Dimension of secant variety

Question

Given $X$ a projective smooth curve of genus $g$, let's $Sec_{d/2}(X)$ be the secant variety defined as the union of $Span(D)$, with $D$ that varies among all the divisor of degree $d/2$. Is it true that $dim(Sec_{d/2})$ is less or equal to $d$? I hope so, because I need this to prove that an opportune moduli space is not empty.

Answer 1

Let's write $\delta$ for $d/2$.

The space of effective divisors of degree $\delta$ on $X$ is just the symmetric product $\Sigma^\delta X$, that is, the quotient of $X^\delta$ by the symmetric group $S_\delta$. This has dimension $\delta$.

First let's assume that the general point of $\Sigma^\delta X$ corresponds to $\delta$ projectively independent points, so their span has dimension $\delta-1$. Then the same is true for a dense open subset $\Sigma^\delta_0 X$.

Let's form the incidence variety

$$ I = \{ (D,x) \mid x \in \operatorname{Span} D \} \subset \Sigma^\delta X \times \mathbb P^{g+d-1}.$$

Projection to $\Sigma^\delta X$ makes it clear that this has dimension $2 \delta-1$. On the other hand it surjects onto the secant variety $\operatorname{Sec}_\delta (X)$.

Now if the general point of $\Sigma^\delta X$ is a divisor whose span has dimension $k<\delta-1$, the same argument shows that the secant variety has dimension at most $\delta+k$, which is even smaller.

Answer 2

@ArthurStuart A general formula for the dimension of a secant variety is the following: Let $X \subset \mathbb{P}^N$ be an irreducible, reduced variety of dimension $n$ over $\mathbb{C}$. Then the $expected$ $dimension$ of $Sec_s(X)$, for any $s \geq 1$, is $$ expdim Sec_s(X) = min\{ sn + s - 1, N \}. $$ It can be prove by using similar arguments as used in the answer by Asal Beag Dubh. The expected dimension is also the maximum possible dimension of the secant variety $Sec_s(X)$.