it change anything? No
Why are you like this lol
There is a difference between voicing your opinion and contributing to an argument
Yes, there is, but you seemed to miss it above
I didn't miss anything. You simply misunderstood in what way your opinion is irrelevant. So I've elaborated, its irrelevant for the discussion
It would be relevant if it were elaborated upon
Do you believe that one can be confused about something, but the very confusion stops one from being able to state what they are confused about? Cases where the resolution of their confusion actually dissolves the confusion, rather than answering a concrete question?
15:55
@Jakobian I would strongly recommend that you let other people be wrong on the internet.
Your comments are starting to cross the line into hostile, and, with my moderator hat on, I'm not going to tolerate it.
@XanderHenderson They're not hostile, you're over-interpreting my responses
@XanderHenderson you got me I suppose
@Jakobian Can you answer my question please
@tigre I think that I am going to insist that this conversation is over. Please let it be.
@tigre I can try, I didn't read it yet. I was taken aback because of Xander's suggestion that I might be hostile
lmao
16:02
@tigre I see your point, yes, I think that's reasonable
I figured something like that was what Xander had in mind, and one finds when they teach the same content over and over again, the same types of errors appear, and students ask the same strange broken questions about those same errors. So the teacher/tutor/lecturer slowly learns how to answer weird broken questions - where in particular, the response they get is 'oh wow, that's exactly what I meant to ask about', etc.
I also know that with strongly decreasing frequency I've asked a question that ended up being nonsense, but the response I got was both why I realised it was nonsense, and resolved my hidden confusion that I couldn't express
A question from 2018 I posted here is beyond nonsense lol
I get that view, but I ask questions myself and I know I didn't put any effort into some of them before asking, and I wasn't particularly confused about them either... I just posted them out of laziness.
The question looks lazy and not that the OP is confused to me
Here is me being beyond confused:

https://math.stackexchange.com/questions/2594017/how-to-treat-isomorphic-objects-in-a-category
The comments are pretty funny
@Jakobian Sorry, not that the OP is not confused, of course they are. Its more that the question is too open ended, too much there can be taken up to interpretation, even if it is about decimals, its not clear at all, and attempts at clarifying it are really adding your own interpretation to the question, so its better if its closed
I mean that's possible. I do think one can have expertise in interpreting such things though. I.e. some people might be able to better associate a hidden question with the asked semi-nonsense
16:13
Its just that I think there's the gap at what could be feasible assumption to make, and the gap here is too large
And people nowadays are going to post their own answers with their own interpretations... its just nonsense that its not closed yet
Wait are we even looking at the same question?
The one with 100% not being able to be divided into three parts? I'm talking about that one
Oh this is embarrassing, I wasn't even looking at that one
Oh this question is interesting
Some of the answers are quite cool
Bad questions can produce great answers
I'm thinking if free commutative semigroup or free commutative monoid fit better in context of unique factorization
I was thinking about that earlier today
I was thinking monoid
16:25
I think I want something that would include both
Well do you care more about unital rings, or arbitrary rings?
I care about unital and non-unital rings, but I won't work with the latter
For semi-groups I think situation is a bit different
Well I was just thinking that usually the factorisation is in regard to the multiplicative monoid underlying a unital ring
yeah, and that would be hard to extract internally if we don't have a monoid
I suppose you're right, lets take monoids just so that the work is easier
Is every monoid the multiplicative monoid underlying some unital ring?
16:32
no, because the latter always has absorption element $0$
Oh, good point
those are called absorption monoids
And is that now enough to endow it with a second operation for which it is an abelian group, with compatibility, so as to give it ring structure?
Hmm I guess not
I'm guessing no but its not immediately obvious to me, I need to think
16:38
Ah yes
The unital ring with $p$ elements is only $\mathbb{Z}/p\mathbb{Z}$, where $p$ is prime
But there's massive amount of unital semigroups with absorbing elements, especially as $p$ gets larger
Any semigroup can be adjoined two points $0$ and $1$ to make it an absorbing monoid
but the number of semigroups of size $n$ raises almost at an exponential pace if not faster
hmm... now that I think about it, I think I did attempt to make a list of which semigroups come from a ring structure, and which come from a category
by the latter I mean that you can give category a semigroup structure by setting $f\circ g = 0$ when the composition is undefined
this family of semigroups seems interesting but isn't unital
I thought this was interesting: https://math.stackexchange.com/a/2626383/519033

But I have to go now, so I'll check back later to see what you have! Enjoy!
