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Xander Henderson
Xander Henderson
00:01
@copper.hat Noice if you can get it. :D
Jakobian
Jakobian
@Thorgott I was probably close with monoids of order $5$ but I'm too lazy to go and check every case. So I've decided to post it
Checking $|I| = 3$ when $I$ is a simple semigroup of order $3$ was a little annoying
I have a feeling this has something to do with Ore conditions
Maybe not have to do, but the cases are similar, so...
Ted Shifrin
Ted Shifrin
00:27
@Thor @Jakobian I've never seen the notation here before. What is $\underset R{\phantom{e}}S$ for rings $R$ and $S$?
Do people write that for $S$ as an $R$-module?
Jakobian
Jakobian
I think I've seen it before
leslie townes
leslie townes
some operator algebra people do that (maybe not with module structure but something equivalent)
Ted Shifrin
Ted Shifrin
I got to the doorstep of 71 years old without seeing it before.
Jakobian
Jakobian
here my guess, for now, would be Hom sets
Thorgott
Thorgott
@Jakobian upvodted
Jakobian
Jakobian
00:35
thanks
Thorgott
Thorgott
@TedShifrin yeah, I think so
Ted Shifrin
Ted Shifrin
Hom sets?
Thorgott
Thorgott
the point is that ${}_RS$ is the left $R$-module structure on $S$ and $S_R$ is the right $R$-module structure on $S$, I guess
don't think it has much use in a commutative setting
Ted Shifrin
Ted Shifrin
If we're talking about $\Bbb Z_{pq} \cong \Bbb Z_p\oplus \Bbb Z_q$, why bother with this crap?
Thorgott
Thorgott
probably just wanted to emphasize the ground ring and did it in a cumbersome way
Ted Shifrin
Ted Shifrin
00:36
No, not p-adics and q-adics.
Thorgott
Thorgott
I'm more terrified by writing $\circ$ for multiplication
Ted Shifrin
Ted Shifrin
Yes, there was that, too. I first wondered how to compose primes.
Jakobian
Jakobian
@Thorgott ah. That's probably it
Ted Shifrin
Ted Shifrin
In all my years of going to algebra lectures, though, I don't recall seeing that notation. People may have specified it in words or made some specific ad hoc notation at the time.
Jakobian
Jakobian
wikipedia lists that notation too
Ted Shifrin
Ted Shifrin
00:39
Ah, that is the definitive ruling.
Thorgott
Thorgott
I wrote things like ${}_RM_S$ to keep track while proving the bimodule version of the Tensor-Hom adjunction once
Jakobian
Jakobian
I just wonder when I've seen it, was it Vakil, Liu or maybe Aluffi... or maybe Lang?
Ted Shifrin
Ted Shifrin
It's a pain in the rump to typeset properly.
Why did Knuth not think of pre-subscripts?
Jakobian
Jakobian
I imagine people define it when writing a document so its not a chore
Xander Henderson
Xander Henderson
@TedShifrin Because anyone use pre-subscripts is either a maniac or a chemist. In either case, they should be STOPPED before they do more damage.
@Thorgott Case in point.
Ted Shifrin
Ted Shifrin
00:43
Yes, one must write macros.
I don't like the look of ${}_RM$. The spacing is wrong.
Jakobian
Jakobian
@Thorgott If $G$ coincides with set of left inverses, why is $S/G$ absolute?
Ted Shifrin
Ted Shifrin
@Xander How was the martini?
Xander Henderson
Xander Henderson
@TedShifrin I went for Safeway gin and Fever Tree elderflower tonic.
It was good.
Ted Shifrin
Ted Shifrin
FeverTree deserves better gin :)
Xander Henderson
Xander Henderson
@TedShifrin Probably, but I only have a very small bottle of St George, and that is a straight up sippin' gin.
Ted Shifrin
Ted Shifrin
00:52
If that’s the one that’s cured in whisky barrels, I hate it.
From Alameda, almost my old home.
Xander Henderson
Xander Henderson
No, though St George does do a reposado that I have been meaning to try. The little bottle that I have is the one with the red label, which is a rye gin, I think.
