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This might be a stupid question, but why is the normalization of the Gell-Mann matrices (basis of the $\mathrm{su}(3)$ Lie algebra) chosen to be $$\mathrm{trace}(\lambda_i\lambda_j)=2\delta_{ij}$$ instead of just $\delta_{ij}$ without the factor $2$? In most of linear-algebra, basis vectors are normalized to $1$ (or not normalized at all). Why not in the context of Lie Algebras? Is there a way of looking at this which makes the factor $2$ seem natural?

On a related note, some physics texts change the normalization by defining "the generators of the $\mathrm{SU}(3)$ group" as $T_i=\frac{1}{2}\lambda_i$. But these just fulfil $\mathrm{trace}(T_iT_j)=\frac{1}{2}\delta_{ij}$ which seems just as unnatural to me. (And the difference between these two normalization conventions just cost me an hour of chasing a missing factor $4$ in a long calculation. Which is why I'm asking this question xD).

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History. The Gell-Mann matrices are an extension/generalization of the Pauli spin matrices for su(2), and $\lambda_{1,2,3}$ identify with these, so obey the same trace relation.

You also understand why the Pauli matrices are further normalized this way by an extra 1/2, so as to then obey the canonically normalized su(2) algebra with structure constant ε, thereby avoiding half-angle exponentials.

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  • $\begingroup$ well, Pauli matrices have the same "problem" of $\mathrm{tr}\sigma_i\sigma_j=2\delta_{ij}$ (also $[\sigma_i\sigma_j]=2i\epsilon_{ijk}\sigma_k$). But changing their normalization would destroy the rather nice property $\sigma_i^2=1$ (which implies Eigenvalues$=\pm 1$ and Determinant$=-1$). These properties even kinda carry over to Gell-Mann, whose eigenvalues are $1,-1,0$ (except for $\lambda_8$). Thanks, Cosmas. $\endgroup$ – Simon Aug 18 '20 at 14:39
  • $\begingroup$ @Cosmas Zachos: Actually, it would be nice if you could expand your answer a bit, because I wonder why only in gauge theory the normalisation Lie-algebra generators is done, but not in representation theory. Why is this so? Why does it matter in gauge theory? Secondly, the trace of 2 generators looks like a killing form $B(x,y) = tr(ad_X, ad_Y)$. Curiously for a compact group like $SU(2)$ I would expect a negative normalisation $tr(t_i, t_j)=-\frac{1}{2}\delta_{ij}$, and in some books it is indeed like this, but why not here ? $\endgroup$ – Frederic Thomas Aug 18 '20 at 21:47
  • $\begingroup$ Sorry, normalization is an endless subject... In gauge theory you care about group identities and Casimirs, anomaly coefficients, etc.. and normalizations matter, e.g. this. In rep theory they don't care... it's beneath them. Note in physics we like hermitian generators with an i on the r.h.side of the Lie algebra, whereas mathematicians naturally go for real structure constants... $\endgroup$ – Cosmas Zachos Aug 18 '20 at 22:01
  • $\begingroup$ By contrast, for the rotation group, where the generators are real, antisymmetric matrices, the structure constants are real without an i... It's all conventions of convenience. $\endgroup$ – Cosmas Zachos Aug 18 '20 at 22:28

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