one way to approach this, perhaps, is that in the usual V-E+F=1 statement, you've got no restrictions beyond it being planar. here you have constraints on the 'external' vertices
Similarly, $\Bbb R$ minus a point is homeomorphic to two copies of $\Bbb R$.
You just need to show that an open ray (such as the interval $(0,\infty)$) is homeomorphic to $\Bbb R$, and the map $f:\Bbb R\mapsto(0,\infty)$ defined by $f(x)=e^x$ provides that homeomorphism.
Is it right that I can cover $\mathbb{RP}^2$ with 2 open sets, each of which sits completely within one of the usual charts of $\mathbb{RP}^2$, ie the set $x_i \neq 0$ for some $i$? And then letting those 2 sets be $U$ and $V$, $U \cap V$ is $ \{ [x_1, x_2, x_3] : x_2, x_3 \neq 0 \}$ (choosing the open sets $x_2 \neq 0$ and $x_3 \neq 0$?
And then, $U \cap V$ is $\mathbb{R}^2$ with one axis removed
Ooooooh wait, there's a mistake there. My 2 open sets don't contain the line $[x_1, 0, 0]$ so I missed something
@BalarkaSen Ok so I have 2 ideas. 1) Take 2 of my chart open sets and union them and the leave the third as it normally is 2) OR make the opens sets all the $[x1, x2, x3]$ with $x1, x2,$ or $x3 < \epsilon$ and then everything else
Do you have any advice about how I should choose between my ideas
But it actually reminds me a lot of doing some calculations in physics
The full calculation is too hard. And there's LOTS of ways of splitting it up into smaller calculations. But an arbitrary way of splitting it won't be very helpful. One has to think about about what a good way is
The kind of decomposition you want your space to have gets quite apparent as you work with singular cohomology. The thing is called a "CW decomposition"
And once you have it, the cohomology calculation becomes a basic busywork. You can program a computer to do it for you
That's why it's a useful tool; it can be computed purely algorithmically
@BalarkaSen This $U$ is real weird. Its an annulus around the equator of a sphere, but antipodal points are identified. So for example rather than rotating by $2 \pi$ to get back where you starter, you can also rotate by $\pi$ and then go down through the equator. Which means if I cut along the equator, and rotate one of the annuli by $\pi$ then glue it back along the cut equator..... that makes it seem like its just $S^1 \times I$?
start with a nematic liquid crystal (order parameter space is a director in RP^2) then find some reason to forbid the director field from pointing too much away from the equator
@BalarkaSen Is it a Mobius Band? Because there you can go all the way round, but you can also go part-way, cross through the equator and be back where you started?
@BalarkaSen So there are no never-zero 2-forms on the Mobius Strip. So that means you need at least 2 2-forms to generate all of them. Because the first will be zero somewhere, so you need a second thats not zero there.
Oh and this is actually an open mobius strip, so its not compact either
Ok we're gonna have to do things recursively. One can split the Mobius strip into another mobius strip and an $S^1 \times I$
Random art ramble: 1. Art is a language of expression, which the audience are free to interpret however they like. It is the only known language without a syntax 2. Mathematics is a form of art 3. Logic is a foundation of mathematics -------------------------------------------- Yet no syntaxless logic exists
@BalarkaSen Is it right that since $H^1_{DR}(\mathbb{R}^2) \cong 0$ and $H^1_{DR}(Mobius) \cong \mathbb{R}$, this map $i^* - j^*$ seems like it has rank 0 since a 1-form on the Mobius loop has to have whatever covector you put at $(0, x)$ the same as what you put at $(2 \pi, -x)$ but $S^1 \times I$ has no such restriction?
I remember that I saw on some users profile a link to his own text about real analysis. But I cannot find it now. (I should have saved it back than.) Does anybody know who this might have been?
Im not sure if this is right but the 1-form that integrates to $2 \pi$ on $S^1$, this guy could be the pullback of a 1-form on the Mobius strip by the inclusion map. But then we can't homotope that $S^1$ to the an $S^1$ in $U \cap V$ because it would have to cross the equator somewhere it seems
Oh, better idea was to look at my browser history and look there for pages containing "math.stackexchange.com/users". After a lot of scrolling I remembered which name rings the bell - it was Aloizio Macedo.
Sorry for disturbing, I hsould have figured this one myself.
