# Exterior Power: Find a orthornormal basis in Hilbert Space

Let $\wedge^{k} H$ with two inner products $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ where $(H, \langle \cdot, \cdot \rangle)$ is a real, n-dimensional Hilbert space with $\langle u_{1} \wedge ... \wedge u_{k}, v_{1} \wedge ... \wedge v_{k} \rangle_{1} = det (\langle u_{i},v_{j} \rangle)_{k\times k}$.

Prove/disprove that there exists a orthonormal basis $\mathcal{B}=\{e_{1},...,e_{n}\}$ of $H$ such that $\mathcal{C} =\{e_{i_{1}} \wedge ... e_{i_{k}}: 1 \leq i_{1} < ...<i_{k} \leq n \}$ is a orthogonal basis of $(\wedge^{k} H, \langle \cdot, \cdot \rangle_{2})$.

I tried to following the ask in this topic:

Show that a Hilbert space with two inner products has a basis that is orthogonal with respect to both inner products

I consider the theorem of Singular Value, but I didn´t get any advance.

Hints or solutions are greatly appreciated.

## 1 Answer

This is false in general as can be seen by dimensional considerations. Define

$$S = \{ (e_1, \dots, e_n) \, | \, (e_i)_{i=1}^n \text{ is an orthonormal basis of } (H, \left< \cdot, \cdot \right>) \}$$

and

$$T = \{ g \colon \Lambda^k(H) \times \Lambda^k(H) \rightarrow \mathbb{R} \, | \, g \text{ is an inner product on } \Lambda^k(H) \}$$

and let $l = \dim \Lambda^k(H) = { n \choose k }$. Consider the map $\varphi \colon S \times (\mathbb{R}_{+})^l \rightarrow T$ which sends $((e_1, \dots, e_n), (\lambda_I))$ to an inner product $g$ on $\Lambda^k(H)$ which is determined uniquely by requiring that $g(e_I, e_J) = \delta_{I,J} \lambda_I$ for all $I,J$ (where $I = (i_1 < \dots < i_k)$ is an increasing multi-index and $e_I = e_{i_1} \wedge \dots \wedge e_{i_k}$). You can readily verify that the map $\varphi$ is a smooth map from a $\frac{n(n-1)}{2} + l$ dimensional manifold to a $\frac{l(l+1)}{2}$ dimensional manifold.

Consider for example $n = 4$ and $k = 2$. Then $\varphi$ is a smooth map between a $12$-dimensional manifold and a $21$-dimensional manifold and so it can't be onto. Hence, there exists inner products $g$ on $\Lambda^k(H)$ for which we cannot find an orthonormal basis $(e_i)$ of $H$ such that the corresponding basis $e_I$ of $\Lambda^k(H)$ is $g$-orthogonal.