Yeah, as the math has gotten harder, I've found I can rely less on my backburner, and must rely more on paper or chalkboard/whiteboard to develop the ideas--something I used to avoid, because I'm good at mental algebra.
I think all but the most self-assured (justifiably self-assured or not...) feel that. I solved that by ignoring it; I'll cross that bridge when I come to it.
There's some math I can do in my head, some I absolutely need paper for.
One thing that annoys me is I'm really slow at some kinds of computations. Like if you do differential geometry you'll look at pullbacks of differential forms, or exterior derivatives, or whatever; these really amount to substitution and partial differentiation. I absolutely cannot do these in my head. I need paper.
And oftentimes a small computation can take me a lot of time. It's killer.
yeah! worst case scenario I'm no good and I go into industry or whatever instead, right? but it's not going to help me today or tomorrow or even when I'm applying to jobs to think about that.
I guess the actual worst case is that a bus comes in through the wall of this bar and runs me over and I die, but I don't lose any sleep on that either
Perhaps that's a failure of the imagination. But, not to toot my own horn, I clearly have some knack for math. And I think that knack goes above and beyond my knacks for other stuff.
Shrug. If I spent as much time studying math as I spend on redundant introspection, I'd be much, much more knowledgeable than I am now.
slowly turns MSE into his personal therapy session
Let me try to be on topic. Man, rings are cool, right? All that addition, and the multiplication too, and whoa, that distributivity, that's some real neat stuff right there. Man.
cool ring-related thing: the set of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with addition being $(f+g)(x) := f(x) + g(x)$ and multiplication being $(f \cdot g) (x) = f(g(x))$ is not a ring because it fails exactly one ring axiom, namely distributivity on the left
well, fibers are orbits of Aut(p). one can just as easily take orbits of Gal(L/K). you were asking about how to translate the notion of fibers into field theory language weren't you?
fundamental group can be defined both as Aut(p) of the universal covering map p : \tilde X \to X and as homotopy classes of loops. Gal(\bar k/k) is suspiciously like the former, but there has been theory unifying these. but what goes with the latter definition?
if the two theories are similar, there should be an analog for the latter definition.
i think points have something to do with algebraic closures and paths have something to do with morphisms between them. explanation : i have done a trivial observation last night - \pi_1(X, x_0) is isomorphic to \pi_1(X, x_1) by conjugating by a path joining x_0 and x_1. in absolute galois groups, choice of basepoint comes from choice of algebraic closure. here, similarly, Gal(\bar k/k) and Gal(\tilde k/k) are isomorphic via conjugation by an isomorphism $\bar k \to \tilde k$
@MikeMiller Is $\pi_1(H)\to\pi_1(G)\to\pi_1(G/H)\to\pi_0(H)$ exact? seems so. that means $B_n({\cal S})\to{\rm Mod}({\cal S},n)$ is an isomorphism precisely when ${\rm Diff}^+({\cal S})$ is trivial right? To what extent are $\pi_1$s of ${\rm Diff}$s of surfaces (compact surfaces possibly with boundary minus a finite number of punctures) classified? Maybe I should just start reading A Primer and fill in the fundamentals later just for fun.
I mean, here I have $G$ being a diffeo group, in which case I only care about the connected component of the identity, which is fine because $\pi_1(G,e)$ only cares about that component too.
what makes you think it isn't the right invariant I want?
@anon: Yes, it is. I guess my answer left out the most important tool: what the homotopy long exact sequence is. When you have a fiber bundle $F \to E \to B$, you get a long exact sequence in homotopy groups $$\dots \to \pi_{k+1}(B) \to \pi_k(F) \to \pi_k(E) \to \pi_k(B) \to \dots$$ Just because the middle map in your bit is an isomorphism doesn't mean the outer terms are zero, just that the maps are zero (but of course you knew this).
I can prove that your guess (the map is an isomorphism iff $\pi_1(\text{Diff}(\mathcal S)) = 0$) is correct with a little more work. I guess you'd prefer to do it yourself than for me to tell you how to do it. Everything you need should be in my answer or its comments.
The homotopy type of $\text{Diff}(\mathcal S,n)$ is probably known for all $\mathcal S,n$. When $n>0$ I don't know it, but I'd be astonished if this wasn't known decades and decades ago. I would be unsurprised if you could prove it without too much work as a consequence of the Earle-Eeles theorem. In any case if you're particularly interested to know it you can surely find out with some googling.
@SohamChowdhury you may be interested in the tree topology - that's basically how you should think about p-adics topology. let me get a relevant link for you.
It doesn't make sense to say that $G$ is a fiber bundle. A map $E \to B$ is a fiber bundle. The issue is a) writing down the sequence in the first place - what's your map $\pi_1(G/H) \to \pi_0(H)$? b) how do you prove exactness in the middle?
Let $G$ be the 2-torus and $H$ be a line with irrational slope. Then $G/H$ is an uncountable set with the indiscrete topology. The exactness of that sequence fails in that center spot.
Besides, @Balarka, I will study NT and algebraic geometry later. But not right now. And class field theory looks forbiddingly hard, maybe I'll think about it sometime later when you start learning geometry.
In your context (as mentioned below my answer) whether or not $H \to G \to G/H$ is a fiber bundle is equivalent to asking whether or not $G \to G/H$ has a continuous local section.
@anon: When $H$ is closed and $G$ is a Lie group it's immediately true, because you get some results about the manifold structure of $G/H$ and the map $G \to G/H$ (it's a submersion). I'm not comfortable saying anything in more generality than that.
