Proving $\mathbb{R}P^n$ is orientable if and only if $n$ is odd.

Question

Prove that $\mathbb{R}P^n$ is orientable if and only if $n$ is odd.

I know this question has been asked many times on this site, but all solutions consist of $n$ forms or homology groups which I can't use. The definition of orientability I can use is: a manifold is orientable if it admits an atlas $(V_\alpha,\phi_\alpha)$ such that the transition maps have a Jacobi matrix with positive determinant.

I was able to prove that the map $\alpha: S^n\to S^n$ defined by $\alpha(x)=-x$ is orientation preserving iff $n$ is odd. Moreover, if $\pi:S^n\to\mathbb{R}P^n$ is the projection, then $\alpha\circ\pi=\pi$. I thought mayble I can somehow use this fact (without using the fac that $\pi$ is a covering map, since it hasn't been taught yet).

Moreover, I have constructed specific atlas for the projective plane: $$\{\varphi_i:U_i\to\mathbb{R}^{n}\},\,U_i=\{(x_0:\ldots:x_{n}):x_i\neq0\}$$ defined by: $$\varphi_i(x_0:\ldots:x_{n})=(\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_{n}}{x_i})$$ and was able to find the transition maps, but calculating the determinant in the general case seemed too hard, and will not disprove the existence of orientation in the even case (maybe this is the way to go). Is there an elementary approach that uses the fact that $\alpha$ is orientation preserving and that $S^n$ is orientable, or given the restriction to use "elementary" methods only I have to calculate the determinant of the Jacobi matrix?

Any help would be appreciated.

Answer 1

If $n$ is odd, then $\Bbb{RP}^n$ is orientable:

To show the transition maps have positive Jacobian determinant, it is enough to consider $0\leq i<j\leq n$, because if $i=j$, the transition it the identity map which has determinant $1$, and if $i>j$, the transition is the inverse (so the determinant will still have the same sign). Now, we have that \begin{align} (\phi_j\circ\phi_i^{-1})(t_1,\dots, t_n)&=\left(\frac{t_1}{t_j},\dots, \frac{t_i}{t_j},\frac{1}{t_j},\frac{t_{i+1}}{t_j},\dots, \frac{t_{j-1}}{t_j},\frac{t_{j+1}}{t_j},\dots, \frac{t_n}{t_j}\right). \end{align} One has to be careful with the indices and ordering (make sure you know where the condition $i<j$ has been used). Clearly, this ordering is not pretty; the factor $\frac{1}{t_j}$ seems out of place, and we have a jump in $\frac{t_{j-1}}{t_j},\frac{t_{j+1}}{t_j}$. So, it would be nice to permute the columns $(j-1)-(i+1)+1=j-i-1$ times so that we get the mapping \begin{align} f_{ij}(t_1,\dots, t_n)&=\left(\frac{t_1}{t_j},\dots, \frac{t_{j-1}}{t_j},\frac{1}{t_j},\frac{t_{j+1}}{t_j},\dots, \frac{t_n}{t_j}\right). \end{align} In other words, $(\phi_j\circ \phi_i^{-1})=\sigma\circ f_{ij}$, where $\sigma$ is a permutation (a linear map on $\Bbb{R}^n$) that makes $j-i-1$ many column swaps. Thus, $\det((\phi_j\circ\phi_i^{-1})’(t))=(-1)^{j-i-1}\cdot\det [f_{ij}’(t)]$. We start by calculating the Jacobian matrix for $f_{ij}$ (hopefully the block structure is clear; I’m not sure how to centre things nicely): \begin{align} f_{ij}’(t)&= \begin{pmatrix} \begin{array}{c|c|c} \frac{1}{t_j} I_{j-1} &-\frac{t_1}{t_j^2}& \mathbf{0}\\ &\vdots & \\ & -\frac{t_{j-1}}{t_j^2}\\ \hline \mathbf{0}& -\frac{1}{t_j^2}&\mathbf{0}\\ \hline \mathbf{0}&-\frac{t_{j+1}}{t_j^2}&\frac{1}{t_j}I_{n-j}\\ &\vdots\\ &-\frac{t_n}{t_j^2} \end{array} \end{pmatrix} \end{align} Here, $I$ denotes the identity matrix of appropriate size, and $\mathbf{0}$ denotes the zero matrix of appropriate size. Now, using the multilinearity of determinants, we get \begin{align} \det[(\phi_j\circ\phi_i^{-1})’(t)]&=(-1)^{j-i-1}\det[f_{ij}’(t)]\\ &=(-1)^{j-i-1}\cdot\left(\frac{1}{t_j}\right)^{j-1}\cdot \left(-\frac{1}{t_j^2}\right)\cdot \left(\frac{1}{t_j}\right)^{n-j}\cdot\det \begin{pmatrix} \begin{array}{c|c|c} I_{j-1} &t_1& \mathbf{0}\\ &\vdots & \\ & t_{j-1}\\ \hline \mathbf{0}& 1&\mathbf{0}\\ \hline \mathbf{0}&t_{j+1}&I_{n-j}\\ &\vdots\\ &t_n \end{array} \end{pmatrix}\\ &=\frac{(-1)^{j-i}}{t_j^{n+1}}\cdot \det I_n\\ &= \frac{(-1)^{j-i}}{t_j^{n+1}}, \end{align} where in second equal sign I pulled out the various powers of $\frac{1}{t_j}$ from the columns, and in the third equal sign, I used the fact that by adding suitable multiples of row $j$ to the other rows (an operation which doesn’t affect determinants due to the alternating nature), we can get rid of the $t_1,\dots, t_n$ in the other rows.