Okay, see you later
The answer seems to be generalization of unique factorization, assuming that the group of units is trivial
Interesting nonetheless
I have a couple of ideas that this brought me to ponder about
17:09
how would you call an analogue of removing signs
absolution? Absolute value?
$\textbf{absolution}$:
ecclesiastical declaration that a person's $\textit{signs}$ have been forgiven.
Can I tell that the matrix $((1, 2), (2, 1))$ is diagonalizable without calculation?
Or, How can I tell it?
one: there's a fairly well known theorem that real symmetric matrices are diagonalizable
Or look for eigenvectors
i'm not sure if you have something more specific in mind about how to "tell"
In this instance $(1, 1)^T$ is a pretty obvious choice
17:15
if someone gives me some random matrix of real numbers, i absolutely do not know how to "tell" without some kind of calculation
@leslietownes Nope, that's it.
but if you give me a symmetric matrix, i do
the other guess would be $(1, -1)^T$
they're both eigenvectors? Great
jakobian is pointing to another kind of idea which is that if for whatever reason you can 'see' a basis of eigenvectors, that is another way to tell, although if the question is "how/why would i be able to do that without calculation" my general answer might be "i don't know"
sometimes for whatever reason it's definitely easier or at least possible to identify eigenvectors without running some general diagonalization algorithm
and if you find enough of them, you know that you have a diagonalization even if you haven't run a general diagonalization algorithm
@leslietownes read the mind of your lecturer
open the third eye
become one with the exercise sheet
Question: The functors $S\mapsto S^1, S\mapsto S^0$ of semigroups obtained by adjoining $1$ and $0$ element respectively are defined so that $S = S^1$ if $S$ is a monoid, correct?
So that they're adjoint to the forgetful functor
17:23
its maybe mildly interesting that for general real symmetric matrices, relying on the theorem is always possible in a way that being able to write down "nice" eigenvalues/eigenvectors is not
sort of how compactification of a compact space is defined to be itself
8
Q: Are these adjoint functors to/from the category of monoids with semigroup homomorphisms?

Thomas KlimpelDo the forgetful functors $G_H:\bf Monoid \to \bf Semigroup^1$ and $G_O:\bf Semigroup^1 \to \bf Semigroup$ have left and/or right adjoints? Here $\bf Semigroup$ is the category of semigroups with semigroup homomorphisms, $\bf Semigroup^1$ is the category of monoids (semigroups with identity) with...

here's something interesting discussion about the category of semigroups, semigroups with unity, and category of monoids, and functors between them
In Vretblad's Fourier Analysis and Its Applications, there is the following exercise.
> Let $f$ be a $2\pi$-periodic function with (complex) Fourier coefficients $c_n, \ n\in \mathbb Z$. Assume that for an integer $k>0$ it holds that $$\sum_{n\in\mathbb Z}|n|^k|c_n|<\infty.$$ Prove that $f\in C^k$.
I'm dumbstruck on this one. The condition above means that the $k$th termwise differentiated Fourier series of $f$ converges absolutely and (I think) uniformly to a function $g$ (maybe $g$ is even continuous, I'm not sure). How do we know that $f=g$?
If I have a map $f:S\to T$ between the two monoids such that $f(1)\neq 1$ then I need to adjoin identity to both to make the morphism unital, so I do need to adjoint it, even if it was a monoid already
and similar thing applies to absorbtion elements
17:38
@psie embarrassing, I meant $f^{(k)}=g$
@psie Yes, uniformly too
And is $g$ continuous?
psie: can you answer this in the case k = 0? note how the question from the other day (about a specific c_n generated by solving some differential equation) maybe fits into this setup
Since Fourier series together with all up to $k$th derivatives converge uniformly...
yeah, very much so
17:41
doesn't that automatically imply that the limit is $C^k$?
but does the limit equal the $k$th derivative of the function?
what are you asking
In the mathematical field of analysis, uniform convergence is a mode of convergence of functions stronger than pointwise convergence, in the sense that the convergence is uniform over the domain. A sequence of functions ( f n ) {\displaystyle (f_{n})} converges uniformly to a limiting function f {\displaystyle f} on a set E {\displaystyle E} as the function domain if, given any arbitrarily small positive...
this is a classic isn't it
good old classic, indeed
ok, it maybe follows from that
Yep. It implies that the function to which the Fourier series converges is in $C^k$
Moreover, there should be some theorem that tells you that since $k > 0$, it follows from uniform convergence that the Fourier series converges to the intended function
hmm, yes, if the function $f$ is continuous I believe
17:45
maybe also buried in there is the ever-useful fact that continuous [or L^1] functions with the same fourier coefficients have to be [a.e.] the same
not 100% sure how you are thinking about this
@psie all they say is that $f$ is $2\pi$ periodic and $c_n$ exist
yes, there probably is a missing assumption (either book's fault or psie's), that $f$ is continuous
Example... $xy$ and $yx$ are invertible but $x, y$ aren't?
