can you calculate $H_n(P^n,\mathbb{Z}_2)$ given that you know that $H_i(S^n,\mathbb{Z}_2) = \mathbb{Z}_2$ for i = n and i = 0 and other wise 0. And that you have the following long exact sequence
I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)$ and $(c)$, which I'll share below; however, I was not sure what to do about the $0<...
@MikeMiller @TedShifrin So, it's not really clear to me how to define the $d : \Omega^0 \to \Omega^1$ for a smooth vector bundle $E/M$. For the cotangent bundle one takes a smooth function $f$ on $M$ and gets the 1-form $x \mapsto df_x$. But how do I define the differential of a function for a general vector bundle?
@BalarkaSen No, I'm asking you, before you say you don't know how to define $d$, what $d$ should map between. What you just told me is not consistent with how $d$ works on 0-forms.
$d$ does not take a function and spit out a thing that (eats functions and spits out functions).
Oh. So maybe I misunderstood something there. Is it right that $d : \Omega^0 \to \Omega^1$ eats a smooth function and spits out the 1-form $M \to T^*M$ given by $p \mapsto d_pf$? ($d_pf$ is a linear functional on $T_pM$).
Ok, so something I am not clear on. Is this $d$ you are talking about a map $\Omega^1 \to \Omega^2$ instead of $\Omega^0 \to \Omega^1$? Because we seem to be starting with smooth sections of $E$ (which are analogous to 1-forms rather than 0-forms).
No, that's not right.
Sections of $E^*$ are 1-forms, not $E$.
Oh, or have we just replaced $T^*M$ itself by $E$? Then I retract my previous "not right" comment.
@MikeMiller So, if I start with a smooth section of $E$, $d$ should take that to a smooth section of $E \wedge E$.
Hmm, so maybe I didn't understand this buisness. Let me ask for a clarification: we replaced the cotangent bundle $T^*M/M$ by an arbitrary bundle $E/M$. $d$ is defined as a map $\Omega^0 \to \Omega^1$ in the cotangent space world. Now you're asking me to look at $d$ for $E/M$, and asking to start with a section of $E$. But in the particular case $E = T^*M$, section of $E$ is a $1$-form, which lives in $\Omega^1$, not $\Omega^0$, which is the domain of $d$. This is where I am confused.
Summary: Why should $d$ eat a section of $E$? In the particular case of $E = T^*M$, a section of $E$ is a $1$-form, whereas $d$ eats a $0$-form.
@BalarkaSen You're getting lost in notation. In the case you're thinking about already, you have sections of the trivial bundle $\Bbb R$ over $M$, and you're taking their derivatives locally to get a 1-form.
Now "taking their derivatives locally" seems like something you could do in any setting. Take a vector bundle $E \to M$. You can define $\Gamma(E) = \Omega^0(E)$ (again - this is just notation).
The "exterior derivative" should take a section of $E$ and send it to... who knows. That's your job to figure out where it would go.
@MikeMiller I see. So the situation is this: sections $M \to M \times \Bbb R$ give smooth functions on $M$ after composing with the second projection $M \times \Bbb R \to \Bbb R$, so these are the same.
The most obvious interpretation of "taking derivative locally" I see is by taking a chart $U$ of $M$, looking at the image of the section sitting in $U \times \Bbb R$ (which sits above $U$) and do best linear approximation. Literally what is happening is that the section in $U \times \Bbb R$ is graph of the smooth function.
This can clearly be done for arbitrary smooth vector bundles, by taking a chart on $M$, shrinking it further to $U$ so that it becomes a trivialization for $E \to M$, then looking at the graph inside $U \times \Bbb R^k$ again. That's what we are doing, right?
I dunno what you're doing, it's your job to figure that out.
It's Ted's question anyway. I was just trying to get you to start with step zero: if the question is "Why is the cotangent bundle/smooth functions special", so you start asking if there's a $d: \Gamma(E) \to ?$, your first step is to solve for ?
Yeah, I understand your question now. It's not clear what kind of beast I get if I apply my "take derivative locally" to some section of $E$. But I should certainly get something.
I need to figure out what it is.
@PVAL I pinged you a proposed counterexample to the question you asked yesterday. Have you seen it/seen the ping?
Why do proofs of things in set theory, for example De Morgan's Laws, assume $x \in A$? What if $A$ is empty? Is it because De Morgan's Laws, for example, hold trivially if any one of the sets is empty?
In the moduli space $\Bbb P^2 \times \cdots \times \Bbb P^2$ ($n$ times) of configuration of $n \geq 3$ points in $\Bbb P^2$, there is an open subset such that picking any configuration from it I get noncollinear points. That's what I broke up there.
A ok..I want to find a code with at most 27 elements and I found one with 9 elements and an other on with 3... But I don't want the equality to hold... @DanielFischer
@PVAL OK, maybe you're right. It's a generic counterexample, I should say. There is a closed set in the moduli space of all configuration of lines and points in $\Bbb P^2$ (whatever that means) where it can be realized.
