Inner Product on $\Lambda^n(E)$

Question

Let $E$ together with $g$ be a inner product space(over field $\mathbb R$) , $\text{dim}E=n<\infty$ and $\{e_1,\cdots,e_n\}$ is orthonormal basis of $E$ that $\{e^1,\cdots,e^n\}$ is its dual basis. Now we define $\omega:=e^1\wedge\cdots\wedge e^n$ as element of volume of $E$.

How can I prove that $\omega(u_1,\cdots,u_n)\omega(v_1,\cdots,v_n)=det[g(u_i,v_j)] \qquad \forall u_i,v_j\in E\qquad\text{and } i=1,\cdots,n\ ?$

Of course, I prove that $\omega(u_1,\cdots,u_n)=det[u_1 \cdots u_n]_{n\times n}$ ($u_i$'s are columns of the matrix)but I do not find out how uses it for my problem.

Answer 1

Hint : (1) Let $$ U=[u_1\cdots u_n],\ V=[v_1\cdots v_n]$$

Then $$ (U^TV)_{ij} = u_i\cdot v_j =g(u_i,v_j) $$

(2) ${\rm det} (U^TV)={\rm det}\ U {\rm det}\ V$

Answer 2

The equation you wish to prove is linear in each of the vectors $u_j$ and $v_j$. Therefore it suffices to show the identity when these vectors are basis vectors. There are $n$ basis vectors from which we now want to choose the $n$ vectors $u_1,\dots,u_n$. If we choose any two to be the same, then both sides of the identity vanish (and the identity is true), so we may assume that they are all distinct. Both sides of the identity change signs in the same way if we permute the vectors $u_1,\dots,u_n$, so way may in fact assume $u_j=e_j$. Similarly, we may assume $v_j=e_j$. So it only remains to check this one case, and now the identity is again trivial: both sides equal one.