Ted Shifrin
Ted Shifrin
Oh, I still hate it :)
Xander Henderson
Xander Henderson
The St George Terroir is my favorite gin---it honestly tastes like the state of California---a little pine-y, with some sage (Artemesia sp., not garden sage).
stgeorgespirits.com/spirits/terroir-gin
Thorgott
Thorgott
@XanderHenderson I don't think I'm a chemist, sir
Xander Henderson
Xander Henderson
@Thorgott By process of elimination, you much be a maniac.
Thorgott
Thorgott
00:59
b-but there's no way I would be a maniac!
Ted Shifrin
Ted Shifrin
A debatable point.
Thorgott
Thorgott
this follows from the relation given yesterday. it suffices to establish that $g(S)$ is an equivalence class. the congruence $\sim$ was the transitive closure of $x\sim'y$ iff $x=a_1z_1\dotsc a_nz_n$ and $y=a_1'z_1\dotsc a_n'z_n$ for some $z_1,\dotsc,z_n\in S$ and $a_1,\dotsc,a_n,a_1',\dotsc,a_n'$ in $g(S)$. if $x\sim e$, we have a chain $x=x_1\sim'\dotsc\sim'x_n=e$.
let's do the first one. if $x_n=e=a_1z_1\dotsc a_nz_n$, then $z_n$ has a left-inverse, hence is invertible by assumption, but $a_n$ is also invertible, so we obtain that $a_1z_1\dotsc a_{n-1}z_{n-1}$ is invertible. by induction…
Xander Henderson
Xander Henderson
@Thorgott Sometimes, the truth hurts. I'm very sorry for your loss.
Jakobian
Jakobian
@Thorgott I see, yeah
Can I include this in my post?
referencing you of course
Ted Shifrin
Ted Shifrin
Credit Sir Thorgott!
Thorgott
Thorgott
01:52
of course
Jakobian
Jakobian
02:19
@Thorgott alright, hope the way I phrased it is good
Juliamisto
Juliamisto
Hello. It is well known that $a < b < c$ is the abbreviation for "$a < b$ and $b < c$." One also writes $a < b = c$ for "$a < b$ and $b = c$". Does one usually write $a < b \geq c$ for "$a < b$ and $b \geq c$," too?
Jakobian
Jakobian
You can write that and it will make sense but I don't think people usually write that because it looks confusing
Juliamisto
Juliamisto
It is true that I do not usually see that; one just says "a is less than b and b is greater than or equal to c."
It does not matter, anyway.
Jakobian
Jakobian
If you have $a < b$ and $c < b$ you might write something like $a, c < b$ though
Juliamisto
Juliamisto
That is true.
By the way, I guess that there exists an area of mathematics in which the word "strictly" is often used for clarity. I have met phrases such as "strictly greater than," "strictly less than," etc.
(Well, I guess that "etc." is not used here.)
Jakobian
Jakobian
02:36
We've already discussed this before
Jan 8 at 2:44, by Juliamisto
Hence there is "strictly positive".
Juliamisto
Juliamisto
Yes.
I just met new phrases with "strictly."
See you. I still have work to do, even on weekends.
Jakobian
Jakobian
yeah, this is the same as for positive/negative
Jakobian
Jakobian
03:18
@Thorgott Eric solved it
@Thorgott If $G$ is a group and $S$ is an absolute monoid, then $G\times S$ has absolution $S$
I think this is what you meant with disjoint union before?
So yes, $|(S\times G)/\{1\}\times G| = |S|$ so the bound is achieved for $|S|\leq (|S|-1)|G|$
we just have to take, say $G = \mathbb{Z}/2\mathbb{Z}$ and $|S| = 2$
Ted Shifrin
Ted Shifrin
03:56
@Jakobian Eric is wise.
