One first picks the smallest vector in the lattice, and rotates + translates to make it $e_1$. Then pick the second smallest vector; call it $a_2$. Now the authors claim "If these did not generate the lattice, then there would be some $z \in \Gamma, z=\lambda_1e_1+\lambda_2a_2$, with $|\lambda_i|<1/2$. Where does this norm condition on the $\lambda_i$ come from?
What is the definition of genus? For a singular curve, you can mean either arithmetic genus or geometric genus (which is the genus of a smooth resolution of the curve).
Riemann-Hurwitz is fun for (smooth) Riemann surfaces with a mapping to $\Bbb P^1$. That's actually -- gasp -- topology for @Balarka, although it can be done via differential forms and the adjunction formula :P
The inverse limit of the groups $\Bbb Z/p^i$s with the pullback maps $\Bbb Z/p^{i+1} \to \Bbb Z/p^i$ defined by modding through $p^i$ is precisely $\mathbf{Z}_p$, right?
Now think of the Cayley graphs $\Gamma(\Bbb Z/p^i)$ of the cyclic groups.
There is also a geometric construction similar to the construction of $p$-adics : Consider the inverse system of $\Gamma(\Bbb Z/p^i)$ which are all quasi isomorphic to $S^1$s with the pullback morphisms being the covering maps $x \mapsto x^{p^i}$ as $\Gamma(\Bbb Z/p^{i+1}) \to \Gamma(\Bbb Z/p^i)$
My question is, what would be the inverse limit of this.
@Balarka the Wikipedia article shows that you won't possibly end up with $\Gamma(\Bbb Z_p)$ as your inverse limit, since solenoids aren't path-connected.
Just let me try to think about it, @Mike. I've let you know about them just because you might be interested. Note that I know nothing of solenoids, I have simply found a construction of something which has a fancy name.
I'm going to think of them about the construction with the topological aspect and the graph theoretic aspect both, and try understanding them. I don't want any facts to be revealed, @Mike. I'd appreciate if you won't reveal them. Hints, if asked, would be most appreciated.
(I haven't yet seen any references about solenoids -- I want to think about it first)
In particular $\Gamma(\mathbf Z_p)$ does not exist. For constructing the Cayley graph you need a finite or at least countable generating set to begin with @Mike. $\mathbf{Z}_p$ is uncountably infinite.
@Balarka The construction works fine with an uncountable generating set. The graph obtained just sucks as a metric space. Anyway, it's very difficult to have a conversation with you when it's not clear what we're talking about at any given point. Let me know when you've got some coherent ideas penned down.
@Alizter Yes. In Cambridge you will start with Analysis I instead of Calculus I, but Calculus I and II cover more than the calculus syllabus in the A levels.
@Alizter Calculus III will be like Cambridge's Vector Calculus course.
@Nick Turns out he had an off-sure account and was directing profits for his own gain. He also has a history of tax avoidance. He was transferred to a mental hospital after he scared some members of the public talking to construction vehicles. When he was released he had further charges of public vandalism.
@Alizter The books got renamed when it split between 2 publishers, Wiley and Springer. So although Springer says it's a 2 volume work it really depends on the Wiley one. There is some overlap but they are essentially different.
@Alizter As you will soon realise, every university math course in the world is different, lol. Cambridge has the best 3 year course in the world to me.
@Alizter However, maybe Warwick has the best 4 year course, lol.
@Jasper: Oh, so it's like a summary that goes before the content. Interesting :D
(I have not read a single math paper with a decent abstract then... maybe I haven't read a decent math paper yet... or even worse, I haven't decently read a math paper yet)
"Doubt" means you don't believe someone or something. Question means you're curious about something.
@Alizter There are a hell of a lot of papers. One can't go through them all. Knowing the results at the start helps one decide if it's worth it for what you're interested in.
Well, I'm off to bed. Ah, my chest hurts like hell. Anyone have any tips to ward off a runny nose... and an annoying persisting cough that sounds like a faulty Volkswagen engine starting?
@Nick Some people pull (diluted!) rose essence through the nose. Burns like hell, and that sort of sniffing fluids makes your head feel like exploding, but after that, you have got hours of peace.
@JasperLoy In the beginning of Nietzsche's Thus spoke Zarathustra, Z. meets an old man and tells him that he wants to give his insights (or whatever, I've forgotten) to the people. The old man said: Give them nothing, better take a bit off their load. (In the sense of "relieve them a bit", you know.)
@Chris'ssis Ok. Let $I(a)=\int_0^1 (\sqrt{x+1}-\sqrt{x-1})^adx$ then $I''(0)$ is the one you want. For $I(a)$ sub $\sqrt{x+1}-\sqrt{x-1}=u$ to get some mess. Then being careful with powers that are being integrated evaluate this simple power rule marathon. I see a lot of rationals logs and pis being generated. CBA to work this all out but setting $a=0$ at the second derivative gives you the value.
Yeah I used mathematica and the answer was not suprising.
