Let $G$ be a $n$ dimensional Lie Group, and let $L_g$ denote the left action of $G$ on itself. I am trying to understand the notion of left invariance of various objects on $G$.
I understand that for vector fields, the notion of left invariance. A vector field is left-invariant if we have that: $$ (dLg)_hX(h) = X(gh) $$ Thus, if we know $X(e)$ where $e$ is the identity of $G$, we can compute $X(g)$ without much trouble, as it is given by: $$ X(g) = (dLg)_eX(e) $$ Now, let's generalize this to differential forms. We say that a $k$ form on $G$ is left invariant if we have that: $$ L^*_gw = \omega $$ when I saw this, it seemed a bit cryptic to me, so I decided to write it out in full. Formally, the above pullback satisfies: $$ (L^*_g \omega)_h(v_1, \cdots , v_k) = \omega_{L_g(h)}(dL_gv_1, \cdots , dL_gv_k) $$ For any $v_1, \cdots , v_k \in T_{L(g)}G$. What does it mean for $\omega$ to be left invariant here? The above equality omits some details such as which points we are evaluating the pullback at etc. And once we have this, how can we show that the form at any point is determined by tensor we get at the identity?
You have most of the right ideas, but the left-invariant piece is that
$$ \omega_{L_g(h)}(dL_gv_1, \ldots, dL_gv_k) \;\; =\;\; \omega_h(v_1,\ldots, v_k). $$
This is where we get the equation $L_g^*\omega = \omega$. The main idea here is that left translation doesn't affect the overall value of the form.