The complex projective plane $\mathbb{C}P^2$ can be thought of as the set of rank one 3-by-3 Hermitian matrices with norm one, i.e., $\mathbb{C}P^2 = \{xx^\* : x \in \mathbb{C}^3, x^\*x=1 \}$. As such, the space $\mathbb{C}P^2$ is a manifold with $4$-real dimensions embedded in the space of all $3$-by-$3$ Hermitian matrices, $H\_3(\mathbb{C})$ (which has 9 real dimensions). 


Suppose $V$ is an affine subspace of $H\_3(\mathbb{C})$ with real dimension $7$, and that $V$ has nonempty intersection with $\mathbb{C}P^2$. Is the intersection simply connected, or is there a simple counterexample? 



Simple connectedness of $\mathbb{C}P^2$ intersected with an affine subspace
Put $V=\{A\in H\_3(\mathbb{C}):a\_{11}=a\_{33}=1/3\}$. I claim that $V\cap\mathbb{C}P^2$ is the set of matrices of the form
$$ P = \frac{1}{3}\left[\begin{array}{ccc} 1 & z & zw \\ \overline{z} & 1 & w \\ \overline{zw} & \overline{w} & 1 \end{array}\right],
$$
where $|z|=|w|=1$. The proof is not hard if you just expand out the equations $P^2=P^\*=P$ and $\text{trace}(P)=1$, looking at the diagonal entries first. We conclude that the intersection is a torus, and so is not simply connected. 


In terms of the more usual picture of $\mathbb{C}P^2$, we just have
$$ \{[x\_0:x\_1:x\_2] : |x\_0|=|x\_1|=|x\_2|\}. $$


