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In the following link:

http://hilbertthm90.wordpress.com/2009/09/09/what-i-talk-about-when-i-talk-about-orientation/

They state that the antipodal map $f: \mathbb{S}^{n} \rightarrow \mathbb{S}^{n}$ is orientation preserving if n is odd.

I'm trying to show this rigorously.

Can we argue as follows?

We can consider the map $h: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+}1$ given by $h(x)=-x$ then $f$ is the restriction of $h$ to $\mathbb{S}^{n}$. Viewing the restriction as the composition of $h$ with the inclusion map $i$ we get that $D(f)(a)=Df(i(p)) \circ Di(p)$. But since the inclusion is essentially the identity map then its derivative is the identity map. So $Df(a) = Dh(a)$ so the representation of the derivative of the map $f$ in local coordinates is the same as the jacobian of $h$. Would it suffice then to observe that the determinant of the Jacobian of $h$ is simply $(-1)^{n+1}$ ?

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That looks fine to me -- but how do you actually define "orientation preserving"? –  Henning Makholm Oct 21 '11 at 23:41
    
Henning Makholm: thank you, good point, that's what confuses me I've seen like $4$ definitions (which maybe they are equivalent?) but the one I'm using is that if $f: M \rightarrow N$ is an diffeomorphism then $f$ is orientation preserving if its derivative $Df$ is orientation preserving (where we use the the notation of orientation preserving for vector spaces, i.e for some fixed bases the change of basis matrix has positive determinant. –  user17182 Oct 21 '11 at 23:44
    
@Henning Makholm: Is this the definition you are used to? –  user17182 Oct 21 '11 at 23:53
    
@Henning Makholm: er I meant "where we use the concept of..." not rather than "notation". –  user17182 Oct 22 '11 at 0:11
    
I'm afraid I'll have to admit that I'm not used to any definition and have hitherto just thought the concept was reasonably obvious. After thinking a bit more, however, I've reached the conclusion that I don't actually know the first thing about orientability and was talking nonsense when I said your argument looked fine. It may be fine, but I couldn't tell you why or why not. Sorry. –  Henning Makholm Oct 22 '11 at 1:19
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2 Answers 2

up vote 5 down vote accepted

This is an elaboration on my comments on the question, and on Jason DeVito's answer.

Firstly, what does it mean to choose an orientation on a manifold $M$?

  1. One way to think of it is that an orientation is a collection of charts covering $M$ such that the Jacobians of the change-of-coordinate map on all overlaps have positive determinants.

  2. Another way to think of an orientation is that we have to choose an orientation for $TM_p$ for each $p \in M$, with the property that given any $p \in M$, there is some chart $U$ containing $p$, with local coordinates $x_1,\ldots,x_n$, such that the orientation on $TM_q$ is the one that contains the basis $\partial_{x_1}, \ldots, \partial_{x_n}$, for each $ q \in U$. (Recall that an orientation on a vector space over $\mathbb R$ is a collection of bases such that all change of basis matrices have positive determinant.)

It's not hard to check that these two notions coincide. Indeed, given a colllection of charts as in the first definition, we can define an orientation on the tangent space $TM_p$ for each $p\in M$ as follows: if $p \in U$ for one chart $U$ in our given collection, and if $x_1,\ldots,x_n$ are the local coordinates on $U$, then we define the orientation on $TM_p$ to be the one containing the basis $\partial_{x_1},\ldots,\partial_{x_n}$. Note that our assumption on the transition maps on the overlaps of the charts means that this really does give a well-defined orientation on the vector space $TM_p$ for each $p$. By construction the resulting set of orientations on the tangents spaces $TM_p$ satisfies the conditions of 2.

Conversely, given a set of orientations on the $TM_p$ as in definition 2, consider the set of charts $U$ whose existence is guaranteed by 2; this collection of charts evidently satisfies the conditions of definition 1.

For the sphere, there is a standard way to choose an orientation: fix a unit normal vector field $\mathbb n$ on $\mathbb S^n$, either the inward pointing normal or the outward pointing normal. Also fix an orientation on $\mathbb R^{n+1}$ as a vector space. If $p \in \mathbb R^{n+1}$, then $T\mathbb R^{n+1}_p \cong \mathbb R^{n+1}$ canonically, and so we get an orientation on $T\mathbb R^{n+1}_p$ for each $p$. (A slightly more long-winded way to describe what I just did, which might nevertheless be helpful, is: I am using $\mathbb R^{n+1}$ as a global chart on itself, and hence defining an orientation on $\mathbb R^{n+1}$ as a manifold as in definition 1. I am then using the procedure described above, of going from 1 to 2, to get an orientation on each $T\mathbb R^{n+1}_p$.)

Now for each $p \in \mathbb S^n$, define an orientation on $T\mathbb S^n_p$ such that the induced orientation on $T\mathbb S^n_p \oplus \mathbb R\mathbb n = T\mathbb R^{n+1}_p$ (induced orientation meaning that we add $\mathbb n$ to any positively oriented basis of $T\mathbb S^n_p$ so as to get a basis for $T\mathbb R^{n+1}_p$) coincides with the given orientation on $T\mathbb R^{n+1}_p$.

Now that we have fixed on orientation on $\mathbb S^n$, we are finally in a position to make a Jacobian computation to compute whether $f$ preserves or reverses orientation.

