Let $\mathscr{H}$ denote a Hilbert space. The projective Hilbert space is defined as $\mathbb{P}\mathscr{H}:=(\mathscr{H}\setminus \{ \mathbf{0}\})/\mathbb{C}$. It is the set of one-dimensional subspaces of $\mathscr{H}$. I have a question about it:
By $(\mathscr{H}\setminus \{ \mathbf{0}\})/\mathbb{C}$ we mean there is an action $\mathbb{C}$ on $\mathscr{H}\setminus \{ \mathbf{0}\}$, i.e. a funtion $\mathbb{C}\times \mathscr{H}\setminus \{ \mathbf{0}\}\to \mathscr{H}\setminus \{ \mathbf{0}\}$ such that $1v=v$ and $(z_1 z_2)v=z_1 (z_2 v)$ for all $v\in \mathscr{H}\setminus \{ 0\}$ and $z_1 ,z_2 \in \mathbb{C}$. I guess this comes from the fact that $\mathscr{H}$ is a $\mathbb{C}$-vector space: we have a scalar product $\cdot :\mathbb{C}\times \mathscr{H}\to \mathscr{H}$ with $1v=v$ and $(z_1 z_2)v=z_1 (z_2 v)$ for all $v\in \mathscr{H}$ and $z_1 ,z_2 \in \mathbb{C}$. But how can we have this? I mean $0\in \mathbb{C}$ and $0v=\mathbf{0}$, so the image of $\mathbb{C}\times \mathscr{H}\setminus \{ \mathbf{0}\}$ under scalar product is not contained in $\mathscr{H}\setminus \{ \mathbf{0}\}$. Am I right? Is there another action $\mathbb{C}$ on $\mathscr{H}\setminus \{ \mathbf{0}\}$?