While solving the problem 4-7 in Spivak's "Calculus on Manifolds", I came up with a hypothesis which can be helpful for solving the problem. The following is the problem which I'm trying to solve:
Every non-zero $w \in \Lambda^n(V)$ is the volume element determined by some inner product $T$ and orientation $\mu$ for $V$.
Also, the following is the hypothesis:
Let $V$ be a vector space such that $\dim V=n$ and $w \in \Lambda^n(V)$, which is non-zero. Then there exists a basis $\{v_1, \cdots, v_n\}$ for $V$ such that $w(v_1, \cdots, v_n)=1$.
My proof of this hypothesis is as follows:
Let $f:\mathbb{R}^n \rightarrow V$ be an orthogonal transformation such that $$f^*w=\det$$ and $e_1, \cdots , e_n$ be the usual basis for $\mathbb{R}^n$. Then we observe that $f$ is invertible and thus an isomorphism by putting $f^{-1}=f^T$; $f^T$ is the transpose of $f$. Hence $f(e_1), \cdots, f(e_n)$ is a basis for $V$. Moreover, according to the definition of $f^*$, $f^* : \mathcal{J}^n(V) \rightarrow \mathcal{J}^n(\mathbb{R}^n)$ is a linear transformation such that $$f^*T(v_1,\cdots,v_n)=T(f(v_1),\cdots,f(v_n))$$ for every $T \in \mathcal{J}^n(V)$ and $v_1,\cdots,v_n \in \mathbb{R}^n$. Therefore, $$\det(e_1,\cdots,e_n)=f^*w(e_1,\cdots,e_n)=w(f(e_1),\cdots,f(e_n))=1.$$
In this proof, I'm not sure that there does exist such $f$; i.e. does there always exist an orthogonal transformation $f$ from $\mathbb{R}^n$ to $V$ such that $f^*w=\det$ for every vector space $V$ such that $\dim V=n$ and $w \in \Lambda^n(V)$? If someone provides some ideas about this question to me, I appreciate it.