I wanted to figure out, how we can define the Inner Product on Exterior Powers in terms of the positively definite attribute on Inner Products.
Let V be a n-dimensional $\mathbb{R}$-Vectorspace with a inner product: $$g_1 = V\times V \to \mathbb{R} $$
For all $k \in \{1,\dots,n\}$ exsists a Bilinear Map
$$g_k := \bigwedge^k V \times \bigwedge^k V \to \mathbb{R}$$, which is uniquely defined by
$$(v_1\wedge\dots\wedge v_k,w_1\wedge\dots\wedge w_k) \mapsto det(g_1 (v_i,w_j)_{1\leq i,j\leq n})\in \mathbb{R} $$
We can assume that for every orthonormal basis $B:=\{e_1,\dots,e_n\}$ there is a induced basis of $\bigwedge^k V$ $\{e_{i_1},\dots,e_{i_k} |1\leq i_1 < \dots < i_k\le n\}$
which is also a orthonormal basis.
My current approach is to construct the Gram-Matrix of $g_k$ with the a orthonormal basis B and have the inner product of $g_1$ become the identity matrix $I_n$ and check for positive definition with the induced basis. Though i am currently having difficulties wrapping my head around on how to use that Induced Basis.
Any help is appreciated.
Best Answer
If you can assume that you have an induced ON basis $\{e_I = e_{i_1}\wedge \dotsm\wedge e_{i_k}\}$, (where I used $I$ as a multi-index for shorthand) then this directly means $$g(e_I,e_J) = \delta_{IJ}$$ so if $v\in \bigwedge^k V$, then $v = \sum_I v^I e_I$ so $$g(v,v) = \sum_{IJ} v^Iv^J g(e_I,e_J) = \sum_I (v^I)^2 \geq 0$$ shows $g$ is positive definite.
But make sure you can prove that the given basis is indeed orthonormal.