# Dimension of projection of projective variety on hyperplane

I have given a closed projective variety $$X$$ of dimension $$k$$ and a hyperplane $$H$$ in $$\mathbb{P}^n$$. When we take a point $$P \notin H$$ we can construct the projection $$\pi$$ by $$P$$ on $$H$$. I managed to show that the map $$\pi$$ is a closed morphism and hence $$\pi(X) \subset H$$ is a closed variety. However, I'm having trouble by proving some dimension claims about $$\pi(X)$$. There are actually three different cases:

(1) $$P \notin X$$: in this case, we have to prove that $$\dim(X) = \dim \pi(X)$$. So far, I managed to show that the dimension of $$\pi(X)$$ is at most $$k$$ because if $$V \subset \pi(X)$$ is a closed subvariety then, $$\pi^{-1}(V)$$ is a closed subvariety of $$X$$. However, when we have a chain of chain of subvarieties $$U_i \subset X$$ then we know that $$\pi(U_i) \subset \pi(X)$$ is also a subvariety, but I think we can't assume that they are distinct. Is there a way to fix this?

(2) $$P \in X$$ but there is a $$Q \in X$$ such that the line $$PQ$$ is not fully contained in $$X$$. I also have to prove that $$\dim(X) = \dim\pi(X)$$. I think I need to 'choose' a specific chain of subvarieties with the help of the point $$Q$$ but I don't know how I can construct this.

(3) $$P \in X$$ but for all $$Q \in X$$, the line $$PQ$$ is fully contained in $$X$$. I now have to prove that $$\dim\pi(X) = \dim(X) - 1$$. It is clear that $$\pi(X)$$ = $$X \cap H$$ so I'm wondering if I can say something about the dimension of the intersection of two projective varieties, but again, I don't really have an idea on how to start.

For two varieties $$X,Y$$ in projective space, we define $$J(X,Y)$$ the join of $$X,Y$$, to be the union of all lines in $$\Bbb P^n$$ connecting distinct points in $$X$$ and $$Y$$. Now I claim that $$\pi(X) = J(X,P)\cap H$$, because both sides represent taking the lines through $$X$$ and $$P$$ and then intersecting them with $$H$$. So by your work in (3), it suffices to determine $$\dim J(X,P)\cap H$$.

We can get rid of the intersection with $$H$$ in the dimension calculation easily, via the projective dimension theorem:

Projective dimension theorem (ref Hartshorne I.7.2): Let $$X,Y$$ be two irreducible closed subvarieties of $$\Bbb P^n$$ of codimensions $$r,s$$ respectively. Then every irreducible component of $$X\cap Y$$ has codimension at most $$r+s$$, and if $$r+s\leq n$$ then this intersection is nonempty.

If we know that $$J(X,P)$$ is irreducible, then as $$P\notin H$$, we see that $$J(X,P)\cap H$$ is a proper closed subvariety of $$J(X,P)$$, so it must have dimension at most $$\dim J(X,P)-1$$. On the other hand, by the theorem, it has dimension at least $$\dim J(X,P)-1$$. So we get $$\dim J(X,P)\cap H = \dim J(X,P)-1$$.

Now all we need to do is to prove that $$J(X,P)$$ is irreducible and determine it's dimension. Here we get a little bit of casework: in case (3), the join variety is just $$X$$ again, so it's irreducible of dimension $$\dim X$$. In cases (1) and (2), the following applies. Let $$J'(X,Y)=\{(x,y,z)\subset X\times Y\times \Bbb P^n \mid x\neq y, z\in [x,y]\}$$ where $$[x,y]$$ denotes the line passing through $$x$$ and $$y$$. Then $$J(X,Y)$$ is the projection of the closure of $$J'(X,Y)$$ to the final factor of $$\Bbb P^n$$. On the other hand, we can consider the projection of $$\overline{J'(X,Y)}$$ to $$X\times Y$$. The fibers of this projection are lines, thus irreducible of dimension 1. As a closed map with irreducible target and irreducible fibers must have irreducible source, we see that $$J(X,Y)$$ is irreducible and of dimension $$\dim X + \dim Y + 1$$. In our case, $$Y$$ is a point which has dimension zero, so $$\dim J(X,P) = \dim X + 1$$.

Subtracting 1 via the projective dimension theorem, we get the desired result in each case.

• Why does it hold that $J(X,Y)$ is the projection of the closure of $J'(X,Y)$. How can we be sure elements on the edge get mapped to $J(X,Y)$. And also why is the second projection a closed map? Don't we need that the image is compact for that? Nov 22, 2019 at 14:37
• $X$ and $Y$ are projective, so $X\times Y$ is projective, thus proper, so the structure morphism of $X\times Y$ is universally closed. As the projection on to the $\Bbb P^n$ factor is the base change of the structure morphism of $X\times Y$ by $\Bbb P^n$, it must also be closed. The closure only matters for the case that $X\cap Y \neq \emptyset$, and it deals with getting the correct tangent lines at the intersection points. This is (more or less) a definition, see page 9 of chapter 0 of 3624 & All That, for instance. Nov 22, 2019 at 18:27
• Why is $J(X,P) = X$ in case (3)? Nov 24, 2019 at 8:26
• $X\setminus P \subset J(X,P)$ and $J(X,P)$ is closed, so $X=\overline{X\setminus P} \subset J(X,P)$. Conversely, $\pi_{\Bbb P^n}(J'(X,P))\subset X$, and $X$ is closed, so $J(X,P)=\overline{\pi_{\Bbb P^n}(J'(X,P))}\subset X$. Nov 24, 2019 at 9:41