TITLE: The class of the diagonal in the symmetric product of a smooth curve
QUESTION [9 upvotes]: Let $C$ be a smooth curve of genus $g$, and let us consider its $d$-th symmetric product $\textrm{Sym}^d(C)$ and its Jacobian $J(C)$. Fixing a point $p_0 \in C,$ there are two maps $$u_d\colon C_d \to J(C), \quad u_d(D):=\mathcal{O}_C(D-dp_0),$$ $$i_{d-1} \colon \textrm{Sym}^{d-1}(C) \to \textrm{Sym}^d(C), \quad  i_{d-1}(D):=D+p_0.$$ Then the following two divisor classes in the Néron-Severi group of  $\textrm{Sym}^d(C)$ are independent on $p_0$: $$\theta:=u_d^*(\Theta), \quad x:=i_{d-1} (\textrm{Sym}^{d-1}(C)),$$
where $\Theta$ is the theta divisor in the Néron-Severi group of $J(C)$. It is well-known that, if $C$ is with general moduli, (i.e., it defines a very general point in the moduli space $\mathcal{M}_g$), then the following holds:

(1) the Néron-Severi group of $\textrm{Sym}^d(C)$ is generated by $\theta$ and $x$;
(2) the class $\delta$ of the diagonal of $\textrm{Sym}^d(C)$ is given by $$\delta = 2((d+g-1)x - \theta).$$

Now, it is not difficult to produce counterexamples (for instances, curves with non-trivial correspondences) showing that (1) does not hold if $C$ is not with general moduli. So let me ask the following

Question. Is (2) true for every smooth curve of genus $g$?

A. Kouvidakis, in Lemma 7 of his paper Divisors on symmetric product of curves (Trans. Amer. Math. Soc. 337), claims that (2) is due to MacDonald, and refers to Arbarello-Cornalba-Griffiths-Harris'book Geometry of Algebraic Curves I, Proposition 5.1 p. 358  for the proof.
Actually, the (much more general) result quoted in the book is apparently stated without assumptions on $C$, but the proof is based on a Lemma whose statement begins with "Let $C$ be a curve with general moduli", and whose content is essentially (1). So I am a bit confused.
My impression is that the answer to the question should be yes and that some kind of specialization argument should do the job. Any reference to the relevant literature will be particularly appreciated. I am primarily interested in the case $d=g=2$, but I would like to know the answer to the general question, too.

REPLY [5 votes]: Let me try to give a slightly different approach, which hopefully complements Jason's answer. Basically, let us compute as much as we can about the natural line bundles on the symmetric power.
First of all, the line bundle $O(-\Theta)$ on $\mathrm{Pic}^{g-1}(C)$ coincides with the line bundle $\det \mathrm{R\Gamma}$ (whose fiber over $\ell\in\mathrm{Pic}^{g-1}(C)$ is $\det\mathrm{R\Gamma}(C,\ell)$). I prefer to work with degree $g-1$ bundles on $C$ here, so that the $\Theta$ divisor is defined canonically, not just up to a shift. Your formula involves the pullback of $\Theta$ (or, equivalently, of this line bundle) under the map $$u_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(D+(-d+g-1)p_0),$$ but I would prefer to work with the map $$v_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(-D+(d+g-1)p_0),$$ which is algebraically equivalent to $-u_d$ (and $\Theta$ is invariant under the inversion anyway).
Now in these terms, we can state the following identity:
$$v_d^*(\det\mathrm{R\Gamma}^{\otimes 2})\simeq O(\Delta-2(d+g-1)H),$$ where following Jason's answer, $\Delta$ is the diagonal in $\mathrm{Sym}^d C$, and $H$ is the reduced divisor whose complement is $\mathrm{Sym}^d(C-{p_0})$. (Note that this is an isomorphism of line bundles, not just an algebraic equivalence.) Indeed, the left-hand side is the line bundle whose fiber over $D\in\mathrm{Sym}^dC$ is $$\det\mathrm{R\Gamma}(C,O(-D+(d+g-1)p_0))^{\otimes 2}\simeq\det(O/O(-D))^{\otimes-2}\otimes (O(-D))_{p_0}^{\otimes 2(d+g-1)}.$$ It now remains to notice that the line bundle whose fiber over $D$ is $\det(O/O(-D))^{\otimes 2}$ is isomorphic to $O(-\Delta)$, while the line bundle whose fiber is $O(-D)_{p_0}$ is isomorphic to $O(-H)$.