Show that the standard atlas for $\mathbb{RP}^2$, ${(U_i,\phi_i)}_{i=1}^3$, is not orienting.

$U_1=\{[x^1,x^2,x^3]\in \mathbb{RP}^2 \ | \ x_1 \neq 0\}$, $\phi_1 ([x^1,x^2,x^3])=(\frac{x^2}{x^1},\frac{x^3}{x^1})$

My attempt:

An atlas $\mathcal{A}$ is orienting if any two charts in $\mathcal{A}$ are orientation compatible. It suffices to prove that there are 2 charts that are not orientation compatible in the standard atlas for $\mathbb{RP}^2$.

Let's check the compatibility of the charts $(U_1,\phi_1)$ and $(U_3,\phi_3)$

$\phi_3 \circ (\phi_1)^{-1} (x^1,x^2)=\phi_3 ([1,x^1,x^2])=(\frac{1}{x^2},\frac{x^1}{x^2})$

$[(\phi_3 \circ (\phi_1 )^{-1})_*]= \begin{bmatrix} 0 & \frac{-1}{(x^2)^2} \\ \frac{1}{x^2} & \frac{-x^1}{(x^2)^2} \end{bmatrix} $

$\text{det} [(\phi_3 \circ (\phi_1 )^{-1})_*] =\frac{1}{(x^2)^3}$

I could not decide weather this is positive or negative in $\phi_1 (U_1 \cap U_3)$. I know that

$U_1 \cap U_3=\{[x^1,x^2,x^3] \ | \ x^1,x^3\neq 0 \}$

$\phi_1 (U_1 \cap U_3)=\{(\frac{x^2}{x^1},\frac{x^3}{x^1}) \ | \ x^1,x^3\neq 0 \}=\mathbb{R}^2 \setminus \{y=0\}$

But still cannot figure it out.

I also calculated $\text{det} [(\phi_2 \circ (\phi_1 )^{-1})_*] =\frac{-1}{(x^1)^3}$, and $\text{det} [(\phi_3 \circ (\phi_2 )^{-1})_*] =\frac{-1}{(x^2)^3}$.

I still do not know weather they're positive or negative. Any help would be appreciated.


1 Answer 1


$\frac{1}{x^3}$ has the same sign as $x$. $U_1 \cap U_3$ has points with positive and negative $x_2$, and so positive and negative determinant, so it’s not orientation preserving.


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