30 votes
Accepted

Understanding the orientable double cover

Almost $2$ years later, I'll give a complete answer to my own question. Step 1 (Topology of $\widetilde{M}$): Take an atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\Lambda}$ such that $\{U_\alpha\}_{\...
rmdmc89's user avatar
  • 9,889
20 votes
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Does the Gauss-Bonnet theorem apply to non-orientable surfaces?

Orientability is not needed. Indeed, one can deduce unorientable version of Gauss-Bonnet theorem from the orientable one : Given a (compact) nonorientable surface, say $M$, with metric $g$, consider ...
Henry's user avatar
  • 3,115
15 votes
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Homology and cohomology of 7-manifold

For any connected manifold $M$, there is a homomorphism $\pi_1(M)\to\mathbb{Z}/2$ which sends a loop to $0$ if going around the loop preserves orientation and sends the loop to $1$ if going around the ...
Eric Wofsey's user avatar
14 votes
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Orientability of $m\times n$ matrices with rank $r$

$M_{m,n,r}$ is an homogenous space under the natural action of the (orientable) Lie group $G=Gl^+(m)\times Gl^+(n)$, through the action $(P,Q) M= PMQ^{-1}$. The isotropy group of the matrix $I_{m,n,r}=...
Thomas's user avatar
  • 7,450
12 votes
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How does one orient a simplicial complex?

As Eric Wofsey says, there are two notions here: An orientation of an $n$-dimensional simplicial complex is a choice of orientation for each $n$-dimensional simplex. The easiest way to find one is to ...
David E Speyer's user avatar
10 votes
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Is orientability needed to define volumes on riemannian manifolds?

You don't need orientability to define the volume of a parametrized region. Your value $vol(R)$ is completely well-defined in terms of a parametrization, and doesn't even depend on the parametrization ...
Aloizio Macedo's user avatar
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9 votes
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Partition of unity and volume form on a manifold

Suppose $p \in U_1\cap U_2$ and consider $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p \in \bigwedge^nT_p^*M$. While $\phi_1^*(\operatorname{vol})_p$ and $\phi_2^*(...
Michael Albanese's user avatar
9 votes
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extended kalman filter equation for orientation quaternion

The biggest issue with dealing with rotation in an EKF is the fact that (so far as I know) there is no sensible way to define the rotation in a vector space. Euler angles, Quaternions and rotation ...
superjax's user avatar
  • 428
9 votes
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Prove that flow map $\theta_t$ is orientation preserving

There is a "coordinates free" way of the above argument. Let $\omega$ be a non-vanishing $n$-form on $M$ which defines the orientation. Then a diffeomorphism $\Phi : M \to M$ is orientation ...
Arctic Char's user avatar
  • 15.9k
9 votes
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In what sense is an orientation continuous?

To see in what sense a continuous orientation for $M$ is continuous, first prove the following easy topological lemma. Lemma. If $f\colon X\to D$ is a function on a topological space $X$ valued in a ...
Alex Ortiz's user avatar
  • 24.5k
8 votes

Partition of unity and volume form on a manifold

A convex combination of volume forms could turn out to be zero, violating the nowhere-vanishing condition. For example, $$\frac{1}{2}dx \wedge dy + \frac{1}{2}(-dx) \wedge dy = 0$$ This is actually ...
Nick Alger's user avatar
  • 18.8k
8 votes

Does the Gauss-Bonnet theorem apply to non-orientable surfaces?

Following the OP's request, I'm posting the details of my computations for the explicit embedding of the Möbius strip $M$ in $\Bbb R^3$. The orientation of (halves) of the boundary curve turns out to ...
Ted Shifrin's user avatar
8 votes
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Seifert surfaces for knots $6_1, 6_2, 6_3$.

Here's how you can get a Seifert surface using Seifert's algorithm for $6_1$'s usual diagram: First you choose an orientation of the knot -- either gives the same result. Then you smooth the ...
Kyle Miller's user avatar
  • 19.2k
7 votes
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Every smooth map $S^4 \to\mathbb{CP}^2$ has degree zero

Note that $f$ induces a homomorphism $f^* : H^4 (\mathbb{CP}^2; \mathbb{Z}) \to H^4(S^4; \mathbb{Z})$. As $\mathbb{CP}^2$ and $S^4$ are orientable and closed, both cohomology groups are isomorphic to $...
Michael Albanese's user avatar
7 votes
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Every oriented vector bundle admits an orientation reversing isomorphism?

