@Daminark Solutions to Laplace's equation are fixed points of the heat equation flow, and you can come to solutions by picking something with given boundary conditions and performing the heat flow.
@MikeMiller Hmm, so you're asking, why are the germinal holonomy either conjugate to germ of the identity or the antipodal map? I guess I haven't worked that out.
@ZachHauk He'll take you from $\varepsilon$-$\delta$ limits to integration in finite terms, even covering a few of the same objects and definitions from real analysis but in more casual tone oc.
Though you often talk about these things called nets in general topological spaces for whatever reason, point is, Hausdorff spaces have enough resolution that a sequence has a unique limit
Well, limit point is defined in terms of a given set, to be precise, but yeah
@PVAL-inactive Look at the proof of the fact for flat G-bundles in Morita's geometry of characteristic classes. It's short and intuitive. The point is that you can define a map from the G-bundle determined by the holonomy representation as $\tilde M \times_{\pi_1} G$ to your bundle by picking a basepoint and doing parallel transport
(Then the flatness in the differential geometric sense is what implies that this map is well-defined; homotopic paths induce the same parallel transport.)
@ZachHauk Consider the cofinite topology on $\Bbb Z$. That is, the open sets are the ones whose complements are finite (also the empty set is open 'cause it has to be)
You convince yourself that foliated bundles over compact manifolds are the smooth analogue of flat G-bundles, a nbhd of the zero section of the normal bundle is a foliated bundle, determined by its holonomy, blah blah blah.
I think that's essentially the main idea of a Hausdorff space — A space is Hausdorff when you can't have something "really close" to two things at the same time.
@MikeMiller I read a paper (admittedly 70's) saying its open whether codim k foliations of compact manifolds by compact leaves have regular neighborhoods of leaves.
@PVAL-inactive Ah that's really interesting. I have this belief that foliations by compact manifolds are fiber bundles over orbifolds but I can only prove it when the foliation is of dimension 1 or codimension 1. I asked Rachel Roberts about it and she was skeptical but didn't know how to prove or disprove it.
By the way, Zach, there's a way to formalize what I said using some extremely weird objects called ultrafilters. But you really don't need to know anything about those to study topology.
Joke 1.3: Suppose $\mathfrak{J}^{\infty}_{\otimes \Bbb Z}$ is a coprime, compact, Hausdorff bundle whose complement is a co-sheaf of the canonical homomorphism from $\mathfrak{T}_{fin} \oplus V^{\star}$ to $\Bbb Z^4$
Sorry, I am confused. How did we establish the existence of a small saturated neighborhood in codim 1 in the first place, then? Don't we need that before we figure out what it is by holonomy
@PVAL-inactive Like they say having good nbhds is the same as having finite holonomy. In dim 1 let's assume for convenience of conversation that the foliation is co-orientable. Then suppose the holonomy of $\gamma$ was nontrivial. Call the germ of a homeo $\Bbb R \to \Bbb R$ it gives $f_\gamma$.
Pick your favorite number $x$ such that $f_\gamma(x) \neq x$. If $f_\gamma(x) < x$ then (because orientation-preserving homeos are order-preserving) $f^{(n)}_\gamma(x) < f^{(n-1)}_\gamma(x)$, and lifting the loop $\gamma$ to the a path in the foliation, we get a map $[0,\infty) \to M$ with $g(n) = f^{(n)}_\gamma(x)$ with image in a single leaf.
I'm confused, give me a minute. The fact that some subsequence of this accumulates should be the problem but it's not obvious why.
If $\|x_n\|_p \to \|x\|_p$, prove that you actually have strong convergence
What we have at our disposal is that $(\ell^p)^* \simeq \ell^q$ where $\frac{1}{p} + \frac{1}{q} = 1$, and that weak convergence in $\ell^p$ is equivalent to convergence in each component
I'm not yet seeing (even though I feel like I should) how to piece this stuff together
Ah yeah. Call that sequence $x_n$. We see that the leaf corresponding to $\lim x_n$ must also be the leaf corresponding to the $x_n$, since some subsequence of $x_n$ converges inside the leaf, and therefore to $\lim x_n$.
But if $x_n > y > x_{n+1}$, then clearly $y_n = f^{(n)}(y)$ also has that same limit.
So the $y_n$ are also in the same leaf as $\lim x_n$. Similarly we see that everything in $[\lim x_n, x]$ is in the leaf, which like Balarka said is nonsense since we chose that to be transverse.
So $f_\gamma = id$. If $f_\gamma(x) > x$ then follow the proof for $f_\gamma^{-1}$.