From definitions this implies there are $a, b$ with $xa = bx = 1$
$a = bxa = b$ of course
so $x, y$ are invertible
18:12
I am reading Hogg, McKean and Craig book for Introduction to mathematical statistics 6th ed. There are 2 two definitions there for continuous random variable (R.V): 1- R.V for which space is continuous (for e.g. space is real numbers) are continuous R.V. 2- R.V is continuous if its cumulative distribution function is continuous. Which one is correct or both are equivalent in some way?
what do you mean by the first definition
@Thorgott are you here?
I need someone that enjoys algebra
18:28
sup
are you interested in a question about monoids
cant guarantee ill have something to say about it, but ill hear it
If $A\subseteq S$, let $S/A = S/\sim_A$ where $\sim_A$ is the smallest congruence for which $a\sim_A a'$ for all $a, a'\in A$. If $S$ is a monoid, let $\mathfrak{g}(S)$ be its group of units and call $\mathfrak{a}(S) = S/\mathfrak{g}(S)$ the absolution of $S$. Call a monoid absolute if $\mathfrak{a}(S) = S$. Is the absolution of a monoid an absolute monoid?
congruence meaning $a\sim a'$ and $b\sim b'$ implies $ab\sim a'b'$?
Yes
In other words, a ring is absolute when its group of units is trivial
@Thorgott That's for $A\neq \emptyset$, lets say $S/\emptyset = S^0$ where $S\mapsto S^0$ is the functor adjoining an absorbing element to a semigroup
18:35
fair, but $g(S)$ is always non-empty for our application
Yeah
I'm thinking about iterating $\mathfrak{a}$ operation, but for this I would probably need to know if it even needs iterating. It seems like there should be an example
Say $S = \{0, 1, 2\}$ where $(\{0, 1\}, *)$ is $\mathbb{Z}/2\mathbb{Z}$, $2^2 = 2$ and $\{0, 1\}*2 = \{0, 1\}*2 = \{0, 1\}$ maybe?
$1*2 = 2*1 = 1$ I mean, I think that works
$0*1 = 1*0 = 1$ and $0*x = x*0 = x$
hi
there is someone?
think of how upsetting it would be if the answer were no
I think the answer is positive in the commutative case
my example is not associative
18:48
consider the relation $\sim_A^{\prime}$ defined by $x\sim y$ if there is an integer $n$, elements $z_1,\dotsc,z_n\in S$ and $a_1,\dotsc,a_n,a_1^{\prime},\dotsc,a_n^{\prime}\in A$ such that $x=a_1z_1\dotsc a_nz_n$ and $y=a_1^{\prime}z_1\dotsc a_n^{\prime}z_n$. this respects the multiplication, is reflexive and symmetric. it is not transitive, but the transitive relation generated by it (so taking sequences of such relations) should yield the congruence $\sim_A$.
now, if $S$ is commutative, then $g(S)$ has the wonderful property of being saturated, i.e. $xy\in g(S)$ implies that $x,y\in g(S)$. then, from the above relation, it is easy to see that $y\sim_{g(S)}1$ iff $y\in g(S)$. so $[x][y]=[1]$ in $a(S)$ iff $xy\sim_{g(S)}1$ iff $xy\in g(S)$ iff $x,y\in g(S)$ iff $[x],[y]=[1]$, i.e. $a(S)$ is absolute.
So you're saying that if $S$ is commutative then $g(S)$ is an equivalence class in $S/g(S)$
yeah, and it has this "saturated" or "prime" property
this doesn't really need commutativity i guess, but rather just that an element has a right inverse iff it has a left inverse (not sure if that property has a name)
Does anyone see why $pu=u^{1/2}p u^{1/2}$?
