I am reading a proof of the following theorem:
For each $\xi \in \mathfrak{g}$, there is a unique $1$-PSG $\lambda_\xi$, such that $\lambda'_\xi(0)=\xi$
Here $\mathfrak{g}$ is the Lie algebra associated to a Lie group G and a 1-PSG is a smooth homomorphism $\lambda: \mathbb{R} \rightarrow G$.
The proof takes the left-invariant extension of $\xi$, which is denoted by $\tilde{\xi}$ and considers the map $\Phi^t:G \rightarrow G$ which is induced by the integral curve of $\tilde{\xi}$ and asserts that
As $\tilde{\xi}$ is left-invariant, $\Phi^t(g_1g_2) = g_1\Phi^t(g_2)$.
This seems like it should be reasonably obvious, however the notes don't specify what such an induced map should do and I haven't been able to find a source for such a map. The closest thing I can find is a map $\phi^t:G \rightarrow G$ which sends each point $p \in G$ to $\gamma_p(t)$, where $\gamma_p$ is the unique integral curve with $\gamma_p(0) =p$.
My guess for what the induced map should be is the following: $$\Phi^t(p) = \gamma(t)p$$ where $\gamma$ is the integral curve of $\tilde{\xi}$. Supposing this is the case I struggle to see how $\Phi^t(g_1g_2) = g_1\Phi^t(g_2)$.
Question:
Is this definition of the induced map correct? If so, how can I show that this map satisfies the required property, I suppose that it should be a consequence of $\tilde{\xi}$ being left-invariant, though I don't see how this applies here. If I haven't defined the correct map, could someone suggest what I should try instead?
You're close: in fact in your notation, $\Phi^t(p) = p\gamma(t)$. I would write this as $\Phi^t(g) = g\exp(t\xi)$. One way of stating this is that the left-invariant vector field $\tilde\xi$ induces a diffeomorphism $\Phi^t:G\to G$ that is just right multiplication by $\exp(t\xi)$ (or $\Phi^t = R_{\exp(t\xi)}$). To see this, note that $$ \frac{d}{dt} g\exp(t\xi) = \frac{d}{ds}\Big\vert_{s=0}g\exp((t+s)\xi) = \frac{d}{ds}\Big\vert_{s=0}L_{g\exp(t\xi)}\exp(s\xi) = T_eL_{g\exp(t\xi)}\xi = \tilde\xi(g\exp(t\xi)), $$ where $L_h:G\to G$ denotes left multiplication by $h\in G$. That is, with our definition of $\Phi^t$ $$ \frac{d}{dt}\Phi^t(g) = \tilde\xi(\Phi^t(g)), $$ and so $\Phi^t(g) = g\exp(t\xi)$ is the diffeomorphism generated by $\tilde\xi$. From this, the condition you want is clear.