Structure Constants and Jacobi Identity in Lie Algebras

1. The Jacobi Identity

The Lie bracket is, in general, not associative. That is, for three elements $X,Y,Z\in \mathfrak{g}$, it is not true that $[[X,Y],Z]$ equals $[X,[Y,Z]]$. The failure of the bracket to be associative is measured by the associator, $[[X,Y],Z]-[X,[Y,Z]]$.

However, the associativity of the underlying Lie group multiplication, $g_1(g_2{g}_3)=(g_1{g}_2)g_3$, does leave a distinct signature on the algebra. This signature is the Jacobi identity:

This identity holds for any three elements $X,Y,Z$ in any Lie algebra. The cyclic permutation of $X,Y,Z$ in the formula is a characteristic feature of many such identities in physics and mathematics.

For Lie algebras composed of matrices, where the bracket is the commutator, we can prove this identity by direct expansion. Let's verify it:
\begin{align*}
[[X, Y], Z] &= (XY - YX)Z - Z(XY - YX) = XYZ - YXZ - ZXY + ZYX \
[[Y, Z], X] &= (YZ - ZY)X - X(YZ - ZY) = YZX - ZYX - XYZ + XZY \
[[Z, X], Y] &= (ZX - XZ)Y - Y(ZX - XZ) = ZXY - XZY - YZX + YXZ
\end{align*}
If you sum these three lines, you will find that every term cancels with another. For example, the $XYZ$ term in the first line cancels with the $-XYZ$ term in the second line. The sum is identically zero. This shows that for any associative algebra (like matrix multiplication), the commutator naturally satisfies the Jacobi identity.

2. The Jacobi Identity as a Product Rule

While the algebraic proof is straightforward, it doesn't provide much intuition. A deeper understanding of the Jacobi identity comes from viewing the Lie bracket in a different light—not as a product, but as a derivative.

Let's define the adjoint action of an element $X\in \mathfrak{g}$ on another element $Y\in \mathfrak{g}$ as the linear map ${\mathrm{ad}}_X$ given by:

You can think of ${\mathrm{ad}}_X$ as an operator that acts on the vector space $\mathfrak{g}$. From this perspective, the Jacobi identity can be rewritten. First, note that by antisymmetry, $[[X,Y],Z]=-[Z,[X,Y]]=-{\mathrm{ad}}_Z([X,Y])$. The Jacobi identity is:

Using antisymmetry on the last term, $[[Z,X],Y]=-[Y,[Z,X]]$, and rearranging gives:

Now, let's write this using the $\mathrm{ad}$ notation:

This shows that the map $X\mapsto {\mathrm{ad}}_X$ is a Lie algebra homomorphism, which we will later recognize as the adjoint representation.

Even more intuitively, if we re-arrange the Jacobi identity slightly differently:

This equation has a remarkable structure. It looks exactly like the Leibniz product rule for derivatives, $d(fg)=(df)g+f(dg)$. The operator ${\mathrm{ad}}_X$ acts like a derivative on the "product" $[Y,Z]$.

This analogy is not a coincidence. The following video beautifully develops this intuition, showing how the Lie bracket can be understood as a type of directional derivative and how the Jacobi identity naturally emerges as its product rule.

3. The Formal Definition of a Lie Algebra

Armed with this intuition, we can now state the abstract, formal definition of a Lie algebra. This definition distills the essential properties we have uncovered from our study of Lie groups.

To summarize the reading, a Lie algebra $\mathfrak{g}$ is a vector space over a field $\mathbb{K}$ (for us, $\mathbb{R}$ or $\mathbb{C}$) equipped with a binary operation $[\cdot ,\cdot ]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that satisfies the following three axioms for all $X,Y,Z\in \mathfrak{g}$ and scalars $\alpha ,\beta \in \mathbb{K}$:

1. Bilinearity:
   $[\alpha X+\beta Y,Z]=\alpha [X,Z]+\beta [Y,Z]$
   $[X,\alpha Y+\beta Z]=\alpha [X,Y]+\beta [X,Z]$

2. Antisymmetry (or Alternating Property):
   $[X,Y]=-[Y,X]$. This immediately implies $[X,X]=0$.

3. The Jacobi Identity:
   $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$

Any vector space with a bracket satisfying these axioms is a Lie algebra, even if it doesn't come directly from a matrix group.

