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Problem of the Week #86 - January 20th, 2014

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Chris L T521

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Jan 26, 2012
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Chris L T521

Well-known member
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Jan 26, 2012
995
No one answered this week's problem. You can find the solution below.

We begin the proof with a lemma (with proof).

Lemma: The reflection $r:\Bbb{S}^n\rightarrow \Bbb{S}^n$ given by $r(x)=-x$ is orientation reversing when $n$ is even and is orientation preserving when $n$ is odd.

Proof of Lemma: Define a continuous normal vector field on $\Bbb{S}^n$ by $n(p) := p$. Then, we can define an orientation on $\Bbb{S}^n$ by stating that a basis $(b_1(p),\ldots,b_n(p)$ of $T_p\Bbb{S}^n$ lies in the orientation class of $T_p\Bbb{S}^n$ provided that $(b_1(p),\ldots,b_n(p), n(p))$ lies in the standard orientation class of $\Bbb{R}^{n+1}$. To see if $r$ is orentation preserving, we need to check whether or not $(dr(b_1(p)),\ldots,dr(b_n(p)),n(r(p)))$ lies in the orientation class of $\Bbb{R}^{n+1}$. However, $dr$ is just the restriction of $d\overline{r}$ to $\Bbb{S}^n$ where $\overline{r}:\Bbb{R}^n\rightarrow \Bbb{R}^n$ is the reflection through the origin of $\Bbb{R}^n$. Since $\overline{r}$ is linear, we have $d\overline{r}=\overline{r}$ and thus $dr=\overline{r}$. Furthermore, note that \[n(r(p)) = n(-p) = -p = -n(p) = \overline{r}(n(p)).\] Thus, we see that \[(dr(b_1(p)),\ldots,dr(b_n(p)),n(r(p))) = (\overline{r}(b_1(p)),\ldots \overline{r}(b_n(p)),\overline{r}(n(p))).\] Since the standard orientation class of $\Bbb{R}^{n+1}$ contains $(b_1(p),\ldots,b_n(p),n(p))$ by assumption, the above basis belongs to the orientation class of $\Bbb{R}^{n+1}$ if and only if $\overline{r}$ is orientation preserving, which is the case only when $n$ is even (or $n+1$ is odd).$\hspace{.25in}\blacksquare$

We now prove the main result.

Proof: Consider the natural mapping $\pi:\Bbb{S}^n\rightarrow\Bbb{RP}^n$ which is a local diffeomorphism given by $\pi(p)=\{p,-p\}$. Now, let $r:\Bbb{S}^n\rightarrow \Bbb{S}^n$ be the reflection through the origin. Then $\pi\circ r=\pi$. If $\Bbb{RP}^n$ is orientable, then we may assume that $\pi$ preserves orientation. Then the above equality implies that $\pi\circ r$ preserves orientation as well. This is not possible only if $r$ preserves the orientation which is the case only when $n$ is odd. Thus $\Bbb{RP}^n$ is not orientable when $n$ is even.


We now seek to show that $\Bbb{RP}^n$ is orientable when $n$ is odd. Let us orient each tangent space $T_{[p]}\Bbb{RP}^n$ as follows: Let $q\in[p]=\{p,-p\}$. Choose a basis of $T_q\Bbb{S}^n$ which is in its orientation class, and let the image of this basis under $d\pi$ determine the orientation class of $T_{[p]}\Bbb{RP}^n$. This orientation is well defined because it is not affected by wheter $q=p$ or $q=-p$. If $(b_1,\ldots, b_n)$ is a basis in the orientation class of $T_p\Bbb{S}^n$ and $(b_1^{\prime},\ldots,b_n^{\prime})$ is a basis in the orientation class of $T_{-p}\Bbb{S}^n$, then it follows that \[(d\pi_p(b_1),\ldots,d\pi_p(b_n)) \quad\text{and} \quad(d\pi_{-p}(b_1^{\prime}),\ldots, d\pi_{-p}(b_n^{\prime}))\]
belong in the same orientation class of $T_{[p]}\Bbb{RP}^n$ (this is because
\[d\pi_p(b_i) = d(\pi\circ r)_p(b_i) = d\pi_{r(p)}\circ dr_p(b_i) = d\pi_{-p}\circ dr_p(b_i)\]
and $r$ preserves orientation; hence we have $(dr_p(b_1),\ldots,dr_p(b_n))$ belonging to the same orientation class as $(b_1^{\prime},\ldots,b_n^{\prime})$). $\hspace{.25in}\blacksquare$
 
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