Source: "An Introduction to Manifolds" by Loring W. Tu, p207, p199.
How do I show that $\omega=-y dx + x dy $ is a left-invariant $1$-form on $S^1$?
I have to show that for all $ g, x \in S^1$, we have $l^*_g(\omega_{gx})=\omega_x$. But, what is $\omega_{gx}$? How do I compute $l^*_g(\omega_{gx})$?
Edit (definition of $\omega_x$): A covector $\omega_x$ at $x$ is a linear function $T_xS^1\to \Bbb R$.
I am not sure of the notations that Tu uses in his book, so you may have to "translate" some of the following arguments into his language.
We are asked to verify that $\omega = -ydx + xdy$ is a left-invariant $1$-form on $S^1$, i.e. that $(l_g)^*\omega = \omega$ for all $g \in S^1$.
Now, in general, if $f \colon M \to N$ is a smooth map of manifolds and $\omega = \sum_i a_i dy_i$ is a $1$-form on $N$, then the pullback $f^*\omega$ is the $1$-form on $M$ given by $f^*\omega = \sum_i (a_i \circ f)\, d(y_i \circ f)$. So, we need to compute the various terms in this expression for our specific $\omega$ and $f$ and check that $f^*\omega = \omega$.
So, fix $g = (\cos t, \sin t)$, and let $(x,y) \in S^1$ be an arbitrary point. We first compute the coefficients ($a_i \circ f$): $$ \bigl((-y) \circ l_g\bigr)(x,y) = (-y)\bigl( (\cos t)x - (\sin t)y, (\sin t)x + (\cos t)y \bigr) = -(\sin t)x - (\cos t)y $$ $$ \bigl((x) \circ l_g\bigr)(x,y) = (x)\bigl( (\cos t)x - (\sin t)y, (\sin t)x + (\cos t)y \bigr) = (\cos t)x - (\sin t)y. $$ Next, we compute $d(y_i \circ f)$. The expressions $y_i \circ f$ are, in our case, just $x \circ l_g$ and $y \circ l_g$. Luckily, we have already computed these expressions right above: $$ d(x \circ l_g) = d\bigl( (\cos t)x - (\sin t)y \bigr) = (\cos t) dx - (\sin t) dy $$ $$ d(y \circ l_g) = d\bigl( (\sin t)x + (\cos t)y \bigr) = (\sin t) dx + (\cos t) dy. $$
We can now write down the expression for $f^*\omega$ as follows: $$ \begin{align*} (l_g)^*\omega &= \bigl( -(\sin t)x - (\cos t)y \bigr) \cdot \bigl( (\cos t) dx - (\sin t) dy \bigr) + \bigl( (\cos t)x - (\sin t)y \bigr) \cdot \bigl( (\sin t) dx + (\cos t) dy \bigr)\\ &= -ydx + xdy = \omega. \end{align*} $$