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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.

Mark Lauer
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1 Answers1

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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.

Keshav
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