@BalarkaSen Oh okay actually no, you can push the right half of the first go-round up, then connect it to the beginning of the 2nd go-round on the bottom, then keep the 2nd go-round on the bottom and push the beginning of the first up of the left to complete the loop
Ok, so the general idea is we want to start with closed forms that integrate to non-zero values on $U$ and $V$ because these guys form a basis for the de Rham cohomology there. And then we want to try to homotope the paths you integrate those guys over to paths that have closed forms that integrate to non-zero values on $U \cap V$.
to every homology class (in this case homotopy classes of loops) there is a dual cohomology class (form (upto an exact form) which integrates to 2pi over a certain loop)
and if the image of the first map is 1-dim, then then kernel of the 2nd is 1-dim, but the 2nd map is also onto, so the $H^2_{DR}(\mathbb{RP}^2) \cong 0$ but its compact so it should be $\mathbb{R}$
@BalarkaSen Youre in luck, Poincare duality was the last thing we covered. Which is why right after I wrote what I wrote i was like "o wait thats wrong"
I was reading some lecture notes earlier and I mustve just misread something
Mayer-Vietoris is the alst thing thats actually on the final. And we didnt have a homework assignment on it, which is why I spent todays tiem doing this stuff
Can't you project on every closed convex set $A$ (linear subspaces are convex) by sending $x$ to $w\in A$ such that $||x-w||=\inf\{||x-v||:v\in A\}$? @brot
@Alessandro I have that theorem for closed subspaces of Hilbert spaces. Not for Banach spaces though. Looking at the proof, it uses the parallelogram identity.
I was the guy who knew no complex analysis a few days ago :P. But yes, I would like to see this sheaf theoretically very soon. Mainly I just want to consider sheaves of smooth, holomorphic and meromorphic functions, and relevant duals hopefully.
I've just started Forster Riemann surfaces today too.
Well that's also another thing I wanted to understand better. Basically if $X$ is a Riemann surface, you can realize it as a branched cover $p : X \to \Bbb{CP}^1$. Now, that gives an inclusion of the field of meromorphic functions $\mathscr{M}(\Bbb{CP}^1) \hookrightarrow \mathscr{M}(X)$
But $\mathscr{\Bbb{CP}^1}$ is really just $\Bbb C(z)$. So that's basically an extension of $\Bbb C(z)$
And if you think about the "holomorphic deck transformations" of this branched cover, that gives a topological way to interpret $\text{Gal}(\mathscr{M}(X)/\Bbb C(z))$
1. There is a maximum, defined to be the unique number that has no successor, and a minimum, defined to be the unique number that has no predecessor. 2. Every number is a successor of some number, except the minimum 3. Every number is either a successor or the minimum, and between any two successor there are no numbers 4. No number can have a sequence of predecessor (if any) that continues indefinitely
This uniquely defines the finite natural numbers without nonstandard elements
But I'm not ready quite yet in many areas. I thought I'd attack my ultra weak point of complex analysis first, and Riemann surfaces seemed a nice excuse to do this.
I'm still dealing with this exercise: Let $\mathcal{F}$ be a sheaf on $X$ and define $\mathcal{F}^w(U):=\prod_{x\in U}\mathcal{F}_x$. Show that this defines a flasque sheaf $\mathcal{F}^w$ on $X$.
It is clear to me that $\mathcal{F}^w$ defines a sheaf on $X$. I am struggling with the flasqueness of it, i.e. to show that the restriction maps $\mathcal{F}^w(X)\to\mathcal{F}^w(U)$ are all surjective.
Given some $s_U=(s_x)_{x\in U}\in\mathcal{F}^w(U)$, I would need to "extend" $s_U$ to some $s_X\in\mathcal{F}^w(X)$ simply by choosing ANY $s_x\in\mathcal{F}_x$ for all $x\in X\setminus U$.
This would be sufficient, because then $s_X$ clearly maps to $s_U$.
However, I do not see why I can always choose such $s_x\in\mathcal{F}_x$ (for all $x\in X\setminus U$). Why can't $\mathcal{F}_x$ be empty in some case?
Or wait... maybe it doesn't matter, cause $\mathcal{F}^w(X) = \prod_{x\in X,\mathcal{F}_x\neq\emptyset}\mathcal{F}_x$.
I see Zariski topology and schemes being useful, but if you're seriously using varieties in proofs, then for most of the results you're just proving special cases
and making commutative algebra useless for number theory
I'm fine with looking at what results imply for varieties
Trolling aside, $\Bbb R$ and $\Bbb C$ is much much more important in "real world" than "geometry over $\Bbb F_p$" :P Literally all of analysis is built over the former two fields
Hence, all of physics is built over them. A wavefunction doesn't take values in a finite field.
It is true there are applications of the finite field theory in a lot of computer scientific worlds, however, true. I don't know much about coding theory but I know encryption
Stuff like ECM/ECPP uses elliptic curves over finite fields