In general no, since such an isomorphism would reverse the sign of the Euler class. Consider $TS^2$ for a simple example where this yields two different isomorphism classes of vector bundles.
Tyrone's user avatar
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7 votes

How does one orient a simplicial complex?

Let $S$ be the set of all the vertices of your simplicial complex. If you choose a total order $<$ on the set $S$, then $<$ is also a total order on any subset of $S$. In particular, the set ...
Eric Wofsey's user avatar
7 votes

Prove that flow map $\theta_t$ is orientation preserving

Here is another answer exploiting the "group" property of flows: $$\theta_{s+t} = \theta_{s} \circ \theta_{t}.$$ Now given $\theta_t$, write it as $$\theta_{t} = \theta_{t/2} \circ \theta_{t/...
Mohith Nagaraju's user avatar
7 votes
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If $M\setminus\{p\}$ is orientable then $M$ is orientable.

The orientation cover of $M \setminus \{p\}$ is the orientation cover of $M$ with $2$ points removed. But the connectedness of a manifold of dimension at least $2$ doesn't change when you delete ...
Max's user avatar
  • 3,927
6 votes
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When is the orientation presheaf a sheaf?

Because of the excision theorem, you can check the sheaf property on a neighborhood of a point homeomorphic to $\mathbb R^n$, in which case it is trivial using the covering space of $R$-orientations ...
Patrick Da Silva's user avatar
6 votes

Are tubular neighborhoods orientable?

Consider $M$ the (open) Moebius band and $N\subset M$ the central circle in $M$. Then the total space of the normal bundle of $N$ is diffeomorphic to $M$ and, hence, is nonorientable. Same for the ...
Moishe Kohan's user avatar
  • 95.7k
6 votes

Does the Gauss-Bonnet theorem apply to non-orientable surfaces?

This is more of an extended comment than a complete answer, but hope that it will close the question. Applying a little Googlomagic (namely, searching for "gauss-bonnet non-orientable") it is ...
Yuri Vyatkin's user avatar
  • 11.2k
6 votes
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Orientable manifold $M$ ,then $\partial M$ is orientable

First, it suffices to consider the case when $M$ and $\partial M$ are both connected. (If $M$ is connected and $\partial M$ contains several components $D_i$, consider manifolds $M_i=(M - \partial M) \...
Moishe Kohan's user avatar
  • 95.7k
6 votes
Accepted

Confusion about the top homology group of a compact manifold.

I know that if the manifold is compact, then all of its homology groups are finitely generated. This is true for homology groups with coefficients in $\mathbb{Z}$. As you have observed, it's ...
Eric Wofsey's user avatar
6 votes
Accepted

Connected intersection of a manifold and orientation

Let $\varphi_i:V_i\to\mathbb R^n$ be the given coordinate maps. Then the map $f:V_1\cap V_2\to\mathbb R$, given by $$f(x)= \det(d(\varphi_2^{-1}\varphi_1)_x)$$ is continuous, and $f(x)\neq0$ for all ...
Aweygan's user avatar
  • 23.1k
6 votes
Accepted

Oriented Atlas for $\mathbb{RP}^3$

The same issue occurs all over with having to choose "arbitrary" orientations when you omit one variable. The negative sign in the second component of the cross product (or the ...
Ted Shifrin's user avatar
5 votes

Prove the sphere is orientable

One way to do it is finding a volume form for $\mathbb{S}^n$. Obviously, using the unit normal assumes the embedding $\iota:\mathbb{S}^n\hookrightarrow \mathbb{R}^{n+1}$. The idea is to take the usual ...
rmdmc89's user avatar
  • 9,889
5 votes

Homology and cohomology of 7-manifold

A quick proof using Stiefel–Whitney classes: a manifold $M$ is orientable iff the first SW class $w_1(M) \in H^1(M;\mathbb{Z}/2\mathbb{Z})$ is zero. But by the universal coefficient theorem, $$H^1(M;\...
Najib Idrissi's user avatar
5 votes

A topological group which is also a (not necessarily smooth) manifold is orientable

The solution by JHF seems correct to me; here is an alternative using the same core idea of paths in the space $M$ yielding homotopies between multiplication maps. The "top then right" leg of the ...
Doeke's user avatar
  • 592
5 votes
Accepted

Orientation twisted on the $2$-sphere

Actually, it's not that simple. Consider a similar example: Let $M$ be the $xy$-plane in $\mathbb R^3$ -- that is, $M$ is the set of all points of the form $(x,y,0)$ for $(x,y)\in \mathbb R^2$. Even ...
Jack Lee's user avatar
  • 46.3k

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