@arctictern Is the problem not that your definition of composition is wrong? Qiaochu commented on your previous question that it's not a fiber product, it's a homotopy limit, or something.
Oh, it's much simpler. I can make $B$ as small as I want around the zero, so small as to make it fit inside a chart inside $W$. Then $f : B - p \to \Bbb R^n - f(p)$ becomes a degree 1 map.
No wait, nobody said $f$ is a local diffeomorphism near $p$.
No, I did pick a regular value in the range of $f$. Near that $f$ is a local diffeo.
@MikeMiller So obviously $f$ is a local diffeomorphism on the zero $p$ of $f$, like I said, and you can make $B$ as small as you want around $0$ to fit it inside a chart, and since $f$ is identity on that chart, $f : B - p \to \Bbb R^n - 0$ has degree 0.
(also, ping to @TedShifrin)
If I have zeroes $p_1, \cdots, p_n$ in general, I can pick small balls $B_1, \cdots, B_n$ around them and do the same argument: $W - \cup B_i$ is a cobordism between $\partial W$ and $\cup \partial B_i$ so $f/|f|$, defined inside that, has the same degree on both ends. $f/|f|$ has degree one on each of $B_i$, so sum of those mod 2 is precisely the degree of $f/|f|$ on $\partial W$, i.e., $n$ mod 2.
This should indeed be true without any mod 2 restriction because any orientation on $W - \cup B_i$ induces the same orientation on the boundary component $\partial B_i$'s. So, at least homologically, the topological degrees are the same.
i mean, even if you don't worry about what they mean, you can still think it's interesting to consider what the (self-adjoint!) operators $i\partial_t$ and $-i\partial_x$ do
i should say, formally self-adjoint. i know there's some stuff in that direction which one shouldn't be careless about
time independent one, I more or less understand because that's a sort of generalization of the wave equation. how does one get to the time dependent one from the time independent one? no idea
@Balarka: Your proof is basically correct. I don't like the sloppiness of saying $f$ is the identity on a chart. You'd have to change coordinates both places, and that alters the unit sphere in $\Bbb R^n$. It's easy enough to fix this.
so while one can definitely ask questions about the assumptions which underpin and motivate the Schrodinger equation, ultimately it's not going to be an absolute physical truth
you know, i feel like that in most of this quantum business. a spinning electron around the nucleus would spiral into the center, and suddenly bohr comes in and says no, it won't, let's take that as an axiom. huh? that clearly contradicts everything everyone did before. but, yeah, funny how it all worked out.
i'd put it a little differently: it's not that he predicted it wouldn't spiral in, it's that the entire story doesn't make sense if you allow that to happen
and given that it doesn't, you end up taking that as a given
@Balarka: So that's a pretty cool way to count roots in higher dimensions, isn't it? (Plus you can fix it all so there's no mod 2, as you already understand.)
Same difference. I'm referring to the arguments that prove Gauss's law. We're always putting a little ball around the charge (or mass, for gravitation).
@Balarka: It all goes much nicer with forms if you realize that a smooth form $f(z)dz$ is closed if and only if $f$ is holomorphic. I am not sure how much of the complex analysis overview lecture ended up on video. I think I did more of it last spring (which wasn't when the video was made).
@TedShifrin Um, write $f = u + iv$. Then $f(z)dz = (u + iv)(dx + idy) = (udx - v dy) + i(udy + v dx)$, and C-R tells the real and imaginary parts are closed, no?
Which makes certain typical CR problems kind've silly. No point in going to the CR equations if you've got $f$ being a real analytic function of $z$ to begin with
I'm currently looking at some of the properties of the Fermi Dirac distribution, given by $f(x) = \frac{1}{e^{(x-a)/b}+1}$. It has some (well, a lot of) significance in physics, but that is not my main reason of interest. What I am wondering about is the following: it is basically a smooth step function, starting at 1 and ending at 0. The parameter a sets where the function takes on the value 0.5, and b determines how steep the step is. But is there a way to quantify this steepness?
What I mean more concretely is, given a certain b, can you determine for which x f(x) is approximately still equal to 1, and for which x it is 0. I know that this is a bit too vague, but perhaps you understand what I am thinking of. Vague in the sense that 0.999 is close to 1, but so is 0.99999, and so is 0.99, depending on who you ask
if $x<a$, then the exponential term on the bottom is small. therefore you can approximate $f(x)$ in powers of that exponential, and therefore obtain the leading term $f(x)\approx 1-e^{(x-a)/b}$
@iwriteonbananas An inductive argument shows that the rational cohomology of $K(G,n)$ is zero for all $n$ when $G$ is finite; now use that there's a CW model for $K(G,n)$ that is finite in all degrees.
@MikeMiller Do you know if Brown ever computes the group cohomology ring of a finite cyclic group? There is a part that bugs me in my current reference.
I'm currently trying to think of something which does not function as a bottle opener to a Scandinavian. I was about to say 'someone's eyebrow' but then recalled I have seen that too (although without success and with blood.)
@TobiasKildetoft Snus is illegal to buy/sell in Denmark if I recall correctly?
Yeah, and the upper lip (so you don't have to spit.) Although some people who use loose snus (i.e. they pack it themselves, it does not come in a small 'teabag') put so much in it kind of goes all over the place.
@AndrewThompson you can't avoid having to spit regardless or where you place it. At least for most people it feels terrible to swallow if you have snus in your mouth
a paper claiming to use TQFT to get representation theoretic results showed up on the arXiv today; I want to know if it's interesting to representation theorists
I've seen this sort of thing show up in what I do, but it's usually a bunch of trig functions flying around for calculating numbers of solutions to PDEs
If X is an n-dimensional CW complex and e is an n-cell in X, then X\e is a subcomplex of X. I'm asked to show that X is homemorphic to an adjunction space of X\e with an n-cell. But since X\e and e are disjoint, then isn't X homeomorphic to the disjoint union of X\e and e? And isn't this an adjunction attaching e to X\e along the empty function?
OKay, I think I see the problem. X\e need to be open. X could be a connected CW complex for example. Then the disjoint union of X\e and e is disconnected, but their actual union is X with is connected, and so they cannot be homeomorphic
So in forming an adjunction space homeomorphic to X, I actually need to make some identifications.
@PedroTamaroff My "example" of any connected CW complex would present a problem, right? Although, I the idea occurred to me by considering [-1,1]. There's a proposition before this exercise in the book that says each n-skeleton is obtained from a n-1 skeleton by attaching n-cells. I think the point of the exercise is to use a similar argument as the proof of that proposition.
