I'm learning "Gauge theory of Elementary Particle Physics Problems and Solutions" by Cheng and Li. In problem 8.4 "$O(n)$ gauge theory" on page 165,

Under infinitesimal $O(n)$ representations a scalar field $\mathbf{\phi}$ transforms in the form $$ \phi_{i}\rightarrow\phi_{i}+\epsilon_{ij}\phi_{j} \quad\text{with}\quad \epsilon_{ij}=-\epsilon_{ji}\tag{8.68} $$ The authors say

"For the covariant derivative we need the adjoint representation of $O(n)$. It is not hard to see that they are just the second-rank antisymmetric tensors," $$ \phi_{ij}\rightarrow\phi'_{ij}=\phi_{ij}+(\epsilon_{ik}\phi_{kj}+\epsilon_{jk}\phi_{ik})\quad\text{with}\quad \phi_{ij}=-\phi_{ji}\tag{8.74} $$ "This gives the global transformation law for the gauge bosons $W_{\mu ij}$"

Question: How can we justify this statement?

Under the definition of covariant derivative of $\mathbf{\phi}$, $$ D_{\mu}\phi_{i}=\partial_{\mu}\phi_{i}+gW_{\mu ik}\phi_{k} \quad\text{with}\, W_{\mu ik}=-W_{\mu ki}\tag{8.75}$$ after some calculations,we get $$ W_{\mu il}\rightarrow W'_{\mu il}=W_{\mu il}-W_{\mu ij}\epsilon_{jl}+\epsilon_{ik}W_{\mu kl} -\frac{1}{g}(\partial_{\mu}\epsilon_{il})\tag{8.81} $$ (There is a misprint in the book,so I've rectified here.) If we set $\epsilon_{il}=$Const. in this expression, we get the above expression for the transformation of $\phi_{ij}$. To be sure.


The thing you add to partial derivative to get derivative covariant with respect to local action of group $G$ is always a one-form (meaning it has one spacetime index) with values in the adjoint representation of $G$. The reason is as follows: for spacetime-independent transformations $\psi \mapsto U \psi$, $U \in G$ we want to have $D \psi \mapsto U D \psi$. This requires that connection coefficients $W$ are in the adjoint representation, $W \mapsto U W U^{-1}$. Adjoint representation of the group is nothing else than its Lie algebra $\mathfrak g$. Geometrically this means that if you want to parallel transport your field along infinitesimal line segment tangent to vector $\xi^{\mu}$, you need to act on it with infinitesimal point-dependent $G$-transformation $-W$.

  • $\begingroup$ Thank you for your answer.I'll ponder with plenty of time. $\endgroup$ – GotchaP Sep 18 '16 at 1:04
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    $\begingroup$ @user369432 If you want more in depth explanation of concepts I mentioned in this very short answer, you might take a look at book "The Geometry of Physics" by Theodore Frankel. $\endgroup$ – Blazej Sep 18 '16 at 9:42
  • $\begingroup$ Thanks. But it's not in my neighbor libraries. And It's a bit expensive in Amazon. $\endgroup$ – GotchaP Sep 19 '16 at 2:19
  • $\begingroup$ Try Russian servers $\endgroup$ – Blazej Sep 19 '16 at 8:59

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