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12:03 AM
i recommend angelo pietro sesame miso dressing for most applications of tofu. if you like sesame and miso.
@dc3rd Some of my friends will tell you my cooking is better than my mathematics :P
i like firm tofu. wildwood makes a good brick of the stuff. pairs very well with curry.
Yeah, I like firm, too. Soft can be used in soups, but even then I prefer firm.
wonders where the math in this room has gone
if you can find a market that makes their own, by all means go with that.
ax^2 + bx + c = 0. math.
Such fascinating math.
Oh, let's solve it in char 2.
12:07 AM
let's not and say that we did.
here's a thought. i think you can recover n from the set of nxn matrices over C as the maximal number of projections (things satisfying p^2 = p) that are orthogonal (pq = qp = 0 for different projections) summing to 1. maybe without even the hypothesis of self adjointness which i love but will remove to make it algebra. is this number more broadly known in ring theory or is it just a matrix thing.
What does this have to do with matrices? It's the dimension of $\Bbb C^n$.
well yeah, but in ring form. i'm wondering what you get if you plug in rings that aren't matrices and maybe don't naturally act on anything. has it been studied. presumably it has. i guess all rings act on something.
There's always the group ring.
anyway that's the math thing in my mind lately. i have some other stuff about eigenvalues but they are too long to fit in the margin.
12:23 AM
You're asking about the maximal number of orthogonal idempotents summing to $1$. If you have such idempotents $x_1,..,x_n$, the map $R\rightarrow\prod_{i=1}^nx_iR$ (note that $x_iR$ is again a ring) given by $r\mapsto(x_1r,...,x_nr)$ is an isomorphism. Conversely, if you have a ring $\prod_{i=1}^nR_i$, then the elements $(0,...,0,1,0,...)$ will constitute a set of $n$ such idempotents. So the maximal number of those should be the maximal number of factors you can decompose the ring into.
that makes sense.
thank you.
12:55 AM
we're coming up on one year since i've left the house for more than my daughter being dropped off at day care, or getting groceries five minutes away. i would love to eat in a restaurant. i'm not going to do that until we're all fully vaccinated. my wife is halfway there. i understand why moronic americans are going crazy.
I'm doubly vaccinated. I'm actually eating outdoors at a restaurant for the first time in almost 13 months tonight.
it doesn't seem to be clear whether vaccination prevents you from spreading the virus, as distinct from having the disease. that's what i'm worried about. i kind of want everybody to be vaccinated.
@TedShifrin must be nice
@leslietownes I don't think it does
that sounds wonderful. please eat something fried. and if you drink, make it a double.
+ new strains to worry about
12:57 AM
They're publishing studies about that now. ... But no question that the official bungling of our politicians (which continues forcefully now) is making this a far greater mess than it had to be.
Not likely fried. I do have heart disease :P
@TedShifrin how would it have gone in Ted's perfect world?
Yes, new mutants are scary. Even more important to vaccinate, but of course the Trompies refuse.
eat something that you wouldn't eat in normal times. it doesn't have to be fried.
Listening to the scientists from day 1, even if they weren't perfect.
LOL, ok, @leslie. I'm trying a new restaurant with good reviews, so I'll let you know.
@TedShifrin playing redcap advocate, scientists said not to wear masks on day 1
12:58 AM
No, this is out of context.
At the beginning, there were no PPE for doctors. And so we were discouraged from stealing them all so that the doctors couldn't function. The lack of PPE was bungling from the top, too, of course.
it's appalling that in the US at least, so much of public health and safety has been reduced to whether you do or do not support the orange man who used to occupy the oval office. and there are so many obstacles to efficient distribution of disease related products due to incoherence from the top and our federal system.
Having a president who modeled decent behavior on all grounds would have been nice, rather than one who incites overthrow of the government.
Yes, politicizing world health has been one of the most appalling things of my lifetime.
And we're not by a long shot through the worst.
@TedShifrin I heard that and I always felt the "no PPE for doctors" line was the noble lie to detract from 1. most foreign nationals + "aware folk" were buying up retail stock of masks and 2. last-minute stocking of hospitals (done for economic reasons, usually a good idea in peacetime) made it so supply was low.
i don't support our democratic governor, he is a scumbag, but the virus doesn't care who my governor is, and it's somewhat annoying that everything seems to be up to people like him, and the decisions of individuals who are reading headlines. the medical guidance has been fairly clear.
