# Computing degrees of projective varieties via Chern classes

I know that the degree of a projective hypersurface $H \subset \mathbb{P}^n$ can be computed in terms of the Chern class of the normal (line) bundle of $H$. Is there a similar formula for the degree of a higher codimension projective variety in terms of Chern classes of the normal bundle?

In general, does degree just depend on the normal bundle of the projective variety in projective space? I feel like the answer is no, which would make it impossible to compute the degree in terms of the Chern class of the normal bundle.

• Just to be clear, what is your definition of degree of a projective variety? – Michael Albanese Mar 26 '16 at 19:35
• @MichaelAlbanese The one taken from en.wikipedia.org/wiki/Degree_of_an_algebraic_variety: the degree of H is the number of points of intersection of H with a linear subspace L with codim L = dim H. – user39598 Mar 26 '16 at 19:41

There is some ambiguity in precisely what you are asking. One issue is that Chern classes are cohomology classes, which means that in general there is no canonical way to compare Chern classes on different varieties, or to turn Chern classes into numbers.

To sidestep this issue, let's just talk about curves. Here the only Chern class of a bundle is $c_1$ and it has a canonically defined degree. So one possible version of the question is the following:

Are there curves $X$ and $Y$ in $\mathbf P^n$ such that $\operatorname{deg} \, c_1(N_X) = \operatorname{deg} \,c_1(N_Y)$ but $\operatorname{deg} X \neq \operatorname{deg} Y$?

The answer to this question is yes. Here's an example.

If $X \subset \mathbf P^3$ is a curve obtained as a complete intersection of hypersurfaces of degrees $a$ and $b$, then $$\operatorname{deg} X=ab \ ; \\ \operatorname{deg}(c_1(N_X)) = \operatorname{deg}(c_1(O(a)\oplus O(b))_{|X}) = ab(a+b).$$

So it is enough to find two pairs of natural numbers $(a,b)$ and $(c,d)$ such that $ab \neq cd$ but $ab(a+b)=cd(c+d)$. The first example I found was $(1,5)$ and $(2,3)$.

• thanks! I later realized that it is possible to recover the degree if we know both $c(TX)$ and $c(\nu)$, where $\nu$ is the normal bundle. Let $i: X^k \rightarrow \mathbb{CP}^n$ be the inclusion map. Then $i^*T\mathbb{CP}^n = TX^k \oplus \nu$ and so $\left< i^*c_k(T\mathbb{CP}^n), [X]\right> = \sum_{i+j = n} \left<c_i(TX) \cup c_j(\nu), [X]\right>$. On the other hand, $c_k(T\mathbb{CP}^n) = (n+1, k)a^k$, where $a\in H^2(\mathbb{CP}^n)$ is the generator, and so $\left< i^*c_k(T\mathbb{CP}^n), [X]\right> = d (n+1, k)$. – user39598 Apr 3 '16 at 22:43
• Therefore $d (n+1, k) = \sum_{i+j = n} \left<c_i(TX) \cup c_j(\nu), [X]\right>$ so we can recover the degree $d$ from $c(TX)$ and $c(\nu)$. – user39598 Apr 3 '16 at 22:43
• Very nice counterexample, Pooh Bear: +1 – Georges Elencwajg Apr 5 '16 at 12:28

Here is a general answer in terms of Chow rings.

Let $i:Y\hookrightarrow \mathbb P^n$ be a smooth closed subvariety of codimension $r$ and degree $d$, so that for the corresponding cycle class we have $[Y]=dH^r\in A^r(\mathbb P^n)=\mathbb Z\cdot H^r$ .
We have (Hartshorne, page431): $$c_r(N_{Y/X})=i^\ast [Y]=i^\ast (dH^r)=dh^r\in A^r(Y)$$ where $h=i^\ast H\in A^r(Y)$ and where $c_r(N_{Y/X})\in A^r(Y)$ is the Chow Chern class
[which, in case the base field is $\mathbb C$, is infinitely more precise than its image in singular cohomology $c_r^\mathbb C(N_{Y/X})\in H^{2r}(Y(\mathbb C),\mathbb Z)$ ].
If $c_r(N_{Y/X})$ is known, we may often extract the degree $d$ of $Y$ from the the above formula $c_r(N_{Y/X})=dh^r$.
However if $h^r=0$, as is the case for curves in $\mathbb P^3$ for example, the equality $c_r(N_{Y/X})=dh^r$ reduces to $0=d\cdot0$, which (in conformity with Pooh Bear's great counterexample) doesn't allow us to compute $d$.