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Let $V$ be a real $n$-dimensional vector space, and let $g,h$ be two inner products on $V$. Fix some $1\le k\le n-1$, and denote by $\text{Gr}_k(V)$ the Grassmannian of $k$-dimensional subspaces of $V$.

Let $\perp_g:\text{Gr}_k(V) \to \text{Gr}_{n-k}(V)$ be the orthogonal complement map operation w.r.t $g$, i.e. $\perp_g(W)=W^\perp$.

Now, suppose that $\perp_g=\perp_h$. Is it true that $g=\lambda h$ for some $\lambda \in \mathbb{R}$?

I guess that the question can be asked more generally, for any two non-degenerate bilinear forms.

One can formulate the question also via linear maps: If $A:(V,g) \to (V,h)$ commutes with taking complements, is it conformal? (Here $A=Id$).

I know that a map which commutes with the Hodge dual operator is conformal, but here we require less.

Asaf Shachar
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Yes, this is true. I'll assume that $g$ is an inner product and $h$ is a symmetric non-degenerate bilinear form (not necessarily positive-definite). Then one can simultaneously diagonalize the forms and find a $g$-orthonormal basis $v_1, \dots, v_n$ such that $h(v_i,v_j) = \lambda_i \delta_{ij}$. We want to show that $\lambda_i = \lambda_j$ for all $i \neq j$. If not, then by reordering the basis we can assume that $\lambda_1 \neq \lambda_2$. Let

$$ W = \operatorname{span} \{ v_1 + v_2, v_3, \dots, v_{k+1} \}. $$

Then a direct calculation verifies that

$$ W^{\perp_{g}} = \operatorname{span} \{ v_{k+2}, \dots, v_n, v_1 - v_2 \} $$ while $$ W^{\perp_{h}} = \operatorname{span} \{ v_{k+2}, \dots, v_n, \lambda_2 v_1 - \lambda_1 v_2 \} $$ and since $\lambda_1 \neq \lambda_2$ we have $W^{\perp_{g}} \neq W^{\perp_{h}}$, a contradiction.

levap
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