In algebraic geometry it is customary to identify locally free sheaves on say a scheme $X$ and vector bundles over the same scheme.

So say I have a nice scheme $X$ (you can assume it to be a variety, over an algebraically closed field $k$), and an exact sequence of locally free sheaves $0\to V'\to V \to V''\to 0$, this exact sequence is exact on the level of stalks, almost by definition, if $x$ is a (closed) point of $X$, $0\to j^*V'\to j^*V \to j^*V''\to 0$ is also exact because $V''$ is locally free, hence flat (here $j$ is the closed immersion $x\to X$).

My question comes to how to practically apply this in the case of normal cones to regular immersion.

Let's look at a simple example, $Z=\{xy=0\}$ regularly embedded in $X=\mathbb{A}^2$. Concerning $Z$ we have the cotangent sheaf. It is not locally free but the normal cone of $Z$ is locally free.

At the origin, we have a geometric tangent cone given by the union the two axis lines. How is this related to the fiber of the tangent sheaf at 0, or to the fiber of the (co)normal cone at 0?

We have an exact sequence $0\to N_{Z/X}^{*}\to j^* \Omega^1_X \to \Omega^1_Z\to 0$. Away form the origin, it is easy to understand what's going on, at a point of the form say $(a,0)$ the stalk of the cotangent sheaf is given by $f(x)dx$, the fiber is simply $f(a)dx$, the conormal sheaf has fiber $k.dy$ (where k is the base field or the residue field of the point).

At the origin it becomes blurry to me, the normal bundle should have a line as a fiber, but which line? The stalk of the tangent sheaf at the point consists of forms of the form $f(x)dx+g(y)dy$ and the fiber is a vector space of dimension 2. Which is also the case for the fiber of $j^{*}\Omega^1_X$, but as the sheafs are not (locally) free I cannot deduce that the fiber of the normal bundle is zero, which is good as it should be a line.

Most importantly, how does this relate to the tangent cone, which is $xy=0$. What is the normal cone? In particular how do i deduce the normal cone from the tangent cone (and not the (co)tangent sheaf and the exact sequence). I was hoping that the normal cone should be related (orthgonal) to something like the vector space generated by the tangent cone, but obviously this is not the case.

I hope my question is somewhat clear, I can clarify it if needed.

Addendum: Algebraically I seem to understand what's going on, it is indeed easy to write down an explicit version of the exact sequence mentionnenned but it is absolutely not clear to me how to interpret the arrow $ N_{Z/X}^* \to\Omega^1_X$, in particular if I understand things correctly, over $(0,0)$ the map on the fiber will fail to be injective (but will be on the stalks of course), as $\Omega^1_Z$ is not free over $\mathcal{O}_{Z,(0,0)}$ this is not problematic but it makes the geometric interpretention of the (co)normal bundle fiber at that point extremely tedious (to me at least). In particular its fiber is not a subthing of $j_{(0,0)}^*\Omega^1_X$.

Is there a way to have a geometric interpretation of that normal line at the origin. By that I mean, on a picture, how would you draw that line and why?


1 Answer 1


Your analysis is essentially correct.

Let me try to explain how to have a geometric interpretation nonetheless.

Let us fix some notations, let $R=k[x,y]/(xy)$, and $R_0=R_{(x,y)}$, and $k_0=R_0/(x,y)$ (which is isomorphic to $k$ of course).

You can do two things that are essentially equivalent. You can look at what happens away from $(0,0)$, as you said everything is clear here you have an exact sequence (here $t=(x-y)$ as the complement of the origin in $Z$ is principal this will make things simpler). $$0\to R_t [xy]\to R_t[dx]\oplus R_t [dy]\to R_t[dx]\oplus R_t[dy]/(x[dy]+y[dx])$$ Notice that $R_t[xy]$ is a free module generated by $[xy]$ (I write it like that to remember that is is a basis and not confuse it with the action of $x$ and $y$ coming form the $R$-module structure).

The map on the left is of course $[xy]\mapsto y[dx]+y[dx]$.

You have the same story replacing $R_t$ with $R$ everything is also free, but the fiber of the exact sequence at $(0,0)$ degenerates (by failure of flatness indeed) and you have an exact sequence $$k_0 [xy]\to k_0[dx]\oplus k_0 [dy]\to k_0[dx]\oplus k_0[dy]/(0[dy]+0[dx])\to 0$$ the map on the left is... zero and the second map is an isomorphism as $0[dx]+0[dy]=0$ generates the null-vector space.

So were you to draw the normal line at the origin in your original picture you would draw... a point. It is simply not visible in the original plane.

However, if you look at everything in $\mathbb{A}^4$, everything becomes much clearer, as $\mathbb{A}^4$ is, the "geometric realization of the tangent bundle $\Omega^1_{\mathbb{A}^2}$.

Here you have $\Omega^1_Z$ sitting in $\mathbb{A}^4$, more precisely you have $\Omega^1_{Z\setminus (0,0)}$ being the locally closed space $\{(x,y,u,v)|xy=0, (x,y)\neq (0,0), xu=-yv\}$, the normal bundle is also here it is given by the subspace $\{(x,y,t,t)|xy=0, (x,y)\neq (0,0) \}$, now we see that the closure of $V(xu+yv)\setminus \{(0,0, u, v)\ u\in k v\in k\}$ inside $Z\times \mathbb{A}^2$ is exactly $V(xu+yv)$ (for instance because it is irreducible and closed) and this IS the normal bundle. The fiber over $(0,0)$ is simply the "limit" of the fiber locally near the singularity.

You have a slightly different way to see it. If you look at $\tilde{Z}=Bl_{(0,0)}Z$ the blow up of your variety at the singular origin, it is sitting inside $\tilde{X}=Bl_{(0,0)}\mathbb{A}^2$ as the strict transform of $Z$. Notice that algebraically it corresponds to exactly the same picture as before but this time you consider $u, v$ as homogeneous coordinates on $\mathbb{P}^1$ instead as bona fide coordinate on $\mathbb{A}^2$.

The strict transform of $Z$ is then the union of two distinct lines and the preimage of the origin is made up of two points $a$ and $b$ say, and you have the differential of the blow down map $\tilde{X}\to X$ which maps the normal line at $a$ and the normal line at $b$ to... $0$ this is the line that is contracted by the blow down after all.

In fact those two lines (the normal lines at $a$ and at $b$) are the same, the exceptional divisor meets $\tilde{Z}$ at these exact two points and this is the normal line at that point (which, if you remember things about strict tranforms as being closure of what happens away from the exceptional locus, is completely coherent with the previous picture).

I hope this makes what happens a bit clearer and why you should look at the blow up to understand what really happens. The normal bundle just so happens to be contracted by the blow down at the origin and thus we cannot see it on the original drawing.


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