I need to expand the function $f(z)=\frac{1}{(z-a)^{k}}$ where $a \in \mathbb{C}$, $a \neq 0$, $k \in \mathbb{Z}$, $k>0$ in a Laurent series in the annuli
(a) $0< |z|<|a|$
(b) $|a|<|z|$
Note that earlier, I found the Laurent series for $\displaystyle g(z) = \frac{1}{(z-a)}$ to be $\displaysty...
all I need is uniform convergence of the sum and each term to be analytic on the domain. The analyticity I have. Just not sure if the uniform convergence is something I have to explicitly show, or I can use the fact that the series we're dealing with is geometric and "it is known" that geometric series are uniformly convergent (if indeed that's true).
@robjohn Hey. I know you're pretty busy all the time. In the near future I'd like to have my book reviewed (the stuff is pretty hard to follow), and I have some people in mind, but I hope not all will be that busy. Also a partial review might be precious.
Here $\ell_1, \cdots, \ell_6$ are cyclically placed so that they form a hexagon. Vertices of the hexagons are $p_1, \cdots, p_6$. Vertices of the stars are $q_1, \cdots, q_6$. Intersection of the parallels at infinity are $r_1, r_2, r_3$ respectively, @PVAL. Then add a line $\ell_7$ which intersects $q_1, q_2$ and $p_5$.
That, I believe, is a fine incidence data which can not be realized as a configuration in $\Bbb{CP}^2$.
If $q_1, q_2, p_5$ were ever colinear, then the whole construction would be degenerate.
The deal with the previous construction (where $\ell_7$ passed through $r_1, r_2, r_3$ instead) was that it was sometimes realizable, like if the hexagon was regular in which case they lie on the line at infinite, but not generically so in the sense that such configurations would only form a closed set in the moduli space.
"I do not want any high-level theorems like Mayer-Vietoris or Seifert-Van Kampen or deformation retracts or \pi_1(S1) or removing points or using covers or other tricks people try to impress me with." Person's a bit narcissistic - what they don't understand is that these aren't "tricks to impress them" but genuine ways one can and do prove things.
@Pallas I don't understand the N column in that graphic, the directions seem to imply the normal vector is (0,0,1) identically. In which case the integral is (4+7-3+8)/4=4.
In Greene's theorem, puu.sh/opQbh/45c5e3a5ee.png this problem has two curves C1 and C2, and i know $\partial D = C_1- C_2$, but how would i actually apply Greene's theorem here?
I just found the Laurent series for $z^{2}e^{1/z}$ for $z = 0$, and now I need to find it at $z = \infty$. (for $z=0$, it was $\displaystyle \sum_{n=0}^{\infty}\frac{z^{2-n}}{n!}$, by the way).
I'm not sure how to approach this problem for $z = \infty$, however...
I came across one method invo...
@arctictern i think it's not an automorphism. Because if we take the first case assuming that $x \in H$ then that would mean that $\phi(x)=\psi(x)$ and since $\psi(x)$ is in $H$ (given that it's an automorphism of $H$) that would imply that the co-domain of $\phi(x)$ is $H$ and hence it's not an automorphism on $G$. Is this reasoning correct or completely wrong?
@Paradox101 Consider the function f(x)=x for real numbers. When x is positive, f(x) is positive, therefore I conclude the codomain of f(x) is positive numbers, hence f(x) is not a bijection. Is this argument valid?
@Pallas I'm guessing the vortex F is conservative, so its curl is zero, so the double integral in Green's theorem is zero, so the two boundary integrals are equal, right?
@Paradox101 so just because phi() sends elements of H to elements of H, doesn't mean phi() doesn't more generally send elements of G to elements of G. In fact phi() is a bijection. It permutes elements of H amongst themselves, and elements of G-H amongst themselves.
@Pallas what?
write down Green's theorem for that domain. you get the difference of two boundary integrals on one side, and zero on the other.
@Pallas you argue the double integral on the other side of Green's theorem is zero, because the integrand is curl(F) dot normal, and curl(F) is zero since F is conservative.
@arctictern if I take the first case where i let $a,b \in H$, then $\phi(a)=\psi(a)$ and $\phi(b)=\psi(b)$, then $\phi(ab)=\psi(ab)=\psi(a) \psi(b)$ (as $\psi$ is an automorphism), similarly the second case also shows $\phi$ to be a homomorphism. What am I doing wrong?
(BTW your there are cases you aren't considering, like when c is in H but d isn't.)
@Paradox101 Consider any G and proper H, let c be any element not in H and consider d=c^-1. The obviously cd is in H. More generally, if c is any element not in H, and h is in H, then d=(c^-1)h is not in H, and cd=h is in H.
you're not using Green's theorem in part (a), you're doing that in part (b) because that's when you're actually enclosing a region with a boundary. the curve in part (a) doesn't enclose anything, it's a line segment.
@arctictern yes you're right i didn't think of considering those. So if i want to disprove homomorphism then for the second case if i define the product of $cd=h$ and state that if $\phi(cd)=cd=h \in H$ which is not possible is ths enough to prove that it's not a homomorphism?
if i want to disprove homomorphism then for the second case if i define the product of $cd=h$ and state that if $\phi(cd)=cd=h \in H$ which is not possible is ths enough to prove that it's not a homomorphism? @arctictern
I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)$ and $(c)$, which I'll share below; however, I was not sure what to do about the $0<...