Does $xy, yx\in g(S) \implies x, y\in g(S)$ not suffice for this?
it would, yeah
I only used one-sided inverse-ness in my argument
oh wait
but what you just said always holds
cause $xy$ having a right inverse implies $x$ has a right inverse and $yx$ having a left inverse implies $x$ has a left inverse, so $x$ is invertible, and vice versa for $y$
19:00
yes I mentioned it above
ah i didn't see that
as you can tell, im too used to thinking in commutative terms
because I thought $[xy] = [yx] = [1]$ would imply that $xy, yx\in g(S)$
@Sha I have no idea what $F(H)$ means. But if $u=pu$, doesn't it follow that $u^{1/2} = pu^{1/2}$?
but since the congruence is being generated, I need a careful argument for this
@TedShifrin $F(H)$ means operators on $H$ of finite rank
19:02
sha: p is the range projection of u. an operator always commutes with the projection onto its range (e.g. if u is positive the operators (u + 1/n)^{-1} u converge strongly to the range projection of u)
aha, I wanted to verify that, but I thought it was not plausible (or I was just lazy)
thanks, I'll do the verification!
well, my construction yields a congruence satisfying $a\sim a'$ for all $a,a'\in A$, but it's also generated as an equivalence relation by relations that are composed of $a\sim a'$ for $a,a'\in A$, reflexivity and compatibility with multiplication, hence contained in the smallest such congruence
let me see
@ShaVuklia you may be thinking/intuiting that the range projection isn't generally in the C star algebra generated by the operator (which indeed might not contain any nontrivial projections). you do need to be able to take strong limits
there should also be a categorical proof, I think
19:08
hmm, okay, I'll come back to this, because I don't see it yet (and I have to go now). thanks already for your comments!
Doesn't it follow from the finite-dimensional spectral theorem that $u^{1/2}$ has the same eigenspaces as $u$ and hence $pu^{1/2}=u^{1/2}$?
Leslie is saying all this fancy stuff over my head.
we don't necessarily have a finite dimensional spectral theorem or eigenspaces, but it does follow from the fact that pu = u that pu^{1/2} = u^{1/2} (e.g. because u^{1/2} is a limit of polynomials in u)
@Thorgott Okay I agree that the transitive closure of $\sim_A'$ is a congruence
Ah, so you need to use some version of Stone-Weierstrass?
No, that doesn't apply.
i wasn't attempting to imply that my approach was the only one, just connecting the question to some ambient operator theory that is used frequently enough to be a common way of thinking about stuff like that
use whatever you'd use to make sense of "u^{1/2}" in general, which you could definitely do using something like stone-weierstrass (meaning, something like identifying 'functions of u' with a function space)
19:14
Yeah, I see that I was being stoooopid. Even though the range of $u$ is f.d., the domain is still all of $H$.
I presume one could apply the Hilbert space spectral theorem, too.
since the range projection of u would commute with u irrepsective of its dimension, 'using' the finite dimensionality would seem kinda odd from some points of view
yeah its all basically "spectral theorem"
This part of functional analysis is something I've never learned.
@Thorgott I think you do need $S$ to be commutative to conclude that $x\sim_A' y$ and $x\in \mathfrak{g}(S)$ implies $y\in \mathfrak{g}(S)$
19:34
If an example exists it must be of size at least four
there are $35$ monoids of size four
$27$ if we consider them up to anti-isomorphism as well
yeah, I agree after all
I'm going to download magma and check the monoids of order four for a counter-example
sounds like a good idea
I mean I don't have it so I'll have to, you know... look around. Or I'll just use GAP
Alternatively, I saw a post by J.E. Pin
6
A: How many monoids of order three are there?

J.-E. PinLet $M$ be a monoid with three elements. Let $G$ be its group of units (elements which have an inverse) and let $I$ be its minimal ideal. Note that if $|I| = 1$, then $M$ has a zero. Let denote by $C_n$ the cyclic group of order $n$. If $M = G = I$, then $M = C_3$. Otherwise, $G$ and $I$ are di...

what he does is consider group of units $G$ and minimal ideal $I$ of the monoid
its probably possible to just check which monoids of order four are there on fingers
all monoids for which $|G| = 1$ or $G = S$ are irrelevant, so that there'll be less checking
20:01
what is "the" minimal ideal
I assume the intersection of all ideals
Since we can take an element from each ideal and consider their product, the intersection is non-empty
A (two-sided) ideal $I\subseteq S$ is a non-empty subset such that $SI\cup IS\subseteq I$
ah, you're right
@leslietownes I'm not sure if I'm understanding you correctly. If $L$ is an operator, and $P$ is the projection onto its range, then we have $L=PL$, and I don't believe that $L=LP$ always holds. E.g., if we take $L:\mathbb C^2\to\mathbb C^2$ such that $Le_1=e_2$ and $Le_2=0$, and if we take $P$ the projection onto $\mathbb C e_2$, then $LP=0\neq L$
sha: okay, here L is positive
sha: in general, P only belongs to the strong closure of the algebra generated by L and L star but if L = Lstar then both will commute with P
i must have missed saying 'positive' somewhere above
20:17
ah, I didn't know that
"here" meaning in your situation above, not "here" in your example
the thing playing the role of L in your example is played by a positive operator u above
Your example doesn't look very self-adjoint.
