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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!

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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