As noted by the OP, $Dh(p)$ has determinant $(-1)^{n+1}$ for any point $p$. On the other hand, $Dh$ takes the unit normal $\mathbb n(p)$ to the unit normal $\mathbb n(f(p))$. (Draw the picture!)

In other words, when we consider $Dh(p): T\mathbb R^{n+1}_p \to T\mathbb R^{n+1}_{f(p)}$, and we decompose this into the direct sum of $Df(p): T\mathbb S^n_p \to T\mathbb S^n_{f(p)}$ and the map $\mathbb R \mathbb n(p) \to \mathbb R \mathbb n(f(p))$ induced by $Dh(p)$, the latter map has the matrix $1$ with respect to the bases $\mathbb n(p)$ in the source and $\mathbb n(f(p))$ in the target. Thus $Df(p)$ also has determinant $(-1)^{n+1}$ with respect to the positively oriented bases on its source and target.

Thus $f$ is orientation reversing/preserving according to whether $n$ is even/odd.

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I think something more needs to be said. The problem is that an orientation preserving diffeomorphism may be orientation reversing on a subspace. For example, Consider the map $h:\mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $h(x,y,z) = (-x,-y,z)$. This is orientation preserving since the determinant of the Jacobian is 1.

However, the yz plane has its orientation reversed. This is because if you restrict to these points, in local coordinates, you get that $(y,z)$ is mapped to $(-y,z)$, which is clearly orientation reversing. If you prefer a compact example, use the unit circle in the yz plane.

I'm not sure if the fact that the sphere is codimension 1 in $\mathbb{R}^{n+1}$ rules this kind of behavior out or not.

Finally, here is an argument that shows the antipodal map is orientation preserving when $n$ is odd.

Lemma 1: Whether or not a diffeomorphism is orientation preserving or reversing can be checked at a single point (at least if the oriented manifold is connected).

Proof: Given a diffeomorphism $f:M\rightarrow M$, the map from $M$ to $\{-1,1\}$ given by taking the sign of $\det(df(p))$ can be shown to be continuous using local coordinates. (The sign is never $0$ since $f$ is a diffeomorphism). If $M$ is connected, this implies the map is constant.

Lemma 2: If $F:M\times[0,1]\rightarrow M$ is smooth and $f_t(p) = F(p,t)$ is a diffeomorphism for each fixed $t$, then all $f_t$ are orientation preserving or they are all orientation reversing.

Proof: By Lemma 1, we can focus on a single point $p$. Then the function taking $[0,1]$ to the sign of $df_t(p)\in\{-1,1\}$ can be shown to be continuous using local coordinates, hence it's constant.

Lemma 3: When $n$ is odd, there is a smooth map $F:S^{n}\times[0,1]\rightarrow S^n$ for which $f_0$ is the identity map and $f_1$ is the antipodal map.

Proof: For $p = (p_1,...,p_{2n})$, Let $F(p,t) = (\cos(\pi t) p_1 + \sin(\pi t)p_2, -\sin(\pi t) p_1 + \cos(\pi t) p_2,... )$, so essentially do a rotation on each pair of coordinates. This is a diffeomorphism for each $t$ because the inverse map is given by rotation each pair of coordinates the opposite direction for the same time.

Finally, $f_0(p) = F(p, 0) = (p_1,...,p_{2n})$ so $f_0 = Id$ and $f_1(p) = F(p,1) = (-p_1,...,-p_{2n}).$ so $f_1$ is the antipodal map.

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Dear Jason, What does $\det(df(p))$ mean? If $A$ is a linear transformation from a finite dimensional vector space to itself, then $\det$ is well-defined. (E.g. choose a basis for the vector space, and compute the det of the matrix of $A$ w.r.t. this basis.) But $df(p)$ is not a linear transformation from one vector space to itself; it maps from one vector space (the tangent space at $p$) to another space (the tangent space at $f(p)$). Thus it doesn't have a well-defined determinant until you choose a basis of each of these vector spaces, and {\em a priori} these two choices have ... –  Matt E Oct 22 '11 at 3:57
    
... nothing to do with one another. However, because $\mathbb S^n$ is orientable, if we choose an orientation on $\mathbb S^n$, then we know what it means to choose a positively oriented basis for the tangent space at each point of $\mathbb S^n$. The sign of $\det df(p)$ is then well-defined independent of the choice of bases for the tangent space at $p$ and $f(p)$, provided we choose them both to be positively oriented. Then we can check (at one point, as you indicate) whether or not $f$ is orientation preserving by computing $\det df(p)$. I'm spelling this out in a comment ... –  Matt E Oct 22 '11 at 4:01
    
... because you don't mention it in your argument, and part of the OP's problem seems to be uncertainty about what it actually means (at a technical level) to have an orientation on $\mathbb S^n$, and how you work with the concept of orientation to check something like orientation preserving. Regards, –  Matt E Oct 22 '11 at 4:03
    
@Matt E: You are absolutely right that I should have been more explicit about the orientation involved. In response to your first comment, I never claimed that $\det (df(p))$ was well defined, only that its sign is defined, and on an oriented manifold, it is. I agree in retrospect that this is not ideal pedagogy. One change I would have made would be saying (in lemma 1) it could be checked in local coordinates using an oriented collection of charts as per your definition (1). –  Jason DeVito Oct 22 '11 at 12:51
    
@Matt E: Since you have supplied the details I was assuming in your own post, I won't edit them in to this post. But thank you for pointing them out to me and the OP. –  Jason DeVito Oct 22 '11 at 12:54
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