TITLE: Incorrect proof that all gauge theories are abelian
QUESTION [0 upvotes]: Consider a gauge field $W_\mu = W_\mu^{a} \tau_a$ where $\tau_a$ are the generators of the Lie algebra and $W_\mu^{a}$ just numbers. Then:
$$
W^2 = W_\mu W^\mu = W_\mu^a\tau_a W^{\mu b} \tau_b = W_\mu^a\tau_a \tau_b W^{\mu b} 
$$
Meanwhile
$$
W^2 = W^\mu W_\mu = W^{\mu b}\tau_b W_{\mu}^a \tau_a = W^{\mu b}\tau_b \tau_a W_{\mu}^a = W_{\mu}^a\tau_b \tau_a W^{\mu b}
$$
Therefore:
$$
W^2 - W^2 = W_{\mu}^a [ \tau_a,\tau_b]W^{\mu b} = 0
$$
So $[ \tau_a,\tau_b]=0$ and all gauge groups are abelian.
Where's the mistake?

REPLY [3 votes]: The mistake is that $W^a_\mu [\tau_a, \tau_b] W^{\mu b} = 0$ does not imply $[\tau_a, \tau_b] = 0$, this is true for any antisymmetric object $A_{ab}$.
In your example, you can see this in the following way:
Let $A_{ab}$ be antisymmetric: $A_{ab} = -A_{ba}$. Then:
$$
\begin{align}
W^a_\mu A_{ab} W^{\mu b} &= W^b_\mu A_{ba} W^{\mu a} \qquad (a\leftrightarrow b) \\
& = W^{b \mu} A_{ba} W^a_\mu  \qquad \text{Raise/lower } \mu \\
& = W^a_\mu A_{ba} W^{b \mu}  \qquad \text{Commute } W \\
& = -W^a_\mu A_{ab} W^{b \mu}
\end{align}
$$
So $W^a_\mu A_{ab} W^{\mu b} = 0$ for any antisymmetric $A_{ab}$. 
This is really a more general fact from linear algebra: For vector $v$ and matrix $A$, $v^T A\ v = 0$ if A is antisymmetric.

REPLY [1 votes]: No, one can only conclude that the symmetrization
$$[ \tau_a,\tau_b] + (a\leftrightarrow b)~=~0$$
is zero, which is indeed true because the commutator is antisymmetric.
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