Insert $\Lambda^\mu{}_{\nu}=\delta^\mu{}_{\nu}+\alpha\omega^\mu{}_{\nu}$ into the <Lorentz group> condition and keep the coefficient of $\alpha$. It gives
$$
\eta_{\sigma\nu}\omega^\sigma{}_{\mu}+\eta_{\mu\tau}\omega^\tau{}_{\nu}=0.
$$
Define $\omega_{\mu\nu}=\eta_{\mu\rho}\omega^\rho{}_{\nu}$. The displayed equation is $\omega_{\nu\mu}+\omega_{\mu\nu}=0$. Raising both indices with the symmetric <Minkowski metric> preserves this antisymmetry:
$$
\boxed{\omega_{\mu\nu}=-\omega_{\nu\mu},\qquad\omega^{\mu\nu}=-\omega^{\nu\mu}.}
$$
These are the generators of the <Lie algebra of the Lorentz group>. In particular $\omega^\mu{}_{\mu}=0$. The mixed-index matrix need not be antisymmetric in the ordinary Euclidean transpose; it is antisymmetric after lowering an index with $\eta$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 4 / iii / Solution

Consider the smooth map
$$
\Psi:M_4(\mathbb R)\to\operatorname{Sym}_4(\mathbb R),
\qquad
\Psi(A)=A^TMA.
$$
Its derivative is
$$
D\Psi_A(H)=H^TMA+A^TMH.
$$
At $A\in O(1,3)$ write $H=AY$. Then
$$
D\Psi_A(AY)=Y^TM+MY.
$$
This derivative is onto the ten-dimensional space of symmetric matrices: for a symmetric $S$, taking $Y=\tfrac12MS$ gives $Y^TM+MY=S$. Hence $M$ is a regular value. Since $M_4(\mathbb R)$ has dimension 16, the <regular level set theorem> gives
$$
\boxed{\dim O(1,3)=16-10=6}.
$$

The tangent vectors are exactly those for which the derivative vanishes:
$$
\boxed{
T_AO(1,3)=\{AY:Y\in\mathfrak S\},
\qquad
\mathfrak S=\{Y:Y^TM+MY=0\}.
}
$$
Writing $Y$ in one-plus-three block form shows explicitly that
$$
\mathfrak S
=\left\{
\begin{pmatrix}
0&v^T\\
v&\Omega
\end{pmatrix}
:v\in\mathbb R^3,\ \Omega^T=-\Omega
\right\}.
$$
The three entries of $v$ and three independent entries of the skew-symmetric matrix $\Omega$ again display dimension six. This is the <Lie algebra of the Lorentz group>.

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 26I / c / ii / Solution

{c}
{title2=$\mathfrak o(1,3)$}

For $M=\operatorname{diag}(-1,1,1,1)$,
$$
\mathfrak o(1,3)=\{Y\in M_4(\mathbb R):Y^TM+MY=0\}.
$$
Every such matrix has the block form
$$
Y=\begin{pmatrix}0&v^T\\v&\Omega\end{pmatrix},
\qquad
v\in\mathbb R^3,\quad\Omega^T=-\Omega,
$$
so the Lie algebra and the <Lorentz group> have dimension six. At $A\in O(1,3)$, the tangent space is $A\mathfrak o(1,3)$.

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