@PVAL: Malcev-Iwasawa says it's diffeomorphic to $G \times \Bbb R^3$, $G$ compact. This should be old knowledge, but at least it was long known that it's a vector bundle over $G$.
@SAWblade, nah, I'm no topologist. I was a differential geometer/complex algebraic geometer. But, yes, I've taught a bunch of topology (and lots of other things).
Proving the product of separable is separable doesn't sound like much of a fight.
But Eric took Guillemin & Pollack, so he has intuition from that, but apparently not enough for the reams of algebra. That's my issue with algebraic topology in general.
Rotman is good. I'm less fond of Halmos. And I've never heard of Carothers.
Thanks for leaving me off your list of likes, mr @Pedro. :D
Hatcher is a mess to read. Very wordy, almost impossible to reference the things you want. There are a lot of important things that are only briefly mentioned inline.
To me, Hatcher wins hands-down because all other alg top books of which I'm aware having terrible exercises — mostly just verifying algebra. Spanier has a few good exercises, but still ...
mr @Pedro, I'm sure people live just to annoy me. Well, I know students have. :P
@TedShifrin Spanier has a few exercises, yes. His exposition is quite slower than Hatcher's, but he's very meticulous. It is also more dry in the applications area. Hatcher gives useful techniques, which Spanier sometimes doesn't.
I think Hatcher's approach is probably fine. He should just define covering spaces (and prove the basic facts about them), if he wants to show it using covering spaces.
Instead of still proving it using covering spaces without ever mentioning them and then referencing this completely arbitrary looking proof when building up the general theory.
Well, @Ted. It says that if you have a set $S$ and two operations $\circ,\ast$ that are compatible in the sense that $(a\circ b)\ast (c\circ d) = (a\ast c) \circ (b\ast d)$ and admit units, then they are the same operation and are in fact commutative.
LOL ... I never remember the Yoneda lemma, even though Balarka mentions it every day. Oh, I haven't seen him in a few days. I hope his health hasn't deteriorated.
I'm OK with that; so do I. (Although more of my own diff geo homework problems are showing up as questions, and I don't like having all the answers to my problems posted.) ... But that doesn't belong on MO.
Yes, @PVAL, looks identical (except for a name) to me.
OK, I've got stuff to do in the kitchen. See y'all later.
Well if no such example existed Yasui's paper about contractible 4-manifolds would have proven there exists a contractible 4-manifold which isn't Stein.
@EricStucky Hey. So after reading over your answer to me, I found that it was quite difficult for me to write my answer in terms of limits in category theory like here: ncatlab.org/nlab/show/limit. I hope that you could clear this up for me.
@MikeMiller Its also a statement which is somehow just about kirby moves and projections of knots, which is really nice. There's no manifold theory necessary to motivate it.
Someone visiting here proved that (maybe errors in my statement of it) that alternating knots with unknotting number 1 had an unknotting crossing in any alternating diagram (using some Heegard-Floer and Donaldson's theorem I think). I really liked that statement because the theorem statement has really no actual "mathematical" content (like its really a statement about pieces of string which are circles rather than about embeddings up to isotopy) though the proof involved heavy mathematics.
I am working on three unrelated problems in topology... I have made some progress on each of them but nothing major yet. Should I keep thinking about all three or force myself to focus on one of them?
Oh, to live on Sugar Mountain With the barkers and the colored balloons, You can't be twenty on Sugar Mountain Though you're thinking that you're leaving there too soon
When you're solving a system of equations with a matrix and you end up with get a row that contains that $1$ and $-1$ with zeroes everywhere else in the row reduced form, why does that mean the system has no solutions?
Given the linear map $T(\mathbf{x}) = \mathbf{Ax}$, what's meant by "the image point $T(3 \mathbf{e}_1-2\mathbf{e}_2 - \mathbf{e}_3)$"? And how do you find this?
@TobiasKildetoft Okay. So I've been searching online, couldn't find anything on it. How do I solve it. Does it help that I've already calculated the null space of the map?
@DanielFischer: I'm not sure about the application of Cauchy-Schwarz in this proof. shouldn't I get a factor $(1+|\eta|^2)^t$ in both integrals after the inequality? the way I understand it, we first use CS to get the norm of the two functions wrt. the $L^2$ norm, which is less or equal to the Sobolev norm which is equivalent to the Sobolev norm defined with the Fourier transform (for general $s \in \mathbb{R}$). this would lead to an exponent $t$ and not $-t$.
@DanielFischer: nevermind, they apply Cauchy Schwarz differently than I thought. I found out how now.
A recent question asked about polynomial solutions to the functional equation $p(x^2)=p(x)p(x+1)$. Subsequently, Robert Israel posted an answer showing that solutions are necessarily of the form $p(x)=x^m(x-1)^m$.
What I had hoped to do myself was provide a solution by interpreting $p(x)$ as a g...
@MikeMiller: I have an argument to generalize Rellich to compact manifolds. if you have some time, can you ping me and I'll write it down? I'm not quite sure if it's correct since it seems a bit too easy.
The given vector $x$
$$
x = 4 e_1+ 4 e_3
$$
can be represented according the first or the second basis too.
\begin{align}
x
&= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)}
= (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\
&= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} ...
Am I misunderstanding the whole concept when I say I'm caught in-between two non-standard basis? Or maybe that wasn't the transition matrix I was supposed to find. Could someone please explain.
hello there. I have a question. As we consider diagonals of bisimplicial objects, can't we consider the diagonal of a double chain complex? If we take the convention that double complex = chain complex in chain complexes (i.e. squares commute, not anticommute), then the definition of the "diagonal" seems straightforward but I haven't seen anyone define it. Am I missing something obvious?
with this diagonal we would have, if I'm not mistaken, that if we start with a bisimplicial abelian group, then applying diagonal then Moore is isomorphic to Moore in both directions then diagonal
@user46225: That seems like a reasonable construction, but in my experience, not something that one would really want to do, given all the information that would be lost. For instance, you've lost your bi grading! And that's a shame.
@MikeMiller well, I mean, sometimes one works with the diagonal of a bisimplicial set. I think it's reassuring to know that one can do the same for bicomplexes
The given vector $x$
$$
x = 4 e_1+ 4 e_3
$$
can be represented according the first or the second basis too.
\begin{align}
x
&= x_1^{(0)} b_1^{(0)} + x_2^{(0)} b_2^{(0)} + x_3^{(0)} b_3^{(0)}
= (x_1^{(0)},x_2^{(0)},x_3^{(0)})^T = [x]_0 \\
&= x_1^{(1)} b_1^{(1)} + x_2^{(1)} b_2^{(1)} + x_3^{(1)} ...
