6
$\begingroup$

In differential geometry and tensor analysis, lower and upper indices appear naturally through covariant and contravariant transformations. One uses the metric tensor and its inverse to lower and raise indices: $V_\mu=g_{\mu\nu}V^\nu$, and $R^{\mu\nu}=g^{\mu\lambda}R_\lambda^\nu$.

In Anthony Zee's Group Theory in a Nutshell for Physicists on page 232, when talking about representations of $SU(N)$, Zee defines the lower index as simply a notation for the complex conjugate representation:

$\psi_i={(\psi^i)}^*$

He calls these indices covariant and contravariant too, because if $\psi^i$ transforms like $\psi^i\mapsto {U^i}_j\psi^j$, then $\psi_i\mapsto{\psi_i}^{'} = \psi_j {(U^\dagger)^j}_i$.

Then he says that these indices can be lowered or raised with the total antisymmetric symbol $\varepsilon^{ij\dots k}$ (the Levi-Civita symbol). For a tensor $\phi_k^{ij}$ in $SU(4)$ with two upper and two lower indices, the tensor $\phi_{kpq}:=\varepsilon_{ijpq}\phi^{ij}_k$ will have three lower indices (page 235).

My question is:

  • Applying the totally antisymmetric symbol $\varepsilon$ to a tensor representation looks like the Hodge star operator to me. Is it the same, or is there a difference? Zee does not write about the Hodge star operator, but it seems he avoids certain names to avoid confusion (which is the cause of much confusion for me - why not just say that it's known as the Hodge star?).
  • Is there a connection between the index raising/lowering in differential geometry (using the metric) and the one of Zee (using $\varepsilon$)?
  • The definition $\psi_i={(\psi^i)}^*$ is useful for complex representations, but for real (or pseudoreal) representations, it does not make much sense. However, the Hodge star operation is defined for all representations, complex and noncomplex. If Zee really talks about the Hodge star operation, why does he define it for complex reps only?
$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.