Yes, and closing down the agency there explicitly to deal with pandemics a year prior was really excellent.
1:02 AM
i bought a shitload of KN95 masks. the chinese standard is good enough for me.
@TedShifrin I do agree with this. something as easy as a mask mandate would have probably saved us from much suffering + won him the presidency
He's less a scumbag than the governors of TX, AL, and even NY.
The one I had broke within an hour, @leslie.
my wife met him once and kind of likes him. that might be why i don't like him.
@TedShifrin "sick people $\in$ nursing homes" NY gov
yeah that sucked pretty bad.
That's a mature reaction, leslie.
I was furious with him for the French Laundry incident. Otherwise, I think he's done way better than most every other governor.
The citizenry refuse to abide by the rules. "Personal freedom" Trompies all.
Anyhow, we're not supposed to politic here.
1:05 AM
he's just a greaseball, politics is full of them. he's not actively malevolent, as are so many others. my wife was on a news broadcast getting a flu vaccine many years ago. the idea was to educate the public as to the flu vaccine. my wife worked for the SF department of public health.
i like how we used to only worry about the flu.
That's because we are all routinely vaccinated for TB and everything else as children. Essentially every single person. To go to public school or college, a student has to present proof of certain vaccinations.
i still have the VHS tape of this appearance. my wife, gavin newsom, it was broadcast on the mandarin language channel in the bay area.
Why the mandarin channel?
they covered it. i have no idea.
Maybe some of the Asian population in SF were not wanting to be vaccinated?
Quite possible.
1:09 AM
i had to do proof of vaccines to go to law school, and i didn't have it because i don't generally use the health system and my doctor was a sole practitioner who retired about 10 years before i went to law school. so they made me get all of them.
yes i think it was that. i don't blame various populations of being skeptical of injections. we do not have a good history of that.
Agreed. I no longer have proof of anything, but I've been getting a lot of new vaccines. Shingles vaccine. Hepatitis A and B vaccines. Flu vaccines. Oh yeah, and COVID.
i had really bad hives once and my doctor at the law school made me get a test, he didn't explain what it was, but it was for syphilis. syphilis can present as almost anything and he was a rigorous clinician. my wife did not high-five me when i tested negative for syphilis.
LOL ... I have several friends who've had it.
my wife was like 'does this test for anything other than syphilis?' no, it ONLY tested for syphilis. it's good not to have it, that's all i can say.
there's no shame in it. bacteria can come from anywhere.
2:00 AM
from manifolds to vd.
1 hour later…
3:01 AM
Hello Guys...
Hi @copper.hat
3:22 AM
i do my best to elevate the tone around here.
3:33 AM
@Thorgott I'm a little scared looking at this
Why do you want to know this theorem
Hi @123
Mathematics is neither discovered nor invented; it is _____
fill in blank
(aiming for humor)
3:57 AM
nobody knows who originally gurgitated it. but it is only regurgitated now
that's a very solid bit @leslie.
Did you know you can multiply two certain $2 \times 2$ integer matrices, each having determinant $1$ and the result is matrix no having determinant $1$?
Q: Why are these determinant $1$ matrices not multiplying to a determinant $1$ matrix?

StudySmarterNotHarderLet $A = \begin{pmatrix} 4 & 3 \\ 5 & 4 \end{pmatrix}$ then $\det(A) = 1$. Let $B = \begin{pmatrix} 6 & 5 \\ 7 & 6\end{pmatrix}$ then $\det(B) = 1$ as well. But $AB = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}$, so $\det(AB) = 3 - 4 = - 1 \neq 1 = \det(A)\det(B)$. Where is my conception going...

the upper left entry looks wrong, as someone has commented. there is no minusing in the matrix multiplication.
the version of multiplication where there is minusing would be potentially interesting to study, if only to kill some time. it might not even be a regurgitation.
by minusing i mean subtraction. it mildly peeves me when people say 'summation' when 'sum' would do. i imagine others feel the same about 'minusing.'
