Is there any part of this which needs more specific info than what's stated, e.g. abelian-ness and such. Mind you, I forgot everything I knew about group theory. I had to learn again that one cannot quotient away just any subgroup
"is taking quotients functorial in both arguments" is probably a more appropriate question
and does it commute with direct sums in both arguments
the concise algebraic way to state what you want to state is that $\dots\rightarrow C_n\rightarrow C_{n-1}\rightarrow\dots$ and $\dots\rightarrow\bigoplus_{i\in I}C_n^i\rightarrow\bigoplus_{i\in I}C_{n-1}^i\rightarrow\dots$ are isomorphic chain complexes, where the chain maps are precisely the isomorphisms you give first
the commutativity of the resulting diagram is the same as saying that $\partial_n$ in the first chain complex corresponds to $(\partial_n\vert_{C_n^i})_{i\in I}$ in the second chain complex
then the result follows since isomorphic chain complexes have isomorphic homology groups
which is an immediate consequence of homology groups being functorial
Yeah the clearest thing to do is to write $f: \bigoplus_{j \in \pi_0 X} C_\ast(X_j) \to C_\ast(X)$ by $f = \bigoplus_{j \in J} (i_j)_*$. Each $(i_j)_*$ is a chain map, since it's the induced map of a continuous map. This is surjective since every singular simplex in $X$ lies in some $X_j$.
$C_*(\sqcup_{j \in \pi_0 X} X_j) \to C_*(X)$ induced by the obvious continuous map $i$ is an isomorphism --- injectivity because $i$ is literally injective, surjectivity because every simplex lies in a path-component
Then use that homology groups of disjoint unions are the disjoint union of homology groups
As for the algebra questions: Yes, taking quotients is functorial in the right sense. You can consider a category where objects are pairs $(N,G)$ of a group $G$ and a normal subgroup $G$ of $N$. A map $(N,G)\rightarrow(M,H)$ is a group homomorphism $f\colon G\rightarrow H$ such that $f(N)\subseteq M$ (so $f$ preserves the subgroup). There is a functor to groups mapping $(N,G)$ to the quotient group $G/N$ and $f$ to the map $\tilde{f}\colon G/N\rightarrow H/M,\,gN\mapsto f(g)M$. This is well-defined, precisely because $f$ maps $N$ to $M$. Note that chain maps commuting with the differentials…
Injectivity and surjectivity are both completely clear. $C_n$ is free abelian on the set consisting of all $n$-simplices in $X$. $C_n^i$ is free abelian on the set consisting of all $n$-simplices in $X_i$. Each $n$-simplex in $X$ lies in one and only one $X_i$ (by path-connectedness), so we have an explicit bijection between bases.
idea is the same if you take coefficients elsewhere than $\mathbb{Z}$, still
the answer I should have given is that singular homology is maps from singular manifolds mod singular bordisms, an both singular manifols and bordisms are path-connected, hence take place in a given path-component
The point is that the data of a cycle --- $\sum \sigma_i$ with $\sum d\sigma_i = 0$ --- is almost exactly the data of an $n$-ddimensional simplicial complex $P$ with orientations of each simplex so that each codim 1 simplex is contained in exactly two codim 0 simplices
This $P$ is a manifold except along points in the (n-2)-skeleton
AKA: "manifold with singularities in codimension 2"
To say that $\sum \sigma_i = d(\sum \eta_i)$ again almost precisely means $P = \partial B$ where $B$ is an (n+1)-dimensional simplicial complex so that each codim 1 simplex is contained in exactly either 1 or 2 codim 0 simplices
orientations etc etc
So that homology groups are exactly what you get when you look at maps from singular manifolds up to singular bordism
not necessarily. Computability was a huge practical and philosophical thing
A model of what a computer can follow, intricately related to how we do mathematics, in the sense of incompleteness basically being a halting problem#
But for biological computations and spaces in which "language" resides etc, things that gromov seeks to describe in his essays, this approach plainly failed
@TedShifrin Apparently it's true that if $X$ is a complex simply connected Kahler manifold of nonpositive curvature then every complex submanifold $Y \subset X$ deformation retract to a half-dimensional CW complex
Or, more practically, training a neural net, or an algorithm that vectorizes words such that "paris - france + germany = berlin", is way easier than doing the same thing with a logical rule system
