differential-formsdifferential-geometrylinear algebramanifolds
This is a very basic definition of orientable and very basic example 20.5 however ı could not understand definition in an good way so ı want you to explain my green writing please 🙂 and my example please help me ı want to learn orientation on manifold if ı could not understand in a good way this ı will not understand in a good way rest of the subject please help me
Before discussing the boundary of a manifold, perhaps it would be best to look at the case with embedded hypersurfaces in general, and for this I will be following Lee's Intro to Smooth Manifolds.
Let $M$ be a smooth, oriented, $n$-manifold and $S$ be an embedded ($n-1$) hypersurface. Let $\omega$ be an orientation form on $M$, so that $\omega$ is just a nowhere vanishing $n$-form. We would like to find a way of tweaking this $\omega$ to define a nowhere vanishing $n-1$ form on $S$. How do we do this? The key is that since $\omega$ is nowhere vanishing then (without loss of generality) it is always positive; that is, for all $p \in M$, $$ \omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_n}\right|_p\right) >0$$ where the partials form a basis for $T_pM$. If we take $p \in S$, then $T_pS$ will give use $n-1$ vectors to plug into $\omega_p$, so our goal is to find a vector field $v_\text{out}: S \to TM$ such that if $B_p=\left\{\left.\frac{\partial}{\partial x_1}\right|_p, \ldots, \left.\frac{\partial}{\partial x_{n-1}}\right|_p\right\}$ is a basis for $T_pS$ then $\{(v_\text{out})_p\} \cup B_p$ is a basis for $T_pM$. In that case we can define an $n-1$ form $\hat \omega_p$ on $S$ as $$ \hat \omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right) = \omega_p \left((v_\text{out})_p,\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right)$$ which will also be positive for every $p \in S$ since $\omega$ is positive for every $p \in S \subseteq M$.
So how do we guarantee that $(v_\text{out})_p$ is always "linearly independent" to $B_p$? We simply demand that $(v_\text{out})_p$ never lies in $T_pS$!
Now if we specifically look at boundaries, our discussion above suggests that we find a vector field $N: \partial M \to TM$ such that $N_p \notin T_p\partial M$ (these are usually called inward-pointing vector fields). The Stokes' orientation is to take the $-N$ and define the $\hat \omega_p$ as above.
The other way of defining induced orientations is to do it in terms of the charts, so that it suffices to look at just the upper half plane $$ \mathbb H^n = \{(x_1,\ldots,x_n) \in \mathbb R^n: x_n \geq 0\}.$$ Diffeomorphisms $T: \mathbb H^n \to\mathbb H^n$ with positive Jacobian induce diffeomorphisms $\partial T: \partial \mathbb H^n \to \partial \mathbb H^n$ with positive Jacobian (this is a straightforward but messy exercise with determinants). The orientation form on $\mathbb H^n$ is the standard form $\omega = dx_1 \wedge \cdots \wedge dx_n$, and the vector $-\frac{\partial}{\partial x_n}$ is outward pointing along the boundary. Furthermore, $$\omega\left(-\frac{\partial}{\partial x_n}, \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_{n-1}} \right) = (-1)^n \omega\left(\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}, \ldots, \frac{\partial}{\partial x_{n}} \right)$$ so this suggests that we should take $\hat \omega = (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}$ to be the induced orientation on $\partial \mathbb H^n$. We then define the induced boundary on $\partial M$ to be $$ [\partial M] = \phi^*[\partial \mathbb H^n]$$ where the square brackets are the equivalence class of orientations and $\phi: M \to \mathbb H^n$ is a coordinate chart of $M$.
First understand what we mean by an orientation for $\mathbb{R}^3$. This is tied to the idea of a right-handed set of vectors: we say $\{v_1,v_2, v_3 \}$ is a right-handed set of vectors if $\text{det}(v_1|v_2|v_3)>0$. Geometrically, this means that $v_3$ is on the same side of the plane spanned by $v_1$ and $v_2$ as $v_1 \times v_2$. We are most familar with the case of an orthonormal right-handed frame where we simply insist $v_1 \times v_2 =v_3$. In two dimensions, the condition $\{ v_1,v_2 \}$ be right-handed again is captured by $\text{det}(v_1|v_2)>0$. Geometrically, the condition in two-dimensions means that the second vector is obtained from the first by a Counter-Clockwise rotation. In all cases, an orientation allows us to decide which way is up once all the other ways are fixed. To be slighly more precise, if we fix $n-1$ coordinates in $\mathbb{R}^n$ then an orientation will provide us a framework in which we can decide what the positive direction is for the remaining $n$-th coordinate. This is equivalent to insisting there be an ordered set of vectors $\{v_1,v_2,\dots v_n \}$ for which $\text{det}[v_1|v_2|\cdots|v_{n-1}|v_n]>0$.
