I don't think we're changing the $J$ for $\overline{E}$. We're using the isomorphism of complex bundles $\Psi : \overline{E} \to E^*$ then using this identification to put a holomorphic structure on $\overline{E}$ because we have one on $E^*$.
... so yeah I thought I could get around the degree computation in the MV proof. So the hard part is computing the map $H_{n-1}(S^{n-1}) \overset{(-i)}{\rightarrow} H_{n-1}(V) \overset{h}{\rightarrow} H_{n-1}(\Bbb RP^{n-1})$, where $h$ is a deformation retract. For odd $n$ this map is 0, and for even $n$ we compute the degree
I don't think so. The fact that $\overline{E}$ has the opposite complex structure is what makes $\Psi$ linear. If $e \in \overline{E}$ then $i\cdot e = -i\ast e$ where $\cdot$ is multiplication for $\overline{E}$ and $\ast$ is multiplication for $E$.
@Astyx OK, try the following. (I : a), the denominator ideal (all things in A, when multiplied by a, goes in I) is also bigger than I, so is principal. I claim if for all a, I + (a) and (I : a) are principal so is I.
Proof: If I + (a) = (b) then for every i in I, i in I + (a) = (b) as well, so i = br. But so r is in (I : b) = (c), so r = ck. So i = bck. I is contained in (bc).
@MikeMiller okay, it really is. Also boils down to the degree computation. I assume you meant the LES $$...\rightarrow \tilde{H}_k(\Bbb R P^{n-1}) \rightarrow \tilde{H}_k(\Bbb R P^n) \rightarrow H_k(S^n) \rightarrow ...$$
@Thorgott nah I want to avoid that but its impossible. I get the local calculation intuitively since it's one identity and one reflection, but the actual proof was brushed away as tedious calculation in local coordinates (in our lecture)
@Astyx Ok, so I want to prove that if $A$ is a Dedekind domain, for any nonzero ideal $I$, $A/I$ is a principal ideal ring. I can boil it down to $I$ a prime power $I = \mathfrak{p}^n$. Then by Balarkayama, we further boil it down to proving $\mathfrak{p}/\mathfrak{p}^n$ is principal.
But dimension of $(\mathfrak{p}/\mathfrak{p}^n)/(\mathfrak{p}/\mathfrak{p}^n)^2$ as a $(A/\mathfrak{p}^n)/(\mathfrak{p}/\mathfrak{p}^n)$-module is the same as $\mathfrak{p}/\mathfrak{p}^2$ as a $A/\mathfrak{p}$-module is the same as dimension of tangent space of $A$ at $\mathfrak{p}$ because $A$ is regular (1dim+integrally closed) is the same as dimension of $A$ is $1$.
@user2103480 Yeah I meant that, though I usually think of it in terms of open sets: $U = \Bbb{RP}^n \setminus \{[0 : \cdots : 0 : 1]\}$ and $V = \Bbb{RP}^n \setminus \Bbb{RP}^{n-1}$. The first projects to $\Bbb{RP}^{n-1}$ in such a way that it gets the structure of a line bundle isomorphic to the tautological line bundle.
Say I have a compact connected smooth manifold. It is known that there is a CW structure on it with a single cell in top dimension. By cellular homology, I then obtain that it is orientable iff this top cell attaches with degree 0 to every cell one dimension lower. Is there some direct geometric interpretation of this?
It means the codimension 1 skeleton is 2-sided, which you should expect in orientable manifolds. Compare with $\Bbb{RP}^2$, where the 1-skeleton $\Bbb{RP}^1$ is 1-sided. Sure enough, the disk attaches with degree 2.
@BalarkaSen Ok, codimension 1 skeleton being 2-sided I buy. So this translates to the attaching by saying that for every time you attach along one side of a cell in the codimension 1 skeleton, you also have to attach along the other side so that these cancel out in degree or something like that?
@Thorgott Something like this. You can probably investigate what it is explicitly; cellular boundary of the top cell gives you some cycle of codimension $1$, namely just pushforward the fundamental class of the boundary sphere $S^{n-1} = \partial D^n$ by the attaching map of the top cell. This is some class in $H_{n-1}(X)$ where $X$ is the codimension $1$ skeleton. What is it? I don't know.
The 2-sided thing you should also be careful to trust me with. $\Bbb{RP}^2$ in $\Bbb{RP}^3$ is 1-sided, but $\Bbb{RP}^3$ is orientable.
This is a famously confusing fact, the covering map $S^2 \to \Bbb{RP}^2$ is $2$-sheeted but if you compose with the collapse $\Bbb{RP}^2 \to \Bbb{RP}^2/\Bbb{RP}^1 = S^2$ then the map $S^2 \to S^2$ is degree $0$ in reality.
for surfaces, the picture works and the condition corresponds to the characterization of having no embedded Möbius strips. what's the right higher-dimensional analogue for this latter characterization?
There is a simple one but this is not relevant to what you want. An $n$-manifold is orientable iff it does not contain embedded $S^1 \times_{\Bbb Z_2} D^{n-1}$'s.
The twisted product is mapping torus of a reflection $D^{n-1} \to D^{n-1}$.
What you describe is something dual, which is also possible to state: A codimension 1 oriented submanifold of an oriented manifold is 2-sided
In my opinion you should understand, for a CW-decomposed $n$-manifold $M$, the space $M/M^{(n-2)}$, and neighborhoods of $M^{(n-1)}$ thereof. $\Bbb{RP}^3/\Bbb{RP}^1$ is already a hard space.
