Derivative of an exact form $\omega_t$ is still exact?

Question

Let $\omega_t$ be an exact form depending smoothly on time $t$.

Then is $\dfrac{d\omega_t}{dt}$ also exact?

$\dfrac{\omega_t - \omega_{t_0}}{t-t_0}$ is clearly exact.

However I am stuck in showing that its limit is exact.

Thanks in advance.

Using Ted's hint

Let $c$ be an any smooth k-cycle. Since $\omega_t$ is exact, $\displaystyle\int_c \omega_t =0$

It suffices to prove that $\displaystyle\int_c \lim_{t \to t_0} \dfrac{\omega_t - \omega_{t_0}}{t-t_0} = \lim_{t \to t_0}\displaystyle\int_c \dfrac{\omega_t - \omega_{t_0}}{t-t_0} $.

However this is on the world of manifolds and diffenretial forms, I am not sure it works well.

Answer 1

HINT: A closed $k$-form is exact if and only if its integral over every $k$-cycle is zero. (Comment: The Leibniz rule for differentiating under the integral sign applies to manifolds as well, giving $$\frac d{dt} \int_M \omega = \int_M \frac{\partial\omega}{\partial t}.$$ The "nice" proof using dominated convergence works just as for one-dimensional integrals.)

Answer 2

An exact form is a differential form $\omega$ such that $\omega = df$ for some differential form $f$. Then $\frac{d\omega_{t}}{dt}$ is exact as it is the derivative of $\omega$.