I saw this question: Reconstructing Lie group globally from the exponential map asking whether we can full cover a Lie group $G$ globally given by the local Lie algebra $t$ away from the identity element (of the Lie group), and extend to write the exponential map $$ \exp(\sum \alpha_a t_a). $$
What I meant for the full cover is that given any group element $g \in G$, we are always able to find a set of $\alpha_a' $ such that the $g = \exp(\sum \alpha_a' t_a)$.
We also should specify certain dimensions of the matrix representations of the Lie algebra.
There the answer seems to be negative in general.
However, I am more simple-minded. I am curious for the case of $SU(n)$. Can we reconstruct the Lie group globally from the exponential map and the local Lie algebra of $\mathfrak{su}(n)$? for each $n \geq 2$.
For $n=1$, it is a trivial Lie group with only the identity.
For $n=2$, if I am not mistaken, we can construct the $SU(2)$ Lie group out of the local $\mathfrak{su}(2)$ Lie algebra. So YES.
To clarify, I think we can construct the $SU(2)$ by the 2-dimensional representaiton of $\mathfrak{su}(2)=\mathfrak{so}(3).$ However, we can construct the $SO(3)$ by the 3-dimensional representaiton of $\mathfrak{su}(2)=\mathfrak{so}(3).$
For $n=3$, if I am not mistaken, we can construct the $SU(3)$ Lie group out of the local $\mathfrak{su}(3)$ Lie algebra? YES or NO?
For general $n$, as above, $\dots$ what is the case?
How about the general $n$, is there a critical $n'$ such that the situation changes from Yes $n<n'$ and No for $n\geq n'$?