# Jacobi identity of Lie algebra intuition

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• lriuui0x0
In summary: Fekete's book is a great read, and it goes into more detail about this.:oldconfused: 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.
lriuui0x0
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?

Last edited:
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/

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.

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.

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:
$\vec a \times (\vec b \times \vec c) + \vec b \times (\vec c \times \vec a) + \vec c \times (\vec a \times \vec b) =\vec 0$
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 $\vec a, \vec b, \vec c$.
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, $\ \vec b \times \vec c\$, $\ \vec c \times \vec a\$, $\ \ \vec a \times \vec b\$.

Using the same correspondence between planes and vectors,
we see that the vectors
$\ \vec a \times (\vec b \times \vec c)\$, $\ \vec b \times (\vec c \times \vec a)\$, $\ \vec c \times (\vec a \times \vec b)\$,
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.

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).

docnet

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

The Jacobi identity of Lie algebra intuition is a fundamental property of Lie algebras, which are mathematical structures used to study symmetries in various areas of mathematics and physics. It states that for any three elements in a Lie algebra, the bracket operation (also known as the Lie bracket) must satisfy a specific algebraic identity.

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

The Jacobi identity is crucial in Lie algebra because it ensures that the Lie bracket operation is well-defined and satisfies certain properties, such as being bilinear and skew-symmetric. It also allows for the use of powerful tools, such as the Baker-Campbell-Hausdorff formula, to solve problems and make calculations in Lie algebra.

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

The Jacobi identity is closely related to the commutator bracket, which measures the failure of two elements in a Lie algebra to commute with each other. In fact, the Jacobi identity can be thought of as a generalized version of the commutator bracket, involving three elements instead of just two.

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

Yes, the Jacobi identity can be extended to higher dimensions in the context of n-Lie algebras. In this case, the Jacobi identity involves n+1 elements and is a necessary condition for the Lie bracket to be well-defined and satisfy certain properties.

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

Yes, the Jacobi identity has numerous applications in mathematics and physics. It is used in the study of symmetries and conservation laws in quantum mechanics, general relativity, and other areas of theoretical physics. It also has applications in differential geometry, Lie group theory, and algebraic topology.

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