I guess the relevant Morse function $f : Y \to \Bbb R$ is $f(y) = d(y, x_0)^2$ where $d$ is the Riemannian distance function, $x_0 \in X$ is a fixed generic point
How much rigor would you assume for someone to show that the klein bottle / the connected sum of two real projective planes is an S^1-bundle over S^1? From a picture, it's pretty clear why this is true, but if anything, I could only imagine smart quotient maps to yield a readable symbolic proof. Some polar coordinates stuff maybe suffices too
If by definition K is RP^2 # RP^2 then by definition this is obtained by pasting two copies of RP^2 \ D to its boundary --- so if you know that's the Mobius band, then you see it's obtained by pasting two mobius bands along their boundary
I am too dumb for this maybe. I want to show if M is a compact Kahler manifold then hol sec. curvature is somewhere positive. I was trying to think about curves in M: let exp : T_p M -> M be the exponential map, and V in T_p M be a complex line, which you can exponentiate everywhere and get a totally geodesic curve in M; this should be some closed, embedded curve right? I feel like it cannot be noncompact.
Yeah, if you think of this inside the sphere all we are considering is top hemisphere vs bottom hemisphere
On the sphere the map I'm discussing is reflection across xy-plane
Now here's the truck
Trick
It follows that kP^n \ D can be represented as the annulus of inradius 1 and outradius 2 where points of outradius 2 are identified iff they are in the same line
It can also be identified as the annulus of inradius 1/2 and outradius 1 where points of radius 1/2 are identified iff they are in the same line
Gluing those together, you get: the annulus of inradius 1/2, outradius 2, where points on the inradius are identified if in same line and points on the outradius are identified if on the same line
There's almost a well-defined projection to the sphere of radius 1, but it doesn't quite work because we identified points on the same line
I have two continuous functions $f\colon \mathbb{R} \to \mathbb{R}$ and $g\colon \mathbb{R} \to \mathbb{R}$ and a dense set $X$ sucht that $f(x)=g(x)$ for all $x\in X$.
I've shown that because $X$ is dense, then for all $z\in\mathbb{R}$ I can find a sequence $(z_n)_n$ in $X$ such that $z_n\to z$.
Now I want to show that $f=g$. Because they are continuous, for every $z\in\mathbb{R}$ we have $\lim{f(z_n)} = f(z)$ for every sequence $z_n\to z$ and the same applies to $g$.
Now how do I show that they are equal? Do I use the $\epsilon-N$ definition? I'll get that
But if we instead project to lines on the sphere of radius 1 you get that there is a fiber bundle kP^n # bar(kP^n) -> kP^{n-1}. (swapped orientation on 2nd) (k, n) = (R, 2) this specifies to what you want
@AttractorNotStrangeAtAll I don't follow what you end with here. If you use the triangle inequality on |f(z) - g(z)| = |f(z) - f(z_n) - (g(z) - g(z_n))| you bound it by eps_1 + eps_2 by your inequalities.
So you know 0 <= |f(z) - g(z)| <= eps for all real numbers eps>0.
Hi! I have a question regarding computing simplicial homology groups. If I have a structure which looks like two faceless triangles connected at a point, with another triangle (that does have a face) composed by drawing an edge between the two next closest, not connected points _ _ e.g. the structure \/_\/, where the middle triangle has a face, how do I find the orientation of the 1-simplicies, and how many distinct 0-simplicies there are? Do we necessarily have to have 3 simplicies which cannot be identified with each other in order to have the 2-simplex (e.g. the face)? And do those fac…
@MikeMiller But why would it be $ 0 \leq \left| f(z) - g(z)\right|$ and not $ 0 < \left| f(z) - g(z)\right|$? Because I need it to be $0$ in order to say that they are indeed equal
Constant sectional curvature is equivalent to saying the curvature tensor $R$ is proportional to $g(KN)g$ where $KN$ is Kulkarni-Nomizu, by just verifying it on $(X, Y, X, Y)$, in which case it's definition; the full tensor is determined by these. I bet holomorphic sectional curvature says but with KM product replaced by a complexified KM product
@AttractorNotStrangeAtAll Yes. You know that $|f(z) - g(z)|$ is either equal to zero or positive. But if it was positive (say, $a$), then taking $\epsilon = a/2$, you would also know that $a < a/2$, AKA, $a < 0$. Contradiction!
