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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?

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    $\begingroup$ You have been given fine answers so I'll only chime in to say that the reason we use hermitian generators is that we are often interested in turning them into quantum operators. Which correspond to observables only if they are hermitian. $\endgroup$ – Bence Racskó Mar 25 '17 at 12:16
  • $\begingroup$ @Uldreth That's actually an excellent point which I was forgetting. Indeed, I can't believe I'd never really brought this thought to the fore, because the physicist's convention has always grated on me a little - until now. Which is weird now I think of it, because I've always found a measurement theoretical view of QM, with a strong emphasis on observables, the most enlightening. $\endgroup$ – WetSavannaAnimal Mar 25 '17 at 12:27
  • $\begingroup$ @Uldreth while you are certainly correct, your observation is often the source of confusion with non-compact groups - particularly Lorentz - when users forget that, in finite dimensional non-unitary irreps, it is not possible to have the boost generators as hermitian if the generators of the compact $so(3)$ part are also hermitian. $\endgroup$ – ZeroTheHero Mar 25 '17 at 12:52
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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.

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Strictly speaking, the first convention is almost an abuse of notation, for there is no Hermitian matrix Lie algebra, at least with the usual matrix commutator as the Lie bracket. To understand this, do the:

Exercise Prove that the Lie bracket of two Hermitian matrices $X,\,Y$ is skew Hermitian, i.e. $[X,\,Y]^\dagger = -[X,\,Y]$

and so the putative Hermitian algebra cannot close under the Lie bracket. However, one can cheat, or abstract (depending on your view) a bit and define:

$$L(X,\,Y) \stackrel{def}{=} -i\,(X\,Y-Y\,X)$$

and this binary operation is indeed skew-symmetric, billinear and fulfills the Jacobi identity, so one can indeed, at this abstract level, define a Lie algebra of Hermitian matrices. Our new Lie bracket $L$ also conserves tracelessness, so that one can regard this algebra as the Lie algebra of $SU(N)$.

This is, in fact, what we are doing here for the sake of the convenience of working with Hermitian matrices.

The Lie algebra of traceless, skew-Hermitian matrices kitted with the wonted commutator bracket is the "correct" Lie algebra when one insists on using the commutator bracket $L(X,\,Y)\stackrel{def}{=}X\,Y-Y\,X$. This would be the usual convention in a mathematics text, for example. fi you do your exercise again, you'll see that the commutator of two skew-Hermitian matrices is again Hermitian, and so now the algebra closes under the wonted Lie bracket.

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