@LukasHeger just to be clear, the closed bounded subset you take could miss some finite points right?
@LukasHeger Oh, there is a statement in by book: Every d-Cauchy sequence of points of S converges to a point of S <=> every geodesic can be extended to a geodesic on the domain \Bbb R
This is what Hopf-Rinow theorem you mentioned right?
@love_sodam you can actually find a closed bounded subset that contains the whole Cauchy sequence. But if you have one that misses only finitely many points, that works for convergence, too
@TedShifrin If I know my book author's email then I will tell him to change everything in his book. Too few examples and explanations and especially those wired terminology he made
yeah an equivalence of categories is fully faithful and essentially surjective, so every dude in M_n(A)-Mod is (naturally isomorphic to) the image of a guy in A-Mod under that functor
Reminds me of the classic "The first author must state that his coauthor and close friend, Tom Trobaugh, quite intelligent, singularly original, and inordinately generous, killed himself consequent to endogenous depression. Ninety-four days later, in my dream, Tom's simulacrum remarked, "The direct limit characterization of perfect complexes shows that they extend, just as one extends a coherent sheaf." Awakening with a start, I knew this idea has to be wrong, since some perfect complexes have a non-vanishing K0 obstruction to extension. I had worked on the problem for 3 years, and saw this…
@stackex33 You're presumably trying to show that the quotient M/G is a manifold. Do you see why there should be problems whenever you have a fixed point?
@MikeMiller well, you're close. I was actually looking at the covering X over X/G and that led me to this. In wiki, they mention free action, and I had heard of it as "fixed point free", so I just wanted to confirm that it meant what I thought it meant.
As for the problems when it's not free, I'm guessing it messes up the Hausdorff condition, an extreme example I can think of is non zero reals acting by multiplication on reals. But in general, I suppose one has certain "pinch points" (I'm thinking of folding the unit disc by conjugation), that may be problematic. I don't know, I'll have to study it
@Thorgott S(g,n) is the surface recursively defined by S(1,0) = T^2, S(0,1) = D^2, S(g,0) = S(g-1, 0) # S(1,0), and S(g,n) = S(g,n-1) # S(0,1)
@stackex33 The Hausdorff condition is actually a different matter. "Freeness" is basically an independent problem. It has more to do with your pinch-points.
Usually one demand that the group action is "proper" in the sense that the map G x M -> M x M, given by (g,m) -> (gm, m), is a proper map. This is what you need to force the quotient to be Hausdorff.
@stackex33 It's a manifold with boundary, but your theorem doesn't say "when M/G is a manifold with boundary", it talks about when M/G is a manifold. If you add boundary points in this way you're out of luck
Consider the action of Z/4 on C by [n] * z = i^n z
If you take the quotient of this action you can prove that what you get is still homeomorphic to C.
But what is the smooth structure? Can there be a smooth structure so that $p: \Bbb C \to \Bbb C/(\Bbb Z/4)$ is a smooth submersion? (No: prove it. Think about $dp_0$.)
@MikeMiller one homeomorphism is given by passing z\mapsto z^4 to the quotient, and there's no submersion because the x and y axes have the same image (so the derivative is not injective), that right?
The proof of existence of Laurent series is cohomological in nature, right? Is it illuminating to rephrase it in that language?
Nevermind
Let $U, V$ be the upper and lower hemispheres of $\Bbb{CP}^1$. You want to write a holomorphic function on $U \cap V$ as a difference of a holomorphic function on $U$ and on $V$; that's the same as saying the $1$-cocycle is a $1$-coboundary. It's saying $H^1(\Bbb{CP}^1; \mathcal{O}) = 0$.
@Thorgott I was having issues with this earlier this month when I was planning lectures. Things like "knot genus" in terms of the minimal genus of a Seifert surface. What made it clearer for me at the time was Conway's ZIP proof for the classification of compact surfaces (with or without boundary): maths.ed.ac.uk/~v1ranick/papers/francisweeks.pdf
Kind of made the whole idea of genus a bit clearer because then you know that for orientable surfaces, they are classified by two pieces of data, the genus, and the number of boundary circles (holes), and these affect the Euler characteristic (taking it originally as 2 for the sphere) in dependable ways (-2g for the genus, and -n for holes).
In any case, the ZIP proof is a great read on its own and is worth remembering.
I did it with a bunch of high school kids and they seemed to like it.
It is, of course, incomplete. He gets to avoid Schoenflies by never explaining why some important operations (zipping, perforation) are well-defined.
