# Is every smooth projective variety contained in a chain of smooth projective varieties of increasing dimension?

Let $$X ⊆ \mathbb{P}^n$$ be a smooth projective variety (over $$\mathbb{C}$$). I think we can find a chain of irreducible varieties $$X = X_0 ⊆ X_1 ⊆ X_2 ⊆ \cdots ⊆ X_k = \mathbb{P}^n$$ whose dimension increases by one at every step by writing $$X = \mathcal{V}(f_1, \dots, f_n)$$ and dropping some of the $$f_i$$ until the dimension of the irreducible component containing $$X$$ increases, and then proceeding by induction on $$k = \operatorname{codim}(X)$$.

Is it possible to a chain where all of the $$X_i$$ are smooth?

• A while ago on math.SE, I showed that every smooth complete intersection is contained in such a chain; see math.stackexchange.com/questions/17157 . Aug 10, 2022 at 12:33

Suppose that $$\operatorname{dim}(X)>1$$ and that such a chain exists. Since $$\operatorname{Pic}(\mathbf{P}^n)\simeq \mathbf{Z}$$, the variety $$X_{k-1}$$ is an ample divisor in $$\mathbf{P}^n$$, and hence by the Lefschetz hyperplane theorem we have $$\operatorname{Pic}(X_{k-1})\simeq \mathbf{Z}$$. So $$X_{k-2}$$ is an ample divisor on $$X_{k-1}$$, and again its Picard group is $$\mathbf{Z}$$. By induction, we obtain that each $$X_{i-1}$$ is an ample divisor on $$X_{i}$$, and then by hyperplane Lefschetz for $$\pi_1$$ we obtain that $$\pi_1(X)\simeq \pi_1(\mathbf{P}^n)$$ is the trivial group. So to conclude, an abelian variety of dimension at least two embedded in $$\mathbf{P}^n$$ gives a counterexample. (N.B. There exist abelian surfaces in $$\mathbf{P}^4$$, constructed by Horrocks and Mumford).

Edit. Of course this contradicts Sasha's answer posted roughly at the same time. I am puzzled as to where the mistake is.

• Indeed, my argument is incorrect (the explanation is given in the edit to my answer). Aug 10, 2022 at 9:15
• I understand that this is outside scope but what about simply connected $X$? Aug 10, 2022 at 9:21
• @მამუკა ჯიბლაძე: No difference, there are smooth simply connected surfaces in $\Bbb{P}^4$ which are not complete intersections.
– abx
Aug 10, 2022 at 9:36

There is already a complete and correct answer. I am just writing another answer since I am having trouble finding the old answer mentioned in my comment above. Update. Thanks to user @MinseonShin for finding the old answer: A Bertini-type result for hypersurfaces containing a subvariety

The question in the original post is a special case of the following question (which I believe was asked in a previous post).

Question. Is every smooth proper closed subvariety $$X$$ of a smooth projective variety $$Y$$ contained in a smooth hypersurface $$Z$$ in $$Y$$?

Proposition. If there exists a hypersurface $$Z$$ in $$Y$$ that contains $$X$$ and is smooth at every point of $$X$$, then the total Chern class $$c(N_{X/Y})$$ equals $$(1+[Z]|_X)\cup \alpha$$ for the total Chern class $$\alpha$$ of a locally free sheaf of rank $$\text{dim}(Y)-(1+\text{dim}(X))$$.

Proof. If there is such a hypersurface $$Z$$, then there is a short exact sequence of locally free sheaves on $$X$$, $$0 \to N_{X/Z} \to N_{X/Y} \to N_{Z/Y}|_X \to 0.$$ Then, by the Whitney sum formula, the total Chern class $$c(N_{X/Y})$$ equals $$(1+[Z]|_X)\cup c(N_{X/Z})$$. QED

This fails in many cases. For instance, for the embedding of a $$2$$-plane $$X$$ in a smooth quadric hypersurface $$Y$$ in $$\mathbb{P}^5$$ (a Schubert subvariety of the Grassmannian $$\text{Gr}(2,4)$$, in other words), this gives, $$1+H+H^2 = (1+mH)(1+nH) = 1+(m+n)H +mnH^2,$$ for integers $$m$$ and $$n$$. Clearly this has no solution in integers, thus there is no hypersurface $$Z$$ in $$Y$$ that contains $$X$$ and is smooth at every point of $$X$$. (Note, the restriction map on Picard groups from $$Y$$ to $$X$$ is an isomorphism in this example.)

• I wonder if you were thinking of this question or this question? Aug 11, 2022 at 22:40
• @MinseonShin. Yes, indeed! My answer above is the same observation as my comment to the MO post that you linked: mathoverflow.net/questions/226230/…. Aug 12, 2022 at 11:36

EDIT. The argument below is incorrect. Indeed, for smoothness of a divisor along $$X$$ one needs the zero locus of a section of $$I_X/I_X^2(mH)$$ to be empty (not just smooth), and this typically is impossible when the rank of this bundle (equal to $$\mathrm{codim}(X)$$) is less or equal than $$\dim(X)$$.

Yes, this follows from Bertini's Theorem.

Indeed, let us check that if $$X \subset Y$$ is an embedding of smooth projective varieties then there is a smooth divisor $$D \subset Y$$ containing $$X$$ (then we will proceed by induction). Indeed, let $$I_X$$ be the ideal of $$X$$ and let $$H$$ be an ample divisor class on $$Y$$. Then for $$m \gg 0$$ the sheaf $$I_X(mH)$$ is globally generated, hence by Bertini's Theorem on Y a general section of $$I_X(mH)$$ is a divisor smooth away from $$X$$.

On the other hand, for $$m \gg 0$$ we have $$H^1(Y,I^2_X(mH)) = 0$$, hence the morphism $$H^0(Y,I_X(mH)) \to H^0(X,I_X/I_X^2(mH))$$ is surjective, and the twisted conormal bundle $$I_X/I_X^2(mH)$$ is globally generated. Therefore, for a general its section (hence for general section of $$I_X(mH)$$) the zero locus on $$X$$ is also smooth (now by Bertini's Theorem on $$X$$), hence it is smooth everywhere.

Now, finally, we apply an inductive argument. First, we consider the embedding $$X \subset X_k := \mathbb{P}^n$$ and construct a smooth divisor $$X_{k-1} \subset X_k$$ containing $$X$$. Next we consider the embedding $$X \subset X_{k-1}$$ and construct a smooth divisor $$X_{k-2} \subset X_{k-1}$$ containing $$X$$. Iterating this procedure we construct the required chain.

• I don't think that any smooth surface in $\mathbb{P}^4$ is contained in a smooth hypersurface, otherwise by Lefschetz theorem it would be a complete intersection?
– abx
Aug 10, 2022 at 8:37
• @abx: You are absolutely right, my argument is incorrect (see the edit). Aug 10, 2022 at 9:14
• Nevertheless, this is an interesting error (and instructive to me as I failed to detect it and probably would have if I had been refereeing a paper). Aug 10, 2022 at 12:02
• This has come up before on MO (I can find a link). One obstruction to existence of a smooth hypersurface of degree $m$ containing $X$ is the image of the top Chern class of the normal bundle of $X$ inside the quotient of the Chow ring of $X$ by the principal ideal generated by $mc_1(\mathcal{O}(1))$. Aug 10, 2022 at 14:55