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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Nov 22, 2019 at 14:37
  • $\begingroup$ $X$ 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. $\endgroup$
    – KReiser
    Nov 22, 2019 at 18:27
  • $\begingroup$ Why is $J(X,P) = X$ in case (3)? $\endgroup$
    – Kamil
    Nov 24, 2019 at 8:26
  • $\begingroup$ $X\setminus P \subset J(X,P)$ 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 $J(X,P)=\overline{\pi_{\Bbb P^n}(J'(X,P))}\subset X$. $\endgroup$
    – KReiser
    Nov 24, 2019 at 9:41

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