证明$ℝℙ^2→ℝ^4$单射

hbghlyj

\begin{gather*}x_1,y_1,z_1,x_2,y_2,z_2\inℝ
\\x_1^2+y_1^2+z_1^2=x_2^2+y_2^2+z_2^2=1
\\\left(x_1^2-y_1^2, x_1 y_1, y_1 z_1, z_1 x_1\right)=\left(x_2^2-y_2^2, x_2 y_2, y_2 z_2, z_2 x_2\right)\end{gather*}
证明 $\left(x_2, y_2, z_2\right)=\pm\left(x_1, y_1, z_1\right)$

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尝试:
$(x_1^2 - y_1^2) z_1^2=(x_1 z_1)^2-(y_1 z_1)^2=(x_2 z_2)^2-(y_2 z_2)^2=(x_2^2 - y_2^2) z_2^2$
若$x_1^2-y_1^2=x_2^2-y_2^2≠0$, 则$z_1^2=z_2^2$.
由$x_1^2+y_1^2+\cancel{z_1^2}=x_2^2+y_2^2+\cancel{z_2^2}$与$x_1^2-y_1^2=x_2^2-y_2^2$得$x_1^2=x_2^2,y_1^2=y_2^2$,矛盾.
故$x_1^2-y_1^2=x_2^2-y_2^2=0$.
由$x_1y_1=x_2y_2$得 $(x_1,y_1)=±(x_2,y_2)$, 代入$x_1^2+y_1^2+z_1^2=x_2^2+y_2^2+z_2^2$得$z_1^2=z_2^2$,
代入$x_1z_1=x_2z_2$得$\left(x_2, y_2, z_2\right)=\pm\left(x_1, y_1, z_1\right)$.

Czhang271828

已知
\begin{align*}
\sin^2\theta_1\cos2\varphi _1&=\sin^2\theta_2\cos2\varphi _2,\\
\sin^2\theta_1\sin2\varphi_1&=\sin ^2\theta_2\sin2\varphi_2,\\
\sin 2\theta_1\cos\varphi_1&=\sin 2\theta_2\cos\varphi_2,\\
\sin 2\theta_1\sin\varphi_1&=\sin 2\theta_2\sin\varphi_2.\\
\end{align*}
证明,\[
(\sin \theta_1\cos\varphi _1,\sin\theta_1\sin\varphi _1,\cos \theta_1)=\pm (\sin \theta_2\cos\varphi _2,\sin\theta_2\sin\varphi _2,\cos \theta_2).
\]
这是显然的.

hbghlyj

For the smooth map $F: \mathbb{R} P^2 \to \mathbb{R}^4$ prove $dF_p$ is injective $\forall p \in \mathbb{R} P^2.$
Injective map from real projective plane to $\Bbb{R}^4$
Show there is a unique solution
Showing injectivity for an immersion $\mathbb R P^2 \to \mathbb R^4$
Show that the function $f: RP^2 \to R^4$ given by $\{x,-x\} \mapsto (x_1^2-x_2^2,x_1x_2,x_1x_3,x_2x_3)$ is injective.
Why a specified chart is smooth
Quotient maps and existence
How do you show this multivariable function is injective?
Understanding an immersion in $\mathbb{R}P^{2}$
Riemannian metric induced by immersion
Show $F(x,y,z) = (x^2-y^2,xy,xz,yz)$ gives an embedding of $\mathbb{R}P^2$ into $\mathbb{R}^4$
Steep of the proof of embedding $\mathbb{RP}^{2}$ into $\mathbb{R}^{4}$
An embedding of $\mathbb{RP}^2$ into $\mathbb{R}^4$
连续性:
Prove that $f$ determines a continuous map $\tilde{f}: \mathbb{R}P^2 \to \mathbb{R}^4$
使用$\Bbb C^2$的回答:
Continuous mapping between $\mathbb{S}^2$, $\mathbb{R}P^2$ and $\mathbb{R}^4$
embedding of $\mathbb{RP}^2$ in $\mathbb{R}^4$

hbghlyj

Difference between Cross-Cap, Mobius Band, and Real Projective Plane
The cross-cap is a model in 3 dimenions, with self intersections, of the projective plane.

The projective plane can also be seen as a Mobius band with a disc glued onto the boundary; but this cannot be done in 3D. A representation of this gluing and its relation to rotations in 3D is shown in part of this presentation Out of Line.