Is a real Lie group one that takes a vector in R^n always to another vector in R^n and a complex Lie group one that takes a vector in C^n always to another vector in C^n? Does that mean that the real Lie group is a subgroup of the complex lie group?
@DIRAC1930 You have in mind a family of Lie groups called matrix Lie groups; these are subgroups of $GL(n, \Bbb R)$ (real) or $GL(n, \Bbb C)$ (complex).
There are Lie groups which are not matrix Lie groups, useful in physics, however.
@shintuku I see, thanks. So if we were to change the question a little and write $x_1$ and $x_2$ to be real numbers, then would the bound on $\frac{x_1}{x_2}$ change or still remain $0$ to $\infty$?
If you know what a Lie group is, then for $G$ a universal covering group $\widetilde{G}$ is a simply connected Lie group such that there is a surjective homomorphism $f : \widetilde{G} \to G$ which induces an isomorphism on Lie algebras
Eg, $SU(2)$ is universal covering group of $SO(3)$, because $SU(2)$ is simply connected, there is a surjective (2:1, in fact) homomorphism $SU(2) \to SO(3)$, and it induces an isomorphism on Lie algebras $\mathfrak{su}(2) \cong \mathfrak{so}(3)$
In fact you just need to know the Lie algebra of a given Lie group to construct the universal covering group, there is a unique such thing, the map can be constructed by hand. These can all be found in textbooks, though I do not know of a handy one to recommend. I am sure physicists have written their own textbooks which will be more accessible to you
if we can represent a secret message by a large prime number $p$, we can transmit over the network the number $r = p · q$, where $ q > p$ is another large prime number that acts as the encryption key.
What is the worst-case time complexity of the above algorithm? Since the input to the algorithm is just one large number $r$, assume that the input size n is the number of bytes needed to store $r$, that is, $n = \frac{(log2 r)}{8}$, and that each division takes time $O(n)$.
So, since $r$ can take 100 bits, we need $\frac{r}{p}$ divisions in order to get the encrypted message. Now the questions is how we can calculate the complexity of the algorithm where we need to guess each and every single possibility, thus I said we need $\frac{r}{p}$ divisions
So, the question is now, I am not sure why the complexity is $O(n \times 2^{4n})$?
$\frac{r}{p} = \frac{r}{2^{100}}$ divisions
I assume the complexity in the question is referring to number of divisions or both divisions and space?
If you guys are talking about Robert Bryant, he’s a superstar of long standing. This note on Chern’s work is more than 5 years old, unless something has changed. He was one of the people whom tried to stop Atiyah ….
i was on kadison's side of the kadison-singer conjecture. he didn't think it was true and i didn't either. he hated the way it was named. it turned out to be true.
@TedShifrin If Robert Bryant declared he's solved the S^6 complex structure problem I'd just say oh damn and move on with the crushing realization that there's another piece of great mathematics from my time that I'll never learn.
Yes, out of frustration, I just searched for it. Seems hardly worth taking the trouble. Seems unlikely that the complex structure and the Riemannian metric don't have to intertwine in some nontrivial way. But the crucial "fact" comes from the same person's earlier work. Shrug.
I'm happy to let you know though that the proof of my little theorem seems correct, after having written down the core (just finished today). It's not finding a complex structure on S6, but still.
Thank you :) still scrutinizing, it's a little surprising this isn't known because it's too simple, the scare of being wrong is always there. But seems to work
my daughter seems to have stopped yelling for the night.
she's not shy, at all. i was very shy at her age. she interrogates people on zoom. i guess it's zoom and not real life, but she will ask them about their business.
she asked a coworker who had not had time to change after spin class, "why is your hair stringy and disgusting."
or not, if it's obviously crank stuff. there was a guy i used to interact with who had a history of interesting results but somehow thought he had solved the invariant subspace problem, and probably hadn't. people wouldn't talk to him at conferences.
he was nice, you just didn't want to talk about the invariant subspace problem with him.
at the beginning of this crap we sued a company that was at work on a covid vaccine, completely unrelated to what we were suing them for, but they dumped it out into the press and suddenly we were getting death threats.
