« first day (4389 days earlier)      last day (58 days later) » 

12:10 AM
Sorry!
 
 
3 hours later…
2:52 AM
I posted this in a MathOverflow chat room but it seems far less active than this room.
I'm seeking a widely used notation for "the basis of S". Something like b(S). I'm dealing with I'm dealing with bases of matroids and I want to say things like b(J1) ⊆ b(J2) where J1 and J2 are the independence sets of matroids M1 and M2. b(J1) = basis of J1, b(J2) = basis of J2.
Does anyone know of a widely used convention for "labeling" a basis like this?
 
Probably not a safe notation, since it suggests a "functional" property for b, and b clearly is multi-valued.
 
@anak Thanks for the quick answer! So your concern is that this looks like "b" is a function(al) that takes "S" as an input and outputs a basis, and since each each S can have multiple bases then "b" is multi-valued?
 
Yes, that's what I mean.
There is worse notation, though. Like - being a symmetric symbol, despite it being an asymmetric relation.
 
@anak you're referring to a specific asymmetric relation that uses a symmetric symbol?
 
Just your standard subtraction. Like a-b
It's a pet peeve of mine to use symmetric symbols for asymmetric relations, or vice versa.
In this context, it's a joke, meant to say you can probably get away with whatever notational sin you want.
 
3:03 AM
Ideally the notation I use for a "basis of S" would be one that's used by others though. I was wondering if there's any notation out there for a "basis of S". I'm a bit surprised at how hard it is to find a notation that's been used for this!
 
In my experience, people just state somewhere near the starting of the discussion, "We fix a basis $\mathcal B$ for ...".
But in your situation it sounds like you are flexibly introducing lots of objects, all which have a basis.
But if you are doing this in a systematic, single-valued way, you can definitely safely use the function notation.
 
you could come up with a notation (e.g. "Bases(__)") for the set of all bases for __ and then say stuff like "for i = 1, 2, .. choose B_j \in Bases(S_j)"
its personally hard for me to envision an application where you'd be doing that so often that it would be worth it, but that would be one approach
 
For matroids in graph theory with the ground set being the edges of a graph, there's a finite number of bases possible (and each matroid can't have very many bases in my work). The bases can all be "ordered" easily. So I guess b1, b2, b3, ... bn could be the bases, and I could say b1_S, b2_S, b3_S, b4_S to denote the first 4 bases of S. This avoids the "functional" notation.
 
the answer to 'is there a widely used notation for this' is probably 'no'
 
@leslietownes I'm even wondering whether there's "any" notation for it that's been used by others!
 
3:07 AM
Be the first, blaze the trail!
 
i just invented one. i license it on generous terms for academic users, $50 per use
 
No, just 50 lesliecoin per use.
 
lesliecoin bubbled and i lost my allowance :(((
 
at the moment we're only accepting fiat currency for the Bases license
which is good news because fiat is worthless so it's almost like we're having a sale
 
Before the topic changes too far into the direction of cryptocurrencies, I'd just like to thank you @anak and @leslietownes for the input!
 
3:15 AM
:DD
 
Have either of you worked with matroids?
 
never, but i saw the definition in a combinatorics class once upon a time
 
Cool! I think combinatorics is where I've seen them used the most :)
What about you @leslietownes?
 
3:42 AM
similar experience to anak's
 
When we say "GL_n(C) is path connected.", what topology is being considered in GL_n(C)? Is an element of GL_n(C) being treated as to have come from $R^{2n^2}$?
I think no, the (metric) topology is induced by the operator norm $||B||=\sup_{||x||=1}||Bx||_2$
 
Usually, people identify $M_n(\Bbb C)$ and $\Bbb C^{n^2}$.
 
3:58 AM
koro: it's the same topology
so, either one
 
You know that different norms on a f.d.v.s. Induce the same topology?
Oh goody. I get to write fdvs.
 
i guess for purposes of proving that from matrix entries, if you don't wanna detour through that result, you might wanna define it to have a topology where it's very close to the definitions that a continuous path in GL_n(C) is the same thing as n^2 continuous paths in the matrix entries
if you think in terms of paths implementing elementary row operations it boils down to C \ {0} being path connected
 
yes, on fdvs norms are equivalent.
leslie: you mean after using Schur's theorem on invertible matrices?
 
i dunno what that is (too many things are called 'schur's theorem'). i was thinking, if you know A is invertible <-> A is reducible to I via elementary row operations
 
i.e., every linear operator on a complex space has upper triangular matrix.
If the operator is invertible, then entries on the diagonal are non zero.
@leslietownes then I could connect two matrices in GL_n(C) with a path lying completely in GLn(C)
@leslietownes oh, that sounds very nice.
 
