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Lie algebra of a isometry group

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^{t X} v ∣ e^{t X} w) = (v ∣ w)

because

e^{t X}

has to be in the isometry group by definition as shown at Section "Lie algebra of a matrix Lie group".

Then:

\frac{d ( e ^{t X} v ∣ e ^{t X} w )}{d t} ∣ ∣ ∣ ∣ ∣_{0} = 0 ⟹ (X e^{t X} v ∣ e^{t X} w) + (e^{t X} v ∣ X e^{t X} w) ∣ ∣ ∣ ∣ ∣_{0} = 0 ⟹ (X v ∣ w) + (v ∣ X w) = 0

so we reach:

\forall v, w \in V (X v ∣ w) = - (v ∣ X w)

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

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Lie algebra of $O (n)$