(Of course every time I've been writing down $\pi_i(X)$ I implicitly mean $\pi_i(X,x)$, having chosen some basepoint $x \in X$. For you your basepoint is just the identity, and you have an obvious basepoint in $G/H$: $[e]$.)
I'm fascinated by this open problem (if it is indeed still that) and every few years I try to check up on its status. Some background: Let $x$ be a positive real number.
If $n^x$ is an integer for every $n \in \mathbb{N}$ then $x$ must be an integer. This is a fun little puzzle.
If $2^x$, $3^x$...
@anon: by the way, for the thing I'm thinking about, it turns out I can stick to irreducible representations: there aren't any nontrivial reducible reps at all. this is probably convenient
"Mathematics is the most beautiful and most powerful creation of the human spirit." - Stefan Banach
@r9m ^^^ maybe it makes more sense ;)
"Mathematics has beauty and romance. It's not a boring place to be, the mathematical world. It's an extraordinary place; it's worth spending time there." - Marcus du Sautoy
@Hippalectryon Yes but if you have collaborated with 2 Erdos collaborators then you must be worth more than if you have collaborated with 1 Erdos collaborator.
It seems interesting because its an exploration of this 0 to 1 log scale, it's a very simple formula, it's gives curve that decelerates from a slope of probably around 1 gradually to 0, and as the value approaches 1 from a value of 0, and it is this interesting itteratie to repeatedly add one and half, but for a non-integer number of times.
@MatsGranvik How would you create the "new" erdos numbers then ? You'd have to make sure that someone with 10000000 collaborations with people who have an erdos number <= 3 still has an erdos number <3
@BalarkaSen tell me your wonderful ideas when I can understand them, okay? I expect you'll write a paper or something on the connections you're finding. :)
@MatsGranvik Uh, work with one Erdos collaborator, and carefully cut off $\sqrt{2}-1$-th of another off. Collaborate with both. Profit.
(Note: I'm not responsible for what happens if anyone follows this :P)
Consider two groups $G$, $H$ and let there be homomorphisms
$$\eta_1, \eta_2: G\to H.$$
1) Let the kernel of both homomorphisms be $K\triangleleft G$. Must $\eta_1 = \eta_2$?
2) Let the kernels of $\eta_1$ and $\eta_2$ be $K_1, K_2\triangleleft G$, with $K_1 \cong K_2$. Does this imply $\eta_1...
$\eta_1 : \Bbb Z \to \Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ given by $\eta_1(x) = (q(x), 0)$ has kernel $2\Bbb Z$. $\eta_2$ is the same with the second factor. $q$ is the quotient map.
Plotting values between 0 and infinity in a space between 0 and 1, like arc-tan but more elegant, I'm afraid is the only one I can think of. The other nifty properties are an added bonus.
@Hippalectryon The questions I propose on channel here are in general very easy problems to say it right. I might show you one day what hard means to me.
Using imaginary numbers with it, for the question "What is the probability of having rolled a given value on a (2 + i) sided complex dice at lease once by 5 rolls, the answer is (0.012... + 0.013... i). Seemingly more rolls of a complex number sided dice can decrease your chance of ever having rolled something at all.
@Rememberme Yes, in much the same way that we talk about functions. Those homological algebra-ish thingies appear a lot in "higher mathematics". Nothing special about algebraic topology per se.
Ammm..... Let A to A closure be a function and complementation from A to X-A(where A is a subset of the topological space X) . If I apply these two operations successively I can form no more than 14 sets @BalarkaSen
@BalarkaSen its there in munkres though .... But I wasn't able to prove it
So my question is there an example of an uncountable infinite set which is a kuratowski set@BalarkaSen
Some progress : getting expression for $I^f_{\omega}(t)$ which we here on simply denote as $I_{\omega}(t)$
Let $$Z_{\alpha}(t) = f(t)\ast\frac{e^{i\alpha t}}{t}\tag{1}$$ and
$$R(\omega) = |F(\omega)|$$ and $$g(\omega) = \int_0^{\omega}R(\alpha)d\alpha$$ then
$$I_{\omega}(t) = \frac{1}{g(\omega...
Are time and energy inversely related? Some mathematicians have a lot of energy but no time, while other mathematicians have a lot of time but no energy.
@skillpatrol In my case my aim is to publish the best book on integrals, series and limits ever (as a collection of problem and solutions), not for the sake of being the best, but since I'm just a self-educated, I wanna prove myself that with the only resources I have from self-education I can reach the sky and more beyond.
Consider two groups $G$, $H$ and let there be homomorphisms
$$\eta_1, \eta_2: G\to H.$$
1) Let the kernel of both homomorphisms be $K\triangleleft G$. Must $\eta_1 = \eta_2$?
2) Let the kernels of $\eta_1$ and $\eta_2$ be $K_1, K_2\triangleleft G$, with $K_1 \cong K_2$. Does this imply $\eta_1...
Some progress : getting expression for $I^f_{\omega}(t)$ which we here on simply denote as $I_{\omega}(t)$
Let $$Z_{\alpha}(t) = f(t)\ast\frac{e^{i\alpha t}}{t}\tag{1}$$ and
$$R(\omega) = |F(\omega)|$$ and $$g(\omega) = \int_0^{\omega}R(\alpha)d\alpha$$ then
$$I_{\omega}(t) = \frac{1}{g(\omega...
