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- Jan 26, 2012

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**Problem**: Prove that $\Bbb{RP}^n$ is orientable if and only if $n$ is odd.

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- Thread starter Chris L T521
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- Thread starter
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- #1

- Jan 26, 2012

- 995

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Jan 26, 2012

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We now prove the main result.

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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