I'm currently facing the problem of computing chern classes for Varieties. More precisely the product of such varieties.

Let $C_i$ be a variety in $\mathbb{CP}^2$ given by the Weierstraß $\wp$-map. I want to construct a product of 3 such varieties. Nothing fancy, just $C_1\times C_2 \times C_3$ and calculate it's chern classes.

I'm specifically interested in the second chern class of the Tangent Bundle and some Vector bundle. However, I'm having real trouble actually starting some calculation.

In the case of a single variety in $\mathbb{CP}^2$ I could have used the splitting principle and used the fact that the Normal bundle is a Line Bundle to calculate the total chern class.

However in the case of 3 such varieties, the first problem that arises is, I don't even know where it lies in? According to Segre Embedding I'd say $\mathbb{CP}^{26}$, but that seems a bit high. Perhaps $\mathbb{CP}^4$ would suffice? $\mathbb{CP}^{2+2+2}$? However, this would only help me with the Tangent Bundle.

Could anyone give me some pointers on how to calculate the total chern class in such a case or some reference where it is done in a similar case? Thanks!

  • 2
    $\begingroup$ I think you are confused about terminology. You say you are interested in a toric variety in $\mathbb{P}^2$ given by the Weierstrass $\wp$ map. The map $\wp$ parametrizes a cubic curve. This is a genus $1$ curve, not a toric variety. (It is, topologically, a torus.) Are you interested in are chern classes of tangent bundles to products of cubic curves? If so, the answer is very easy -- they're all zero. This is because the tangent bundle is a trivial bundle. If you do care about toric varieties after all, there are good answers, but I'm not sure they're what you want. $\endgroup$ Commented May 30, 2012 at 19:59

3 Answers 3


It appears that you are assuming that your varieties $C_{i}$ are smooth (you seem to assume that since you are talking about the tangent bundle). In this case each $C_{i}$ is an elliptic curve (I guess, this is what you meant by "toric") and so $C_{1}\times C_{2}\times C_{3}$ is a three dimensional abelian variety. Since it is a group its tangent bundle is trivial and all of its Chern classes are zero. For any other bundle, it will depend on how the bundle is defined. If you have more information about your other bundle it should be easy to figure out the answer.


For the product of three varieties in $\mathbb {CP}^2$, the Segre map gives an embedding into $\mathbb {CP}^{26}$. This may be unsatisfying. Since the variety in question has dimension $3$, an argument similar to the one that shows all curves embed in $\mathbb {CP}^3$ should give an embedding into $\mathbb {CP}^7$.

But that's not the right way to go about this. Instead, you should use the fact that the tangent bundle of the product of three varieties/manifolds is the direct sum of the pullback of the tangent bundles of the varieties. Total chern classes are multiplicative on direct sums, so you only need to consider the tangent bundles of the varieties in $\mathbb {CP}^2$. But it seems like you understand those.

Aren't all toric varieties in $\mathbb {CP}^2$ except $\mathbb {CP}^1$ singular?

How to calculate the Chern class of the vector bundle depends on what form you have that vector bundle in. I don't believe there's a simple theory of vector bundles on toric varieties, the same way there is a simple theory of line bundles - otherwise there would be a simple theory of vector bundles on $\mathbb {CP}^n$, when in fact there are open problems!

  • $\begingroup$ @Will Although the theory of vector bundles on toric varieties is not as simple as that of line bundles, there is a very good theory. See the expository sections of arxiv.org/abs/math/0605537 $\endgroup$ Commented May 30, 2012 at 20:45

As I say in my comment above, I don't think you are actually interested in Chern classes of toric varieties. But, if you are:

For a smooth proper toric variety $X$, the $k$-th Chern class of the tangent bundle is represented by the sum of the codimension $k$ toric subvarieties. For example, if $X = \mathbb{P}^2$, then $c_1$ is $3 \cdot [\mathrm{line}]$ and $c_2$ is $3 \cdot [\mathrm{point}]$. I'm having trouble finding you a reference for this, because everyone wants to prove more complicated things! But it isn't hard to prove directly. Let coordinates on the torus be $(t_1, \ldots, t_n)$, for $t_i \in \mathbb{C}^{\ast}$. Let $\theta_i$ be the tangent vector field $t_i \partial/(\partial t_i)$. These extend to sections of $T_{\ast} X$. We'll explicitly take $n-k+1$ sections of $T_{\ast} X$, written as linear combinations of the $\theta_i$ and compute where they become linear dependent.

Let's do $k=n$ first. So we have one section of $T_{\ast} X$, written as $\sum a_i \theta_i$. Let's work in the neighborhood of a fixed point of $X$. Without loss of generality, we may assume that the $t_i$ are local coordinates at that fixed point. So we want to know where $\sum a_i t_i (\partial/\partial t_i)$ vanishes. Well, it's exactly where all the components vanish, so where $a_1 t_1 = a_2 t_2 = \cdots = a_n t_n =0$. Assuming all the $a_i$ are nonzero, this is precisely where $t_1 = \cdots = t_n = 0$. In other words, at the fixed point. The local computation goes the same at every fixed point -- so $c_{n}(T_{\ast} X)$ is represented by the sum of the torus fixed points. Sanity check: The top Chern class of the tangent bundle is the Euler characteristic, and the Euler characteristic of a toric variety is equal to the number of torus fixed points.

Now let's do $k=1$. So we want to take $n$ sections, of the form $\sum_{i=1}^n a_{ij} \theta_i$, and figure out where they are linearly independent. Again, compute in the neighborhood of a torus fixed point. We want $\det (a_{ij} t_i) =0$, or $\det (a_{ij}) t_1 t_2 \cdots t_n$. If we have chosen the constants $a_{ij}$ generically, this is the same as $t_1 t_2 \cdots t_n$, so the union of the coordinate planes through the fixed point. Again, taking the union over all charts, the sections between linearly independent on the union of the toric divisors.

I won't do the general case, but it isn't much harder than these two.


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