I don't like the triangulation proof, but his proof is not much different than doing a handle-decomposition style proof, so I'd rather just do that instead..
I think my favorite at this point is (assuming you know that every surface can be written by iteratively attaching discs along portions of the boundary, which is the hard part) showing that the collection of standard surfaces is closed under handle-attachment. This reduces to some local computations of handle attachments on discs and annuli.
The point is that if you do a 1-handle attachment to S, it all happens local to two of the boundary components, and you can split off boundary components with S = S' # (something simple like a disc or annulus)
Since you have presumably already given a careful proof that the connected-sum is well-defined, which of course everyone has seen a careful proof of /s, it reduces you to simple computations
Still not my favorite (my favorite cuts down instead of assembling up). But the cutting down proof requires a more geometric understanding of homology than I can usually present
There's a simple proof that the only simply connected closed smooth surface is $S^2$; it's definitely $D^2 \cup_{S^1 \times I} \cup D^2$ by Morse theory, so it suffices to compute $\pi_0 \text{Diffeo}(S^1)$ which we can do. What about simply connected noncompact smooth surface?
I guess get a proper Morse function
Ok, you can definitely prove an increasing union of $\Bbb R^2$'s is homeomorphic to $\Bbb R^2$
Not as easy is it? If $f$ is say a non-negative proper Morse function, $f^{-1}[0, n)$ can be obtained from $f^{-1}[0, n-1)$ by handle-attachment, they handles just get capped off
It in general will show that any noncompact surface w/o boundary and H_1(X;F_2) = 0 is an ascending union of discs and hence R^2 after some Schoenflies / annulus thm arguments
Here's my understanding of the 2D situation. Let $X_n = f^{-1}[0, n]$. We know $X_{n+1}$ is obtained from $X_n$ by attaching handles. $X_n$ is a standard handlebody for any $n$, and eventually all the plugs (there are only finitely many) in it get filled so that it becomes a disk. This is true for each $n$, so this gives my surface is an increasing union of disks
I am not seeing where this is failing in 3D. The handles are $I \times D^2$ or $D^3$
If you attach $I \times D^2$ you get some contribution of the fundamental group, the core circle has to die later
So the "plug" obtained from attaching $I \times D^2$ eventually gets filled (by a $D^2 \times I$)
So why am I not able to make an ascending union of 3-balls following the same strategy?
But maybe this is wrong. $X_n$ will look like some genus $g$ handlebody anyway
There are $g$ many circles that should die in some $X_m$, $m \gg n$
No idea why the same idea doesn't work
Lol nevermind it won't work for the simple reason that there are many ways to kill $\pi_1$ by attaching handles in 3D
Ehh but then again there's not too many simply connected $3$-manifolds with boundary
the boundary components are all spheres by classification of surfaces, half-lives-half-dies. So capping them would give a simply connected closed $3$-manifold which is $S^3$ by PC
so the guy has to be a 3-sphere with balls taken out
ok so instead of an increasing union of D^3's you get increasing union of "pants", D^3 \ (D^3_1 U ... U D^3_k) for some k varying
I'm honestly too stupid for homology, sad state of affairs: I try to show that that reduced homology group of X is isomorphic to the homology of (X,*) for any point *. I go via the appropriate long exact sequence induced by inclusion and projection and for n>=2 i get 0 - > reduced H_n(X) -> H_n(X,*) - > 0 so it's easy, and for 0 I can just calculate.
But for n = 1 I get 0 -> H_1(X) - > H_n(X,*) -> Z - > ..., where I'm stuck. Do I need to do algebra here, or think about this topologically?
Actually I need to run, let me tell you a sketch. (1) $H_n(X, A) \to H_{n-1}(A)$ is the map which sends the homology class represented by the relative cycle $\xi \in Z_n(X, A)$ to the homology class represented by $\partial \xi \in Z_{n-1}(A)$. Chase the proof of the snake lemma to see this (2) A relative cycle in $Z_1(X, pt)$ is actually just a cycle, because $\partial [0, 1] = \{0, 1\}$, and these boundaries have to cancel in a way that the remaining boundary points lie in $pt$ to make such a singular cycle, forcing everything to cancel actually.
The KL Divergence is quite easy to compute in closed form for simple distributions -such as Gaussians- but has some not-very-nice properties. For example, it is not symmetrical (thus it is not a metric) and it does not respect the triangular inequality.
What is the reason it is used so often in M...
