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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 $x$.

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  • $\begingroup$ Do you want the equality $\pi_L^{-1}(\pi_L(\sigma_d(x)))=\sigma_d(x)$ to be set-theoretical or scheme-theoretical? $\endgroup$
    – Sasha
    Apr 6 at 5:33
  • $\begingroup$ @Sasha I would prefer the equality to be scheme-theoretic, but set-theoretic equality is also OK. From the work of Joel Roberts (Theorem 1 of "Generic projections of algebraic varieties") it seems that a scheme-theoretic equality is possible for a generic point of $X$. But, I am not sure a similar result exists for all points of $X$ i.e., if my question has a positive answer. $\endgroup$
    – user43198
    Apr 6 at 8:11
  • $\begingroup$ For the scheme-theoretic equality the answer is negative, see my answer. $\endgroup$
    – Sasha
    Apr 6 at 8:58
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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.

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  • $\begingroup$ Perhaps it is not clear from the question, I do not expect $\pi_L^{-1}(\pi_L(\sigma_d(x))=\sigma_d(x)$ for each point $x$ and fixed $L$. Of course, I do not expect $\pi_L$ to be an isomorphism. My question was, given an $x$, can I find an $L$ such that $\pi_L^{-1}(\pi_L(\sigma_d(x))=\sigma_d(x)$ . For different $x$, I will need to choose different $L$. I have edited the question to make this point slightly clearer. $\endgroup$
    – user43198
    Apr 6 at 10:23
  • $\begingroup$ @user43198: If you want the condition to hold only at one point, then the answer is evidently positive; see the second paragraph in my answer. $\endgroup$
    – Sasha
    Apr 6 at 10:31
  • $\begingroup$ Thank you. That was my question (at only one point). Sorry for the confusion. $\endgroup$
    – user43198
    Apr 6 at 10:34

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