Proof of antisymmetry of the infinitesimal Lorentz parameters

A finite Lorentz transformation satisfies the equation

(1)
\begin{align} \Lambda \cdot \eta \cdot \Lambda^T &= \eta \ . \nonumber \\ {\Lambda^\mu}_\nu \ \eta^{\nu{ \rho}} \ {\Lambda^\sigma}_{ \rho} &= \eta^{\mu\sigma} \end{align}

Remark: Note that in the second equation above, the transposition is implicitly present in the second $\Lambda$.
Expanding $\Lambda$ as follows where $\lambda$ parametrises infinitesimal Lorentz transformations:

(2)
\begin{align} {\Lambda^\mu}_\nu ={\delta^\mu}_\nu +{\lambda^\mu}_\nu + \mathcal{O}(\lambda^2)\ . \end{align}

For instance, for a Lorentz boost with velocity $\mathbf{u}=(u_1,u_2,u_2)$, one has (see here)

(3)
\begin{align} \lambda(\mathbf{u}) = \begin{pmatrix} 0 & -\tfrac{u_1}{c} & -\tfrac{u_2}{c} & -\tfrac{u_3}{c} \\ -\tfrac{u_1}{c} & 0 & 0 & 0 \\ -\tfrac{u_2}{c} & 0 & 0 & 0 \\ -\tfrac{u_3}{c}& 0 & 0 & 0 \end{pmatrix}\ . \end{align}

Exercise: Obtain the matrix ${\Lambda^\mu}_\nu$ for the simple case of a boost in the $x_1$ direction i.e., with $u_2=u_3=0$ by computing the matrix exponential. Hint: Compute $\lambda^2$ and see that it is proportional to the identity matrix.

On substituting Eq.(2) in Eq.(1), to first order in $\lambda$, we get

(4)
\begin{align} {\lambda^\mu}_\nu \ \eta^{\nu\rho} \ {\delta^\sigma}_\rho+{\delta^\mu}_\nu \ \eta^{\nu\rho} \ {\lambda^\sigma}_\rho &= 0\ ,\\ \implies {\lambda^\mu}_\nu \ \eta^{\nu\sigma} \ + \eta^{\mu\rho} \ {\lambda^\sigma}_\rho&=&0\ , \\ \implies \lambda^{\mu\sigma}+\lambda^{\sigma\mu}&=&0\ . \end{align}

We thus see that Eq.(1) implies that $\lambda^{\mu\nu}:={\lambda^\mu}_\rho \eta^{\rho\nu}$ is antisymmetric as claimed in the lecture. In fact, one has the stronger statement:

(5)
\begin{align} {\Lambda^\mu}_\nu =\exp\left({\lambda^\mu}_\nu\right)\ . \end{align}

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