Remember a topology is just a thing $(X,\mathcal{J})$ such that the union of any things from $\mathcal{J}$ is in $\mathcal{J}$ and so forth. We just call the things in $\mathcal{J}$ "open sets"
I encourage people to give concepts stupid names they know not to be true when they meet a concept with a familiar name, (a,b) is KNOWN to be an open interval, rather than thinking of open-ness (because of the name) it can be useful to think of it as potato in the topology induced by (metric)
@Soham In the first algebraic topology class in the uni, SB was giving a few examples of topological spaces. He asked "does everyone know what a graph is?" and when everyone said yes he asked us to define what it is. When nobody replied, I tentatively said "um, a 1-dimensional simplicial complex?" :P
"I am a software developer by trade, and while I can't say I've taken great interest in category theory, category theory seems to have taken great interest in my field recently." -- from the reviews of that book :P
I guess the grad students I started to learn math with knows more math than I do now. The course also included singular cohomology, but I dropped out because of schoolwork, so I didn't get to sit on that.
If someone had given me some kind of "simplified Princeton Companion" or told me a little about what "serious math" was like, I would've started far earlier.
The Basel problem is a problem in mathematical analysis with relevance to number theory, first posed by Pietro Mengoli in 1644 and solved by Leonhard Euler in 1734 and read on 5 December 1735 in The Saint Petersburg Academy of Sciences (Russian: Петербургская Академия наук). Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given...
he factorizes the function in terms of the roots etc.
the real analogy lies in realizing $\mathsf{Spec} \, \Bbb Z$ as a 3-manifold and embeddings $\mathsf{Spec} \, \Bbb F_p \to \mathsf{Spec} \, \Bbb Z$ (which looks like embedding of circles in $S^3$)
it's actually more than a topological space : at every point of $\mathsf{Spec} \, R$ (i.e., a prime ideal $\wp$ of $R$), there is an associated field $R/\wp$. so every point has a field attached to it. you can do the same with every open set (localization).
$\mathsf{Spec} \, R$ is defined to be collection of prime ideals of $R$ and the topology on it is defined by setting the closed sets to be collection of prime ideals which contain some certain ideal $I$ of $R$.
after doing the torus thing, I again got interested in polynomial equations. I used to be mad about all those. analytical geometry is still fun for me.
like plotting higher analogues of ellipses and stuff.
At the very end of the line, there is $(0)$ sitting in it. Note that every ideal of $\Bbb Z$ contains $(0)$, so every neighborhood of $(0)$ intersects $(p)$ for any $p$.
So this topology in non-Hausdorff.
@SohamChowdhury Yeah, the fuzz comes from what I said above.
oh, btw, you should also note that the line sticking out of $(x^2 + 1)$ forms something like a 2-sheeted covering space over the line sticking out of $(x)$.
that's something formalized by Grothendieck, I think.
@BalarkaSen: I have two definitions: Firstly, $M,N$ are locally isometric, if for all $p \in M$ exists an open nbhd $U$ of $p$ and a $q \in V$ open in $N$ such that $U$ is isometric to $V$. Secondly, a local isometry between $M$ and $N$ is a map $f:M \to N$ such that for all $p \in M$ and all $X,Y \in T_p M$ we have $$\langle df_p X, df_p X \rangle = \langle X, Y \rangle.$$ I find those definitions rather weird. Are they standard?
@BalarkaSen: Also, just looking at the notion, I'd guess that being locally isometric should be equivalent to the existance of a local isometry. But looking at the definitions, this doesn't seem to be the case. Is this just bad notion or am I misunderstanding something?
@Huy: It is not true that these two notions are equivalent. If $M$ and $N$ are compact, a local isometry $M \to N$ has to be a covering map. In particular this means that the relation "there is a local isometry $M \to N$" is not symmetric: take, say, $M$ and $N$ to be genus 2 and genus 3 surfaces, respectively; there is a covering map $\Sigma_3 \to \Sigma_2$ but not the other way around. These are locally isometric.
