@MikeMiller So in this nice case, I have $0\to \Bbb Z^d\to \Bbb Z^d\to 0$ and underneath that, another complex $0\to \Bbb Z\to \Bbb Z\to 0$, and the map from the left $\Bbb Z^d\to \Bbb Z$ just send every $1$-cell to the same $1$-cell. The generator of $H_2(\Delta_d)$ on the top can be taken as $e_1+e_2+\dots+e_n$, which is sent to $de_1$ on the bottom, i.e. $H_2(S^1)\to H^2(S^1)$ is the degree $d$ map
I do the dimension count but forget it every time. If $E/M$ is rank $n$, and you take $r <= n$ sections, that gives a section of $E^{\otimes r}$ whose fibers are $M_{r \times n}(\Bbb R)$. Make this section transverse to the substratified space of matrices of rank less than $r$, which is of dimension $(r - 1)(r + n - r + 1) = (r - 1)(n + 1) = rn + r - n - 1$, so codimension $n - r + 1$.
My retardation knows no bounds, can't even get a sign right
So for $T^2\to T^2$, I can take $S^1\times S^1$ to have the cell structure given by $\Delta_{d_1}\times\Delta_{d_2}$ in Mikes notation. This has $d_1d_2$ $0$-cells, $1$-cells and $2$-cells. The fundamental class in $H_2(T^2)$ is given by $\sum_{i=1,j=1}^{i=d_1,j=d_2}e_i^1\times e_j^1$, where every one of these terms is taken by the cellular map to $e\times e$ (the single $2$-cell in $T^2$ with the other model, i.e. $\Delta_1\times\Delta_1$), so that the map is degree $d_1d_2$
@BalarkaSen It's a bit late for me to start learning geometry and algebra :( some advice I got was to not start over, but rather to build on what I've done. And the way I am thinking now is to add rigorous Analysis to my work, at the PhD level. Stuff like functional analysis, harmonic analysis, and stick with mathematical modeling.
Oh, I misread "I don't do algebra" as "I do algebra", that's what confused me
Probably because I don't read such a horrible sentence often
Next I want to find for $(z_1,\dots,z_n)\mapsto (z_1^{d_1},\dots,z_n^{d_n})$ the induced map on the second homology. Probably I can keep using this model, but now the description of the generators of $H^2(T^2)$ is hard, since the dimension is potentially big, i.e. $n\choose 2$ I think
@TedShifrin the remann theorem on removable singularity says if $f$ is bounded on $\Omega-\{z_0\}$, then $z_0$ is a removable singularity. I don't see what I need to show
(Small comment: You wrote the Laurent series with finitely many negative term. It might as well have infinitely many, because a-priori the singularity at $z_0$ could be essential instead of being a pole - you have to prove none of this happens of course)
I'm having trouble working out the induced map $H_2(T^n)\to H_2(T^n)$ still, coming from $(z_1,\dots,z_n)\mapsto (z_1^{d_1},\dots,z_n^{d_n})$. I have worked out (in theory) that there are $d_1\dots d_n{n\choose 2}$ $2$-cells in $\Delta_{d_1}\times\dots\times \Delta_{d_n}$ and ${n\choose 2}$ in $\Delta_1\times\dots\times \Delta_1$.
So in the even case this is like $T^2\times\dots\times T^2\to T^2\times\dots\times T^2$, and in hte odd case I just get an extra factor of $S^1\to S^1$, right
This is easiest to argue with cohomology and cup product but you can think dually... understanding a basis of $H_2(T^n)$ is the first step, as Ted said.
Give me an explicit formula for the cellular basis of $H_k(T^n)$
Consider given integers $A,B$ such that $AB \neq 0$.
Consider a given polynomial $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ of degree $n > 1$ with rational coefficients $a_i$.
Now I wonder about solving the diophantine equations of type :
$$ A(x+y_1)(x+y_2)...(x+y_n) + B(x+z_1)(x+z_2)...(x+z_k) = f...