My picture goes something like this. *Locally* the codimension 1 skeleton is two-sided, so while you're attaching, you have to traverse it twice to "attach both sides". If you always traverse in opposite directions, you get the 0 map in homology and have orientability. In the other case, you traverse it twice in the same direction (so in homology you get a multiple of two, which is consistent with cellular homology), but have "flipped sides". In a surface, this corresponds to an embedded Möbius strip, but in higher dimensions I have no imagination.
The space is S3 wedge S2 upto homotopy equivalence not upto homeomorphism
But RP3/RP1 is not a manifold either :)
It's just some wild idea. The problem in your interpretation was coming from the fact that the 2-sidedness is only captured in the (n-1)-simplices, not further below
So why not try and quotient all simplices below dimension n - 1?
So M/M^(n-2)
Who knows what that looks like
it's going to be D^n with the boundary attached to a wedge of (n-1)-spheres
It's the n-dimensional analogue of a 2-complex really
yeah, and restricting D^n to S^{n-1} and projecting onto the wedge summands of course returns the maps whose degree we're interested in, but I'm not sure where orientability lurks in this picture anymore
But all points on the (n-1)-spheres away from the collapsed wedge point are manifold points
So projecting to the wedge summands will give you 2-sheeted maps
2-sheeted everywhere except the wedge point
Where junk collapses
Lol I guess if M is nonorientable, then $D^n \cup_{\times 2} S^{n-1}$ is a cellular quotient of M. Proof: There is some (n-1)-sphere where the 2-sheeted map is degree 2 and not 0. Collapse the rest of the spheres
It'd be awesome if you could say RP^n is a cellular quotient of M
Yeah I dunno what else to say. I hope it makes sense that there is a cellular quotient M -> D^n U_2 S^(n-1)
Lmao this is dual to there exists an embedding D^(n-1) x_2 S^1 -> M
@Thorgott If $W = D^n \cup_2 S^{n-1}$, then $H^{n-1}(W) = \Bbb Z_2$. If there is a cellular quotient $M \to W$, then I assume this gives an injection $H^{n-1}(W) \to H^{n-1}(M)$, which forces nonorientability of $M$; one of the equivalent ways to state nonorientability is to say $H^{n-1}(M)$ has no torsion.
but tbf if we only care about detecting torsion in H^{n-1}, then this is immediate directly from the the cellular boundary formula even without passing to the quotient
cellular boundary with single top cell gives the direct bridge between "top homology Z" and "codimension 1 homology torsion-free", this is a point Hatcher makes at some point
I can give a geometric interpretation of "top homology Z" by drawing a picture of a triangulation as generator of the top homology, is there a direct geometric interpretation of "codimension 1 homology torsion-free"? probably more complicated
i grew up in ireland in the 60's and it was all about who you know. i thought the us would be different mainly because of the vastly larger population.
@Thorgott Codimension 1 homology classes are hypersurfaces (literally). The connected ones are either 1-sided or 2-sided. If it is 2-sided it either splits M or it doesn't. In the first case the homology class is zero. In the second case it's non-torsion: there is the natural pairing of H_1 on H_{n-1} by intersection numbers. If the complement of a 2-sided hypersurface is connected it represents a non torsion class because you can construct a loop which has intersection number 1 with it.
The 1-sided ones have double covers which are a boundary; the complement wants to have boundary the orientation double cover of the 1-sided submanifold.
Homology classes are always represented by oriented submanifolds. If you're in an oriented manifold you only have 2-sided oriented submanifolds and unoriented 1-sided submanifolds.
But only the oriented ones show up in homology; so H_{n-1} contains no torsion.
In the unoriented case you have both oriented 2-sided and 1-sided submanifolds. So you get some Z in some Z/2 in your top homology.
I believe only a single copy of Z/2 but that seems more complicated to explain.
How can I show a nonzero vector space is faithfully flat? I already know all vector spaces are flat and my definition of faithfully flat module M is flat and $M\otimes N = 0\Rightarrow N=0$.
If $N'\to N\to N''$ is an exact sequence of $B$-module and if there is a ring homom $f:A\to B$, then $N'\to N\to N''$ is an exact sequence of $A$-module?
@MikeMiller The intersection pairing requires orientability to work though, right? Otherwise they'd have to be mod 2, but then this only gives non-triviality of the homology class, but not non-torsion.
Something like this, but the details escape me without working it out again
There's a notion of twisted orientation, because for any line bundle L, Aut(L) still has two path components whether or not L is trivial
A twisted orientation ought to be something like a preferred representative L of the line bundle coming from the oriented double cover and an isomorphism det(TM) ~ L; you then ought to define twisted orientations on submanifolds (naturally you might guess an orientation of their normal bundle but you want to come up with a notion that will work fine even when your maps aren't immersions)
@Sarabsrimt Yes. When we want only to select things we use Cr and when we want to select +arrange we use Pr. In your example as you reasoned , we don't need to arrange.
yeah, there's a hom $F_2\rightarrow\mathbb{Z}\times\mathbb{Z}$ that's induced by $a\mapsto0$ in the first and $b\mapsto0$ in the second factor, your subgroup is the preimage of $2\mathbb{Z}\times2\mathbb{Z}$
so very much a subgroup
I guess better look at the map $F_2\rightarrow\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
your thing is precisely the kernel and the map is surjective
There are algorithms for these kinds of questions, look at the original work by Nielsen or Schreier, @AkivaWeinberger; maybe also any good book on combinatorial group theory
