differential-geometrygeneral-topologygeometrymobius-band
I am trying to understand the notion of an orientable manifold.
Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1}))> 0$ (where defined). My question is:
Using this definition of orientation, how can one prove that the Möbius strip is not orientable?
Thank you!
I just want to flesh out something Ronaldo said
Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go "continuously" from one to another.
To this end, consider the vector space $\mathbb{R}^n$. Let $Gl = Gl_n(\mathbb{R})$ be the set of all invertible linear maps from $\mathbb{R}^n$ to itself and let $M = M_n(\mathbb{R})$ be the set of all linear maps (possibly noninvertible) from $\mathbb{R}^n$ to itself. The space $M$ is naturally a vector space and is naturally isomorphic to $\mathbb{R}^{n^2}$. Use this identification to topologize $M$. The important thing about this topology is that polynomials in the entries of a matrix are continuous - this is all we'll use about it.
We have a map (the determinant) $det:M\rightarrow \mathbb{R}$, which is a continuous homomorphism (it's given as a polynomial in the entries of the matrix).
We can identify $Gl$ with $det^{-1}(\mathbb{R}-\{0\})$, i.e., $Gl$ is an open subset of $M = \mathbb{R}^{n^2}$. Give $Gl$ the subspace topology.
Now we get to the main claim: With this topology, $Gl$ has precisely two path components. I'll prove this in a second. In the mean time, notice that this makes what Ronaldo said precise: If we start in the component of $Gl$ containing the identity matrix, then a path of elements in $Gl$, starting at the identity matrix, gives rise to a path of basis vectors starting with whatever choice of basis you've started with - it gives you a way to continuously go from one basis to another. But you'll never be able to get to things of opposite orientation because that would require a path from one component of $Gl$ to another, which is impossible.
Now I'll establish the claim that $Gl$ has precisely two components. First notice that is must have at least two components because $det:Gl\rightarrow \mathbb{R}^n-\{0\}$ is continuous, surjective, and the range has two components. Let $X^+\subseteq Gl$ be those matrices of positive determinant and $X^-$ those of negative determinant, so we have $Gl = X^+\cup X^-$.
Next, I claim that $X^+$ and $X^-$ are homeomorphic. To see this, let $A\in X^-$ be anything at all and consider the map $f:X^+\rightarrow X^-$ given by $f(B) = BA$. Since matrix multiplication is given by polynomially in the entries of the matrices, $f$ is continuous. It has inverse $f^{-1}(B) = BA^{-1}$, which is continuous for the same reason $f$ is. Thus, $f$ is a homeomorphism.
All the following maps and paths will be continuous for the same reason: they'll be given as polynomials in the entries of the matrices involved.
Finally, I just need to show $X^+$ is path connected. So, let $A\in X^+$ be arbitrary. I want to connect it to the identity matrix by a path. Let $E_{ij}$ denote the matrix of all 0s with a 1 in the $ij$th slot. Let $F_{ij}(t) = I + tE_{ij}$. Notice that for $i\neq j$, $F_{ij}$ is triangular, so has determinant 1. Left multiplying by $F_{ij}(t)$ corresponds to the row operation of adding $t$ times the $j$th row to the $i$th row, which doesn't change the determinant if $i\neq j$.
Since the $F_{ij}(t_0)$ are just row operations, we know we can multiply by several of them (each with a different appropriate $t_0$) to bring $A$ to a diagonal form without changing the determinant. By multiplying by $F_{ij}(t)$ and letting $t$ go from $0$ to $t_0$, we make a path out of this. Doing this for all the $F_{ij}(t_0)$ simultaneously gives us a path from $A$ to a diagonal matrix where for all time, the determinant is constant.
We've now connected $A$ to a diagonal matrix $D$ by a path in $X^+$. It remains to connect the diagonal matrix to the identity. We know all the entries of $D$ are nonzero because $D$ has positive determinant. If an entry $d$ of $D$ is positive, use the path $t + (1-t)d$ in that slot to continuously change $d$ to $1$, all the while keeping the determinant nonzero. If $d$ is negative, using the same trick, we can change $d$ to $-1$. At this point, the determinant must be $1$ - it's positive because we're in $X^+$ and it's a product of $\pm 1$s.
We've now reduced $D$ to a new diagonal matrix $E$ whose entries are plus and minus 1s. If all are positive, we're done, so assume we have a $-1$ somewhere. Since $E$ is in $X^+$, there must be another entry which is $-1$. For notation, let's assume one $-1$ is in the $i$th position and the other is in the $j$th position. Think of the two $-1$s as $cos(t\pi)$ and think of the $ij$th entry and $ji$th entries as $sin(t\pi)$, all with $t=1$. Now, let t go from 1 to 0. One can easily check the determinant stays $1$, and this gives a path which changes the two $-1$s to $1$s. Doing this finitely many times eliminates all the $-1$s and changes $E$ to the identity matrix.
Thus, we've connected $A$ to the identity matrix by a path, so $X^+$ is connected.
(Kudos to anyone who reads the whole thing - it ended up being WAY longer than I anticipated, but it was a good exercise for me to go through).
Let $M\subset \Bbb R^3$ be a $2$ dimensional submanifold with boundary (say the Möbius strip) and let $p\in M$ be a boundary point. Let $U$ be an open ball of $\Bbb R^3$ centered on $p$, like in your definition of embedded submanifold. Then $U\cap M$ will be an open neighbourhood of $p$ in $M$, diffeomorphic to an open half-disk $D$ in $\Bbb R^2$ ("diameter" included). Finally, observe that there is no open subset of $\Bbb R^2$ diffeomorphic to $D$, so there is no $V\subseteq \Bbb R^3$ open such that $\phi(U\cap M)=V\cap \Bbb R^2$.
Note that the difference between a manifold with an one without boundary lies exactly in the fact that every point of the former has a neighbourhood diffeomorphic to $\Bbb R^n$, while this fails for some points of the latter, which admit open neighbourhoods diffeomorphic to an open half-ball in $\Bbb R^n$.
Best Answer
If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)
The point is that the positivity condition you wrote gives you that the normal at any $p'$ is independent of the specific $(U_{\alpha'},\phi_{\alpha'})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.
(This is just a formalization of the intuitive argument.)