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?

  • $\begingroup$ 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 $\endgroup$ – Dmitri Panov Feb 7 '10 at 1:49
  • $\begingroup$ 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. $\endgroup$ – Enrique Acosta Feb 7 '10 at 2:00
  • $\begingroup$ 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. $\endgroup$ – Harry Gindi Feb 7 '10 at 2:06
  • $\begingroup$ 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. $\endgroup$ – Deane Yang Feb 7 '10 at 3:39
  • $\begingroup$ 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". $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.