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We can almost reach the <Lie algebra> of any <isometry group> in a single go. For every $X$ in the <Lie algebra> we must have:
$$
\forall v, w \in V, t \in \R (e^{tX}v|e^{tX}w) = (v|w)
$$
because $e^{tX}$ has to be in the isometry group by definition as shown at <Lie algebra of a matrix Lie group>{full}.

Then:
$$
\evalat{\dv{(e^{tX}v|e^{tX}w)}{t}}{0} = 0
\implies
\evalat{(Xe^{tX}v|e^{tX}w) + (e^{tX}v|Xe^{tX}w)}{0} = 0
\implies
(Xv|w) + (v|Xw) = 0
$$
so we reach:
$$
\forall v, w \in V (Xv|w) = -(v|Xw)
$$
With this relation, we can easily determine the <Lie algebra> of common isometries:
* <Lie algebra of O(n)>

Bibliography:
* <An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)> page 151

Lie algebra of a isometry group

Lie algebra of O(n)

Ciro Santilli @cirosantilli 37

 Incoming links: Lie algebra of $O (n)$

Ciro Santilli @cirosantilli 37

 Incoming links: Lie algebra of O(n)

 Incoming links: Lie algebra of $O (n)$