That's the stupid part though. Exponentiating the normal bundle we get a bundle with a foliation transvere to the fibers (for small enough time). That's all we need.
To be more precise about the co-oriented case: Not only is every $f_\gamma$ order 2, in fact the image is precisely two elements. For if $f_\gamma$ and $f_\eta$ are both nonzero, $f_\gamma f_\eta = 1$ by the above argument, so $f_\gamma = f_\eta$.
Somehow in the codim 1 case you can stay very close to the leaf if you go along this transverse germ at any point, just by conjugating this path to other points.
In general if $g$ is a homeomorphism $[0,\infty)$ to itself, then $f$ is always of the form $f(x) = - g(x)$, for $x \geq 0$ and $g^{-1}(-x)$ for $x < 0$. Set $h(x) = x$ for $x \geq 0$ and $h(x) = g(-x)$ for $x < 0$. Then $hfh^{-1} = -1$.
So the whole thing is conjugate to a rep to $\Bbb Z/2$.
@MikeMiller @BalarkaSen even with a small neighborhood transverse to the fol'n I STILL don't understand why the parallel transport on a leaf (besides the original itself) must stay within that neighborhood.
Maybe PVAL is thinking about phenomenon like foliations of the form $xy = c$ and $y = 0$, near a neighborhood of the leaf $y = 0$? If you cut in by a tubular neighborhood you may get leaves like that
Suppose we consider the ring $R = M_n(B)$ where B is some ring. Then, I claim that there is one to one correspondence between left ideals of R and left ideals of B.
First if J of B is a left ideal. I proved that $J \mapsto M_n(J)$ is an ideal of R.
Conversely if $M_n(I)$ is an ideal of $M_n(B)$ I proved that $M_n(I) \mapsto \{c \in B: c = x_{11} \ for \ X \in M_n(I)\}$ is a left ideal of B.
well, you said "if $M_n(I)$ is an ideal of $M_n(B)$..." - aren't you supposed to say something like "let blah be an ideal of $M_n(B)$" without assuming it's of the form $M_n(I)$?
If C is a ideal of $M_n(B)$, then we have that $\phi(C)$ is composed of all elements of B for which the elements is the first coordinate $x_{11}$ for some matrix inside C.
First of all if we have $M \in (\psi \circ \phi)(C)$. Then M is a matrix whose first coordinate is $m_{ij}$ coordinate is first coordinate of some matrix X inside C. Then, we want to construct a matrix J whose coordinate agrees with M. We can do that by by first multiplying by switching of rows matrix and adjusting that right @arctictern ?
Show that a finite union of compact subspaces of a topological space $X$ is compact.
The proof goes as follows : "Let $A_1, \ldots, A_n$ be compact subspaces of a topological space $X$. Let $\mathscr{B}$ be a collection of open sets of $X$ which covers $\displaystyle{\bigcup_{i=1}^n A_i}$. Then, $\mathscr{B}$ covers $A_i$ for each $1 \leq i \leq n$. Since each $A_i$ is compact, we can choose a finite subcover $\mathscr{B}_i$ of $A_i$. But then, $\displaystyle{\bigcup_{i=1}^n \mathscr{B}_i}$ forms a finite subcover of $\displaystyle{\bigcup_{i=1}^n A_i}$"
If I have region R on r = 2 sin(theta) + 4cos(theta) from [0,2] in the first quadrant, how could i find the volume of the solid generated by revolving R about the y-axis using disks and washers
Karim: I'm not paying attention to what you're discussing with tern, but remember that row operations correspond to left multiplication by elementary matrices.
@Ted I'm going to look over some of those chapter 2 problems and then start where I left off tomorrow. right now i'm going to watch "The Shining" for the first time
@TedShifrin Hmm, I can't see why it doesn't matter. Can you give me a hint, because currently I'm thinking there could be some open cover of some $A_i$ consisting only of open sets with respect to the subspace $A_i$ and in which case some finite subcovering is given by a finite collection of open sets with respect to the subspace $A_i$ (wherein all of those open sets with respect to $A_i$ need not be open with respect to $X$)
@TedShifrin If I turn it into cartesian coordinates, I get (x-2)^2 + (y-1)^2 = 5. I guess it's a little confusing for me because region R is part of a circle, so I'm not sure how to do disk/washer with that
@MathisLife: You have to break it up into intervals. On one, you get disks when you slice perpendicular to the $y$-axis. On the other, you get washers.
I got bored and inspected the in- and circumcircles of regular polygons. Deriving the radii from the polygonal side length is obvious for even number-sided figures
