I've been stuck for quite a while on what is probably a trivial problem. Let $X\subset\mathbb{P}^n$ be a smooth projective curve, and let 
$$\mathcal{I}=\{(p,q,r):p,q\in X,p\neq q,r\in\overline{pq}\}$$ 
(where $\overline{pq}$ is the line that joins $p$ and $q$) and let 
$$\mathcal{J}=\{(p,r):p\in X,r\mbox{ lies on the tangent line to }X\mbox{ at }p\}.$$
It is easy to see that $\mathcal{I}$ is a complex 3-manifold and $\mathcal{J}$ is a complex 2-manifold. Let $\alpha:\mathcal{I}\to\mathbb{P}^n$ so that $(p,q,r)\mapsto r$, and let $\beta:\mathcal{J}\to\mathbb{P}^n$ so that $(p,r)\mapsto r$.


Why is it that if $n\geq 4$ then there is a point in $\mathbb{P}^n$ that is not in the image of either function? I can see that the image of $\alpha$ has complex dimension at most 3 and the image of $\beta$ at most 2, but I can't see why their images can't cover all $\mathbb{P}^n$.


For reference, this was taken from Algebraic Curves and Surfaces by Rick Miranda, page 101.


Thanks.



Projecting projective curves
$\alpha(\mathcal I)$ is the *secant variety* of $X$. I think it is enough if you understand this one, the other is similar. Of course the actual **proof** that $\alpha(\mathcal I)\subsetneq \mathbb P^n$ is exactly the dimension count you're mentioning. So, I assume you're looking for intuition and/or a heuristic explanation of what's happening.


I suspect your problem is really that you can't see in dimension $4$. Don't take this as a criticism, I can't **see** in dimension $4$ either and whoever claims they can, they don't really mean it that way.


$r\in\alpha(\mathcal I)$ means that $r$ lies on a secant line of $X$, in other words, there is a line through $r$ that intersects $X$ in two points, say $p,q\in X$. Now let's fix $p$ for a moment and turn around the observation: which points in $\mathbb P^n$ will lie on a secant line through $p$? Well, the points of the secant lies. Duh. So, you can look at all the lines through $p$ and observe that the ones that are contained in $\alpha(\mathcal I)$ are the ones that connect other points of $X$ to $p$. Let's call the union of these lines $A\_p$, that is, let $\mathcal I\_p=\{(a,b,c)\in\mathcal I|a=p\}$ and let $A\_p=\alpha(\mathcal I\_p)$


Let's take a (random) hyperplane $H\simeq \mathbb P^{n-1}$. It is enough to prove that $H\not\subseteq \alpha(\mathcal I)$. What is $H\cap\alpha(\mathcal I)$? First of all 
$$\alpha(\mathcal I)=\bigcup\_{p\in X} A\_p.$$
Second of all, $H\cap A\_p$ is the projection of $X$ to $H$ and hence it is an *algebraic* curve. As $p$ runs through the points of $X$ we get a $1$-parameter family of curves, but that cannot fill a space of dimension $n-1\geq 3$.


Of course, this is the same proof as counting the dimension of the secant variety, but perhaps something that you can imagine better.


