# If some of the vectors in a dual basis are orthogonal then so are the original vectors?

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.

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 $\partial_{n+1},\dots,\partial_{2n}$?

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).
• 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. – Asaf Shachar Aug 17 '18 at 9:34