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?

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    $\begingroup$ Could you define $V$ by $a_{11} = 0$, $a_{22} - a_{33} = 0$ to obtain a circle in $CP^2$? $\endgroup$ Jul 17, 2014 at 9:01

1 Answer 1


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 $$ \{[x_0:x_1:x_2] : |x_0|=|x_1|=|x_2|\}. $$


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