So one of the few ways I know how to recognize a complex manifold is that intersection numbers (of distinct submanifolds) must be nonnegative, @MikeM. Can that work?
@Karim I don't know much functional analysis (actually none except basic ideas about measures and few more stuff which is not the real functional analysis stuff). Though I like the idea of functional analysis
A manifold M admits an almost complex structure iff the principal GL(n,\Bbb R) bundle associated to the frame bundle of TM admits a reduction to a GL(n,\Bbb C). All the classical homological and homotopical obstructions come from this. IIRC, for any dimension $c_1=w_1$ mod 2 and $c_1$ must be a class which satisfies an integral "Wu formula". In \Bbb C-dim 2 that is a complete obstruction,but I do not know what the relations are in higher dimensions or what other obstructions exist.
@MikeMiller No, I imagine they got quite complicated in very high real dimensions (I don't think I've seen the real version ($\Bbb Z/2$) written out for real dimension higher than 4). You assume there exist some Wu-classes which satisfy something to do with squaring the sw-classes and see what that implies.
@PVAL: For the case we care about $X = \Bbb{CP}^4 \# \Bbb{CP}^4$. So all cohomology is in even degrees. So we only care about the odd homotopy groups. One can show by hand (+ bott periodicity) that $\pi_1, \pi_3, \pi_5$ are zero.
So the obstruction must live in $H^8(X;\pi_7(SO(8)/U(4))$.
I have absolutely no idea how I would want to calculate the obstruction.
I assure you I don't want to start by building an AC structure on the 7-skeleton...
The variety is not an open subset of affine space.
As Milne does the way you should model an algebraic variety in full generality is as something with charts that are isomorphic to algebraic subsets, and transition maps that are regular. The "open subset of affine space" thing is probably unnatural.
Not sure if you can actually do it. You can for the one case I checked. This is not particularly good enough.
@MikeMiller $w_2(CP^4)=1$ and $w_6(CP^4)=0$ So by naturality $w_2(P^4#P^4)=(1,1)$ and $w_6(P^4#P^4)=(0,0)$, so $u=(x,y)$ both odd $v=(z,w)$ both even. The Euler characteristic is 18=2 mod 4, and clearly $u.v=0 $ mod 4. Which contradicts part c of the paper I linked.
Assuming I did not make a mistake in computing the binomial coeff or $\Chi$/
@MikeMiller but that's bad news. I wanted to cover $V$ by open sets isomorphic to $\Bbb A^n$, so that you choose a finite subclass $\{U_i\}$ by compactness, and you build the map $V \to \Bbb A^n$ by gluing all the isomorphisms $U_i \to \Bbb A^n$. This map is clearly surjective, and has finite fibers, as the class is finite.
That's how I wanted to realize $V$ as a branched cover of $\Bbb A^n$
@MikeMiller Do the cohomology isomorphisms for the connect sum coming from Mayer-Vietoris respect the direct sum structure? I probably used that while computing (read guess) the sw-classes.
@PVAL: Yah. The cohomology ring of $M \# N$ is the direct sum of the cohomology rings of $M$ and $N$, except that you identify the units and the volume forms
@PVAL: Can we think about this in terms of maps into BSO(8)? We know what the map is when restricted to each half. By cellular approximation I think that tells us that p_1 splits as (5,5).
But p_2 isn't going to be as easy probably.
What were the last obstructions again @PVAL? I don't have the paper up
@PVAL: If we can calculate $p_1$ correctly the Hirzebruch signature formula gives us $p_2$. But I think it's not $(5,5)$ because then the Hirzebruch formula would make $p_2$ not an integer.
@PVAL: Mevermind. It all works fine. The formula gives $p_2=20$.
@PVAL: OK putting it all together and taking the last equation mod 2 gives a contradiction. I will write this all up later.
? The isomorphisms $U_i \to \Bbb A^n$ can have wildly different images for each $i$. How can you guarantee it to be continuous by making $U_i \to \Bbb A^n$ and $U_j \to \Bbb A^n$ to agree on the intersection $U_i \cap U_j$?
ah. i don't know what you mean by standardness in that context. :P
@Rememberme That's right. Btw, you should right vectors as $(a_1, \cdots, a_n)$ instead of as $\{\cdots\}$. Vectors are ordered, while sets are unordered.