Said more clearly: Every positive real number has another positive real number smaller than it. Therefore there is no positive real which is smaller than every other positive real.
Since $|f(z) - g(z)|$ is either zero or positive... but it's also smaller than every positive real number... it must be zero.
Ok this is too hard for me. Next exercise is if $X, Y \subset \Bbb C^N$ are complex submanifolds with $\dim X + \dim Y \geq N$ and $\{(x, y) \in X \times Y : d(x, y) \leq c\}$ is compact for every $c$, then $X$ and $Y$ intersect
"Hi! I have a question regarding computing simplicial homology groups. If I have a structure which looks like two faceless triangles connected at a point, with another triangle (that does have a face) composed by drawing an edge between the two next closest, not connected points _ _ e.g. the structure \/_\/, where the middle triangle has a face, how do I find the orientation of the 1-simplicies, and how many distinct 0-simplicies there are? Do we necessarily have to have 3 simplicies which cannot be identified with each other in order to ha…
That's a simplex, but I still need to know what a simplicial complex is (the input to simplicial homology) and what the simplicial chain complex / boundary operator are defined as. Easiest to tell me the reference.
So a Delta-complex comes equipped with characteristic maps for each simplex, maps from the standard simplex.
That's the standard orientation
Actually let me just do this eplicitly
The point is that when you give a Delta-complex you start by orienting every simplex (not necessarily compatibly)
You have five 0-simplices, five one-simplices, and one 2-simplex
Sorry, OK, let me start from the topo. I got thrown off by your discussion of orientation.
I have to assume your structure looks like a triangle with two "wings". When you say "do those faceless triangles nee to have a 1-simplex..." it's kind of unclear what you mean --- you give me the space
In a Delta-complex you can identify different sides of a triangle to the same side, so those 3 sides do not actually have to be distinct
But each side of the triangle does need a 1-simplex to be glued to (they can just all be the same if you like)
Maybe dist(x, y)^2 : X x Y -> R is a Morse function after perturbing X and Y a little bit in C^N. Suppose $(x_0, y_0)$ is a critical point such that $x_0 \neq y_0$, then intuitively the index is $\dim_{\Bbb C} X$ (there are $\dim_{\Bbb C} X$ direction you can nudge $x_0$ in so that the secants emanating out of those clutter at $y_0$) + $\dim_{\Bbb C} Y$ (same logic but with nudging $y_0$ now, $x_0$ fixed)
So if $\dim_{\Bbb C} X + \dim_{\Bbb C} Y > N$ was a strict inequality you get screwed because dimension of $X \times Y$ has dimension at most $2N$, so half dimension at most $N$
And my guy is Stein so deformation retracts to a half dimensional object, which having a Morse function with more than half dimensional index critipoint will contradict
@MikeMiller
I'm going to sleep now but to elaborate a little more on the index, at a critical pair of points (x0, y0) it should be the number of allowable directions you can move the secant L = x_0y_0 while staying in the secant variety so that you "move towards" L. Fixing y_0, there are (dim_R X)/2 = dim_C X choices: For every complex 1D submanifold of X, distance to x_0 never attain local min or local max by Liouville so out of every 2 dimension only 1 is available to you
Similarly dim_C Y choices the other way around. So at most dim_C X + dim_C Y > N index I think
Also, could anybody inform me how to make a covering projection that doesn't evenly cover the whole space (Topology)? For the figure 8, say, if I have one sheet that maps to the one half, and another sheet that maps to the other, that evenly covers the whole figure 8.
I don't really understand that question. If you're looking for a simple example of a non-trivial covering space I suggest f_n: S^1 -> S^1 with f_n(z) = z^n.
