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For $v \in \mathbb{R}^n$ , $v^*\in (\mathbb{R}^n)^* $is given through \begin{equation} v^*:\mathbb{R}^n \rightarrow \mathbb{R}, u\mapsto v^*(u):=\langle u,v\rangle . \end{equation}

i)Show that $v_1,...,v_k$ is linearly independent if and only if $v_1^*\wedge...\wedge v_k^*=0$

ii)Let $ \gamma:\mathbb{R} \rightarrow \mathbb{R}^n \in C^2$ and \begin{equation} \omega:\mathbb{R}\rightarrow\Lambda^2(\mathbb{R^n})^*, t\mapsto\gamma(t)^*\wedge\gamma'(t)^*. \end{equation} Show that:

a) If there is a function $\lambda:\mathbb{R}\rightarrow\mathbb{R}$ with $\gamma''(t)=\lambda(t)\gamma(t)$ than $\omega$ is constant.

b) If $\omega$ is constant and $\gamma''(t)\neq 0 $ for all $t$, so exists a function $\lambda:\mathbb{R}\rightarrow\mathbb{R}$ with $\gamma(t)=\lambda(t)\gamma''(t)$.

I got i), but for ii) a) and b) i have no idea how to start. This subject is still slightly confusing to me.Any help or a hint is much appreciated. Thanks!

edit: sorry, forgot some details

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    You mean $\gamma\colon\Bbb R\to\Bbb R^n$ (with $n>1$)? And what does $\gamma(t)^*$ mean? – 2017-01-31

1 Answers 1

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For a) i have:

$ d(\omega)=d(\gamma(t)^*\wedge\gamma'(t)^*=d\gamma(t)^*\wedge\gamma'(t)^*-\gamma(t)^*\wedge d\gamma'(t)^*$

$=d\sum_{i=1}^n\gamma(t)_ix_i\wedge\gamma'(t)^*-\gamma(t)^*\wedge d\sum_{i=1}^n(\gamma'(t))_ix_i $

$=\gamma'(t)^*\wedge\gamma'(t)^*-\gamma(t)^*\wedge\gamma''(t)^*$

$=0-\gamma(t)^*\wedge(\lambda(t)\gamma(t))^*=\lambda(t)(\gamma(t)^*\wedge\gamma(t)^*=0$

Is this correct? Any hints/solution for b)?