Compute the induced homomorphism on deRham cohomology

Question

Consider the smooth map on the torus to itself $f: S^1 \times S^1 \rightarrow S^1 \times S^1$ defined by $f(z_1, z_2) = (z_1^2z_2, z_1^{-1}z_2)$ (here we identify $S^1$ as the unit circle on the complex plane). Compute the homomorphism $f^*: H^2_{dR}(S^1 \times S^1) \rightarrow H^2_{dR}(S^1 \times S^1)$. Hint: Compute the degree of $f$.

My attempt:

So I want to first compute the degree of $f$. There are three ways I know to compute the degree:

(i) compute $f_*:H_2(S^1 \times S^1) \rightarrow H_2(S^1 \times S^1)$ (taking $[x] \mapsto \deg(f) [x]$); or

(ii) compute $\int_{S^1 \times S^1} f^* \omega = \deg(f) \cdot \int_{S^1 \times S^1} \omega$; or

(iii) find a regular value $y$ of $f$ and compute $\mathrm{deg}(f) = \sum\limits_{x \in f^{-1}(y)} \mathrm{sgn}(x)$.

It seems like (ii) is the easiest so I work out the following computation:

$$\int_{S^1 \times S^1} f^*(dz_1 \wedge dz_2) = \int_{S^1 \times S^1} d(z_1^2 z_2) \wedge d(z_1^{-1} z_2) = 3 \int_{S^1 \times S^1} z_2 dz_1 \wedge dz_2$$

It seems like the $2$-form I chose led to an inconclusive result because the integrand has a factor of $z_2$. Would I have to explicitly compute $\int_{S^1 \times S^1} z_2 dz_1 \wedge dz_2$ to retrieve the degree of $f$? (Maybe I should've done (iii) instead)

Also, thinking ahead, after I compute the degree of $f$, how does that tell me what the induced homomorphism on deRham cohomology is? I am guessing that it would be multiplication by $\frac{1}{\mathrm{deg}(f)}$ as it is the "inverse" to the homology case written in (i).

Answer 1

Thanks to @Ted Shifrin on the comments, I figured out the degree.

So the volume form on $S^1 \times S^1$ is $d\theta_1 \wedge d\theta_2$ over angles. Therefore, to figure out the volume form in terms of $z_1$ and $z_2$, we apply a change of coordinates using the relations $z_1 = e^{i\theta_1}$ and $z_2 = e^{i\theta_2}$.

The relations above will give you that $\theta_k = -i \ln(z_k)$ and $d\theta_k = -ie^{-i\theta_k}dz_k$. So working this out gives us that: $$d\theta_1 \wedge d\theta_2 = - \frac{1}{z_1z_2} dz_1 \wedge dz_2 =: \omega$$ and this is now the appropriate volume form to compute the degree of $f$.

Thus, $$\int_{S^1 \times S^1} f^*(\omega) = \int_{S^1 \times S^1}-\frac{1}{z_1z_2^2} (3z_2 dz_1 \wedge dz_2) = 3 \int_{S^1 \times S^1} \omega$$ so the degree is officially $3$!

Using this volume form, we can say that $H^2_{dR}(S^1 \times S^1) = \langle \omega \rangle \cong \mathbb{R}$. Therefore, the induced map $f^*:H^2_{dR}(S^1 \times S^1) \rightarrow H^2_{dR}(S^1 \times S^1)$ sends the generator $\omega \mapsto k \omega$ to a constant multiple of the generator. Consequently, we can now justify that $$3 \int_{S^1 \times S^1} \omega = \int_{S^1 \times S^1} f^* (\omega) = \int_{S^1 \times S^1} k\omega$$ so $k= 3=\mathrm{deg}(f)$. That is, the induced homomorphism is a multiplication by $\mathrm{deg}(f)=3$ map.