# How many points can the projection of a variety to a line omit?

Let $$V$$ be a closed subvariety of $$\mathbb{A}^n$$. Let $$\pi:\mathbb{A}^n\to \mathbb{A}^1$$ be the projection map that forgets the last $$n-1$$ coordinates (say). Assume that the Zariski closure of $$\pi(V)$$ is $$\mathbb{A}^1$$. Then $$\pi(V)$$ must be of the form $$\mathbb{A}^1\setminus S$$, where $$S$$ is a finite set of points in $$\mathbb{A}^1$$. What sort of useful bound can we give on the number of points $$|S|$$ in $$S$$?

I am looking for a bound of the form $$|S|\leq (\deg V)^D$$, where $$D$$ is a function of $$n$$. It would be very nice if $$D$$ were bounded by a polynomial on $$n$$. (I think I already know how to get a horrific $$D$$.)

• Question arising from work with my former postdoc Jitendra Bajpai and my former student Daniele Dona. The formulation is Daniele's. Commented Nov 26, 2020 at 14:22
• Sure, that's what I meant. (Let's not get distracted, but there must be some difference here depending on your foundations; if you are just using classical foundations and define a variety as a closed algebraic subset of $\mathbb{A}^n$ or $\mathbb{P}^n$, there's of course no difference. You mean an embedding of an abstract variety in affine space?) Commented Nov 26, 2020 at 14:43
• Ah, right, good point. Changed the wording. Commented Nov 26, 2020 at 14:49
• If $V \subset \mathbb{A}^2$ I would say that the maximum number is $\deg V$. Compactifying, we get a projection $\pi \colon \bar{V} \to \mathbb{P}^1$. If there are missing points in the image of the original projection, they come from the points at infinity that I add to $V$. If these points are $x_1, \ldots, x_d$, where $d=\deg V$, then the maximum number of missing points occurs when $\pi$ is the projection from $x_1$ (say) and all the secant lines $x_1x_j$ are distinct (included the case $j=1$, giving a tangent line to $\bar{V}$). In this case we have $d$ missing points. Commented Nov 26, 2020 at 14:53
• uhm...not sure anymore of what I have said. If the line $x_1x_j$ intersects $\bar{V}$ in a further point $y \in V$, then there is no missing point corresponding to this line... Commented Nov 26, 2020 at 15:06

It seems to me that one can bound $$|S|\leq (n-1) \deg(V)$$.

First, note that we can work projectively, that is, we will be able to work with the projective closure $$\overline{V}\subset \mathbb{P}^n$$. In the end, the points of $$\overline{V}\setminus V$$ will only contribute a point at infinity in $$\mathbb{P}^1$$, and we are not counting that point anyhow. We will write $$V$$ instead of $$\overline{V}$$ henceforth.

We can define a map $$\pi_n:\mathbf{P}^n\setminus P_{0,n}\to \mathbf{P}^{n-1}$$ by $$\pi_n(x_0:x_1:\dotsc:x_n) = (x_0:\dotsc:x_{n-1})$$, where $$P_{0,n}\in \mathbf{P}^n$$ is the point $$(0:0:\dotsc:0:*)$$. As pointed out in How many holes may a projection of an algebraic variety have?, either (a) $$\dim(\overline{\pi_n(V)})=\dim(V)$$ and $$\pi_n(V)$$ contains $$\overline{\pi_n(V)}\setminus W$$, where $$W$$ is a variety of dimension $$\leq \dim(V)-1$$ and degree $$\leq \deg(V)$$, or (b) $$V$$ is a cone whose vertex contains $$P_{0,n}$$, and so $$\pi_n(V)$$ is closed and of dimension $$\dim(V)-1$$. Clearly $$\deg(\overline{\pi_n(V)})\leq \deg(V)$$.

We iterate: we define $$\pi_{n-1}:\mathbf{P}^{n-1}\setminus P_{0,n-1}\to\mathbf{P}^{n-2}$$ just as above. If we are now in case (a), we have $$\dim(\overline{\pi_{n-1}(\pi_n(V))})=\dim(\overline{\pi_n(V)})$$, and $$\pi_{n-1}(\pi_n(V))$$ contains $$\pi_n(\pi_{n-1}(V))\setminus (W' \cup \overline{\pi_{n-1}(W)})$$, where $$\deg(W')\leq \deg(V)$$ and $$\dim(W')\leq \dim(\overline{\pi_n(V)})-1$$, and $$W$$ is as above (and is empty if we were in case (b) before). If we are in case (b), then we need not remove a new variety $$W'$$, and we also notice that what we must remove from $$\pi_{n-1}(\overline{\pi_n(V)})$$ is the variety consisting of the points of $$\pi_{n-1}(W)$$ whose preimage under $$\pi_{n-1}$$ is contained in $$W$$. That variety is either empty or of dimension $$\leq \dim(W)-1$$; its degree is presumably $$\leq \deg(W)$$.

We iterate further, and are done.

• I think you even have $|S|\leq \deg(V)-1$. This follows from your argument and the observations that, assuming $V$ is not a cone over the point of projection, we have $\deg W + \deg \pi(V)=\deg V$. So the total amount of degree you pick up in the "variety of possibly missing points" by your sequence of projections is $\deg V -1$. (Thanks to Aaron Landesman for pointing out that the degree of $V$ should go down by projecting from a point on $V$!) Commented Dec 4, 2020 at 1:20
• Humor my slow brain: I take that $\deg(W) + \deg \pi_n(V) = \deg V$ follows from the observation that $\deg(W)$ equals the multiplicity of $V$ at $P_{0,n})$? (Thus, a space of dimension complementary to that of $V$ going through $P_{0,n}$ contains in fact $\deg(W)$ copies of $P_{0,n}$, and thus can have at most $\deg(V)-\deg(W)$ points of intersection with $V\setminus P_0$? Commented Dec 4, 2020 at 22:43
• Right, you need the relation between the multiplicity and the degree of $W$. A reference is Harris' "Algebraic geometry" p. 259, Ch. 20 in the part titled Multiplicity. The main point is that the class of the proper transform of $V$ is given by $[\widetilde{V}]=\deg V [L]-m[\Lambda]$, where $L$ is a linear subspace of $\mathbb{P}^n$ and $\Lambda$ a linear subspace of the exceptional divisor (both with the same dimension as $V$). Commented Dec 5, 2020 at 2:20