4:13 AM
Crap, yeah, was using subtraction because the determinant formula was stuck in my head
do you know about permanents? mind will be blown.
en.wikipedia.org/wiki/Permanent_(mathematics) i shouldn't be promoting this stuff. i have gone on the record as being against this sort of thing.
@leslietownes that does look interesting
the permanent is to the determinant kind of what regular multiplication is to your minusy version of matrix multiplication. not quite, but kind of.
it's harder to compute in some formal sense. i gave a talk about this once.
I'm more concerned about the subgroup of $S_2(\Bbb{Z})$ generated by $T = \begin{pmatrix} n & p \\ q & n \end{pmatrix}$ since if that subgroup is not finitely generated, then there have to be infinitely many twin primes, right?
Since each element of $T$ defines a twin prime pair
what is $S_2(\mathbb{Z})$?
4:18 AM
it isn't clear to me when that matrix will have determinant 1. i mean, it will some of the time, but the relation between those instances and twin primes is not clear to me.
$n^2 - pq = \det(A) = 1 \iff n^2 - 1 = pq \iff (n-1)(n+1) = pq \iff$ $p = n-1, q = n-1$ for sufficiently large $n$ $\iff$ $p,q$ are a twin prime pair
Thus twin prime pairs are in 1-to-1 correspondence with elements of $T$.
Up to permutation
of $p, q$
So they are in 1-to-2 correspondence actually, but the statement still holds
finite is finite
@leslietownes see now?
I mean $q = n + 1$ not $n-1$
i sort of do. i don't think in integers.
the 'for sufficiently large n,' is that to rule out stuff where n is 2? is something else going on there?
Actually, they are in 1 to 3 correspondence if you include $-p, -q$ possibility
wouldn't that make 4?
(p,q), (q,p), (-p,-q), (-q,-p)
4:25 AM
Yes where $n$ is 2, so you could just say $n \neq 2$
Yes, 4 actually, you are write if you include $-\Bbb{P}$ allowed
but only do that if there is some mathematical purpose
Thus a proof that the subgroup is not finitely generated is a proof of twin primes. This is not the same proof as the one I mentioned yesterday, but a new one. The one I mentioned yesterday failed miserably
the trouble i'd have is that writing the twin prime conjecture in an equivalent form, doesn't mean it's any easier to answer in that form
Since if $T$ is finite and $\langle T \rangle$ is not finitely generated, then contradiction
I think it's not equivalent though, which might make it not work
as a restatement
so you're looking at that set of matrices where n isn't 2, and p and q are prime, and the result lands in SL_2(Z). i guess from your calculation that will happen when pq is a product of (n+1) and (n-1), which by numbery stuff forces one of p, q to be n-1 and the other to be n+1. i would certainly doublecheck the numbery stuff. i live in $\mathbb{C}$ where none of this is possible. but it seems OK
i guess i don't see how working with that subset of SL_2(Z) is easier or more manageable than working with twin primes as twin primes. but i haven't thought about it.
Well, when are teh twin prime pairs themselves elements of an algebraic structure, not just the individual primes. Ask yourself that.
if p and q aren't prime it seems like you could land in SL_2(Z) in some other way, with the factors aggregating together to form n+1 and n-1. and ruling that out seems nontrivial.
4:30 AM
$n^2 - pq = 1$ naturally asks to be a determinant equation
Let me think on it for a bit, I'm betting I'm wrong here
i prefer not to ask myself that although it's an interesting idea. i have gone on the record as against determinants.
Also the symmetricity of that matrix is astounding
Not that it's a symmetric matrix
but that the inherant symmetry that you can't quite define but can see, $n = n$ both non diagonal elements are odd primes
@leslietownes are we talking determinants in general, or determinants for characteristic polynomials
i think of matrices of integers as really, really hard. i took a complex analysis class where we studied modular forms, and it was just, so much harder than what i was used to.
You could express it as symmetric though with $n, n$ not on the diagonal, see
4:32 AM
i think i went on the record as just generally being down on determinants.
determinants as expressing volumes is geometrically nice enough that i can't dislike them
Determinants are just multilinear maps o the entries
Or homomorphisms of the whole matrix
i get why people don't like them as the route to characteristic polynomials tho
They go hand in hand. Dets are a nice example of a hom there
i tend to just work in QM land tho
4:34 AM
there's a meme of a crazy looking guy in front of a board with images connected by twine. he looks like he's been deranged by a conspiracy. that's me, and determinants are on the board.
so my matrices are almost always finite and hermitian
Just think of them as multilinear maps satisfying a few properties
are determinants that hard
which eliminates a lot of the weird jordan normal form possibilities
up with permanents.
