a) Let $V$ be an oriented $k$-dimensional vector subspace of $\mathbb{R}^N.\,$Prove that there is an alternating $k$-tensor $T\in\bigwedge^k (V^*)$ such that $T(v_1,\ldots,v_k )=1/k!$ for all positive oriented ordered orthonormal base. Furthermore show that $T$ is unique is called the volume element of $V$

b) In fact, suppose that $ϕ_1,\ldots,ϕ_k\in V^*$ is an ordered basis dual to some positively oriented ordered orthonomal basis of $V$. Show that the volume element of $V$ is $ϕ_1\wedge\cdots\wedge ϕ_k$.

Here is how I understand the problem

a) By the determinant theorem $A^*T=(\det A)T$ for every $T\in \bigwedge^k(V*)$ meaning if $\phi_1,\ldots,\phi_k \in \bigwedge^1(V*)$ then $A^*\phi_1 \wedge \cdots \wedge A*\phi_k= (\det A) \phi_1 \wedge\cdots \wedge \phi_k$.

Note that a basis $v_1,\ldots,v_k$ of $V$ is orthonomal if

$$v_i.v_j= \begin{cases} 1 & \text{if $i=j$};\\ 0 & \text{if $i \not =j$}.\end{cases} $$

From the result of one of previous exercise, we have

$$\phi_1 \wedge\cdots\wedge(v_1,\ldots,v_k)=\frac{1}{k!} \det[\phi_i(v_j)]$$

where $[\phi_i(v_j)]$ is a $k \times k$ matrix. From the definition of orthonomal, we can say that $[\phi_i(v_j)]$ is orthogonal, so $\det[\phi_i(v_j)]=\pm 1$. If we feed all positive base into $T$ we will have

$$\phi_1 \wedge\cdots\wedge(v_1,\ldots,v_k)=\frac{1}{k!} (1)=\frac{1}{k!}$$

the uniqueness is trivial from the determinant theorem.

for b)

I know that any oriented manifold has non vanishing top form. But I feel like I need something more.


It seems like you have (a) well in hand. For (b), let $x_1, \dots, x_n$ denote the basis dual to the $\phi_i$, and let $v_1, \dots, v_n$ denote an aribtrary oriented, orthonormal basis of $V$. Let $A:V \to V$ denote the map with $A(x_i) = v_i$. Since the $v_i$ are orthonormal, we have $$(Ax_i, A x_j) = (v_i, v_j) = \delta_{ij}.$$ It follows from linearity that $A$ preserves the inner product on $V$; that is, $A\in O(V)$. Since $A$ takes an oriented basis of $V$ to another oriented basis, we have $A\in SO(V)$. In particular, $\det A = 1$. Thus \begin{align*} (\phi_1 \wedge \cdots \wedge \phi_n)(v_1, \dots, v_n) &= (\det A)\left(\phi_1 \wedge \cdots \wedge \phi_n\right)(x_1, \dots, x_n) \\ &= \left(\phi_1 \wedge \cdots \wedge \phi_n\right)(x_1, \dots, x_n) \\ &= \frac{1}{k!} \end{align*} by your computation in part (a). Hence $\phi_1 \wedge \cdots \wedge\phi_n$ is the required volume element by part (a).


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