Ok, so we’re almost there. Unfortunately, these charts are not the oriented ones. Consider $\psi_i=(-1)^i\phi_i$. Then, the transition map is (exercise) \begin{align} (\psi_j\circ\psi_i^{-1})(t_1,\dots, t_n)&=(-1)^j\left(\frac{t_1}{t_j},\dots, \frac{t_i}{t_j},\frac{(-1)^i}{t_j},\frac{t_{i+1}}{t_j},\dots, \frac{t_{j-1}}{t_j},\frac{t_{j+1}}{t_j},\dots, \frac{t_n}{t_j}\right). \end{align} In words, we’re multiplying the transition $\phi_j\circ\phi_i^{-1}$ by $(-1)^j$ overall, and multiplying one of the components by $(-1)^i$. Hence, \begin{align} \det[(\psi_j\circ\psi_i^{-1})’(t)]&=(-1)^{nj}\cdot(-1)^i\cdot\det[(\phi_j\circ\phi_i)’(t)]\\ &= (-1)^{nj}\cdot(-1)^i\cdot \frac{(-1)^{j-i}}{t_j^{n+1}}\\ &=\frac{(-1)^{(n+1)j}}{t_j^{n+1}}. \end{align} Thus, for odd values of $n$, this determinant is positive, and hence for odd $n$, $\Bbb{RP}^n$ is orientable, and the $\psi_i$’s provide an oriented atlas.

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If $\Bbb{RP}^n$ is orientable then $n$ is odd:

I don’t have a purely elementary proof, so for now, I’ll just mention the quick proof using differential forms (which is actually pretty easy and how I would have preferred proving the above implication as well). Suppose $\Bbb{RP}^n$ is orientable; then is has a nowhere-vanishing continuous $n$-form $\zeta$. The pullback $\mu:=\pi^*\zeta$ is then a continuous $n$-form on $S^n$, and this is nowhere-vanishing since $\pi$ is a local diffeomorphism. Now consider \begin{align} \omega=\sum_{j=1}^n(-1)^{j-1}x^j\,dx^1\wedge\cdots, \wedge \widehat{dx^j}\wedge\cdots\wedge dx^{n+1}. \end{align} This is the local-coordinate formula for the usual volume form on the sphere. From here, it is evident that $\alpha^*\omega=(-1)^{n+1}\omega$, where $\alpha(x)=-x$ is the antipodal map.

With this, we note that we can write $\mu=f\omega$ for some function $f:S^n\to\Bbb{R}$; in fact since $\mu$ is nowhere-vanishing, so is $f$, and since $\mu,\omega$ are continuous, so is $f$. So, $f$ being a continuous, nowhere-vanishing function on $S^n$, which is a connected space, implies $f$ has constant sign on all of $S^n$. Finally, we observe that $\pi=\pi\circ \alpha$ implies that \begin{align} f\omega&=\mu=\pi^*\zeta=(\pi\circ\alpha)^*\zeta=\alpha^*(\pi^*\zeta)=\alpha^*(\mu)=\alpha^*(f\omega)=(f\circ \alpha)(-1)^{n+1}\omega, \end{align} and thus $f=(-1)^{n+1}f\circ \alpha$, and so $(-1)^{n+1}=\frac{f}{f\circ \alpha}$. This quotient makes sense since $f$ is nowhere-vanishing, and now since $f$ maintains constant sign, the numerator and denominator have the same sign and hence $(-1)^{n+1}$ must be positive, i.e $n$ is odd.