Jakobian
Jakobian
this also gives me an example that I can't characterize monoids with unique factorization as "$\mathfrak{a}(S)$ is free"
were I define the former as monoids isomorphic to $G\times S$ where $S$ is free and $G$ is a group
this question arised because I was looking at internal characterizations of this property
nickbros123
nickbros123
04:12
@AlessandroCodenotti there's probably a good combinatorics problem hiding in there
Jakobian
Jakobian
Split exact sequence
In mathematics, a split exact sequence is a short exact sequence in which the middle term is built out of the two outer terms in the simplest possible way. == Equivalent characterizations == A short exact sequence of abelian groups or of modules over a fixed ring, or more generally of objects in an abelian category 0 → A → a B →...
pretty sure this is related
So I think I need to require that there is a retraction $h:S\to \mathfrak{g}(S)$
Jakobian
Jakobian
04:54
Suppose there is a retraction $h:S\to\mathfrak{g}(S)$. Then $f:S\to \mathfrak{g}(S)\times \mathfrak{a}(S)$ defined by $f(x) = (h(x), [x])$ is a homomorphism. If $u\in \mathfrak{g}(S)$ and $[x]\in \mathfrak{a}(S)$ then $f(u(h(x))^{-1}x) = (u, [x])$ so $f$ is surjective. Not sure about injective though
 
7 hours later…
nickbros123
nickbros123
11:31
math.stackexchange.com/q/4848137/1118236 this question has many things going wrong with it in terms of meeting the guidelines, but is "needs more focus" the correct one?
i would rather close it under homework types
Sine of the Time
Sine of the Time
12:12
@TedShifrin sorry, I was busy yesterday. I did not see the message :(
Jakobian
Jakobian
12:24
Pronto
user 85795
user 85795
12:52
:)
Xander Henderson
Xander Henderson
13:22
@nickbros123 "Needs Context" (under site specific close reasons) is more appropriate.
 
1 hour later…
Thorgott
Thorgott
14:38
@Jakobian ah, very nice
@Jakobian no, I just meant the genuine disjoint union
Jakobian
Jakobian
ah then yeah, if its just disjoint union you get $\mathfrak{g}(G\sqcup S) = G$ so $\mathfrak{a}(G\sqcup S) = \{1\}\cup S$
so $|S/G| = |S|-|G|+1$ in this case
John Zimmerman
John Zimmerman
15:14
I think I can show that $J(x)$ is continuous now. Differentiability will be more difficult
for continuity my argument takes advantage of the fact that piecewise linear functions on compact intervals must have finite collection of intervals which the function is affine. Then I use this to argument to show a contradiction, that passing to the limit shrinks the piecewise linear function to a continuous function
 
1 hour later…
Obliv
Obliv
16:32
"A paradox occurs in an axiom system when both a statement and its negation are deducible from the axioms. This in turn implies (by an exercise in elementary logic) that every statement in the system is true, which is hardly a very desirable state of affairs." I don't understand this
If a paradox ensues from axioms, why does that imply that every statement is true?
i.e when an "if P, then Q" statement for when P is true, but Q is both true and false.
that somehow implies that any implication statement starting with P true, is true?
@TedShifrin How do you feel about hungerford's abstract algebra? That's the text we're using for this intro abstract algebra course.
psie
psie
I'm paraphrasing from Bridges' Foundations of Real and Abstract Analysis.
Let $w(t)$ be a nonnegative continuous function on $I=[a,b]$ such that if $f\in C(I)$ and $\int_I w(t)f(t) dt=0$, then $f=0$ (the codomain of $f$ is either $\mathbb R$ or $\mathbb C$). Then $$\langle f,g\rangle=\int_a^b w(t)f(t) \overline{g(t)} dt,$$ defines an inner product on $L_2(I,\mathbb F)$ (here $\mathbb F$ is the codomain of $f$ and I assume this is the same space as $L^2$, although the author prefers to use subscripts).
Thorgott
Thorgott
for any statement Q, we have that Q is equivalent to "(P and (not P)) or Q", because "P and (not P)" is true by assumption, but it is also equivalent to "(P or Q) and ((not P) or Q)", but "P or Q" is true because P is true and "(not P) or Q" is true because not P is true, so this statement is true
Xander Henderson
Xander Henderson
Principle of explosion
In classical logic, intuitionistic logic and similar logical systems, the principle of explosion (Latin: ex falso [sequitur] quodlibet, 'from falsehood, anything [follows]'; or ex contradictione [sequitur] quodlibet, 'from contradiction, anything [follows]'), or the principle of Pseudo-Scotus (falsely attributed to Duns Scotus), is the law according to which any statement can be proven from a contradiction. That is, from a contradiction, any proposition (including its negation) can be inferred; this is known as deductive explosion.The proof of this principle was first given by 12th-century French...
