3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $-W$.

$\endgroup$
  • $\begingroup$ Thank you for your answer.I'll ponder with plenty of time. $\endgroup$ – GotchaP Sep 18 '16 at 1:04
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.