Let $X\subset\mathbb{P}^n$ be a smooth irreducible variety (over $\mathbb{C}$) and $p\in\mathbb{P}^n$ a point. Let $\pi:\mathbb{P}^n\setminus\{p\}\to\mathbb{P}^{n-1}$ be the linear projection with center $p$ and denote by $Y$ the Zariski closure of $\pi(X\setminus\{p\})$. I would like to know:

When is $Y$ smooth and the restriction $X\setminus\{p\}\to\pi(X\setminus\{p\})$ of $\pi$ an isomorphism?

If $p\not\in X$, then I think this is equivalent to $p$ not lying on the secant variety of $X$. But I am mainly interested in the case when $p\in X$. In this case we can rephrase the question as:

When is $Y$ the blow-up of $X$ at $p$?

  • $\begingroup$ If you require $\pi$ being generically one-to-one and the projection center on the variety, then $X$ has degree 2. I think it only happens when $X$ is a smooth quadric hypersurface. $\endgroup$ – AG learner Aug 4 at 18:31
  • $\begingroup$ No, I don't think so. For example if $X$ is the rational normal curve of degree $n$, then $Y$ is the rational normal curve of degree $n-1$ and the restriction of $\pi$ is an isomorphism onto its image. Maybe there was an ambiguity in the question which I edited now. $\endgroup$ – Hans Aug 5 at 7:18
  • $\begingroup$ Sorry, you’re right. I was thinking about codimension one case. $\endgroup$ – AG learner Aug 5 at 15:12

The exact statement for surfaces is given in Beauville's book: Complex Algebraic Surfaces. Lemma 4.4. I imagine that this generalises easily to higher dimensions.

The condition is different whether or not the point is contained in the variety (as you mentioned), I will state both cases.

I quote exactly from the book, to avoid messing it up:

Lemma: Let S be a surface in $\mathbb{P}^N$ and $p \notin S$ (respectively $p \in S$). Let $f: S \rightarrow \mathbb{P}^{N-1}$ (respectivly $f: \hat{S} \rightarrow \mathbb{P}^{N-1}$) be the restriction of the projection away from $p$. Then $f$ is an embedding $\iff$ there is no line through $p$ meeting $S$ in at least $2$ (respectively at least $3$) points, counted with multiplicitly.

A nice corollary of this statement (given immediately after in Beauville's book) is that every smooth projective surface is isomorphic to a surface in $\mathbb{CP}^5$.

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  • $\begingroup$ Every smooth projective surface is isomorphic to a surface in $\mathbb{P}^5$? $\endgroup$ – red_trumpet Aug 5 at 8:49
  • 1
    $\begingroup$ Yes, this is Proposition 4.5 "Complex Algebraic Surfaces". A. Beauville. $\endgroup$ – Nick L Aug 5 at 8:52

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