Thorgott
Thorgott
ah, I always forget the name
psie
psie
@psie how can one verify that the integral is well-defined for $f\neq g$?
Obliv
Obliv
16:41
Oh, thank you. @XanderHenderson @Thorgott
Xander Henderson
Xander Henderson
@psie Everything is continuous on a closed interval $I$. There should be no problem with integrability, no? I don't understand your question I think...
psie
psie
@XanderHenderson but $f, g\in L^2$ may not be continuous, right?
Xander Henderson
Xander Henderson
No, but they are square integrable.
Obliv
Obliv
this book defines classes, is that something specific to godel-bernays?
psie
psie
@XanderHenderson right, I guess one wants to bound $f(x)\overline{g(x)}$ from above somehow...
Xander Henderson
Xander Henderson
16:46
Doesn't this just follow from Cauchy-Schwarz? Then you are weighting by something continuous?
Or maybe it's Hölder that I'm thinking of?
Which is the one about $\|fg\|_r \le \|f\|_p\|g\|_q$ with $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$, with $p=q = 2$?
That's Hölder, isn't it.
Yeah, Hölder: en.wikipedia.org/wiki/H%c3%b6lder%27s_inequality
 
1 hour later…
robjohn
robjohn
17:57
@XanderHenderson It seems to be SE chat that is screwing that up
Xander Henderson
Xander Henderson
Yeah, it does that, doesn't it.
robjohn
robjohn
There, that works.
It needs to be 2 byte UTF-8: C3B6
Xander Henderson
Xander Henderson
18:15
@robjohn Cool? I mean, I just copy-pasted from wikipedia. Works on my machine...
Ted Shifrin
Ted Shifrin
It loads fine on my ipad, but I had issues opening this page.
robjohn
robjohn
@XanderHenderson The link works, but other programs don't include ö in URLs. To make it acceptable here, we need to change it to %C3%B6
Xander Henderson
Xander Henderson
@robjohn Like I said, I just copy-pasted from Wikipedia. It seems like their problem if the URLs they give me are malformed. :/
robjohn
robjohn
@XanderHenderson Indeed, they should not include a ö in a URL. They substitute %27 for '.
@TedShifrin is the URL blue only up to the "H" in Hölder?
Now that I've edited it, it won't
Here is the bad link: en.wikipedia.org/wiki/Hölder%27s_inequality
That takes you to a page about "H"
Sine of the Time
Sine of the Time
the page about Holder is more interesting
Ted Shifrin
Ted Shifrin
18:25
No, it was fine for me, @robjohn
robjohn
robjohn
@TedShifrin what browser are you using?
Firefox, Safari, and BBEdit don't include ö in URLs
Obliv
Obliv
18:37
isn't $\mathbb{Z}_n = \{-n+1,-n+2,...,0,1,...,n-1\}$
robjohn
robjohn
@Obliv I thought it was $\mathbb{Z}/n\mathbb{Z}$
but I might be mistaken here
Obliv
Obliv
I think they're the same thing, not 100%. I was just confused because I wanted to double check that the set of integers under congruence modulo n was that set. like negative and positive numbers
but couldn't find a simple example online. google seo just ruined at this point
Jakobian
Jakobian
@robjohn for me it is (not correct)
and I'm on firefox
Ted Shifrin
Ted Shifrin
@robjohn Safari on ipad.
I’ll go try the Mac with Firefox.
@Obliv You’re double-listing, so no.
Jakobian
Jakobian
@Obliv depends on who you ask
different people use $\mathbb{Z}_n$ for different things
Ted Shifrin
Ted Shifrin
18:45
You all are missing the point. Lots of undergraduate algebra texts (including mine) introduce $\Bbb Z_n$ (before quotient groups). But Obliv did not list the elements correctly.
@robjohn The original link opens fine on Firefox on my Mac, too.
And fine on Safari, as well, on the Mac.
Jakobian
Jakobian
I'm on windows
Ted Shifrin
Ted Shifrin
@Obliv What you must understanding is that in $\Bbb Z_6$, for example, $-2 = 4$. Why?