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Edit: An atlas-based proof of the above implication.

We begin with a preliminary lemma about compatibility of a given chart with a given oriented atlas (essentially how you prove that there are only two possible orientations).

Lemma (Orientation compatibility):

Let $M$ be a smooth manifold which has an oriented atlas $\mathcal{A}_+$, and let $(U,\phi)$ be any chart belonging to the maximal atlas. If $U$ is connected, then exactly one of the following occurs:

• for each chart $(V,\psi)$ in $\mathcal{A}_+$, we have $\det (D(\phi\circ \psi^{-1}))$ is positive on $\psi[U\cap V]$ (and hence $\det D(\psi\circ\phi^{-1})$ is positive on $\phi[U\cap V]$)
• for each chart $(V,\psi)$ in $\mathcal{A}_+$, we have $\det (D(\phi\circ \psi^{-1}))$ is negative on $\psi[U\cap V]$ (and hence $\det D(\psi\circ\phi^{-1})$ is negative on $\phi[U\cap V]$).

As a corollary, we deduce that if $(U_1,\phi_1)$ and $(U_2,\phi_2)$ are charts from the maximal atlas such that $U_1$ and $U_2$ are both connected, then the Jacobian determinant $\det D(\phi_1\circ\phi_2^{-1})$ has constant sign on $\phi_2[U_1\cap U_2]$.

Here is the proof:

Define a function $\sigma:U\to \{-1,1\}$ as follows: given a point $x\in U$, pick a chart $(V,\psi)\in \mathcal{A}_+$, such that $x\in V$, and define $\sigma(x)$ to be the sign of $\det (D(\phi\circ\psi^{-1})_{\psi(x)})$. This is a well-defined quantity since is we had taken a different chart $(V_0,\psi_0)$ from the oriented atlas, then by the chain rule, \begin{align} \det D(\phi\circ \psi_0^{-1})_{\psi_0(x)}= [\det D(\phi\circ \psi^{-1})_{\psi(x)}]\cdot [\det D(\psi\circ\psi_0^{-1})_{\psi_0(x)}]. \end{align} Since the second term is positive (by definition of belonging to an oriented atlas), the well-definition of $\sigma(x)$ follows. Also, by continuity of $\det D(\phi\circ\psi^{-1})$, it follows that in a neighborhood of the point $\psi(x)$, this determinant maintains the same sign, meaning that the function $\sigma:U\to \{-1,1\}$ is locally constant. Therefore, $\sigma$ is actually a constant function. If this constant is $+1$, we get the first bullet point, while if the constant is $-1$, we get the second bullet point.

The corollary is pretty easy to deduce: for each $x\in U_1\cap U_2$, fix a chart $(V,\psi)\in\mathcal{A}_+$ from the oriented atlas such that $x\in V$. Once again, by the chain rule, \begin{align} \det D(\phi_1\circ\phi_2^{-1})_{\phi_2(x)}=[\det D(\phi_1\circ \psi^{-1})_{\psi(x)}]\cdot [\det D(\psi\circ \phi_2^{-1})_{\phi_2(x)}], \end{align} and so the sign of the LHS is $\sigma_1\sigma_2$, where $\sigma_i$ is the constant sign guaranteed by the lemma when applied to the chart $(U_i,\phi_i)$. Thus, the proof of the corollary is complete.

Now, let’s apply this to $\Bbb{RP}^n$, by supposin it admits an oriented atlas $\mathcal{A}_+$. First, we note that the standard charts $(U_j,\phi_j)$ all have connected domains because $U_j=\{[x_0:\dots, :x_n]\,|\, x_j\neq 0\}$ is the image under the (continuous) projection $\pi$ of the $j^{th}$ upper hemisphere $\{(x_0,\dots, x_n)\in S^n\,:\, x_j>0\}$, which is a connected set (of course, $U_j$ is also the image of the $j^{th}$ lower hemisphere). Now, fix any $0\leq i<j\leq n$ as above (take $(i,j)=(0,1)$ for concreteness if you want), we saw that the chart transition $\phi_j\circ\phi_i^{-1}$ is defined on $E=\{t\in\Bbb{R}^n\,:\, t_j\neq 0\}$ and has Jacobian determinant $\frac{(-1)^{j-i}}{t_j^{n+1}}$, so in order for this to maintain the same sign on all of $E$, $n$ must be odd.