(Or, of course, positive.)
yea, I interpreted Leslie too broadly, but that got clarified
Interpreting Leslie at all is usually a big error.
if u is not self adjoint then following the same idea you might only get that its range projection is a strong limit of functions of u ustar [or something else that is not guaranteed to commute with u]
20:28
You are working way too hard to avoid ChatJax and mathematical symbols. You're typing ustar now?
If $S$, $|S| = 4$ were to be absolute we need to have $|G| = 2$ and $|I| = 1$ or $|I| = 2$
@leslietownes I think this argument works for me now :D
this is actually an instance where it makes more sense than usual to avoid chatjax, because when there is more than one literal asterisk in a line, it can be interpreted by the chat as wanting to put something in emphasis, even if the asterisks appear separately in different math modes
one can use \ast though (not that I mind you avoiding chatjax)
there's no time for that
:D
20:31
But if you're typing plain text, wtf do you type ustar instead of u*?
mm, let's see how it handles me writing a u* here and a u* here
Most of the time I find his avoiding chatjax beyond annoying.
okay, that looks OK
i'm writing for the audience of people using this chat from old flip phones and nokia bricks
In other words, you're living 15 years ago.
anyway, it's right to be generally paranoid about how this chat formats asterisks if you're doing star-algebra
20:35
That seems like just another reason not to study those things...
And we can't do contravariant functors or pullbacks in general.
Who needs 'em?!
you're the one with diagrams in your thesis, probably you do
@leslietownes My diagrams aren't that annoying. Though I think that i might have a pullback measure in there somewhere that uses a star.
There aren't any convolutions, though.
oh, that's "just" the biproduct in the category of freyd-lawvere regurgitants
why didn't you say so
20:38
@leslietownes Oh, I see. It's all so clear now.
@robjohn An ancient answer of yours just appeared here. Forgive my rude comment, but I cannot follow at all.
@Thorgott Another thing is that if $|S/G| = |S|-|G|+1$ then $S/G$ must be absolute, so we have to only check those monoids for which $|S/G|\leq |S|-|G|$
@Jakobian This might interest you, if you decide to go down the AOC rabbit hole. I actually have no recollection of Radon-Nikodym's depending on AOC.
@TedShifrin I have just added a more complete explanation of $(1)$, I will add how that supports $(2)$, but I am leaving for PT soon, so it will have to wait until I get back.
I'm not one of those people that go crazy about axiom of choice
20:52
I have no issue with (1). But then there was, I thought, a monstrous leap.
I'm all for global axiom of choice
Do well at PT. I have to do my exercises shortly.
Ah, OK, Jakobian.
@TedShifrin $(2)$ is breaking up $[0,\infty)$ into pieces of size $\frac1n$, then applying $(1)$. I will add a description of that when I get back.
@TedShifrin Thanks.
The appearance of the $n$ seems beyond unexpected. And no discussion of limits as $b\to\infty$ seems very unlike you :P
@TedShifrin $a=\frac{k}n$ and $b=\frac{k+1}n$. I will add that to the answer when I get back.
20:56
Oh.
Anyhow, serious amplification is required.
PT well!
21:11
btw @Leslie, can I bother you with a funana Q that's been bugging me for a while now?
21:41
@Thorgott The case of monoids of order four goes like this. Since $|S/G| = |S|-|G|+1$ implies $G$ is an equivalence class, for $|S| = 4$ we only consider $|S/G|\leq 4-|G|$. Then $|G| = 2$ and $|S/G| = 2$ is the only case. If $|I| = 1$ then $S/G$ has a zero, so is absolute. If $|I| = 2$ then letting $S = \{1, a, b, c\}$ with $a^2 = 1$, $1\sim b$ then $bc = b$ or $cb = b$ so $c\sim b$ as well, contradiction.
So $S/G$ is absolute for monoids of order four
so the monoid I'm looking for must be of order $5$ or above
If $|S| = 5$ then $|G| = 2$ or $|G| = 3$, if $|G| = 3$ then $|S/G| = 2$, the cases of $|I| = 1$ and $|I| = 2$ are like for monoids of order four above
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