Let $V$ be a real finite dimensional vector space. $V\oplus V^*$ has a natural symmetric bilinear form: $$\langle X+\xi,Y+\eta\rangle=\frac{1}{2}(\xi(Y)+\eta(X)) $$ If $B\in \wedge^2V^*$, and we identify it with the transformation $V \to V^*$ given by $X \mapsto B(X,\cdot)$,
$$\tilde{B}:=\left(\begin{array}{cc} 1 & 0\\ B & 1 \end{array}\right)$$ defines an orthogonal (w.r.t the bilinear form above) operator on $V \oplus V^*$ with the special property that it preserves projections to $V$, i.e., if $\pi:V \oplus V^* \to V$ is the projection, $\pi \circ \tilde{B}=\pi$.
My question is: are all the orthogonal operators preserving projection to $V$ like this?
If $T:V \oplus V^* \to V \oplus V^*$ is such operator, we can use the direct sum to write it like this: $$T=\left(\begin{array}{cc} T_1 & T_2\\ T_3 & T_4 \end{array}\right).$$ To preserve projection to $V$ translates to $T_1=Id$ and $T_2=0$. With that information, orthogonality boils down to $$T_3(Y)(X)+T_4(\eta)(X)+T_3(X)(Y)+T_4(\xi)(Y)= \xi(Y)+\eta(X).$$ $T_4=Id$ and $T_3$ an skew-symmetric operator is an obvious solution, but how can I show it is the only one? Is it the only solution?
After some thinking, I got it. It is the only solution. Let $\{X_1,\dots,X_m\}$ be a basis of $V$ e let $\{\xi_1,\dots,\xi_m\}$ be the dual basis. Orthogonality means that $\langle T(X+\xi),T(Y+\eta)\rangle=\langle X+\xi,Y+\eta\rangle $ must be true for any $X+\xi,Y+\eta \in V\oplus V^*$. Particularly, we have $$\langle T(X_i+0),T(0+\xi_j) \rangle=\langle X_i+0,0+\xi_j\rangle \Rightarrow T_4(\xi_j)(X_i)=\delta_{ij}\Rightarrow T_4=Id$$ and $$ \langle T(X_i+0),T(X_j+0) \rangle=\langle X_i+0,X_j+0\rangle\Rightarrow T_3(X_i)(X_j)+T_3(X_j)(X_i)=0$$ $$\Rightarrow\ T_3\ \mbox{is skew-symmetric}.$$