4:35 AM
Well every time you see a determinant equation then, just work on the 2x2 or 3x3 case and let your brain do the extrapolation to higher n x n
believe me, you don't want to leave anything up to my brain.
Let $X$ be the monoid generated by all 2x2 matrices $(a_{ij}) = A$ where $a_{11} = a_{22}$ and the off-diagonal entries are odd prime numbers
Then $X \cap SL_2(\Bbb{Z})$ not finitely genrated $\implies$ twin prime conjecture is true, is the statement
However, I don't think we immediately have equivalence with twin primes and that's why it's interesting because close-together logical equivalences are no-brainers and make no difference
So if the intersection is finitely generated we get nowhere is what I'm saying
that's the downside
However, people have shown that general linear groups are not finitely generated.
This is an approach I just came up with an hour ago, so not sure if it's any good
So, assume, by way of induction that the intersection is not generated by one element
And assume also that we already proved that somehow
That's probably also a hard problem
Then assume it's not generated by any $n \geq 1$ elements
Now, how would we work with matrices and prove that it can't be generated by $n + 1$ elements
$GL_2 (\Bbb{Z})$ is finitely generated, so I mispoke
however, intersections are offten not also
4:50 AM
i think of matrices of integers as a scarier world than the world of integers. like, my map has integers on land and then there's water and pictures of dragons and then there's SL_2(Z) on the edge of it.
Q: Generating the special linear group of 2 by 2 matrices over the integers.

IsomorphismOur Number Theory professor claimed that the special linear group $\text{SL}_2(\mathbb{Z})$ is generated by just two matrices: $$ M_1=\begin{pmatrix} 0& -1\\ 1& 0 \\\end{pmatrix} $$ $$ M_2=\begin{pmatrix} 1& 1\\ 1& 0 \\\end{pmatrix} $$ He commented that the proof is outside the scope of the ...

so the special linear group of matrices is finitely generated, that might make the proof easier
Since the direct way would be to take general product of matrices of the a $n, p, q$ form and show that it's not decomposable as powers of a finite set of such matrices
Crap, if you take the monoid generated by the $n, p, q$ matrices $X$ then the statement implication as is might not hold
So this won't work for now
my cat is enjoying her new toy with the ball inside that she can bat around. we are hoping it will improve her behavior in the middle of the night. she's gotten into a habit of climbing on my wife's head and swatting at her face and meowing multiple times a night. just for the attention.
slight change of subject, it's just on my mind right now.
if anyone has thoughts on how to manage feline attention seeking behavior in the middle of the night, i'm all ears.
5:20 AM
she walks over my head to get to my wife's head. it's funny how i don't qualify for late night attention seeking.
anyway it ought to be easier than the twin prime conjecture.
5:51 AM
are we still on perpetual motion machines?
surely there's a patent on that?
that might be the only time when the patent office will issue a 112 rejection. they should do it all of the time, but in practice they don't do it unless you're like, "here's my spaceship for traveling back in time."
there are also grounds for a 101 rejection in that instance. the patent office guidance is, do a word search in previously issued patents and ask a grown-up for any other ground of rejection. it's very lazy, but it's administrable.
and lots of $$$ into not having the patent office examine stuff on the front end. i wonder if there's a patent on a perpetual motion machine. it's certainly novel and non obvious.
my memory fails me, but i believe there were a few over the years.
i do a lot of biotech litigation and the human genome would have been sequenced in 5 seconds for 5 dollars in 1990 according to a lot of the patents i've seen.
hyperbole is the patent's gamble.
i've also seen it go the other way. an inventor who did not have english as his native language said some stuff in a provisional application that was clearly just broken english. a court and an appellate court said it was a deliberate attempt to capture something in the prior art, invalidating the patent. it was just broken english.
i think the courts are so used to dealing with abuses of the system that sometimes they come down too hard on non-abuses of the system. i don't have a solution for it.
it's greedy to put perpetual motion in the title of the patent. the smart approach would be to claim it but not use the words perpetual or motion. the examiner may not pick up on what you're doing.