4. The Jacobi Identity for Structure Constants

The Jacobi identity must hold for all elements of the algebra, which means it must hold for the basis vectors $T_i$. This, in turn, imposes a condition on the structure constants $f_{ij}^k$.

Let's substitute the commutation relation $[T_i,T_j]={\sum }_m{f}_{ij}^m{T}_m$ into the Jacobi identity for the basis vectors $T_i,T_j,T_k$:

Let's expand the first term using bilinearity:

Expanding the other two terms similarly, and grouping all terms by the basis vector $T_l$, we get:

Since the basis vectors $T_l$ are linearly independent, the coefficient of each $T_l$ must be zero. This gives us the Jacobi identity in terms of structure constants:

This quadratic relationship is a powerful constraint. You cannot choose arbitrary numbers for the structure constants; they must satisfy this condition to define a valid Lie algebra. This is a non-trivial check that is sometimes required when constructing new theoretical models.

For further reference on this derivation, you can consult the text by Pfeifer.

5. Example: Verification for $\mathfrak{so}(3)$

Let's make this concrete by verifying the identity for our familiar example, $\mathfrak{so}(3)$, where the structure constants are given by the Levi-Civita symbol, $f_{ij}^k={ϵ}_{ijk}$. The condition on the structure constants becomes (relabeling the summed index to $m$ and the free index to $l$):

This is a famous identity for the Levi-Civita symbol, which you may have encountered in a course on electromagnetism or classical mechanics.

There's a beautiful way to see this identity physically. As we noted in the last lesson, the algebra $\mathfrak{so}(3)$ with the Lie bracket is isomorphic to the algebra of 3D vectors ${\mathbb{R}}^3$ with the cross product. Let the basis vectors $X_i$ of $\mathfrak{so}(3)$ correspond to the Cartesian unit vectors ${\mathbf{e}}_i$. Then the commutation relation $[X_i,X_j]={\sum }_k{ϵ}_{ijk}{X}_k$ corresponds to the cross product relation ${\mathbf{e}}_i\times {\mathbf{e}}_j={\sum }_k{ϵ}_{ijk}{\mathbf{e}}_k={\mathbf{e}}_k$ (for cyclic i,j,k).

Under this correspondence, the Jacobi identity for the Lie algebra, $[[X_i,X_j],X_k]+\cdots =0$, translates directly to an identity for the cross product:

This is precisely the Jacobi identity for the vector cross product, which can be written more generally for any three vectors $\mathbf{A},\mathbf{B},\mathbf{C}$ as:

So, the somewhat abstract Jacobi identity for Lie algebras is something you've likely used implicitly in vector calculus. It is the fundamental algebraic property that the cross product satisfies.

Conclusion

In this lesson, we completed our construction of the Lie algebra. We introduced the crucial third axiom, the Jacobi identity, which elevates the algebra from a simple space with a bracket to a structure with deep connections to the associative nature of the group from which it originates.

Key Takeaways:

• A Lie algebra is a vector space with a bilinear, antisymmetric bracket that satisfies the Jacobi identity: $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.
• For matrix Lie algebras, the Jacobi identity is an automatic consequence of the associativity of matrix multiplication.
• The Jacobi identity can be intuitively understood as the Leibniz product rule for the adjoint action, ${\mathrm{ad}}_X(Y)=[X,Y]$, where ${\mathrm{ad}}_X$ acts as a derivative.
• This identity imposes a quadratic constraint on the structure constants, ${\sum }_m(f_{ij}^m{f}_{mk}^l+f_{jk}^m{f}_{mi}^l+f_{ki}^m{f}_{mj}^l)=0$, ensuring the consistency of the algebra.
• The algebra of $\mathfrak{so}(3)$ provides a physical realization of these principles, with the Jacobi identity manifesting as the well-known identity for the vector cross product.

We now have a complete abstract definition of a Lie algebra. In the next lesson, we will focus on the map that takes us from the algebra back to the group: the exponential map. We will formalize its definition and explore its properties, including the full Baker-Campbell-Hausdorff formula which allows us to reconstruct the group product from the algebra's structure.