while writing that example, I had $\frac 1{\sqrt 2}$ and $2\pi$ in mind that is the example certainly doesn't have any of the aforementioned periods. Then, you asked why that function is not periodic so I started thinking of proving it in rigour.
but that seems very complicated :-(
@love_sodam: it can be proved that the above function is indeed non-periodic (without graph) like this:
Let $T\gt 0$ be a period of $f$ so we must have: for all x, $f(x)=f(x+T)\implies \{\sqrt 2x\}\sin x=\{\sqrt 2 (x+T)\}\sin (x+T)$ and in particular this must be true at $x=0$
whence we get: $\{\sqrt 2T\}\sin T=0$, hence either $\{\sqrt 2 T\}=0$ or $\sin T=0$ and from here you can verify that neither is possible.
This establishes that the function $f(x)=\{\sqrt 2 x\} \sin x$ is non-periodic over $\mathbb R$.
@leslietownes using cdc numbers for the bimax test (for symptomatic) and the albany case rate (from alameda county) i get that the prob. of having covid given that one tests positive is about 41% (assuming no mistakes).
@love_sodam if $T=\pi$ then you'll have for all real $x$, $\{\sqrt 2x\}\sin x=-\{\sqrt 2(x+\pi)\sin x$ and it follows that $\sin x(\{\{\sqrt 2x\}+\{\sqrt 2(x+\pi)\})=0$(will this hold for all $x$?)
How about this: If f,g are periodic functions with period $T(f), T(g)$ and if we suppose $T(fg)$ is a period (assume $fg$ is periodic) then $T(fg) = nT(f) = mT(g)$ for some $n,m\in\Bbb N$. Hence $T(f)/T(g)\in\Bbb Q$.
@love_sodam To show why neither of these two cases is possible, you may consider two cases: 1)$\{\sqrt 2 T\}=0$ , 2) $\sin T=0$ and in both cases, $f(x)=f(x+T)$ will not hold for all $x\in \mathbb R$.
that's what I hinted at earlier when I said neither is possible
@love_sodam To show why neither of these two cases is possible, you may consider two cases: 1)$\{\sqrt 2 T\}=0$ , 2) $\sin T=0$ and in both cases, $f(x)=f(x+T)$ will not hold for all $x\in \mathbb R$.
the behaviour of periods is strange specially when periods are irrational. If both $f$ and $g$ have rational periods then $fg$ has "LCM of period of $f$ and that of $g$" as one of its periods.
Using the notation found in Vick's Homology Theory.
Let $f: X \to Y$ be a homeomorphism. Let $\sigma_n = \{ (s_0, s_1, \dots, s_n) \in \Bbb{R}^{n+1} : \sum_{i=0}^n s_i = 0$ and $s_i \geq 0, \ \forall i \}$. That is the standard $n$-simplex.
Then if $\phi : \sigma_n \to X$ is singular $n$-simple...
Shows that I understand some of the basics of AT going by Vick's and from what I already knew before.
Reason I'm excited is because this question was left as an exercise in Vick's and I figured it out all on my own knowledge, without googling for hints.
@robjohn: Thank you very much. I actually need his help. I want to know where I can download the past entrance exam of Nanyang Technology University (Singapore). He is living in Singapore.
Thanks a lot for the answer @robjohn. My only question to that is: since $\frac{\log (n)^2}n$ does not tend to $0$, how was the final conclusion made? Here, by $\log (n)^2$, I mean $(\log n)^2$.
If $\frac{\log (n)^2}n$ tended to $0$, the following lim $b_n$ would tend to $0$, solving my problem instantaneously.
padic numbers go off infinitly to the left i.e. $....31525$ real numbers go off infinitly to the right i.e. $3.1415....$ are there any numbers that go off infinitly in both? i.e. $....3125.31415...$ i remember seeing something like this before but i cant find it
Using the notation found in Vick's Homology Theory.
Let $f: X \to Y$ be a homeomorphism. Let $\sigma_n = \{ (s_0, s_1, \dots, s_n) \in \Bbb{R}^{n+1} : \sum_{i=0}^n s_i = 0$ and $s_i \geq 0, \ \forall i \}$. That is the standard $n$-simplex.
Then if $\phi : \sigma_n \to X$ is singular $n$-simple...