6:10 AM
patents are big company games.
although the only point of getting a patent is to sue, or threaten to sue, infringers. i don't know how you'd prove that someone invented a perpetual motion machine. sneaking garbage through the patent office is an easier game than that.
one click
oh, we had a good one. the thing that puts (1) next to your inbox when you have one unread message. patented and sued upon.
i guess when you have nothing else in your guns...
i know i am not a real mathematician when i struggle to apply Zorn's lemma withiut guidance.
(just reading through something which chatting...)
that's interesting. i think a lot of applications of AC coincide roughly with intuition. the thing about chains in partially ordered sets is maybe counterintuitive but the idea of "just pick something" is fairly natural.
"just pick infinitely many things so that something converges," ok, that's harder.
i think my dissertation depends on the axiom of choice. i use "banach limits." i don't think they exist but AC says they do, so i used them.
6:18 AM
i would not hesitate to use AC. all of my related work is 'practical' optimization so niceties of AC are rarely of concern.
i do think it's helpful to actually construct things. i have taken heat for this in the past. someone on my thesis committee wanted to strike a portion of it because i could actually construct some of what i was talking about, when existence was guaranteed from AC. he won and i lost.
i would think constructing where possible to be a reasonable approach.
people on math.SE are also not always sympathetic to constructive arguments, or acknowledgment that constructions are, at least to some people, separate from existence. i don't dig deep into foundations but i do see the value of really writing down what you're talking about, if you are able.
in the 'real' world, existence alone is rarely of interest.
and sometimes, as with some of the problems i dealt with that did not make it into my thesis, the known constructions don't converge! but AC tells you that there's a solution anyway! i'm fine with that. but i hate pulling anything out of a black box unless it is absolutely necessary
i am surprised that more mathematicians are not focused on constructions and avoiding AC. math more than other fields, you don't care what the answer is, you want the argument. i can understand why a chemist would be happy with an oracle saying "this reaction will produce these results" if it could be relied upon. mathematicians don't want the answer, they want the argument. they should be more sympathetic to construcivism, ultrafinitism, etc. but it is something of a sideshow.
it doesn't help that a lot of people who at least provisionally reject AC are doctrinaire. i have dealt with a few statistics people who worship at the temple of Bayes and get very hostile to even basic frequentist modeling. it's not a fight, it's not a power struggle, nobody has to win. there isn't winning. let's just write down what we know and see what we learn from it.
6:28 AM
choice is used an many areas that people do not think about. Arzelà Ascoli for example.
"Academic Politics Are So Vicious Because the Stakes Are So Small"
in my subfield it was the hahn banach theorem. people just go to it, they don't think. sometimes you can construct the functional you would want. or can describe aspects of its properties.
it is like religion, or coding standards. it brings out the worst...
my wife is still an academic. the smaller the stakes, the worse it gets. it's really true. although i know a lot of very pleasant academics.
our intellect evolved not to think deep thoughts but to be better team players.
i have many good friends in academics, nothing otherwise implied above.
it was a comment on your 'nobody has to win'.
my wife's field, sociology, is toxified by relation to the real world. it is mostly people arguing about what words mean. like people arguing whether leibniz or newton notation was better for calculus. who really cares. but that's where the action is. and much of it is very much a game of king of the hill. math is slightly better than that, although not immune from the general prejudices that infect academia.
6:35 AM
much is fashion
i tell my wife to introduce new theories into the field. she tells me that i'm talking out of my ass. it's a very functional relationship.
relationships are delicate things
best at long distances
we lived in different time zones for a while. it did work better then. having a kid has forced us to localize at a point.
kids are rough on relationships. or perhaps kids reveal the relationship.
6:42 AM
our toddler is a miniature version of me. she looks like me, she acts like me, she bothers my wife the same way i do. everybody loves it, except for my wife.
i love kids. i like their honesty
i am going to hit the sack. i might get up early for a ride in the trails.
my daughter is 2 and a half years old. she likes being pushed onto her bed. my wife hates doing it, because who knows, she might bite her lip or something. you don't want to push her kid around. last night, after two or three half hearted pushes onto the bed, my daughter told my wife: "you aren't very good at this."
enjoy the ride.
