A question on linear projection of a smooth projective variety

Question

Let $X$ be a smooth, projective $\mathbb{C}$-variety of dimension $n$. Fix a closed point $x \in X$ and an embedding of $X$ in $\mathbb{P}^m$ for some integer $m$. For a given $d$, denote by $\sigma_d : \mathbb{P}^m \to \mathbb{P}^{N_d}$ the $d$-tuple embedding. My question is: for $d \gg 0$, does there exist a linear subspace $L_x \subset \mathbb{P}^{N_d}$ (depending on $x$) of dimension $N_d-n-2$, not intersecting $\sigma_d(X)$ such that for the linear projection from $L_x$ (sometimes called projection with centre $L$): $$\pi_{L_x} : \sigma_d(X) \to \mathbb{P}^{n+1}$$ we have $\pi_{L_x}^{-1}(\pi_{L_x}(\sigma_d(x)))=\sigma_d(x)$ i.e., the preimage of $\sigma_d(x)$ is only $\sigma_d(x)$ for the chosen closed point $x$? Any hint/reference is most welcome.

EDIT Note that, the choice of $L_x$ depends on the choice of $x$.

Answer 1

It is definitely not possible to expect the existence of a linear projection $\pi_L$ such that the equality $$ \pi_L^{-1}(\pi_L(\sigma_d(x)))=\sigma_d(x)\tag{*} $$ holds scheme-theoretically for each point $x \in X$. Indeed, this equality means that $\pi_L$ defines an isomorphism of $X$ onto its image in $\mathbb{P}^{n+1}$, which is thus a smooth hypersurface. But for $n \ge 3$ a smooth hypersurface in $\mathbb{P}^{n+1}$ has Picard group isomorphic to $\mathbb{Z}$, while the Picard group of $X$ could be arbitrary.

On the other hand, for a given point $x \in X$ it is easy to find a subspace $L$ such that $(*)$ holds. Indeed, the closure of the union of lines joining $x$ with other points of $X$ has dimension at most $n + 1$, and any $L$ disjoint from this union works.