4
$\begingroup$

Let $N = \{ (z_0,z_1,z_2) \in S^5 \mid z_0^3+z_1^3+z_2^3 = 0 \}$, where we consider $S^5\subset\mathbb{C}^3$. The circle $U(1)$ acts on $N$ by $$e^{i\theta} \cdot (z_0,z_1,z_2) = (e^{i\theta}z_0,e^{i\theta}z_1,e^{i\theta}z_2) ,$$ and the quotient $E = N/U(1)$ is an elliptic curve in $S^5/U(1) = \mathbb{C}P^2$.

The resulting bundle $N \to E$ is just the restriction of the Hopf bundle to the elliptic curve $E$.

How can I calculate the first Chern class of this bundle?

I'm told the answer is $-3$, which I believe must have something to do with $$f(e^{i\theta}\cdot(z_0,z_1,z_2)) = e^{3i\theta} f(z_0,z_1,z_2) ,$$ which is suggestive of a map $S^1\to S^1$ of degree 3 (I'm writing $f$ for the polynomial defining the elliptic curve).

$\endgroup$

1 Answer 1

4
$\begingroup$

The circle bundle you get is the restriction to $E$ of the unit circle bundle of the tautological line bundle $\mathcal{O}(-1)$ over $\mathbb{CP}^2$. We will compute the Chern class of this line bundle. It is dual to $\mathcal{O}(1)|_E$, the restriction of the hyperplane bundle to $E$. Since a hyperplane section of $E$ is a divisor of degree $3$, we have $c_1(\mathcal{O}(1)|_E) = 3$. Taking duals gives the answer you are after.

$\endgroup$

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.