:-). thanks!
windows is bad for me
I need to switch to linux
windows is not great
6:53 AM
which linux distro should I use?
I think ubuntu would be good but I am not sure
i have oscillated between slackware and BSD, which probably isn't really linux.
there's got to be something more recent that is better
i need to buy a new phone because it crashes when people from work call me. the phone is over five years old. don't listen to me.
7:13 AM
@leslietownes My phone is 6 years old (bought mid 2015) and it is going strong. I did have to update the battery, however.
nice. the samsung s5 was my first smartphone. i have put i think two aftermarket batteries into it. it replaced a nokia brick which i had since 2008. before then, nobody could reach me. i regarded that as a holy time.
hello you guys.
working on this problem,"How many 8-letter words contain exactly 5 vowels (a,e,i,o,u)? What if repeated letters were not allowed?"
I solved the first part (8C5)*5^5*21^3
the second part I have you can choose 5 words that in the 8 spots.
do words have to be english words? or just symbol strings from the alphabet {A, . . . Z}
strings from the alphabet.
7:36 AM
@leslietownes My first cell phone was a Motorola brick from 1990.
when cell phone service was first introduced to my hometown, my dad was given a phone for a week. the top of it was normal but it connected to something the size of a briefcase. i remember running outside and phoning the neighbors across the street and then waving to them through the window.
Good morning everyone,
any one knows a new printed version for Elements of Quaternions by Hamilton?
@CroCo almost morning here.
@robjohn here 7:40 am
here is 2:42 am.
7:43 AM
@CroCo 11:40 PM
@CroCo Near the prime meridian, then
@robjohn then I'm ahead of you in time. What would you like to know about your future?
@CroCo I would like the stock page from your paper ;-)
and here I am at 1:16 pm
@robjohn just telling NO giving. :)
8:06 AM
@CroCo :-p
2 hours later…
9:46 AM
If $R$ is a commutative ring with unity and also $k$-algebra where $k$ is a field, then I guess the set of $k$-algebra homomorphisms $k[x,x^{-1}]\to R$ and $R^\times$ are bijective where $R^\times$ is a set of units of $R$
Am I right?
I thought the map would be $f\mapsto f(x)$
10:30 AM
@love_sodam yes
&\text { Let } W_{1}, W_{2} \text { be subspaces of finite dimensional inner product space } V \text { such that } W_{1} \subset W_{2} \text { . Prove that then }\\
&W_{2}^{\perp} \subset W_{1}^{\perp}
Let $W_{1}, W_{2}$ be subspaces of finite dimensional inner product space $V$ such that $W_{1} \subset W_{2}$. Prove that then $W_{2}^{\perp} \subset W_{1}^{\perp}$
I already proved this if we not consider proper subspace but how to prove this if W1 is proper subspace of W2 ?
that $W_{2}^{\perp} \subset W_{1}^{\perp}$
@LeakyNun Any idea.
just unfold definitions
@LeakyNun so lets say we have some $w_{2} \in W_{2} $ such that $w_{2} \notin W_{1}$ .
Now we need to show there is some $z \in W_{1}^{\perp}$ that not lies in W2 perp ?
right @LeakyNun.
what's the definition of perp?
W1 perp is element in V such that it is perpendicular to all vectors in W1 .
<z,w1> = 0 for all w1 \in W!
10:41 AM
use that definition
@LeakyNun I want tos how it is proper subspace ? I can show it is subspace but how to show proper ?
I doubt they mean proper subspace
some people use $\subset$ to mean $\subseteq$
Can you give me some counter example such that W1 is proper subspace of W2 but their orthogonal complement are not proper.
you can prove it by considering dimensions
Ohh Ya right
10:53 AM
in infinite dimensions this can happen
2 hours later…
12:51 PM
Why is the continuity of $f^{-1}$ follows from compactness of $C_1\times C_2$?
