# Why is “Severi variety” a variety?

According to the book "moduli of curves" by Harris. He defined $$U_{d,g}$$ as the variety in $$\mathbb{P}^N$$ (coefficient space) of (irreducible reduced) nodal curves with genus $$g$$ and degree $$d$$. We first ignore the other difficult issues about the irreducibility. On page 30 theorem 1.49, he claimed:

$$U_{d,g}$$ is smooth of dimension $$3d+g-1$$.

First, let's simplify the argument of one node. He proved like this: Take $$\Sigma=\{(C,p)|p \text{ is singular on }C\}\subseteq \mathbb{P}^N\times\mathbb{P}^2$$. The proof is a bit mixed with differential geometry: write down the polynomials $$F(X,Y)=0\\F_x(X,Y)=0\\F_y(X,Y)=0$$

and apply constant rank theorem. These are all fine, and we can show that $$\Sigma$$ is smooth of dimension $$N-1$$ at those curves with only one nodes.

Now, he takes projection $$\pi:\Sigma\to\mathbb{P}^n$$. In order to the desired image we must take away those diagonals in order to forbid repeating nodes. Then the map is an immersion at each $$(C,p)$$ and the image is exactly $$U_{d,g}$$. Here comes the problem. We can only be sure that the image is a constructible set. Even though $$\pi$$ is an injective immersion, we can not conclude that the image is smooth (manifold) because of this.

Do I miss something? Actually, I tried to prove that this map is an embedding. But the map can not be proved to be a homeomorphism.

• I don’t have the book to reference, but if $\Sigma$ is constructed as a closed subset of a product of projective spaces then it’s proper, and thus it’s image under any morphism is closed, not just constructible. – Devlin Mallory Nov 12 '19 at 0:10
• @DevlinMallory First, the condition "only one node" can not be obtained by a closed relation. Also for more nodes we need to carve out the diagonals. The variety is extremely ill-behaving at the boundaries. I dont think it will still be smooth if consider the closed set. – Upc Nov 12 '19 at 0:35
• Having exactly one node is neither open nor closed. But having only nodal singularities is open. – Phil Tosteson Nov 12 '19 at 4:18
• I think this is a matter of definition. I don’t have the book on hand, but I think of the Severi variety as the closure of $U_{d,g}$. – Samir Canning Nov 12 '19 at 4:38
• @PhilTosteson Yes! But the the problem is we want to project it onto $\mathbb{P}^N$ so it will trouble some if the condition is not closed. – Upc Nov 12 '19 at 12:56