Is $e^{B}$, where $B$ is a member of the $SU(2)$ matrix algebra, a member of the $SU(2)$ group?

Question

I am doing some work in quantum field theory, and I have a question regarding the representation of the $SU(2)$ group. My understanding is that the $SU(2)$ group can be represented as $e^B$, where $B$ is a member of the $su(2)$ algebra. $B$ can be written as $igB^\alpha t_\alpha$, where $t_\alpha$ are the generators. However, how does this exponential expression give a member of the $SU(2)$ group? Expanding it as a Taylor Series, the determinant of $e^B$ is not going to remain $1$. Do I have incorrect reasoning for any part of this explanation?

Answer 1

It is shown here that $$\det(\exp\mathbf A)=\exp(\operatorname{tr}\mathbf A)$$ But $\mathbf A\in \mathfrak{su}(2)\implies \operatorname{tr}\mathbf A=0$ which shows that $$\det(\exp \mathbf A)=1$$ As expected. There is no problem.

Answer 2

Lie product formula implies that if $\exp t A$, $\exp t B$ are in a (closed) matrix group for all $t$, then so is $\exp(A+B)$. That should ease some of your concerns.