The last sentence
No never mind
1:14 PM
@MikeMiller No particular reason tbh. A special (and easier to prove) case appeared in the construction of Milnors $\lambda$ invariant. Then I read the property apparently holds in general and so I thought it'd be interesting to learn why.
here's a fun fact though: up to multiplication with a constant in each dimension, signature is the only invariant of oriented diffeomorphism type of compact smooth manifolds that is additive with respect to gluing along boundaries
ok, that actually isn't quite right, but I cba to formulate the correct statement
Something like that seems about right
To be honest I have never read this proof
Seems like pain
Seems like the better approach is to prove that the eta invariant attached to signature negates under orientation reversal or something
@Thorgott You may enjoy the first couple pages of "Spectral Asymmetry and Riemannian Geometry I", the atiyah patodi singer paper
Their discussion sort of implicitly assumes Novikov additivity already (it was an important ingredient in them coming up with their theorem)
But the theorem itself does not assume that
1:52 PM
why is it that if $\mu,\nu$ are probability measures on $\mathcal{B}(\mathbb{R})$ s.t. for all open intervals $(a,b)$ we have $\mu(a,b) + \frac{\mu\{a\}}{2} + \frac{\mu\{b \}}{2} = \nu(a,b) + \frac{\nu\{a\}}{2} + \frac{\nu\{b \}}{2}$ that they are equal?
2:16 PM
it didn't help me understand Novikov additivity, but I did enjoy reading the introductory section of that paper. what they discuss seems very interesting, too bad it's still beyond me.
I think the issue might just be that I'm too bad at linear algebra, so let me rephrase everything as a linear algebra problem:
I have a block matrix
Why is the the signature of $M$ the sum of the signatures of $B_1$ and $B_2$?
what is signature?
you can find an invertible matrix $S$ such that $S^tMS$ is diagonal, then it's number of positive entries on the diagonal minus number of negative entries on the diagonal
the intrinsic characterization is dimension of largest subspace on which the form is positive-definite minus dimension of largest subspace on which the form is negative-definite
@Thorgott I'm about to write a linear algebra lecture so it only seems fair I give this the good old college try
2:31 PM
@Thorgott M sends e1 to e4 and e4 to e1
Is $a_{11}$ a number or a map $\Bbb R^n \to \Bbb R^n$
so it sends e1+e4 to itself and e1-e4 to its negative
sorry replace 4 by n, I forgot it's a block matrix
so those two 1's add 1-1 to the signature
that's why $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ has signature $0$, yeah, but in this case we have additional $\ast$s in the way
oh i'm blind
No control over bottom-right entry at all?
2:39 PM
not as far as I can tell
and also not as far as Atiyah and Singer indicate
Leaky seems like he's giving the right proposal. If this matrix acts on $V \oplus V_1 \oplus V_2 \oplus V$, and $V_1^+, V_2^+$ are positive definite subspaces of $V_1, V_2$, it seems like $\{(v, v_1^+, v_2^+, v)\} \subset V \oplus V_1^+ \oplus V_2^+ \oplus V$ ought to be a positive definite subspace of our large vector space. In explicitly writing this out I get stuck with a term $v^\top c v$ coming from the bottom-right (say the bottom-right entry is matrix $c$)
Which is the only term I can't prove is positive
Actually no the outer terms are a whole damn mess
Oh OK let's try to do that then
B_1 and B_2 are indefinite though right
Or not nec def
yeah, can be whatever
they're the (non-degenerate quotients of) the signature forms on $Y,Y^{\prime}$
Oh so they are nondegenerate == definite
2:43 PM
can I assume the matrix is symmetric?
if that's what definite means to you
"definite" to me means "either positive definite or negative definite"
that's not implied by non-degeneracy
Yeah of course not lol
2:44 PM
there can be isotropy
Still, nondegenerate is what I needed
I think, geometrically, isotropy in this context should come from Poincaré duals of submanifolds without self-intersection or sth
I get the sense this would be doable within 30 minutes or an hour of messing around with matrices for me, since I got very good at that for a while
But I don't think I have those 30m or 1h
So I'll leave it to you and Kenny
Nondegeneracy ought to be very important here
where's the topology man
is the punchline to this theorem really some stupid linear algebra computation
Yes, this is why I never read the proof!