4:11 AM
yeah, i was thinking in terms of proving as a lemma that if A and B are in GL_n(C) and related by an elementary row operation, then there is a path in GL_n(C) from A to B
 
So I'll try to construct a path phi from [0,1] to GL_n(C) such that phi(0)=A, phi(1)=I such that phi "records" row operations and since there is finitely many of them, I'll eventually get I.
 
i wouldn't bother trying to 'write down' the path (just the notation for that would be horrible) but proving that you can get a path from a sequence of row operations, yeah
 
That is, for each step i, I define: phi_i on [0,t_i],say. Then pasting these phi_i together should give me the desired path.
(i+1)st path would be on $[t_i, t_{i+1}]$, etc.
@leslietownes but there seems to be one problem in this.
 
yeah, but in the spirit of a topologist, don't actually set up notation for how you'd piece the path together. just focus on proving that you can do it
 
I could use the same argument in GL_n(R) , which is not path connected.
 
4:20 AM
well, you couldn't, actually, and i didn't say that you could
8 mins ago, by leslie townes
yeah, i was thinking in terms of proving as a lemma that if A and B are in GL_n(C) and related by an elementary row operation, then there is a path in GL_n(C) from A to B
it's not an accident that it's C there, and not a random field
16 mins ago, by leslie townes
if you think in terms of paths implementing elementary row operations it boils down to C \ {0} being path connected
R \ {0} isn't path connected. e.g. no path in R \ {0| connecting -1 to 1
but there is a path in C \ {0} connecting -1 to 1
so e.g. if A' is the result of scaling a row of A by a nonzero scalar k, and f(t) is a path in C \ {0} from 1 to k (which exists by path connectedness of C \ {0}), then "the result of scaling that row of A by the nonzero scalar f(t)" is a path connecting A to A' and you can check that it lies in GL_n(C)
if you can come up with a similar argument for the other elementary row operations, then you can deduce that row operations keep you in the same path component of GL_n(C)
so being able to row reduce something to I means you have a path from something to I
there must be some nicer abstract theory way of seeing it
that's the cue for someone to give a one-line proof
 
@leslietownes yeah :).
thanks a lot, Leslie :). I'll think about this more.
 
4:38 AM
@leslietownes Whereof precisely?
 
whomst?
 
One-line proof of whom?
 
4:56 AM
that GL_n(C) is path connected
i can envision stuff based on row operations, and maybe also an induction proof focusing a lot on matrix entries
 
@Koro You could start with the Jordan form $A=V^{-1}JV$, then there is a straight line path that removes the off diagonal elements of $J$, then there is a path for each diagonal element to $1$ that does not pass through $0$, hence we have a continuous path in $GL(n,\mathbb{C})$ that maps $A$ to $I$ .
 
i guess most of the proofs i am familiar with about connectedness or components of matrix groups do rely on stuff like canonical forms or at least reductions to various matrix forms, maybe it's not so unnatural
 
Personally, I prefer the Schur decomposition. I like a Schur thing.
But the slickest proof is the $\log$ one.
 
yeah, that's a good one. works in infinite dimensional hilbert spaces too
although anything relying on a functional calculus in the infinite dimensional setting is maybe not truly a one liner
 
Or maybe $z \mapsto \det (A+z(I-A))$ and find a path from $0$ to $1$ that avoids the zeros of the polynomial.
certainly the functional calculus is on my intellectual boundary. i got stuck in nets way before that.
I have a full house again.
 
5:08 AM
@copper.hat that’s what I also said.
But I like what Leslie said. Using elementary operations.
 
I always found elementary operations to be so, well, mundane.
 
Sorta like the fundamental theorems of whatever.
 
The Oracle Fundamental Theorem of Calculus. A marketing opportunity there.
 
built upon other Oracle's
\o all
 
\o back
 
5:22 AM
= o/
@copper.hat How's life in the refilled house?
Did you ever get those math exam files out of your computer?
 
5:40 AM
Apr 13 at 5:52, by copper.hat
one of my projects was to write them up in latex, but i guess that will take second place to actually opening the stupid files
math exam
Apr 13 at 5:30, by copper.hat
i wrote lots of notes & problems for my kids on microcrap onenote and now i cannot access them.
 
