But the argument from before does not work. What do you get when you take germs at infinity? The space of germs of homeomorphisms of R^n that preserve 0 and are the diffeomorphism away from 0. Which is, what?
Question, what does $(\mathbb{Z}/n\mathbb{Z})^{\times}$ mean notation-wise? I know that $\mathbb{Z}/n\mathbb{Z}$ are the integers modulo $n$, but I'm not sure what the $\times$ notation is supposed to denote
"you have three pulleys inside a roller-coaster ride which are in turning falling from outer space with 981 cm/s acceleration, each of them carrying a spring of variable length. suppose they are connected parallely. find the spring force constant"
and all kinds of extremely unrealistic situations
as a bonus, after you do the math, you find that's not at all what the problem meant. the roller-coaster ride was moving in 981 cm/s^2 acceleration, not the whole system... great
@Balarka Ok. I was going to say I can offer other things to work on if you don't. But if you do you do. I thought you were done with school for a while?
School's off but there's still a bunch of schoolwork. I can hear what you suggest though, spectral sequences was just for a refresher in algebraic topology and I am not going to read much more than Serre spectral sequences anyway
@BalarkaSen Foliations are near and dear to my heart, and you have the background. I don't know that much about them, so at a point we'd be caught up and learning at the same time.
@MikeMiller Great idea. Earlier the differential geometry I started studying with you was too hard for me because it takes a lot of work for me to think in terms of symbols, so I hope there's more topology element in this.
@BalarkaSen Yes. You get over the symbols in differential geometry eventually; I did. (I hate symbol-pushing exercises.) But you only get over that by pushing symbols until you understand the symbols.
@MikeMiller Fair enough, I agree. It took me a lot of work even to push through Ted's stuff (which I am reasonably comfortable with now). Nope, I don't.
If for some reason I make the redefinition $\psi(x)\to e^{i\lambda(x)}\psi(x)$, I can recover something that looks mostly like the Schrodinger equation: $-\frac{1}{2m}\left(\frac{d}{dx}+i\lambda\right)^2\psi(x)+V(x)\psi(x)=E\psi(x)$
(not confident I've done the algebra right on that. working somewhat from memory)
@BalarkaSen A foliation is a decomposition of a smooth manifold into smooth submanifolds (not necessarily embedded) so that locally, the partition looks like the partition of $\Bbb R^n$ into parallel $\Bbb R^k$s.
Note that in this decomposition, there's no reason to believe that one of the smooth submanifolds we started with corresponds to a single $\Bbb R^k$ - it might even correspond to eg $\Bbb Q \times \Bbb R^{n-1}$
but each path component in the chart is one of the $\Bbb R^k$s
The standard example is the decomposition of $T^2$ into irrational lines of the same slope.
For instance, take $[0,1] \times \Bbb R$, the left side and the right side are submanifolds, and the submanifolds in the middle look like parabolas pointing downwards, stacked on top of each other. Does this picture make sense so far?
Then you've got two circles and a bunch of lines in your foliation of $S^1 \times [0,1]$. If you want this on a closed manifold, double that to get a foliation of the torus with two circles and a bunch of lines.
Oh, I was think of the "foliation" by torii of increasing diameter, which is not really a foliation (well, it's a foliation of the complement of the Hopf link). Yeah, with circles that gives one.
Now thinking of this distribution, let's just think about distributions in general. We say a distribution is integrable if locally it's equivalent to, well, standard thing on $\Bbb R^n$. Equivalently, that at every point, there's a submanifold $M$ passing through that point and everywhere tangent to the distribution.
Let's suppose it's codimension 1, so that locally it's the kernel of some nowhere zero 1-form $\lambda$. OK?
Ah, right, given a point $p$, just draw the subamanifold passing through it tangent to the distribution and then take their union for all $p$'s (some of them might coincide but w/e). It should just be that simple.
Suppose $\lim\limits_{n\to\infty}n\left(f(1/n)-f(0)\right) = 1$ and $f$ is continuous at $0$. Can you conclude that the right-hand derivative $f'_+(0)=1$?
@BalarkaSen I'm not super sure where to start with this, I haven't given a foliations lecture in quite some time. There are many things one can study. One can study the set of closed leaves; the dynamics of where leaves accumulate; whether there are any closed foliated subsets other than individual leaves or the whole thing... Whether manifolds can support certain kinds of foliations, or foliations at all.
No. Set $k = 1$. You're asking for the existence of a nonzero vector field.
For $k = n-1$, taking normal spaces you're asking for a line field as well. It's due to Thurston that you support a codimension 1 foliation iff you have $\chi(M) = 0$.
Instead of talking about the categorical way of defining monomorphism. I wanted to see if I have the correct intuition regarding monomorphic function. A function $f : A \rightarrow B$ is said to be monomorphic if for all sets $Z$ and for all functions such that $\alpha_1,\alpha_2 : Z \rightarrow ...
I can throw a question at you I liked working out (but I suspect at this point you're not going to be able to find the full answer to). Suppose $M$ is foliated by circles. Is the foliation the fibers of some fiber bundle?
Or I can say things if you have questions. Basically I suggest picking up Foliations I, Candel and Conlon.
By the way, @BalarkaSen, note that in the example I gave all bundles have trivial total space, so you can pick a nontrivial bundle and the trivial bundle and you can't do any tricks about automorphisms of the base.
I'm a bit sleep-deprived now but you mean that Milnor's theorem says more strongly that the map [X, O] --> [X, TOP] is zero instead of just not being injective, yep? So every bundle (of appropriate rank) has total space homeomorphic to the trivial one.
@MikeMiller I've been rather frustrated looking for modern expositions of Novikov's work. Calegari seems to give up on some of the harder ones (e.g. the consequence in my answer to your older question: a closed leaf of a taut foliation is $\pi_1$-injective).
Injectivity is good as a first intuition but you want to keep in mind that morphisms in a general category need not be functions. So I don't know if there is a "non-algebraic way" to think about them.
You should really think of monomorphisms as morphisms that can be "left-canceled". In particul...
@MikeMiller My only interaction with him, was handing him his packet at a conference in UT. I remember that he insisted that he was given a receipt for the $20 he had to put down for the dinner at that conference.
I think I have been at a few conference with a nametag that he never picked up.
The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space.
In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely relat...
@MikeMiller Isn't exercise 1.1.3 in C-C precisely the Ehresmann's theorem?? "If $\partial M = \emptyset = \partial B$ and $B$ is connected prove that a submersion $f : M \to B$ with compact level sets is a fiber bundle" O_o
Namely: "Show that the bullet-riddled square $ [0,1]^2 \setminus {\bf Q}^2$, and set of bullets $[0,1]^2 \cap {\bf Q}^2$, both have inner Jordan measure zero and outer Jordan measure one. In particular, both sets are not Jordan measurable."