Stupid linear algebra computations are reasonably common tho
2:57 PM
actually, I'm fine with stupid linear algebra computations
but this is a stupid bilinear algebra computation
That happens in life too
yeah, but now I have to deal with the fact that I never understood those
There are a few places in my thesis where I assert things like "If you have a block diagonal matrix of this form... and the unambiguous terms have operator norm less than 1/5 the smallest eigenvalue of the diagonal terms ... then this matrix has signature [blah]"
Kind of pain like this
pain indeed
It gets fun
When you do it a lot
3:06 PM
so I can assume the B1 and B2 are diagonal and then algebraically there's no difference if I assume the whole B1 and B2 block is just 1x1, so M becomes x2^2 + (2x1+Ax2+Bx3)x3. The extra term (2x1+Ax2+Bx3)x3 always contributes 1-1=0 to the signature, using the trick y1y2 = (1/4)[(y1+y2)^2-(y1-y2)^2].

e.g. x2^2 + 2x1x3 + 18x2x3 + x3^2 = x2^2 + (x3+9x2+x1)^2 - (9x2+x1)^2
@Thorgott very low tech proof ^
If you develop M-XI over the first collumn, you find $X(B_1-X)(B_2-X)(c-X) - (B_1-X)(B_2-X)$, which factors through (X^2-cX-1), which has either a positive and a negative root, or no real root
what does develop mean?
Unless I'm breaking some linear algebra determinant rule
it always has real roots because $\Delta = c^2+4 > 0$
Oh yeah duh
$\det M = \sum_{i=1}^n (-1)^im_{i,1}\det M_{i,1}$
is what I mean by "develop"
3:09 PM
Wait but why do we care about determinants
det(M-XI) is the char poly
Assuming everything is diagonalizable, you're counting the positive and negative eigenvalues right ?
Which is encoded in the char poly
3:10 PM
I'm not able to parse what either of you are saying so I'm going to dip
@MikeMiller I'm basically looking at the quadratic form. The extra terms becomes $(2x_1 + \sum_{i=2}^n a_i x_i)x_n$, which has signature 0 because $y_1 y_2 = \frac14 (y_1+y_2)^2 - \frac14 (y_1-y_2)^2$, one positive and one negative
@Thorgott ^ this is probably easier to read than my mess above
I've made sign errors above but I think it works
@Astyx if I interpret your solution correctly, you're assuming the * are zero and the blocks B1 and B2 are 1x1 each?
Oh shoot you're right
could someone help me figure out why this is true? if $\mu$ and $\nu$ are probability measures on $\mathcal{B}(\mathbb{R})$ such that $\mu(a,b) + \frac{1}{2}(\mu\{a\} + \mu\{b\}) = \nu(a,b) + \frac{1}{2}(\nu \{a\} + \nu\{b\})$ for all $a<b$ , then their distribution functions agree at all points of continuity
*at all common points of continuity
which are points s.t $\mu(x) = \nu(x) = 0$
hmm, I guess we can say $\mu(-N,x) \leq \mu(-N,x) + \frac{1}{2}\mu\{-N \} \leq \mu[-N,x)$ and letting $N \rightarrow \infty$ this means $\lim_{N} \mu(-N,x) + \frac{1}{2}\mu \{-N \} = \lim_{N} \nu(-N,x) + \frac{1}{2} \nu \{-N \} = \mu(-\infty,x] = \nu(-\infty,x]$
but my course notes omitted this explanation so im guessing this is supposed to be stupidly obvious
now how does two probability distribution functions agreeing on the complement of a countable sense imply they are the same everywhere...
3:31 PM
@porridgemathematics you changed your question; which one is the correct one?
never mind, I worked it out, we need to use right continuity
what does it mean for a diffeomorphism to not respect addition?
@LeakyNun I don't think I understand. It's not like the signatures just add when you do pointwise sum of two quadratic forms.
@Thorgott yeah but my vectors are linearly independent
like they form a new basis
the new vectors are $x_1 + \sum_{i=2}^n \frac12 a_i x_i + \frac12x_n$ and $x_1 + \sum_{i=2}^n \frac12 a_i x_i - \frac12x_n$
they both have $x_1$ terms, so they're outside the span of $x_2, \cdots, x_{n-1}$
and their $x_n$ terms are different, so their difference is also outside the span of those vectors
so I can just add them together
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