I personally use notation like $-\bar 2 = \bar 4$, so that people don't think they're working with actual integers. Some books write $[4]_6$, to make it clear that it's mod 6 we're working.
robjohn
robjohn
@TedShifrin Here is the original link: en.wikipedia.org/wiki/Hölder%27s_inequality and here is the repaired link: en.wikipedia.org/wiki/H%c3%b6lder%27s_inequality You're saying that both work?
Ted Shifrin
Ted Shifrin
Xander's original link did not appear like that to me. The whole thing was (still is) blue.
Did you change it in here?
robjohn
robjohn
That is because I repaired it
Ted Shifrin
Ted Shifrin
18:54
Oh, then this was all a colossal waste of time :D
robjohn
robjohn
That is what I like to hear ;-)
Obliv
Obliv
19:10
user image
I can't even decipher what this exercise is
"The set $\{\sigma \in S_n\mid \sigma(n) = n\}$ is a subgroup of $S_n$ which is isomorphic to $S_{n-1}$."
It doesn't have any instructions.. it's just a statement
I guess I have to prove that it's a subgroup of $S_n$ and is isomorphic to $S_{n-1}$
Ted Shifrin
Ted Shifrin
Correct.
Thorgott
Thorgott
if an exercise just gives a statement, the exercise should be understood as "Prove <statement>."
Ted Shifrin
Ted Shifrin
I'm inclined to agree that one should ordinarily give an instruction. I am fond of "Prove or give a counterexample" for everything :D
Obliv
Obliv
so it's the set of bijections in $S_n$ of $S$ which leave the $n^{th}$ element of whatever $S$ is unchanged.
yeah that's how I'll interpret it from now on, just not used to this type of text.
Sine of the Time
Sine of the Time
Hi, does anyone have an idea on how to prove this: if $\int f(x)x^ne^{-x^2}dx=0$ for all $n\ge0$, then $f=0$
without using complex analysis
Ted Shifrin
Ted Shifrin
19:20
Can you prove it without the $e^{-x^2}$?
Obliv
Obliv
user image
Not sure I understand this definition. $H$ being closed under the product in $G$ part specifically.
Ted Shifrin
Ted Shifrin
$H$ must itself be a group.
So the product and inverses of elements must be in there.
Obliv
Obliv
right, but I'm trying to discern which one is "bigger"
like subsets
Ted Shifrin
Ted Shifrin
It says $H$ is a nonempty subset.
Huh?
Obliv
Obliv
oh i'm tired, didn't catch the subset part.. read it as set
Ted Shifrin
Ted Shifrin
19:25
Even when I typed it?
Obliv
Obliv
YUP
Ted Shifrin
Ted Shifrin
The point is that you do not have to check associativity and the presence of the identity if you check $H$ is closed under products and inverses. The rest follows from the fact that $G$ is a group.
Obliv
Obliv
Right, so I just have to show that $\{\sigma \in S_n\mid \sigma(n) = n\}$ is closed under function composition and the inverse of that to show it's a subgroup of $S_n$
Sine of the Time
Sine of the Time
@TedShifrin no
Ted Shifrin
Ted Shifrin
Right.
@SineoftheTime First, my question would be on a closed interval. I assume your integral is over all of $\Bbb R$?
Sine of the Time
Sine of the Time
19:29
yes
also, $f\in L^1_{\text{loc}}$ and $\int |f|^2e^{-x^2}dx<+\infty$
Ted Shifrin
Ted Shifrin
The standard question usually is proved using Stone-Weierstrass. But that does not apply on non-compact intervals.
You should think about defining an inner product with weight $e^{-x^2}$. So $f$ will be orthogonal to all the $x^n$.
Sine of the Time
Sine of the Time
the problem arises from proving that the Hermite polynomials form a complete orthogonal system
Ted Shifrin
Ted Shifrin
Well, you can apply Stone-Weierstrass if $f$ is compactly supported. Maybe that's the right approach.
Sine of the Time
Sine of the Time
I don't know that theorem
On wikipedia it's proven using complex analysis and Fourier transform, maybe that's the best way
Xander Henderson
Xander Henderson
Mesquite flour seems to be a lot less dense than wheat flour... I wonder what it is going to do to my bread?
