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.
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.
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.
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.
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.
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.
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.
@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.
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.
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 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.
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 $1$?
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.'
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
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.
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
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.
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.
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.
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.
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
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.
Our 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
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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."
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.
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.
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$
\begin{aligned} &\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} \end{aligned}
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 ?
@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
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?
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 $$M=\begin{pmatrix}0&0&0&1\\0&B_1&0&\ast\\0&0&B_2&\ast\\1&\ast&\ast&\ast\end{pmatrix}.$$ Why is the the signature of $M$ the sum of the signatures of $B_1$ and $B_2$?
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
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$)
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]"
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].
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
@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
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...
Let $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...
Our 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 ...