@Rememberme Right, so you want informations about extension of $B$ by $A$. Not a lot can be said, we have discussed this before. That is, it's an open problem to classify all groups that sits in '?', where $A \to ? \to B$ is exact. Maybe you forgot or didn't pay attention when I told you this.
Btw, you should not think about the set of all extensions of $B$ by $A$. Rather, you should think about the equivalence class of all extensions of $B$ by $A$, where the equivalence between two short exact sequences $N \to G \to H$ and $N \to G' \to H$ is defined to be a large commutative diagram with rows being the two sequences and with maps in the middle are identity for $N$ and $G$ and an isomorphism for the middle groups.
(I am not being very explicit about the equivalence relation here, mostly because I am being lazy about drawing a commutative diagram, ask me if you are curious about it)
Anyway, what I mean is roughly that you want to think about extensions of $B$ by $A$ modulo isomorphism of the middle group (isomorphism in a way that is coherent with respect to the arrows in the short exact sequence). That said, we know the cardinality of this group.
How would we prove this result by real methods ?
$$\int_0^{\infty } \frac{\sin \left(\pi x^2\right)}{x+2} \, dx=\frac{1}{4} \left(\pi-2 \pi C\left(2 \sqrt{2}\right)-2 \pi S\left(2 \sqrt{2}\right)+2 \text{Si}(4 \pi ) \right)$$
As you can easily see, Fresnel integrals are involved. What are yo...
I have other things to do before I have time to write something else.
If you want to understand h-cobordism, first understand handlebody diagrams and handle moves. Gompf-Stipsicz, Kirby Calculus and 4-manifolds, is a good place to learn this. Then read the proof on wikipedia. In precisely one step you'll need to know a certain version of the Whitney embedding theorem, which is where the dimensional constraint appears.
@Rigor how might we explain the fact that on a site with a lot of mathematicians I don't receive an answer for this question math.stackexchange.com/questions/1412761/…? I've been looking forward to it for 4 hours.
Maybe teacher @anon can outline a proof, the brilliant idea that produces a solution.
@Rigor my guess is that there should be a very fast way and simple at the same time.
seeing a sine with a rational function, especially with pi*x^2 inside the sine, my first instinct would of course be to use complex methods. I'm not experienced with using real methods on many nontextbook integrals.
not sure why you're pinging me anyway, as I haven't exactly been known for that stuff for a few years
@anon It seemed to me that a while ago, not too far, you had some opinions on a book about integrals by Paul Nahin, and then I realized that you might be very good in this area.
@PVAL: Posted an answer persuant to our previous conversation. This is the wrong approach if one wanted to prove it in general. Of course I don't know what the right approach would be. I am satisfied with this case and won't work on this anymore.
@Khallil Norm is an abstract way to define "lengths "of vectors in vector spaces. A metric is an abstract way to define "distance" between two points in an arbitrary set.
@MikeMiller I may ask Dan Freed if he knows the answer (I think Ted and Georges are probably the only two people who might know who post here regularly.).
@MGA Not quite. I would start by narrowing down the dimension of the vector space. (Technically not required, but I think it avoids some complications.) And then use Mike's hint on swapping in scalar multiples into the basis.
@MikeMiller I've thought about it a little, and I think if one chases through the usual computation of the (co)homology of $M#N$ and uses the naturality of characteristic classes from the natural maps $M#N \to M$ and $N$, it should be pretty easy to see that the sw-classes and the pontryagin classes factor as $(w_i(M),w_i(N))$ for all but the top dimension and then as $(w_i(M)[M]+w_i(N)[N])[M#N]$, though I haven't written out the computation carefully.
@MikeMiller How do you know that the classifying map is homotopic to $f \vee f$?
The classifying map for a direct sum should nicely depend on the classifying maps for the summands and I think that comes from comparing the pullbacks of the tangent bundles from the maps $M#N \to M$ and $N$ and the tangent bundle of $M#N$.
@PVAL: You an construct the connected sum so that the classifying map is literally the connected sum of the classifying maps, in particular so that it restricts to $f \vee f$.
But I think this pullback thing you're suggesting proves it perfectly well too.
How would we prove this result by real methods ?
$$\int_0^{\infty } \frac{\sin \left(\pi x^2\right)}{x+2} \, dx=\frac{1}{4} \left(\pi-2 \pi C\left(2 \sqrt{2}\right)-2 \pi S\left(2 \sqrt{2}\right)+2 \text{Si}(4 \pi ) \right)$$
As you can easily see, Fresnel integrals are involved. What are yo...