(I'm excited to find out!)
Ted Shifrin
Ted Shifrin
19:44
Will the bread be smoky?
Xander Henderson
Xander Henderson
@TedShifrin No. Mesquite flour is kind of sweet.
Ted Shifrin
Ted Shifrin
I have no experience with it.
Xander Henderson
Xander Henderson
Oh... wait.... you were making a joke.
I missed it. :/
Sorry.
Ted Shifrin
Ted Shifrin
Mix in a little mezcal for good measure.
Xander Henderson
Xander Henderson
@TedShifrin I feel like that might kill the yeast, and kind of ruin the bread...
Ted Shifrin
Ted Shifrin
19:46
I assume it would go in after the yeast is highly active.
Xander Henderson
Xander Henderson
I'm honestly not sure how that would work...
I've never tried to put alcohol in bread in that way.
Ted Shifrin
Ted Shifrin
I guess I haven't either.
Sha Vuklia
Sha Vuklia
20:16
user image
@leslietownes do you agree that the assumption that $a$ is self-adjoint is not necessary?
user image
user image
I think $\phi(a)$ is automatically positive from $\langle \phi(a)x,x\rangle\geq 0$ for all $x\in H$ by the remark and theorem above
leslie townes
leslie townes
sha: i don't read that as assuming that phi(a) is self-adjoint? it might just be commenting on it, on the way to saying that phi(a) is positive?
e.g. self adjointness may just be part of how the author defines "positive", and they are ticking one of the boxes for that
Sha Vuklia
Sha Vuklia
I agree that being positive includes being hermitian, but doesn't Theorem 2.3.5 imply that $u$ is positive, which inculdes $u$ being hermitian?
so the assumption that $a$ is self-adjoint was not necessary, since $\phi(a)$ has been shown to be positive, and hence self-adjoint by Theorem 2.3.5
leslie townes
leslie townes
when you say "i think that phi(a) is automatically positive" i say "i don't see the author as implying otherwise"
i don't know about the rest of it
sorry i'm trying to talk to my daughter at the same time, bbl
Sha Vuklia
Sha Vuklia
okay, priorities first :)
peek-a-boo
peek-a-boo
20:49
@SineoftheTime the Weierstrass approximation theorem (special case of/simpler than Stone-Weierstrass) is a relatively simple (in hindsight) result about uniform approximation of continuous functions by polynomials. There are several proofs of it, for example via convolution with the heat kernel or using Bernstein polynomials. Also, this is a cool result so I would definitely learn about it.
then, letting $d\mu=e^{-x^2}\,dx$ (a finite measure on the line) we can use the uniform approximations to show that for any $g\in C_c$, $\int_{\Bbb{R}}fg\,d\mu=0$, and so $f=0$ a.e by your favorite variant of the fundamental lemma of calculus of variations.
Sine of the Time
Sine of the Time
@peek-a-boo hi! It's definitely in my to do list, however, since the ex I asked before was is a past exam, I think there's a different method to solve it, in fact we did not study the Stone-Weierstrass theorem during the course
peek-a-boo
peek-a-boo
I see, what type of course is it for?
/what results do you already have at your disposal
Sine of the Time
Sine of the Time
it's like an introduction to functional analysis.
