# Exterior Algebra: Inner Product and Cones

Let $\mathbb{R}^{n}$ the real vector space with canonical inner product $\langle \cdot, \cdot \rangle$.

Define $\varphi \colon \operatorname{GL}(\mathbb{R}^{n}) \rightarrow \operatorname{GL}(\wedge^{k}\mathbb{R}^{n})$ by $A \mapsto \wedge^{k}A$,

$T_{1}=\{h:\mathbb{R}^{n}\times \mathbb{R}^{n}|$ $h$ is a inner product in $\mathbb{R}^{n}\}$,

$T_{2} = \{g \colon \wedge^{k}\mathbb{R}^{n}\times \wedge^{k}\mathbb{R}^{n} \, | \, g \text{ is a inner product on } \wedge^{k}\mathbb{R}^{n}\}$, where $\dim (\wedge^{k} \mathbb{R}^{n})= \begin{pmatrix} n \\ k \end{pmatrix}=l$.

We know that all inner product in vector space $\mathbb{R}^{n}$ is determinated by a positive definite matrix in $M_{n}(\mathbb{R}) \cong \mathbb{R}^{n^{2}}$, and it´s a cone in $\mathbb{R}^{n^{2}}$. Define

$S_{n++}=\{A \in M_{n}(\mathbb{R}): A$ is positive definite $\}$

$S_{l++}=\{B \in M_{l}(\mathbb{R}): B$ is positive definite $\}$

So, $S_{l++}$ is open cone in $\mathbb{R}^{l^{2}}$.

Consider $\psi \colon \operatorname{GL}(\wedge^{k}H) \rightarrow T_{2}$ given by $J \mapsto\langle J(\cdot),J(\cdot)⟩$, where

$$\langle u_{1}\wedge...\wedge u_{k},v_{1}\wedge...\wedge v_{k} \rangle \colon= \det(\langle u_{i},v_{j}\rangle)_{k\times k}.$$

and set $R:=\psi \circ \varphi(S_{n++})$.

What is geometrically $R$ in cone $S_{l++}$?

We expect that $R$ is an open cone of dimension $\frac{n(n+1)}{2}$ in $S_{l++}$ s.t. $\partial R \subseteq \partial S_{l++}$. (So, if $x_{n} \to x$ with $x_{n} \in R$ and $x \in S_{l++}$, then $x \in R$).

Related:

Inner Products on Exterior Powers

• What is $\varphi$? – levap Jan 22 '17 at 16:34
• Hi, in the second line. Thanks – Alladin Jan 22 '17 at 19:35