# Difference between infinitesimal parameters of Lie algebra and group generators of Lie group

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!

• 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? Sep 24 '18 at 2:56
• 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. Sep 30 '18 at 14:02
• ... but 4-vectors constitute the defining rep of this group, hence the prominence of your first expression. Sep 30 '18 at 14:03

## 1 Answer

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.