So I've basic results about Hilbert spaces
and to prove $V=\{H_n\}$ is a complete system, the hint is to prove that $V^{\perp}=\{0\}$
maybe the best idea is the one proposed on wikipedia
leslie townes
leslie townes
@ShaVuklia i agree that the hypothesis that a is self-adjoint is not necessary. but, i would guess that the author either (1) views the fact that no non-self-adjoint element is going to satisfy "tau(a) >= 0 for all positive linear functionals tau" as apparent enough to the reader to not be worthy of comment, or (2) doesn't want the reader to be focusing about what they do or do not yet know about the full space of positive linear functionals on Cstar algebras at this point in the exposition
e.g. re (1), you don't really get a "more general" theorem if you relax that hypothesis, and re (2), maybe they haven't yet proved that a Cstar algebra has "enough" positive linear functionals, or don't want to recall all of that stuff
peek-a-boo
peek-a-boo
@SineoftheTime yea perhaps… Fourier transofrm and complex analysis are great tools for dealing with all these classical collections of functions
leslie townes
leslie townes
21:02
i guess since he's talking about the universal representation he has proved that stuff already, and maybe just doesn't want to focus on it
Sine of the Time
Sine of the Time
thank you :)
Sha Vuklia
Sha Vuklia
cools, good to know I haven't missed something then! it's an exercise in Conway where $a$ is not assumed to be self-adjoint, so I wanted to double-check here that I can just copy the proof in Murphy for self-adjoint $a$, in case it shows up in the exam. and also, less assumptions makes theorems easier to remember :)
leslie townes
leslie townes
yeah, it's generally possible to present this material so that a tons of consequences about order structure and self-adjointness precede (sometimes by a lot) any knowledge that something like a universal representation exists, and if you do it that way, you might have results that have "redundant" hypotheses whose redundancy only becomes clear later
it doesn't sound like conway takes that approach generally, but some authors do
Ted Shifrin
Ted Shifrin
You mean Conway calls non-self-adjoint operators positive? Ugh.
leslie townes
leslie townes
no not at all
not remotely
Ted Shifrin
Ted Shifrin
21:08
Oh, good.
leslie townes
leslie townes
he's weird, but not that weird
Ted Shifrin
Ted Shifrin
I've noticed in my time on this site that that is a particular cause of ambiguity and confusion.
leslie townes
leslie townes
i associate that with PDE people for some reason (maybe just my own prejudices), i don't know anybody conway-adjacent who does that
Sha Vuklia
Sha Vuklia
the only big criticism I have for Conway's book is that he assumes his (C*) algebras to be unital for a big chunk of the theory...
Ted Shifrin
Ted Shifrin
Give my regards to munchkin. Have we used golf balls for $2\times 3=3\times 2$ yet? :D
Sha Vuklia
Sha Vuklia
21:09
luckily Murphy doesn't do that
Conway only proves the Gelfand isomorphism for unital abelian C-star algebras...
but it's okay, since Murphy got me covered there :D
Ted Shifrin
Ted Shifrin
Luckily for me, this is mathematics about which I have not just zero — but negative — knowledge.
leslie townes
leslie townes
ted: she's not quite there yet. still just addition and subtraction of pairs of numbers (and now, and more frustratingly for her, attempting to read)
Ted Shifrin
Ted Shifrin
Screech decided earlier to turn off the power switch on the surge protector strip. Down went the computer, the internet, and everything else in my office :)
I was a late reader, but once I started I was way beyond grade level.
Xander Henderson
Xander Henderson
@TedShifrin Same here. I didn't read until second grade, but by the end of second grade, I was reading Jules Verne and Ursula LeGuin.
Sha Vuklia
Sha Vuklia
@TedShifrin Luckily for me, Leslie is here and is a funana expert :D I couldn't follow the lectures (due to other obligations), so I never really got to meet fellow students for this course
Ted Shifrin
Ted Shifrin
21:13
Wow, that really is late. I wasn't that bad :D
Xander Henderson
Xander Henderson
@TedShifrin My mother has always chalked it up to a complete lack of interest in the kinds of books that little kids are taught to read with. Who the heck cares about Spot?
leslie townes
leslie townes
sha: one expository divide you will notice is between people who define cstar algebras such that they have faithful representations on hilbert space by definition, vs. people who define them such that it isn't clear (maybe for a while) why they ought to have any, or whether it's only specific examples that have them
Ted Shifrin
Ted Shifrin
My parents read me Winnie the Pooh, so I wanted to read that ...
Xander Henderson
Xander Henderson
@TedShifrin My father read us ethnographies. Because, I am told, that is what I asked for.
Ted Shifrin
Ted Shifrin
But there was also Goodnight Moon, etc. I don't remember the first thing I actually read.
Xander Henderson
Xander Henderson
21:15
And who doesn't want to be read to from Driver's Indians of North America?
Ted Shifrin
Ted Shifrin
And the time to have asked my mom such details passed 25 years ago.
