# Inner Product on Exterior Powers. Proving that the inner product is positively definite.

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.

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.