The $2$-cells in the base for say $n=3$ are of the form $e^1\times e^1\times e^0, e^1\times e^0\times e^1,e^0\times e^1\times e^1$. All are mapped to zero by the chain map
i read about multivariable calculus but i still feel confused and ignorant about math with many variables. Noting specific, just a feeling of " missing something ". maybe some advice , or is my question too broad and vague ??
I mean, if we a priori know the homology is free of rank ${n\choose 2}$ and I count my $i$-cells in this CW structure to have ${n\choose 2}$ (which is cheating) but still
ok
But describing the $2$-cells upstairs seems horrible
Let's write this a bit precisely, because what you say is correct but messy to the eyes. $T^n = \prod_{\alpha = 1}^n S^1_\alpha$. Then the cellular chain complex is $C_\bullet(T^n)$, where $C_k(T^n)$ is the free abelian group generated by $k$-cells of the form $\prod_{\alpha \in I} e^1_\alpha \times \prod_{\alpha \notin I} e^0_\alpha$ where $I \subset \{1, \cdots, n\}$, $|I| = k$.
I have no idea what a generator in the upstairs copy of $\Bbb Z^{n\choose 2}$ looks like, that is, it's some horrible sum of things of the form downstairs?
I'm confused because if I pick a representative in $C_2(T^n_{upstairs})$ for a class in $H^2(T^n)$, then it has some horrible form, not like the form of a downstairs representative
Like for even the circle you get $\sum_{i=1}^d e_i^1\in H^1(\Delta_d)$ for $\Delta_d\to \Delta_1$
after reading this post, math.stackexchange.com/questions/1284316/…, I think I can do $|(z-z_0)f|\leq A|z-z_0|^{\epsilon}$. $(z-z_0)f$ is bounded in a open set center at z_0 and holomorphic. the function $(z-z_0)f$ has a removable singularity, z_0.
What I wanted you to say was $f$ isn't a cellular map with the same cell structure on $T^n$ in both the domain and target copy. It does not induce a map on the cellular chain complexes.
I don't think you get it from that, @Simple. I'm astonished at how stubborn you are about trying to change your approach. You really need to work on that.
This does seem to agree with what you said @BalarkaSen, since I have additivity, so I'll just get $d_ad_b$ copies of $e^1_{a}\times e^1_b\times \prod e^0$ downstairs
IMO you should not argue by cells at all. Let $e^1_\alpha, e^1_\beta$ be two $1$-cells in $T^n$. By (diagram chase) the homology class in $H_2(T^n)$ correspond to the cell $e^1_\alpha \times e^1_\beta$ is given by image of $1 \in \Bbb Z = H_2(T^2)$ by the inclusion $T^2 \to T^n$ as the $\alpha\beta$-copy.
The image also lands inside this embedding of $H_2(T^2)$, and the map $H_2(T^2) \to H_2(T^2)$ is multiplying by $d_\alpha d_\beta$ (this you can already prove)
This describes the action of $f_*$ on each generator of $H_2(T^n)$ completely.
Top cell gives $1 \in \Bbb Z = H_2(T^2)$, which lands into the homology class corresponding to the cell $e^1_\alpha \times e^1_\beta$ in $H_2(T^n)$, as promised.
If you have a map $M\to N$, such that it induces $H_*(M)\to H_*(N)$, then if you can find maps $L\to M\to N$ such that you can map to any generator of $H_*(N)$ from a map $H_*(L)\to H_*(N)$ factoring through $H_*(M)$, then you can completely describe the map
Well, I don't know about category theory but I was trying to point at the fact that we're secretly saying there's some algebra structure on the total homology of $T^n$ which is natural under maps, and thus describing it on degree 1 describes it everywhere
Because higher degree generators are product of degree 1 guys
Yeah this isn't usually visible on homology. The point is homology and cohomology of torus are the same, because they are free (dual of fg free is the same guy)
Consider given integers $A,B$ such that $AB \neq 0$.
Consider a given polynomial $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ of degree $n > 1$ with rational coefficients $a_i$.
Now I wonder about solving the diophantine equations of type :
$$ A(x+y_1)(x+y_2)...(x+y_n) + B(x+z_1)(x+z_2)...(x+z_k) = f...
