Let $M$ be a von Neumann algebra and $\varphi: M_+\to [0, \infty]$ be a normal, faithful semifinite weight. Consider its associated semi-cyclic representation $$\pi_\varphi: M\to B(\mathfrak{H}_\varphi)$$ (see Takesaki's second book, chapter VII for details).
Consider the associated map $$\eta_\varphi: \mathfrak{n}_\varphi\to \mathfrak{H}_\varphi.$$
Is it true that $\eta_\varphi(\mathfrak{n}_\varphi\cap \mathfrak{n}_\varphi^*)$ is norm-dense in $\mathfrak{H}_\varphi$?
Attempt: If $\xi \perp \eta_\varphi(\mathfrak{n}_\varphi\cap \mathfrak{n}_\varphi^*)$, then for $x,y\in \mathfrak{n}_\varphi$ $$0 = \langle \xi, \eta_\varphi(x^*y) \rangle = \langle \pi_\varphi(x)\xi, \eta_\varphi(y)\rangle$$ so that $\pi_\varphi(\mathfrak{n}_\varphi)\xi=0$. Maybe this is sufficient to conclude that $\xi= 0$? I feel like non-degeneracy of $\pi_\varphi$ is relevant.
Thanks in advance for your help/input/comments!
The key is that $\mathfrak n_\phi$ is weak$^\ast$ dense in $M$, which follows from the semifiniteness of $\phi$: First, since $\mathfrak n_\phi$ is a left ideal of $M$, its weak$^\ast$ closure is of the form $Me$ for some projection $e\in M$. Thus $\mathfrak m_\phi\subset eMe$. Hence also the von Neumann algebra generated by $\mathfrak p_\phi$ is contained in $eMe$. As $\phi$ is semifinite, the von Neumann algebra generated by $\mathfrak p_\phi$ equals $M$ by Takesaki's definition. This implies that $e=1$, that is, the weak$^\ast$ closure of $\mathfrak n_\phi$ is $M$.
To conclude, use that $\pi_\phi$ is normal, hence the weak$^\ast$ closure of $\pi(\mathfrak n_\phi)$ contains $\pi_\phi(M)$. It follows from what you have shown that $\pi_\phi(M)\xi=0$. As $\pi_\phi$ is non-degenerate, $\xi$ must be zero.