# Local versus global embedding dimension

Fix a complex projective scheme $$X$$ and a closed point $$x\in X$$.
Let $$d_x$$ denote the dimension of the Zariski tangent space at $$x$$.
This is the local embedding dimension of $$X$$ at $$x$$ -- the minimal dimension of a smooth scheme containing an open neighbourhood of $$x$$.

In a paper I blithely asserted that $$d(X):=\max_{x\in X}d_x$$ is the global embedding dimension -- the minimal dimension of a smooth scheme containing $$X$$.

I had assumed we could embed $$X\subset\mathbb P^N$$ and then take an intersection of $$N-d(X)$$ generic sufficiently positive hypersurfaces containing $$X$$. But since a troublesome referee has quite unreasonably asked me for a proof (I'm joking) I checked more carefully and saw this construction does not work everywhere at once (even when $$N-d(X)=1$$ and $$X$$ is smooth!).

Can anyone suggest another construction, maybe by suitable projections, or a reference, or...?

• For a reduced $k$-scheme $X$ of pure dimension $n$ and local embedding dimension $n+1$, one obstruction to finding a global embedding into a smooth $(n+1)$-fold with invertible conormal sheaf $\mathcal{I}$ is the connecting map $$H^0(X,\mathcal{Ext}^1_{\mathcal{O}_X}(\Omega_{X/k},\mathcal{I})) \to H^2(X,\mathcal{Hom}_{\mathcal{O}_X}(\Omega_{X/k},\mathcal{I})).$$ If there exists an embedding, the kernel has a section that everywhere locally generates the domain $\mathcal{O}_X$-module (which is locally principal). Jan 28 '20 at 9:52
• Wonderful, thanks Jason. At first I thought we could kill this obstruction by choosing $\mathcal I$ to be a sufficiently positive line bundle, so the $H^2$ vanishes. But then the global sections of $\mathcal Ext^1(\Omega_X,\mathcal I)$ would have zeros and so not generate (assuming the singular locus has dimension $\ge1$). So I think you've probably give a method to debunk my claim. I.e. using this one should be able to find an example of a variety with at worst hypersurface singularities which is nonetheless not a hypersurface (in something smooth)? Jan 28 '20 at 13:31
• I am trying to work this out explicitly for a threefold obtained from a $\mathbb{P}^1$-bundle over a surface where a bisection is glued to itself. There is a class in $H^2$ of the bisection that I need to compute . . . Jan 28 '20 at 14:42
• @Jason Starr: Your formula above (from the local-global-ext-ss) and the argument looks like it is part of a general theory of embeddings of algebraic varieties into each other. What would be a good article or book to look for, where this is explained in more detail? (I tried with Google already, but found nothing especially pertaining). Jan 28 '20 at 14:49
• @JürgenBöhm: Jason is describing the Kodaira-Spencer class that classifies first order thickenings of $X$ by a sheaf $\mathcal I$ (that becomes the square-zero ideal of $X$ inside the thickening). So he's describing the (obstruction to the existence of the) first order part of the ambient space I was after. "Kodaira-Spencer class" is probably the thing to google. Jan 28 '20 at 15:08

It seems there is a counterexample. This is based on Jason Starr's suggestion in the comments.

If we have a surface $$S$$ with two smooth disjoint curves $$C_1$$ and $$C_2$$, which are isomorphic, and let $$X$$ be obtained by gluing $$C_1$$ and $$C_2$$ along that isomorphism $$i: C_1\to C_2$$, then $$X$$ is projective if there is an ample line bundle on $$X$$ whose restrictions to $$C_1$$ and $$C_2$$ are equal (under $$i$$).

$$X$$ has singularities locally isomorphic to a nodal curve cross a smooth curve, thus has local embedding dimension $$3$$. Can $$X$$ be embedded as a hypersurface in a smooth $$3$$-fold? If so, then by (part of) Jason Starr's obstruction, the sheaf

$$\mathcal{Ext}^1_{\mathcal O_X} (\Omega_{X/k}, \mathcal I)= \mathcal{Ext}^1_{\mathcal O_X} (\Omega_{X/k}, \mathcal O_X) \otimes \mathcal I$$ must be globally generated, where $$\mathcal I$$ is the conormal line bundle. This sheaf is clearly supported on the glued curve $$C$$, and we can calculate that it is isomorphic to $$\mathcal I$$ tensored with the normal bundle of $$C_1$$ and the normal bundle of $$C_2$$ there. (It suffices to work, carefully, locally in $$k[x,y]/xy$$, where $$\Omega$$ is generated by $$dx$$ and $$dy$$ with relation $$xdy+ ydx=0$$ and the generator of the $$\mathcal{Ext}^1$$ is precisely the linear map that sends $$xdy+ydx$$ to $$1$$, which the automorphism group acts on the same way it acts on the tensor rpoduct of the normal bundles.)

So for this sheaf to have a nonvanishing section, the conormal bundle $$\mathcal I$$ of $$X$$, restricted to $$C$$ must be isomorphic to the tensor product of the conormal bundle of $$C_1$$ to the conormal bundle of $$C_2$$.