6:00 AM
I have several questions relating to here win.tue.nl/~aeb/codes/Andw.html
I know S() stands for Steiner system, but can't find what TD(), GD() and all the rest mean
 
6:22 AM
@user4539917 It is mixed. I grew up in a very interactive/affectionate household but my current family eschews contact. But it is nice to have them around.
I almost gave up on the One Note thing, and quite by accident discovered that I had a CD with the 2003 version on it, so I was able to make PDFs of the documents. The point was that my daughter was going to give them to here tutor at Oxford, but she never bothered so it was all a waste of time.
 
You could share them on math ed.se?
or other relevant stackexchange's
Students studying for the JEE would greatly appreciate it :-)
 
These problems are for 7-9 year olds :-)
 
Ok, I didn't know that.
 
students who get a peculiarly early start on studying for the JEE would greatly appreciate it
 
I completely agree with that^
But, stackexchange's age restriction of 13 years old stands in the way.
 
6:42 AM
A few of the problems introduce concepts I think are useful for them.
But definitely not MSE material :-)
 
@leslietownes how do you know jee?
 
imho, the only way to really stand a chance of making it in the jee is with peculiarly early preparation.
 
koro: mostly from people asking questions relating to JEE prep on the internet. questions they would not be asking if, as 7-9 year olds, they had access to copper.hat's materials
 
i have had mixed results
 
martinis? what is it tonight?
 
6:55 AM
one graduate, yet to earn, one sophomore , has never earned
 
we could enrich copper.hat's materials...
 
just some New Zealand pinot
 
as they say, teach to the test
 
Mixed Results would be a good name for a cocktail bar
 
Is anyone interested in a jee preparation site proposal on area 54?
 
7:07 AM
a cocktail bar for mathematicians
 
and sciencists with mixed results
 
Prime Time
Sum of all beers
 
The Oracle
of
Truth
 
good night/morning!
 
cya
 
7:25 AM
@user4539917 chemistry is very scoring in jee though.
 
yup, essentially biochemistry that requires rote memorization
 
@leslietownes :)
@user4539917 biochemistry? I recall three: inorganic,organic and physical chemistry.
 
it's a branch of organic chem
 
Imho, inorganic and organic require lot of memorisation.
If one writes jee, I think they find it ‘easy’ to answer chemistry questions. ‘Easy’ because they don’t take much time if one has memorised some things.
 
students who like biology don't mind that approach
 
 
1 hour later…
8:36 AM
Is there any real reason why mathematicians approximate integrals and then estimate the accuracy of their approximation?
say we wanted to integrate sin(t^3) using partial sums, why not just dump it into sage or smth?
 
8:49 AM
@bumblebee Because sage or smth doesn't provide a proof of what it computes, and as computer assisted math programs are not always right, their output cannot be used as proof.
 
@robjohn do you know of any theorem that goes like so: if $\sum_{n=1}^{\infty}(-1)^n b_n$ converges to S, then $|S-S_n| < b_{N+1}$
Where S is the integral sin(t^3).
 
@bumblebee If $b_n\ge0$ then look at the alternating series test
 
Any ideas on the problem
2
Q: Are the determinants of these matrices always negative under these conditions?

BAYMAXSuppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$). Thus we have $B = A+ \xi e_{1}^T$, where $\xi$ is a $n \times 1$ column ve...

 
Got it, thanks.
 
 
1 hour later…
10:20 AM
Maybe we can show that if sucha condition on the eigenvalues are imposed then the two matrices are indeed not that different
 
 
3 hours later…
1:48 PM
@leslietownes it's irreducible in the Zariski topology as it is open in affine space, thus by some nontrivial AlgGeo it's connected in the classical topology
 
2:43 PM
I think the result that a Zariski irreducible set is analytically connected is in volume 2 of Shafarevich's Basic Algebraic Geometry
I'm nuking a mosquito of course
 
3:22 PM
Only way to be sure.
 
3:53 PM
Hey @Xander
 
Hay is for horses.
(But good morning.)
 
what is the metric on the Adele ring that you consider? is it something like a weighted sum of the p-adic metrics and the archimedean metric?
 
Yeah, that's basically the idea.
There is a reasonable set of slides which go over the idea here.
 
thanks!
 
Basically, you want something like a weighted $L^2$ metric. This gets you the topology you want, and, perhaps more importantly, a reasonable Fourier transform.
 
4:11 PM
this looks interesting. I have actually encountered Bruhat-Schwartz functions and adelic Fourier transforms before due to applications in number theory
 
 
8 hours later…
11:55 PM
Wow! The room is on its deathbed!
 

« first day (4389 days earlier)      last day (58 days later) »