What's worse I can find you two hyperbolic 3-manifolds $M,N$ that don't have a local isometry $f: M \to N$ or $f: N \to M$; but they're both hyperbolic, hence locally isometric.
I think it's fairly clear notation. It is a bit annoying that nobody points out that "locally isometric" does not imply "exists a local isometry", though.
@MikeMiller: I usually just assume that implication to be true when reading definitions, like being homeomorphic means there exists a homeomorphism, etc. I'm not sure whether there are further counterexamples though?
Fair enough. The definition of homeomorphic is 'exists a homeomorphism'. If a definition doesn't say ___ means exists a ____, just assume they're not equivalent. (Or try to prove/disprove.)
@TedShifrin: As an exercise, I'm supposed to show that any smooth curvature scalar $k(s) > 0$ and torsion scalar $\ell(s)$ determine a curve in $\mathbb{R}^3$ uniquely up to rigid motion. As a hint, I was told to compute $$\frac{d}{ds} \begin{pmatrix} \tau\\N\\B \end{pmatrix}.$$ I found $$\frac{d}{ds} \begin{pmatrix} \tau\\N\\B \end{pmatrix} = \begin{pmatrix} 0&k&0\\-k&0&\ell\\0&-\ell&0 \end{pmatrix} \begin{pmatrix} \tau\\N\\B \end{pmatrix}.$$ I feel like I should be almost done, what's the
the group would have to itself be a space in order to even call the map a homeomorphism, in which case the homeomorphism doesn't care about the group structure it only cares about the topological structure and so your question reduces to asking how to visualize a homeomorphism between two spaces.
Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?
Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{...
"how do I think about X" is not a precise question. you learn how to think about mathematical objects by knowing features about it that allow you to mentally visualize it (so to speak)
you should be able to get them by considering higher-genus surfaces minus discs
if $\Sigma_{g,n}$ is a genus $g$ surface minus $n$ discs, figure out what its fundamental group is, and therefore what $\Sigma_{0,2k}$ it could possibly cover
then try those
(actually this tells you precisely what the covers of $\Sigma_{0,2k}$ are, since all covers of it are surfaces, and because it has free fundamental group, has covers of all degree...)
if we take A to be the category of finite graphs with valence-preserving surjections, and B the category of closed, compact orientable surfaces with homotopy-classes of covering maps, then I think there is a functor A->B which makes a sphere for every vertex on a graph, then connects the spheres via tubes (after deleting discs and identifying boundaries). which is why I ask.
@anon A sphere minus k open disks is homotopy equivalent to a wedge of k-1 circles. In particular there is a 1:1 correspondence of coverings of a wedge of k-1 circles and a coverings of a sphere minus k open disks. These also correspond to subgroups of the free group in which there is a lot of work trying to understand these (see the end of Hatcher chapter 1 or Stillwell's "...Combinatorial group theory").
For example, if you remove 2 disks from $S^2$, the space is homotopy equivalent to $S^1$ and the coverings (up to covering isomorphism) are from the map $z\to z^n$ or $z\ to -z^n$ thinking of the unit disk as the deleted disk.
$f: M \to N$ is a local homeomorphism if for every $p \in M$, there is an open $p \in U \subset M$ and open $f(p) \in V \subset N$ such that $f(U) \subset V$ and such that $f\big|_U: U \to V$ is a homeomorphism.
@Huy $f : X \rightarrow Y$ is a local homeomorphism if for every $x \in X$, there is an open neighborhood $U$ of $x$ such that $f(U)$ is open in $Y$ and $f|_{U} : U \rightarrow f(U)$ is a homeomorphism.
I'm a grad student at UTexas at Austin. I think if one is careful about the correspondences one can probably find out the genus of any finite index covering from any finite index subgroup of the free group (when one looks at the graph). I don't know how to do this, but I wouldnt be surprised if it was in Stillwell's book.
Actually, I think the cover can be chosen to be holomorphic and the Riemann-Hurwitz formula explicitly calculates the genus.
The ramification index should nicely correspond to the cover induced on $S^1$ from the cover of graphs. Thinking about each $S^1$ in the wedge as a boundary curve in the surface, and yes I do know Tye.