# How many holes may a projection of an algebraic variety have?

Let $$V$$ be a closed subvariety of $$\mathbf{P}^n$$. (We work over an algebraically closed field.) Define $$\pi:(\mathbf{P}^n\setminus P_0)\to \mathbf{P}^{n-1}$$ by $$\pi(x_0:x_1:...:x_n) = (x_0:x_1,...:x_{n-1})$$, where $$P_0$$ is the point $$(0,0,...,0,*)$$ in $$\mathbf{P}^n$$.

If only $$\pi$$ were defined in all of $$\mathbf{P}^n$$, $$\pi(V)$$ would be a closed subvariety of $$\mathbf{P}^{n-1}$$. It isn't, and $$V$$ need not be the a closed subvariety of $$\mathbf{P}^{n-1}$$. (Easy example: $$V:x_0^2 = x_1 x_2$$.) Can one still say that $$\pi(V)$$ contains $$\overline{\pi(V)}\setminus W$$, where $$W$$ is a closed subvariety of positive codimension in $$\overline{\pi(V)}$$ and degree $$\leq \deg(V)$$, say? How?

• It seems natural to guess that one should start by blowing up $P_0$, but I don't know where to go from there. Nov 30, 2020 at 10:06
• In the projective space $(x_0:x_1:\dots:x_n)$ is a usual practical notation for elements.
– YCor
Nov 30, 2020 at 11:50

Blow up to get a morphism $$\Pi: Bl_{P_0}\mathbf P^n \rightarrow \mathbf P^{n-1}$$. Let $$\widetilde{V}$$ be the proper transform of $$V$$ in $$Bl_{P_0}\mathbf P^n$$. Then $$\overline{\pi(V)}=\Pi(\widetilde{V})$$.

Now we can write $$\widetilde{V}=V \setminus \{P_0\} \ \cup \mathbf P(C_{P_O}V)$$ where $$C_{P_0}V$$ is the tangent cone of $$V$$ at $$P_0$$.

So $$\pi(V \setminus \{P_0\})$$ (which in your notation is $$\pi(V)$$) contains $$\Pi(\widetilde{V}) \setminus \Pi (\mathbf P(C_{P_O}V))$$.

As noted above, $$\Pi(\widetilde{V})$$ equals $$\overline{\pi(V)}$$. Moreover, $$\mathbf P(C_{P_O}V))$$ is a closed subset of the exceptional divisor $$E$$, and $$\Pi_{|E} \colon E \rightarrow \mathbf P^{n-1}$$ is an isomorphism.

So we get that $$\pi(V)$$ (in your notation) contains $$\overline{\pi(V)} \setminus W$$ where $$W \subset \mathbf P^{n-1}$$ is a closed subset isomorphic to the projectivisation of the tangent cone of $$V$$ at $$P_0$$.

The closed set $$W$$ has dimension $$\operatorname{dim}(V)-1$$. On the other hand, $$\pi(V)$$ has the same dimension as $$V$$ unless $$V$$ is a cone whose vertex contains $$P_0$$, but in that case $$\pi(V)$$ is a closed set .

As for degree, the degree of $$\mathbf P(C_{P_O}V))$$ as a subscheme of $$E$$ equals the multiplicity of $$V$$ at $$P_0$$, hence is bounded above by $$\operatorname{deg}(V)$$. Since $$W$$ is (isomorphic to) the underlying closed subset of this scheme, its degree is not greater than that of the scheme. So we have $$\operatorname{deg}(W) \leq \operatorname{deg}(V)$$ as required.

• Ah yes (and two colleagues just explained the same to me). Thanks! I'm a bit unsure of what happens if $V$ is singular at $P_0$ - I take you are addressing that at the end? Nov 30, 2020 at 12:07
• Never mind - everything works out. Nov 30, 2020 at 12:19