I am reading Gathmann's notes on algebraic geometry and am doubting his proof that "every smooth cubic surface in $\mathbb{P}^{3}_{\mathbb{C}}$ has 27 lines."

We work in the manifold topology. Let $U\subset \mathbb{P}^{19}$ be the parameter space of smooth cubic surfaces in $\mathbb{P}^{3}$. We have shown that the line counting function on $U$ is locally constant and wish to show that $U$ is connected. Hence Gathmann argues:

We know that $U$ is the complement of a proper Zariski closed subset in $\mathbb{P}^{19}$. But as such a closed subset has complex codimension at least 1 and hence real codimension at least 2, taking this subset away from the smooth and connected space $\mathbb{P}^{19}$ leaves us again with a connected space.

My doubt: $\mathbb{P}^{19}\setminus U$ is a projective variety that may have singular points. So in what sense do we talk about codimensions given that $\mathbb{P}^{19}\setminus U$ is not necessarily a smooth submanifold? How can Gathmann's argument be made precise?

Assuming it is a submanifold, his argument holds by "if $N\subset M$ is a codimension 2 submanifold of a connected manifold, then $M\setminus N$ is connected." But I can't imagine how to show it's a submanifold. It's also possible that he's talking about algebraic codimension. But then we'd be leaving the realm of manifolds, so the linked lemma can't be used.

I suspect this is a problem of complex geometry, of which I know nothing. Any sources or explanations are appreciated!

  • $\begingroup$ Mod details, the zero set of a single polynomial is locally a branched cover over a polydisk of codimension one, which is enough to deduce the complement is connected. Commenting to avoid providing details, and hoping this allays your doubt until someone posts a proper answer. $\endgroup$ Commented May 13 at 0:28
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    $\begingroup$ Although this is not actually needed, let me emphasise that an open subset of a smooth variety is always smooth. $\endgroup$ Commented May 13 at 21:04
  • $\begingroup$ @AndrewD.Hwang Could you point me to some theorems so I can fill in the details? $\endgroup$
    – user816709
    Commented May 16 at 13:23
  • $\begingroup$ If memory serves, Principles of Algebraic Geometry by Griffiths and Harris discusses this sort of issue (e.g., under the "Weierstrass preparation theorem," admittedly much more than needed here). Loosely, echoing Lazzaro's and KReiser's points, the smooth locus is a complex hypersurface (non-separating because real codimension two), and the singular locus is "even lower-dimensional" (so the entire locus is non-separating). $\endgroup$ Commented May 16 at 13:58

1 Answer 1


The result is true, ignoring any conditions on $Z=\Bbb P^{19}\setminus U$ being a manifold. Here is the outline of a proof.

Let $k=\Bbb R$ or $\Bbb C$. Then for any affine $k$-variety $X$ and closed subvariety $Y\subset X$, there exists a triangulation of the pair $(Y,X)$, that is, a simplicial complex $\Delta_X$ with a simplicial subcomplex $\Delta_Y\subset\Delta_X$ (ref, originally due to Hironaka in 1974 - there are more modern references like here which improve this slightly).

On any of the standard affine opens covering $\Bbb P^{19}$, we may apply this result to get a pair of triangulations of $(Z\cap\Bbb A^{19}) \subset \Bbb A^{19}$, and all of the simplices in the triangulation of $Z\cap\Bbb A^{19}$ are of codimension at least 2 (i.e. in our case they're all 17-simplices or smaller - this is what codimension two means). Now we can apply subdivision techniques to get a tubular neighborhood, and then use the standard arguments once we have this tubular neighborhood and this bound on the dimension of the simplices. This will show that $U\cap\Bbb A^{19}$ is connected, and then by a topological argument you can get that $U$ is itself connected.


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