Xander Henderson
Xander Henderson
It's a stone cold children's classic.
leslie townes
leslie townes
munchkin likes stories about sharks, whether they come from the childrens section of the library or the uss indianapolis
Ted Shifrin
Ted Shifrin
LOL ... I'll stick to The Cat in the Hat.
Does she have shark envy? She wants to swallow you all?
Xander Henderson
Xander Henderson
@leslietownes I read bunch of books on European explorers in the Pacific when I was in the 6th Grade (Bligh, Cook, etc). There ere some sharks there, I guess.
leslie townes
leslie townes
21:18
ted: there's something psychological about it. she likes the general concept of wild animals, precisely because of what they are able to do irrespective of the wishes of human beings
hence her occasional alliance with the cat
Xander Henderson
Xander Henderson
@leslietownes I assume the alliance is not permanent only because cats are chaotic neutral, at best?
leslie townes
leslie townes
indeed
Ted Shifrin
Ted Shifrin
Olivia and Screech need to form an alliance. Screech can teach biting techniques.
user 85795
user 85795
22:16
Some teachers are against teaching children how to read too early because they say it is like watering a garden before a rain storm.
user 85795
user 85795
22:30
> The authors conclude that the effects of early reading are like "watering a garden before a rainstorm; the earlier watering is rendered undetectable by the rainstorm, the watering wastes precious water, and the watering detracts the gardener from other important preparatory groundwork".
Re: wikipedia
user 85795
user 85795
23:06
Not so sure the same could be said about arithmetic.
psie
psie
23:17
Let $w(t)$ be a nonnegative continuous function on $I=[a,b]$ such that if $f\in C(I)$ and $\int_I w(t)f(t) dt=0$, then $f=0$. Then $$\langle f,g\rangle=\int_a^b w(t)f(t) \overline{g(t)} dt,$$ defines an inner product on $L^2$. I struggle with verifying the property $$\langle f,f\rangle =0\implies f=0.$$
I know that for continuous non-negative functions, if the integral vanishes, then the function is identically $0$. Here, however, $f$ need not be continuous, right?
Jakobian
Jakobian
23:31
@psie $|f|^2 = 0$
ah
Show that $\int w(t)f(t)dt = 0$ implies $f = 0$ a.e. for any integrable $f$
psie
psie
ok
Jakobian
Jakobian
@psie not for non-negative functions
for continuous functions of any sign
psie
psie
@Jakobian ok, where do you start on this one?
the assumption is $\int w(t)|f(t)|^2dt=0$, and we want to show $f=0$ a.e.
Jakobian
Jakobian
@psie actually yeah there's an assumption missing here
$f$ should be non-negative
psie
psie
ok, it's from Douglas Bridge's Foundations of Real and Abstract Analysis, probably a typo then
Jakobian
Jakobian
23:38
About how to prove this, well, start with simple functions
oscarmetal break
oscarmetal break
Hi, define tensor product $(x_1,x_2)\to x_1\cdot x_2$ from $V_1\times V_2\to W$, where $x_1\cdot x_1$ is linear in each component. How can I show $W$ is a vector space? I tried to find an $(x_1'', x_2'')$ corresponds to $x_1\cdot x_2+x_1'\cdot x_2'\in W$, but I failed to find it.
Ted Shifrin
Ted Shifrin
You should fail.
You need to read your book/exercise much more carefully.
oscarmetal break
oscarmetal break
23:55
oh it is the collection of the map $\tau:(x_1,x_2)\to W$ is a vector space
Jakobian
Jakobian
@oscarmetalbreak define tensor product?
oscarmetal break
oscarmetal break
Typo should be $\tau: (x_1,x_2)\to x_1 \cdot x_2$ from $V_1\times V_2\to W$
Jakobian
Jakobian
Tensor product of $V_1$ and $V_2$ is a pair $(V_1\otimes V_2, \otimes)$ of a vector space $V_1\otimes V_2$ and a bilinear map $\otimes$ from $V_1\times V_2$ to $V_1\otimes V_2$
satisfying a certain property
Not sure what you mean by "tried to find an $(x_1'', x_2'')$"
it seems to me like you're trying to show every element of $V_1\otimes V_2$ is of the form $x\otimes y$, but thats just not true

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