# Normal bundle to a curve in P^2

Let $C$ be a smooth curve of degree $d$ in $\mathbb{P}^2$ over $\mathbb{C}$. Say $C$ is defined by $p(x,y,z)=0$, with $p$ a homogeneous degree $d$ polynomial.

In vector calculus one learns that the gradient of $p$ is normal to $C$ at every point of the curve.

In algebraic geometry, the invertible sheaf associated to the normal bundle $N_{C|\mathbb{P}^2}$ to $C$ in $\mathbb{P}^2$, is given by $\mathcal{O}_{\mathbb{P}^2}(d) _{|C}$.

Is there any relationship between the gradient and the bundle or the sheaf?

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Enrique, you should define what is a gradient. Usually you need a metric to define it, so you should specify the metric on CP^2 or on C^3. Once you do it, it will get more clear what this question is about – Dmitri Feb 7 '10 at 1:49
I simply mean the vector <p_x,p_y,p_z> which one uses in vector calculus. It is normal to the level set p=0 at every point. – Enrique Feb 7 '10 at 2:00
The gradient is the Jacobian matrix of a map R^n->R at a point. Equivalently, it is the matrix realization (after choosing bases) of the total differential of a map from a f.g. real vector space into R. – Harry Gindi Feb 7 '10 at 2:06
It has a been an extremely long time since I have thought about this stuff carefully, but surely the differential $dp$ is a section of the conormal bundle, which is naturally defined without any need for a metric and is dual to the normal bundle. – Deane Yang Feb 7 '10 at 3:39
I think the confusion here is that "gradient" is usually taken to mean a vector field. As Matt points out, without a metric what you naturally get is a 1-form. If you have a metric you can identify this with a vector field, and this then gets called the "gradient". – Kevin McGerty Feb 7 '10 at 16:58

Yes, there is a strong relationship between the two.

First, let's work locally in affine space rather than in projective space (it makes more sense to work locally just because we are dealing with a sheaf, which is defined locally). So I will consider a non-homogen

Working without a metric (as one does in at least the algebraic aspects of algebraic geometry), it is perhaps better to talk not about the gradient of $f$, but its exterior derivative $df$, given by the same formula: $df = f_x dx + f_y dy.$ Since this is differential form valued, we will compare it with the conormal bundle to the curve $C$ cut out by $f = 0$.

Now the exterior derivative can be thought of simply as taking the leading (i.e. linear) term of $f$.

On the other hand, if $\mathcal I$ is the ideal sheaf cutting out the curve $C$, then the conormal bundle is $\mathcal I/\mathcal I^2$. (If $f$ is degree $d$, then $\mathcal I = \mathcal O(-d)$, and so this can be rewritten as $\mathcal O(-d)\_{| C}$, dual to the normal bundle $\mathcal O(d)\_{| C}$.) Now $f$ is a section of $\mathcal I/\mathcal I^2$ (over the affine patch on which we are working), so we may certainly regard it as a section of $\mathcal I/\mathcal I^2$; this section is the (image in the conormal bundle to $C$ of) the exterior derivative of $f$.

The formula $\mathcal I/\mathcal I^2$ for the conormal bundle is thus simply a structural interpretation of the idea that we compute the normal to the curve by taking the leading term of an equation for the curve.

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