I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$, then $T_x(V)=V$ if $x\in V$.

If $x\in V$, then since $V$ is a manifold, there is a local parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$. Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.

An arbitrary element of $T_x(V)$ looks like $$ d\phi_0(v)=\lim_{t\to 0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t} $$

but this doesn't seem very useful to show $T_x(V)=V$. What is the right approach?

  • 2
    $\begingroup$ You can parameterize $V$ by a linear function $\phi$ whose image under $\mathbb{R}^k$ is just $V$. Since $d\phi=\phi$... $\endgroup$ – Alex Youcis Sep 30 '13 at 1:18
  • $\begingroup$ thanks for the hint @AlexYoucis $\endgroup$ – Geovanna Anthony Sep 30 '13 at 1:36

Hint: Fix a basis $\{v_1,...,v_k\}$ of $V$ and consider the following parametrization: $\Phi:\mathbb R^k\to V$ given by $(a_1,...,a_k)\mapsto \sum_{i=1}^ka_iv_i$

  • $\begingroup$ Thanks for the hint azarel. $\endgroup$ – Geovanna Anthony Sep 30 '13 at 1:36

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.