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?
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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).



Chern class of Hopf fibration over elliptic curve
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.


