In the O'neill's differential geometry text, there are following problems. enter image description here

The projective plane $P$ is defined as follows. enter image description here

As you know, this $P$ is an abstract surface. So, we have to define the tangent vectors on an abstract surface.

But, this text book just gives following definitions.

Def) Let $\alpha:I \rightarrow M$ be a curve on the abstract surface $M$. The velocity vector( or tangent vector) $\alpha'(t)$ is defined by $\alpha'(t)=(f\alpha)'(t)$ for each differentiable function $f:M\rightarrow R$.

I think the answer of (a) is {$v_p, -v_{-p}$} where $v_p$ is a tangent vector on $\Sigma$.

How can we evaluate this problem?


Let $x$ in $P$, $x=F(y)=F(-y), y\in \Sigma$. Let $v\in T_xP$, since $dF_y$ and $dF_{-y}$ are isomorphism since $F$ is a local diffeomorphism, there exist a unique $v_y\in T_y\Sigma$ and a unique $v_{-y}\in T_{-y}$ such that $dF_y(v_y)=dF_{-y}(v_{-y})=v$.

Since $\Sigma$ is compact and connected and $F$ continue, $F(\Sigma)=P$ is compact and connected.

Suppose that $P$ is orientable and $\omega$ a volume form on $P$, $F^*\omega$ is invariant by $i:x\rightarrow -x$, this implies that $i^*(F^*\omega)=F^*\omega=-F^*\omega$. This implies that $F^*\omega=0$ contradiction.

  • $\begingroup$ Could you explain why $F$ is a local diffeomorphism? $F$ is defined by $F(p)=${$p,-p$}. $\endgroup$ – Chris kim Jun 1 '16 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.