"Totally real" linear transformations

Question

Identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ via the equality $$(z_1, z_2, \ldots, z_n)=(x_1, \ldots, x_n, y_1, \ldots , y_n)$$ Where $z_j=x_j + iy_j$.

We call a linear invertible map $A: \mathbb{R}^{2n}\to \mathbb{R}^{2n}$ totally real if the image of every complex line is not a complex line (i.e a totally real plane). Equivalently, if $Gr(2,2n)$ denotes the Grassmannian of planes in $\mathbb{R}^{2n}$ then $\mathbb{C}P(n-1)$ lives inside $Gr(2,2n)$, and $A$ being totally real means that $\mathbb{C}P(n-1)$ intersects its image trivially. Note that the (real) dimension of $\mathbb{C}P(n-1)$ is half the dimension of $Gr(2,2n)$.

As an example, for $n=2$, the linear map $$B:(x_1, x_2, y_1, y_2)\to (-y_2 , x_1, y_1, -x_2)$$ Is totally real and satisfies $\det(B)=1$. Since $B$ is isotopic to identity through linear maps, then $B(\mathbb{C}P(1))$ is isotopic to $\mathbb{C}P(1)$ which shows that the self intersection number of $\mathbb{C}P(1)$ (inside $Gr(2,4)$) is zero.

Similarly, the previous example can be generalized to all $n=2k$ even.

My question: is it possible to find a totally real linear (invertible) transformation $A$ for $n\geq 3$ odd? (we allow $A$ to have negative determinant).

Note that if the self intersection number of $\mathbb{C}P(n-1)$ (inside $Gr(2,2n)$) is non-zero for $n$ odd, then $A$ cannot have positive determinant. But I don't know if this is the case and I would like also to know what is the self intersection number in this case.

Thanks in advance.

Answer 1

I argue that being totally real reduces to a problem about polynomials.

In OP's identification of $\Bbb{R}^{2n}$ with $\Bbb{C}^n$, multiplication by ${\rm{i}}$ amounts to $J:=\begin{bmatrix} O_n & -I_n\\ I_n & O_n \end{bmatrix}_{2n\times 2n}$. Now let's see how $A$ can fail to be "totally real". This means that there exists a non-zero vector $v\in\Bbb{R}^{2n}$ for which $AJv\in{\rm{Span}}\{Av,JAv\}$, i.e. the image under $A$ of the complex line in $\Bbb{C}^n$ determined by $v$ is invariant under multiplication by ${\rm{i}}$. Write $v\in\Bbb{R}^{2n}$ as $v=\begin{bmatrix} v_1\\ v_2 \end{bmatrix}$ where $v_1,v_2\in\Bbb{R}^n$, and $A$ as
$$A=\begin{bmatrix} A_1 & A_2\\ A_3 & A_4 \end{bmatrix}$$ where $A_i$'s are $n\times n$ matrices. Then one can easily check that $AJv\in{\rm{Span}}\{Av,JAv\}$ amounts to the existence of scalars $\lambda,\mu\in\Bbb{R}$ for which $$ -A_1v_2+A_2v_1=\lambda(A_1v_1+A_2v_2)-\mu(A_3v_1+A_4v_2),\\ -A_3v_2+A_4v_1=\lambda(A_3v_1+A_4v_2)+\mu(A_1v_1+A_2v_2). $$ These equations can be written as $$ (\lambda A_1-\mu A_3-A_2)v_1=(-\lambda A_2+\mu A_4-A_1)v_2,\\ (\lambda A_3+\mu A_1-A_4)v_1=-(\lambda A_4+\mu A_2+A_3)v_2. $$ Notice that it is impossible for both $$ \det(-\lambda A_2+\mu A_4-A_1),\det(\lambda A_4+\mu A_2+A_3)\in \Bbb{R}[\lambda,\mu] $$ to be identically zero. Because then there should exist non-zero vectors $w,w'\in\Bbb{R}^n$ with $A_2w=A_4w=A_1w=\mathbf{0}$ and $A_4w'=A_2w'=A_3w'=\mathbf{0}$. But then $A^{\rm{T}}\begin{bmatrix} w\\ w' \end{bmatrix}=\mathbf{0}$, contradicting the assumption that $A$ is non-singular. WLOG, suppose
$\det(-\lambda A_2+\mu A_4-A_1)\not\equiv 0$. When $-\lambda A_2+\mu A_4-A_1$ is invertible, one can solve the first equation for $v_2$. Substituting in the second equation, one should have $$ \small \left((\lambda A_3+\mu A_1-A_4)+ (\lambda A_4+\mu A_2+A_3)(-\lambda A_2+\mu A_4-A_1)^{-1} (\lambda A_1-\mu A_3-A_2)\right)v_1=\mathbf{0}. $$ Therefore, assuming that $\det(-\lambda A_2+\mu A_4-A_1)\not\equiv 0$, then $A$ is not totally real iff the rational function $$ \small \det\left((\lambda A_3+\mu A_1-A_4)+ (\lambda A_4+\mu A_2+A_3)(-\lambda A_2+\mu A_4-A_1)^{-1} (\lambda A_1-\mu A_3-A_2)\right)\in\Bbb{R}(\lambda,\mu) $$ vanishes at a point of $\Bbb{R}^2$.

One possible way to show that $A$ cannot be totally real when $n$ is odd is to show that the function above should have a zero. One approach is to analyze its sign as $\lambda\to\pm\infty$ but $\mu$ is fixed.