So here's what we're going to do. We will take $$E_1$$ and $$E_2$$ two distinct, but isomorphic, elliptic curves in $$\mathbb P^1$$. In fact, we will take them to be two isomorphic curves appearing in the Dwork family, so their intersection points will be $$3$$-torsion. We will blow up all $$9$$ intersection points, plus two points $$P_1, Q_1$$ on $$E_1$$ and two points $$P_2, Q_2$$ on $$E_2$$. We choose $$P_1, Q_1, P_2, Q_2$$ very general, subject to the condition that $$i(P_1) + 2i(Q_1) = P_2 + 2 Q_2$$ in the group law on $$E_2$$.

To make our ample class, we'll just take a sufficiently high multiple of the hyperplane class, minus the sum of the exceptional divisors at all $$9$$ intersection points, minus the exceptional divisors over $$P_1$$ and $$P_2$$, minus twice the exceptional divisors over $$Q_1$$ and $$Q_2$$. Because of our assumption on the group law, this restricts to the same line bundle on $$E_1$$ and $$E_2$$, as each exceptional divisor corresponds to that point in the Picard group.

However, the Picard class of the tensor product of the two conormal bundles on $$E_2$$ will be some multiple of the hyperplane class, plus twice the sum of all the $$3$$-torsion points, plus $$i(P_1) + i(Q_1) + P_2 + Q_2$$. If this class comes from a global line bundle, then it must come from a sum of hyperplane classes and exceptional divisors, which means (projecting to Pic) it must come from a sum of $$3$$-torsion points, $$P_2$$ and $$Q_2$$. The exceptional divisors over $$P_1$$ and $$Q_1$$ don't contribute because they don't intersect $$E_2$$. Thus, it can only happen if we have some relation that $$i(P_1) + i(Q_2) = a P_2 + b Q_2$$ for $$a,b\in \mathbb Z$$, up to $$3$$-torsion. But there are countably many such relations, and none of them is forced by our condition on $$P_1,P_2, Q_1,Q_2$$, so none of them will hold for our very general choice.

• Magnificent, thanks Will and Jason. Can I summarise the overall method as follows? Make $X$ by gluing a smooth surface $S$ along two disjoint curves $C_1\cong C_2$. If $X$ lived in a smooth 3-fold $Y$, we know by elementary geometry that the normal bundle $N_{X/Y}$ must restrict on $C_1\cong C_2$ to the line bundle $L:=N_{C_1/S}\otimes N_{C_2/S}$. But you found an example where $L$ is not the restriction of any line bundle on $X$ (though of course it would be on its normalisation $\overline{X}=S$). Jan 28 '20 at 21:18
• @RichardThomas Yes, with the caveat that the line bundle that is that restriction on the normalization is not the tensor product of $N_{C_1/S}$ and $N_{C_2/S}$, but rather the line bundle that is $N_{C_1/S}$ on $C_1$ and $N_{C_2/S}$ on $C_2$. Jan 28 '20 at 22:09

I wonder if the following example, adapted from section 18 of Kollár's Links of complex analytic singularities, would also work: Let $$E_i := V(x_i^3 + y_i^3 + z_i^3) \subset \mathbb{P}_i^2$$ for $$i = 1, 2$$, let $$\tau: E_1 \to E_2$$ be an isomorphism corresponding to a translation of the elliptic curve $$V(x^3 + y^3 + z^3)$$ and use it to glue the 2 copies of $$\mathbb{P}^2$$, to get $$X(\tau) := \mathbb{P}_1^2 \cup_{\tau} \mathbb{P}_2^2$$. Let $$E \subset X$$ denote the common image of $$E_1, E_2$$.

Then using $$\mathcal{O}_{\mathbb{P}_1^2}(1)|_{E}$$ as a basepoint, we can make the identification $$\mathrm{Pic}^3(E) \simeq \mathrm{Pic}^0(E) \simeq E$$. Under this identification $$\mathcal{O}_{\mathbb{P}_2^2}(1)|_{E} = \tau^*(\mathcal{O}_{\mathbb{P}_1^2}(1)|_{E} = \tau \in \mathrm{Pic}^3(E)$$, and more generally $$\mathcal{O}_{\mathbb{P}_2^2}(d)|_{E} = \tau^d \in \mathrm{Pic}^{3d}(E)$$ for $$d \in \mathbb{Z}$$. So $$X(\tau)$$ is projective if and only if $$\tau$$ is torsion, in which case $$\mathcal{O}_{\mathbb{P}_1^2}(d), \mathcal{O}_{\mathbb{P}_2^2}(d)$$ glue to form a line bundle on $$X(\tau)$$ if and only if $$\tau^d = 1$$.

On the other hand, $$N_{E \subset \mathbb{P}_i^2}^\vee = \mathcal{O}_{\mathbb{P}_i^2}(-3)|_E$$ for $$i = 1, 2$$ so that $$N_{E \subset \mathbb{P}_1^2}^\vee \otimes N_{E \subset \mathbb{P}_2^2}^\vee = \mathcal{O}_{\mathbb{P}_1^2}(-3)|_E \otimes \mathcal{O}_{\mathbb{P}_2^2}(-3)|_E$$, corresponding to $$\tau^{-3}\in \mathrm{Pic}^{-18}(E)\simeq E$$. Hence $$X(\tau)$$ is an snc divisor if and only if $$\tau^3=1$$.

• Same idea but simpler, I like it, thanks. Jun 27 '20 at 21:14