#### Problem:

Consider $M = \mathbb{T}^2$ with local coordinates $(\theta, \phi)$ and $N = \mathbb{S}^3$ with local coordinates $(a, b, c)$. Consider the two form $\alpha = 3 a \, db \wedge dc$ on $N$. Compute $f^* \alpha$.

#### Solution:

Let $f: M \to N$ be given in local coordinates by

$a = \cos(2 \theta + \phi)$

$b = \sin(\theta + 3 \phi)$

$c = 4 - 5 \phi$.

Then, $f(\theta, \phi) = (\cos(2\theta + \phi), \sin(\theta + 3\phi), 4 - 5\phi) = (a, b, c)$

and, for the two form $\alpha = 3 a \, db \wedge dc$, the pullback is:

$f^*\alpha = f^*(3a) f^*(db) f^*(dc) = 3(a \circ f) d(b \circ f) d(c \circ f)$

$= 3\cos(2\theta + \phi) d(\sin(\theta + 3\phi)) \wedge d(4 - 5\phi)$

$= 3\cos(2\theta + \phi) (\cos(\theta + 3\phi) (d\theta + 3d\phi)) \wedge (-5d\phi)$

$= 3\cos(2\theta + \phi) [(\cos(\theta + 3\phi) d\theta + 3\cos(\theta + 3\phi) d\phi))] \wedge (-5d\phi)$

But, $d\phi \wedge d\phi = 0$, therefore, $3\cos(\theta + 3\phi) d\phi) \wedge (-5d\phi) = 0$

So,

$f^*\alpha = -15\cos(2\theta + \phi) \cos(\theta + 3\phi) d\theta \wedge d\phi$.

Scott agrees with the final answer, although there appear to be some wedges missing from the work that's shown.