<An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)s> page 146.

Every closed subgroup of $GL(n, \C)$ is a Lie group

Every closed subgroup of GL(n,C) is a Lie group

{tag=Tensor}

<An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)> shows that this is a <tensor> that represents the <volume of a parallelepiped>.

It takes as input three vectors, and outputs one real number, the volume. And it is linear on each vector. This perfectly satisfied the definition of a tensor of <order of a tensor>[order] (3,0).

Given a basis $(e_i, e_j, e_k)$ and a function that return the volume of a parallelepiped given by three vectors $V(v_1, v_2, v_3)$, $\varepsilon_{ikj} = V(e_i, e_j, e_k)$.

Levi-Civita symbol as a tensor

{c}
{wiki}

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

Recommended from <Physics from Symmetry by Jakob Schwichtenberg (2015)> page 92:
* <An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)>
* <Naive Lie theory by John Stillwell (2008)>
* <Lie Algebras In Particle Physics by Howard Georgi (1999)>

Ciro Santilli @cirosantilli 34

 Incoming links: An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)