Jacobi identity of Lie algebra intuition

lriuui0x0 · Jan 9, 2022

My intuition about the Lie algebra is that it tries to capture how infinitestimal group generators fails to commute. This means ##[a, a] = 0## makes sense naturally. However the Jacobi identity ##[a,[b,c]]+[b,[c,a]]+[c,[a,b]] = 0## makes less sense. After some search, I found this article https://www.hashpi.com/lie-groups-intuition-and-geometrical-interpretation, which explains how Jacobi identity of the commutator is a consequence of the associativity of the underlying group. Further on the intuition, if Jacobi identity tries to capture the associativity of the group, then we should be able to derive associativity from Jacobi identity, but the proof doesn't seem to easily go in this reverse direction. Do people know if this is the right way to think about the Jacobi identity?

Another result from the search is that Jacobi identity can be rewritten in the Lebniz form, so that ##[a, \cdot]## becomes a derivation operator over the Lie bracket itself. See https://en.wikipedia.org/wiki/Jacobi_identity#Adjoint_form for more details. However, I don't understand why should Lie algebra be related to derivation. Can somebody comment on this?

fresh_42 · Jan 9, 2022

lriuui0x0 said:

However, I don't understand why should Lie algebra be related to derivation. Can somebody comment on this?

The Lie algebra is the tangent space at the identity element of the Lie group. It is per definition related to derivatives and thus the Leibniz rule.

For more details see (section B)
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

strangerep · Jan 10, 2022

fresh_42 said:

The Lie algebra is the tangent space at the identity element of the Lie group. It is per definition related to derivatives and thus the Leibniz rule.

The Leibniz rule for commutators doesn't depend on whether you're dealing with a Lie algebra. $$[A,BC] = ABC - BCA = ABC - BAC + BAC - BCA = B[A,C] + [A,B]C ~.$$ It just relies on the definition of a commutator, and associativity.

fresh_42 · Jan 11, 2022

strangerep said:

The Leibniz rule for commutators doesn't depend on whether you're dealing with a Lie algebra. $$[A,BC] = ABC - BCA = ABC - BAC + BAC - BCA = B[A,C] + [A,B]C ~.$$ It just relies on the definition of a commutator, and associativity.

This is the wrong order. Lie algebras are primarily tangent spaces, and that they can be realized by an algebra with the commutator rule given an associative matrix algebra is second, a theorem to be exact (Igor Dmitrievich Ado (char 0), Kenkichi Iwasawa (char p)). The adjoint representation ##X \longmapsto (Y\longmapsto [X,Y])## is independent of the commutator rule, but it is still the Leibniz rule.

robphy · Jan 11, 2022

The Jacobi identity arises from the lack of associativity.

Here is an interesting passage from Fekete's Real Linear Algebra.
https://www.amazon.com/dp/0824772385/?tag=pfamazon01-20

Search for jacobi:
https://www.google.com/books/edition/Real_Linear_Algebra/3_AIXBO11bEC?hl=en&gbpv=1&bsq=jacobi

Fekete said:

The cross product of vectors is "nicely" nonassociative, however. We
have the "next best thing" to the associative law, which is the Jacobi identity:
[itex]
\vec a \times (\vec b \times \vec c)
+
\vec b \times (\vec c \times \vec a)
+
\vec c \times (\vec a \times \vec b)
=\vec 0
[/itex]

Fekete then has (bolding mine):

Fekete said:

GEOMETRIC INTERPRETATION OF THE JACOBI IDENTITY
The Jacobi identity is one of the nontrivial, nonetheless highly important,
formulas of linear algebra. Therefore, it will be useful to find its geometnc interpretation.
A three-dimensional analog of the triangle is the trihedron, i.e.., the
figure formed by three noncoplanar vectors [itex] \vec a, \vec b, \vec c [/itex].
These vectors correspond to the vertices of the triangle;
what will correspond to its sides?
To the sides of the triangle there correspond the faces of the trihedron.
The faces of the trihedron are planes
for which we may substitute their normal vectors,
i.e., the vectors perpendicular to them, [itex]\ \vec b \times \vec c\ [/itex], [itex]\ \vec c \times \vec a\ [/itex], [itex]\ \ \vec a \times \vec b\ [/itex].

Using the same correspondence between planes and vectors,
we see that the vectors
[itex]\ \vec a \times (\vec b \times \vec c)\ [/itex], [itex]\ \vec b \times (\vec c \times \vec a)\ [/itex], [itex]\ \vec c \times (\vec a \times \vec b)\ [/itex],
correspond to the altitudes of the trihedron,
i.e., the planes containing an edge and perpendicular to the opposite face.
If the sum of three vectors is the null vector, the three vectors must be coplanar.
The normal vectors of three planes having a point in common are coplanar
if, and only if, the planes also have a line in common.
Hence
the geometric interpretation of the Jacobi identity:
The altitudes of the trihedron are three planes having a line in common.
This is a generalization of the familiar theorem from plane geometry asserting that
the altitudes of the triangle are three lines having a point (the orthocenter) in common.

fresh_42 · Jan 11, 2022

The cross product only works for one special Lie algebra in 3 dimensions. The Jacobi identity remains to be the Leibniz rule in any dimension. There is no better way to describe it than by the formula
$$
\operatorname{Ad}(\exp(A)) = \exp(\mathfrak{ad}(A))
$$
which directly connects the inner derivations of the tangent space (Jacobi) with the inner automorphisms of the group (conjugation).

Jacobi identity of Lie algebra intuition

1. What is the Jacobi identity of Lie algebra intuition?

2. What is the importance of the Jacobi identity in Lie algebra?

3. How does the Jacobi identity relate to the commutator bracket?

4. Can the Jacobi identity be generalized to higher dimensions?

5. Are there any real-world applications of the Jacobi identity in Lie algebra?

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