group theory - Lie Algebra Conventions: Hermitian vs. anti-Hermitian

Consider the Lie algebra of $SU(2)$.

To find the infinitesimal generators we linearise about the identity $$U=I+i\alpha T$$ where $\alpha$ is some small parameter. To find the form of $T$ use the condition $\textrm{det}(U)=1$ to find $\textrm{Tr}(T)=0$ and also $U^{\dagger}U=I$ to give $T=T^{\dagger}$ Hermitian.

But instead linearising as $$U=I+\alpha T$$ we would find the conditions $\textrm{Tr}(T)=0$ and $T=-T^{\dagger}$ anti-Hermitian, which seemingly results in a different Lie algebra. I think the former approach is the one usually used (and results in a nicer answer). Is there some rule that determines whether the factor of $i$ should be used in this process, or is it just a matter of convenience?

Answer

The factor of $i$ is generally a matter of convention. Essentially, it boils down to choosing what constant you'd like sitting in front of the defining equation,

$$[T^a,T^b] = f_{abc} T^c$$

of the structure constants $f_{abc}$ of the Lie group. We could have instead a factor of $i$ or any constant in our definition and it is a matter of convention.

There is also some freedom in choosing the normalisation of the 'inner product' $\mathrm{Tr}(T^a T^b)$ though there are restrictions depending on if the group is compact for instance.

In my own experience, physicsts keep a factor of $i$ explicit and in the mathematical literature it is usually omitted.