The Tangent Space at the Identity
The geometric idea behind Lie theory is that a curved object — a matrix Lie
group \(G\) — can be almost completely captured by a flat one: the
tangent space at the identity. Just as a smooth surface is approximated
near each of its points by a tangent plane, a matrix Lie group is approximated
near \(I\) by a vector space of velocity vectors of curves passing through \(I\).
This vector space — the focus of this page — is the Lie algebra
of \(G\). Among the smooth curves through \(I\), the
one-parameter subgroups
\(\gamma(t) = \exp(tA)\) play a privileged role: as we will see, the exponential
map is the bridge that lets us pass freely between matrices in the algebra and
smooth curves in the group.
The Lie Algebra of a Matrix Lie Group
Definition: Lie Algebra of a Matrix Lie Group
Let \(G\) be a
matrix Lie group.
The Lie algebra of \(G\) is the set of velocity vectors of
smooth curves in \(G\) through the identity:
\[
\mathfrak{g} = \{ \gamma'(0) : \gamma : \mathbb{R} \to G
\text{ is smooth and } \gamma(0) = I \}.
\]
Equivalently, \(\mathfrak{g}\) consists of all matrices
\(A \in M_n(\mathbb{C})\) satisfying \(\exp(tA) \in G\) for every
\(t \in \mathbb{R}\). We call \(\mathfrak{g}\) the tangent space of
\(G\) at the identity and write \(\mathfrak{g} = T_I G\).
The notation \(\mathfrak{g}\) uses Fraktur (Gothic) script, which is
the standard convention for Lie algebras. The Lie algebra of a group denoted by an
uppercase Roman letter is denoted by the corresponding lowercase Fraktur letter: the
Lie algebra of \(G\) is \(\mathfrak{g}\), of \(H\) is \(\mathfrak{h}\), and so on.
For named groups, we use the corresponding lowercase name: the Lie algebra of
\(GL(n, \mathbb{R})\) is \(\mathfrak{gl}(n, \mathbb{R})\), of \(SO(n)\) is
\(\mathfrak{so}(n)\), etc.
The two characterizations in the definition — via the exponential map and via tangent
vectors of smooth curves — coincide, but the two inclusions are of very different
depth. One direction is immediate: if \(\exp(tA) \in G\) for all \(t\), then
\(\gamma(t) = \exp(tA)\) is itself a smooth curve in \(G\) with \(\gamma(0) = I\) and
\(\gamma'(0) = A\). The reverse inclusion — that every velocity vector
\(A = \gamma'(0)\) of a smooth curve \(\gamma\) in \(G\) through \(I\) satisfies
\(\exp(tA) \in G\) for all \(t\) — is non-trivial and follows from Cartan's Closed
Subgroup Theorem
(Cartan's Theorem):
the theorem makes \(G\) a smooth embedded submanifold of \(GL(n, \mathbb{C})\) with a
well-defined tangent space at \(I\), and its proof identifies this tangent space
precisely with the set \(\{A : \exp(tA) \in G \text{ for all } t\}\). Throughout this
page we take the smooth-curve characterization as our operative definition: it is the
natural one for structural arguments — closure of \(\mathfrak{g}\) under sums and
brackets, and the identification of each classical \(\mathfrak{g}\) by
differentiating the defining equation of \(G\). The exponential characterization
enters as a powerful computational tool, used in the converse direction of each
classical-Lie-algebra identification: given \(A\) satisfying the linear condition,
\(t \mapsto \exp(tA)\) supplies the smooth curve in \(G\) realizing \(A\) as a
tangent vector.
Theorem: The Lie Algebra is a Real Vector Space
Let \(G\) be a matrix Lie group. Then \(\mathfrak{g}\) is a
real vector subspace of \(M_n(\mathbb{C})\). That is,
\(\mathfrak{g}\) is closed under real scalar multiplication and addition.
Proof:
We use the smooth-curve characterization: \(X \in \mathfrak{g}\) means
\(X = \gamma'(0)\) for some smooth curve \(\gamma : \mathbb{R} \to G\) with
\(\gamma(0) = I\). The proof exploits the product structure of \(G\)
directly, lifting it to algebraic operations on \(\mathfrak{g}\).
Closure under scalar multiplication. Let
\(X \in \mathfrak{g}\) and \(r \in \mathbb{R}\). Choose a smooth curve
\(A(t)\) in \(G\) with \(A(0) = I\) and \(A'(0) = X\). Define
\(D(t) = A(rt)\). Then \(D\) is smooth, \(D(0) = A(0) = I\) (so \(D\) is a
smooth curve in \(G\) through \(I\)), and by the chain rule
\[
D'(0) = r\,A'(0) = rX.
