I am getting myself confused regarding the differences between the infinitesimal generators of Lie group and the elements of the Lie algebra, likely due to the fact that I am studying from a physics perspective where authors frequently use the group and the algebra interchangeably.

Consider a specific example from QFT, the Lorentz group $SO(3,1)$. Let $\Lambda^{\mu}_{\ \nu}$ be a Lorentz transformation that is infinitesimally close to the identity. i.e. $ \Lambda^{\mu}_{\ \nu} = \delta^{\mu}_{\ \nu} + \epsilon^{\mu}_{\ \nu} $ with $\epsilon^{\mu}_{\ \nu}$ "small".

From the definition of the Lorentz group, we can show that $\epsilon^{\mu}_{\ \nu}$ is antisymmetric, the usual defining property of the Lorentz algebra. This leads me to believe that $\epsilon^{\mu}_{\ \nu}$ is an elements of the lie algebra $\mathfrak{so}(31)$. However, most authors then go on to write a general Lorentz transformation as $$ \Lambda = \text{exp}\left(i\epsilon_{\mu \nu} M^{\mu \nu} \right) $$ with $$ [M^{\mu \nu}, M^{\rho \sigma}] = i(g^{\nu \rho} M^{\mu \sigma} - g^{\mu \rho} M^{\nu \sigma} - g^{\nu \sigma} M^{\mu \rho} + g^{\mu \sigma} M^{\nu \rho}) $$ which is usually how the algebra $\mathfrak{so}(3,1)$ is defined. This leads me to believe that the elements $M^{\mu \nu}$ are the elements of the lie algebra and the $\epsilon^{\mu \nu}$ are simply the parameters that define the transformation (i.e. the angle $\theta^{12}$ of the rotation generated by $M^{12} \equiv J^3$)

The use of these two definitions confuses me, especially because in the latter example, the tensor $\epsilon^{\mu \nu}$ is still referred to as antisymmetric. If anyone could provide some clarification is would be much appreciated!

  • $\begingroup$ The set of infinitesimal generators is usually identified with the elements of the Lie algebra. I don’t think there should be any differences to speak of. What difference do you expect? $\endgroup$
    – rschwieb
    Sep 24 '18 at 2:56
  • 1
    $\begingroup$ The generators $M^{\mu\nu}$, up to signs and is, are antisymmetric in μ,ν . In the quartet representation (acting on a 4-vector), their matrix elements are $M^{\mu\nu}_{ab}\propto\delta^\mu_a\delta^\nu_b -\delta^nu_a \delta^mu_b$, again, cavalierly in signs and is. The tensor parameter $\epsilon_{\mu\nu} $ saturating them only has meaningful antisymmetric components: it is antisymmetric. Acting on 4-vectors, then, it yields $\lambda~x=x+i\epsilon\cdot M ~x+...$ whose components are $x^\mu +i2\epsilon^{\mu\nu}x_\nu$, up to normalizations, similar to your first expression, for 4-vectors. $\endgroup$ Sep 30 '18 at 14:02
  • $\begingroup$ ... but 4-vectors constitute the defining rep of this group, hence the prominence of your first expression. $\endgroup$ Sep 30 '18 at 14:03

I think that I answered my own question, with the help of the commenters. My confusion stemmed from the fact that I didn't see why $\epsilon^{\mu \nu}$ had to be antisymmetric, and mistook the antisymmetry to mean that $\epsilon^{\mu \nu}$ must be an element of the lie algebra. In fact, $\epsilon^{\mu \nu}$ doesn't have to be antisymmetric, but since it is being contracted with $M^{\mu \nu}$, which is antisymmetric, any symmetric part of the tensor will vanish. Thus, we might as well take $\epsilon^{\mu \nu}$ to be antisymmetric by definition.


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.