# Set of orthogonal complements to open set in $Gr(k,\mathbb{C}^n)$ open in $Gr(n-k,\mathbb{C}^n)$?

$$\DeclareMathOperator\Gr{Gr}$$Consider $$\mathbb{C}^n$$ endowed with the Hermitian inner product $$\langle u,v\rangle=u^*v$$, and let $$U \subseteq \Gr(k,\mathbb{C}^n)$$ be a Zariski open dense subset of the Grassmannian of $$k$$ planes in $$\mathbb{C}^n$$. Is the set \begin{align} V=\{u^{\perp} | u \in U\}\subseteq \Gr(n-k,\mathbb{C}^n) \end{align} of orthogonal complements (under $$\langle\cdot,\cdot\rangle$$) open dense in $$\Gr(n-k,\mathbb{C}^n)$$? Or does it at least contain an open dense subset of $$\Gr(n-k,\mathbb{C}^n)$$?

If the bijection $$\Gr(k,\mathbb{C}^n)\leftrightarrow \Gr(n-k,\mathbb{C}^n)$$ given by $$u \leftrightarrow u^\perp$$ were an isomorphism of algebraic varieties then this would be obvious, but unfortunately it appears to only be an isomorphism when these are viewed as varieties over the reals.

Another idea is to somehow use Chevalley's theorem, although this result doesn't seem to hold over the reals.

• Judging by the question you link to, your $u^{\perp}$ is orhogonality with respect to the Hermitian inner product. If it were orthogonality by the standard complex-linear inner product, then you would have an isomorphism of varieties as described that question. But the Hermitian orthogonal complement is the composition of the complex linear orthogonal complement and complex conjugation! Both are automorphisms of the Zariski topology, so the answer is yes! – David E Speyer Feb 14 at 22:13
• @DavidESpeyer Ahh... so simple. Thank you! – doremifasolatido Feb 15 at 15:14