Editing Jacobi identity (section)

==Commutator bracket form==
The simplest informative example of a [[Lie algebra]] is constructed from the (associative) ring of <math>n\times n</math> matrices, which may be thought of as infinitesimal motions of an ''n''-dimensional vector space. The × operation is the [[commutator]], which measures the failure of commutativity in matrix multiplication. Instead of <math>X\times Y</math>, the Lie bracket notation is used:
:<math>[X,Y]=XY-YX.</math>

In that notation, the Jacobi identity is:
:<math>[X, [Y, Z] ] + [Y, [Z, X] ] + [Z, [X, Y] ] \ =\  0</math>
That is easily checked by computation.

More generally, if '''{{math|A}}''' is an associative algebra and '''{{mvar|V}}''' is a subspace of '''{{math|A}}''' that is closed under the bracket operation: <math>[X,Y]=XY-YX</math> belongs to '''{{mvar|V}}''' for all <math>X,Y\in V</math>, the Jacobi identity continues to hold on '''{{mvar|V}}'''.<ref>{{harvnb|Hall|2015}} Example 3.3</ref> Thus, if a binary operation <math>[X,Y]</math> satisfies the Jacobi identity, it may be said that it behaves as if it were given by <math>XY-YX</math> in some associative algebra even if it is not actually defined that way.

Using the [[anticommutativity|antisymmetry property]] <math>[X,Y]=-[Y,X]</math>, the Jacobi identity may be rewritten as a modification of the [[associativity|associative property]]:

:<math>[[X, Y], Z] = [X, [Y, Z]] - [Y, [X, Z]]~.</math>

If <math>[X,Z]</math> is the action of the infinitesimal motion '''{{mvar|X}}''' on '''{{mvar|Z}}''', that can be stated as: 
{{blockquote
 | The action of ''Y'' followed by ''X'' (operator <math>[X,[Y,\cdot\ ] ]</math>), minus the action of ''X'' followed by ''Y'' (operator <math>([Y,[X,\cdot\ ] ]</math>), is equal to the action of <math>[X,Y]</math>, (operator <math>[ [X,Y],\cdot\ ]</math>).
|sign=|source=}}

There is also a plethora of [[Lie superalgebra#properties|graded Jacobi identities]] involving [[anticommutator]]s <math>\{X,Y\}</math>, such as:

:<math>
[\{X,Y\},Z]+ [\{Y,Z\},X]+[\{Z,X\},Y] =0,\qquad
[\{X,Y\},Z]+ \{[Z,Y],X\}+\{[Z,
X],Y\} =0.
</math>
{{See also|Lie bracket of vector fields|Baker–Campbell–Hausdorff formula}}