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Let $(V,g)$ be a $2n$-dimensional real inner product space. Let $v_i$ be a basis for $V$, and let $\theta^i$ be its corresponding dual basis. ($1 \le i \le 2n$).

The metric $g$ induces a metric $g^*$ on $V^*$ in the standard way:

Every $\theta \in V^*$ is identified with a vector $V_{\theta} \in V$, and we define $ g^*(\theta_1,\theta_2):=g(V_{\theta_1},V_{\theta_2})$.

Now, suppose that $\theta_1,\dots,\theta_n$ are $g^*$-orthogonal to $\theta_{n+1},\dots,\theta_{2n}$. How to prove that $v_1,\dots,v_n$ are $g$-orthogonal to $v_{n+1},\dots,v_{2n}$?

Note that I do not assume that $(v_i)$ is a $g$-orthonormal basis for $V$. (In that case the dual basis is also orthonormal). I assume only the orthogonality relations stated above.

Comments:

  1. I am not sure if it's really important that the dimension is even, and that we consider exactly "half" of the basis vectors.
  2. In manifold language: Suppose that $dx^1,\dots,dx^n$ are orthogonal to $dx^{n+1},\dots,dx^{2n}$. Is it true that $\partial_1,\dots,\partial_n$ are orthogonal to $\partial_{n+1},\dots,\partial_{2n}$?
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The basis and the dual basis satisfies, under the standard duality pairing $$ \langle \theta^i, v_j \rangle = \delta^i_j$$

The musical isomorphisms work as follows: given an inner-product space $(V,g)$, by Riesz representation for every $\theta\in V^*$ there exists a unique $V_\theta \in V$ such that $g(V_\theta, w) = \langle \theta, w\rangle$ for all $w\in V$. The metric $g^*$ on $V^*$ is defined to be $g^*(\theta, \eta) = g(V_\theta, V_\eta)$.

Putting this together, we have that $g(V_{\theta^i}, v_j) = \delta^i_j$.

By linearity $$ g(V_{\theta^i}, v_j) = \sum_{k} g( \langle \theta^k, V_{\theta^i} \rangle v_k, v_j) = \sum_k g^*(\theta^i, \theta^k) g(v_k, v_j) = \delta^i_j$$ At which point what you want reduces to the linear-algebraic fact that block-diagonal matrices have block-diagonal inverses (and if you wish, you can just copy the same proof over).

(As as you can see, you correctly surmised that the dimensions of the orthogonal subspaces do not matter.)

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  • $\begingroup$ Thanks! For a moment I forgot that $\langle dx^i,dx^j \rangle=g^{ij}$. From there this is indeed immediate after you realize the "matrix meaning" of the orthogonality relations. A very elegant solution. $\endgroup$ – Asaf Shachar Aug 17 '18 at 9:34

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