Details of the choice: ok, suppose $v_1,\dots , v_{n-1}$ are fixed and you are given $a,b$ vectors in $\mathbb{R}^n$ then $\text{det}[v_1|v_2|\cdots|v_{n-1}|a]>0$ whereas $\text{det}[v_1|v_2|\cdots|v_{n-1}|b]<0$. Then, in terms of the given orientation $\{v_1,v_2,\dots v_n \}$ we see $a$ points in the upward-direction whereas $b$-points in the downward-direction. Of course, this comment is relative to the coordinate system defined by the span of the orientation.
Relation to wedge product: $$ v_1 \wedge v_2 \wedge \cdots \wedge v_n = \text{det}[v_1|v_2|\cdots|v_{n}] e_1 \wedge e_2 \wedge \cdots \wedge e_n. $$ Where $(e_i)_j = \delta_{ij}$ defines the standard basis. The coefficient of the $n$-form will be positive iff the set of vectors $\{v_1,v_2, \dots, v_{n}\}$ shares the same orientation as the standard basis. Moreover, all such related bases form the so-called standard orientation of $\mathbb{R}^n$.
Now, for the case of a manifold it is much the same, however, we have to consider the derivations which fill $T_pM$ at various $p$. The natural orientation given by a coordinate chart $(x^i)$ at $p \in M$ is simply the ordered $n$-tuple (using $\partial_i = \frac{\partial}{\partial x^i}$) $$ \{ \partial_1|_p, \partial_2|_p, \dots, \partial_n|_p \}$$ dual to these $dx^1, dx^2, \dots dx^n$ form the volume form: $$ vol=dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n$$ Given this, we can judge if $v_1,v_2, \dots v_n$ is a coherent orientation to the one which is naturally induced from the coordinate derivations. Simply check: $$ vol(v_1,v_2, \dots , v_n) > 0 ? $$ Now, if there is another coordinate system $(y^j)$ also defined at $p$ then we can ask if $$ \{ \partial/\partial y^1,\partial/\partial y^2, \dots, \partial/\partial y^n \}$$ forms a coherent orientation with the one we already induced from $(x^i)$. The chain rule connects the $x$ and $y$ coordinate derivations and when we feed that to the volume form we find the determinant of the transition function appears. Therefore, two coordinate systems provide coherent orientations if their transition functions have Jacobians with positive determinant.
All of this said, it seems that the phrase you quote:
If we can assign an orientation to each point on a manifold M in such a way that the orientations as any two sufficiently near points on M are coherent, . .
could apply to what I describe above. But, the larger idea concerns a curve so almost certainly the discussion above about coordinate-induced orientations is not the real answer to the question.
What about a curve? Morita talks about choosing an orientation along the curve. I'm not entirely sure which construction he has in mind. But, here's a possibility: given a curve we can frame the curve using the tangent vector, and the change in the tangent vector etc.. well, in the abstract, how to frame a curve? I leave to your imagination, but, in many cases it can be done. So, there is a natural way to assign $n$-vectors $f_1(t), \dots , f_n(t)$ to a given curve at point $\gamma(t)$. Suppose the curve is closed such that after time $T$ the curve returns to the initial point $p=\gamma(0)=\gamma(T)$. Then, we can compare the orientations of $f_1(0), \dots , f_n(0)$ and $f_1(T), \dots , f_n(T)$ and see if they are compatible (coherent). In particular, feed both to the volume form at $p$ and see that they share the same signed result. If you did this for the tangent, normal frame field for a curve on the Mobius band then you'd find the orientation carried by the frame of the curve was not coherent when you go around the band once.
Best Answer
First you have to have a good notion of orientation on a vector space $V$. Since the transition matrix from one basis to another is an invertible matrix it has non-zero determinant. Hence, it's either positive or negative. Two bases of $V$ determine the same orientation if the transition matrix has positive determinant.
So, an orientation on $V$ is an equivalence class of bases for $V$. You can think of the group homomorphism $$ \begin{align} GL(V) \overset{\Large\epsilon}{\longrightarrow} &\{-1, +1\} \\ A \longmapsto & \frac{\det A}{|\det A|}. \end{align} $$
Two bases whose transition matrix lives in $GL^+(V) = \epsilon^{-1}(1)$ determine the same orientation, and two bases whose transition matrix lives in other coset $\epsilon^{-1}(-1)$ determine opposite orientations.
On a smooth manifold, each tangent space is a vector space $T_pM$ of the same dimension. A non-vanishing top form determines a basis in each tangent space in a way that varies smoothly as $p$ varies about $M$.
It's instructive to think about a non-orientable manifold, say the open Möbius strip $$ [0, 1] \times (0, 1) / \sim $$ where $(0, y) \sim (1, 1 - y)$. The left and right edges of the square are identified with a half twist. Any consistent choice of basis at each point in the interior of the square cannot be extended to the left (= right) edge.
For example, the standard basis $\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\}$ which is dual to the $2$-form $dx \wedge dy$ cannot be extended continuously (let alone, smoothly) to the left/right sides. The transition matrix would be $$ \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ on one side or the other.