\]
Therefore \(rX \in \mathfrak{g}\).
Closure under addition. Let \(X, Y \in \mathfrak{g}\).
Choose smooth curves \(A(t), B(t)\) in \(G\) with
\(A(0) = B(0) = I\), \(A'(0) = X\), \(B'(0) = Y\). Define
\[
C(t) = A(t)\,B(t).
\]
Since \(G\) is closed under products, \(C(t) \in G\) for every \(t\); since
\(A\) and \(B\) are smooth, so is \(C\); and \(C(0) = I\). Therefore \(C\)
is a smooth curve in \(G\) through \(I\), so \(C'(0) \in \mathfrak{g}\).
By the product rule,
\[
C'(0) = A'(0)\,B(0) + A(0)\,B'(0) = X \cdot I + I \cdot Y = X + Y.
\]
Therefore \(X + Y \in \mathfrak{g}\).
With \(\mathfrak{g}\) established as a real vector subspace of
\(M_n(\mathbb{C})\), the next step is concrete identification: for each classical
matrix Lie group, we ask which matrices \(A\) lie in its Lie algebra. The answer
is a linear condition extracted by differentiating the group's defining equation
along a smooth curve through \(I\).
The Classical Lie Algebras
In The Matrix Exponential,
we established which linear conditions on \(A\) ensure that \(\exp(A)\) lands in each
classical group. We now identify these conditions as defining the corresponding Lie
algebras. For each classical group, the proof that the stated set equals
\(\mathfrak{g}\) proceeds in two directions. The "if" direction uses the
exponential characterization: if \(A\) satisfies the linear condition, then
\(\exp(tA) \in G\) for all \(t\), and \(t \mapsto \exp(tA)\) is itself a smooth curve
in \(G\) through \(I\) with velocity \(A\), so \(A \in \mathfrak{g}\). The
"only if" direction uses the smooth-curve characterization directly:
given \(A \in \mathfrak{g}\), choose any smooth curve \(\gamma\) in \(G\) with
\(\gamma(0) = I\) and \(\gamma'(0) = A\), and differentiate the defining equation of
\(G\) at \(t = 0\) to extract the linear condition on \(A\).
Definition: \(\mathfrak{gl}(n, \mathbb{R})\)
The Lie algebra of \(GL(n, \mathbb{R})\) is
\[
\mathfrak{gl}(n, \mathbb{R}) = M_n(\mathbb{R}),
\]
the space of all \(n \times n\) real matrices, with no additional constraint.
This follows immediately from the fact that
\(\exp(tA) \in GL(n, \mathbb{R})\) for all \(t\) and all
\(A \in M_n(\mathbb{R})\), since
\(\det(\exp(tA)) = e^{t\,\mathrm{tr}(A)} \neq 0\). The dimension is \(n^2\).
Definition: \(\mathfrak{sl}(n, \mathbb{R})\)
The Lie algebra of
\(SL(n, \mathbb{R})\)
is
\[
\mathfrak{sl}(n, \mathbb{R}) = \{ A \in M_n(\mathbb{R}) : \mathrm{tr}(A) = 0 \},
\]
the space of traceless real matrices. The dimension is \(n^2 - 1\).
Proof:
(If) If \(\mathrm{tr}(A) = 0\), then
\(\det(\exp(tA)) = e^{t\,\mathrm{tr}(A)} = e^0 = 1\), so
\(\exp(tA) \in SL(n, \mathbb{R})\) for all \(t\). Hence
\(A \in \mathfrak{sl}(n, \mathbb{R})\).
(Only if) Suppose \(A \in \mathfrak{sl}(n, \mathbb{R})\).
Choose a smooth curve \(\gamma\) in \(SL(n, \mathbb{R})\) with
\(\gamma(0) = I\) and \(\gamma'(0) = A\). The defining equation of
\(SL(n, \mathbb{R})\) is \(\det \gamma(t) = 1\) for all \(t\). Writing the
Taylor expansion \(\gamma(t) = I + tA + O(t^2)\) and using the standard
first-order expansion of the determinant (an entry-wise calculation from
Leibniz's formula),
\[
\det\bigl(I + tA + O(t^2)\bigr) = 1 + t\,\mathrm{tr}(A) + O(t^2),
\]
the constraint \(\det \gamma(t) = 1\) yields
\(t\,\mathrm{tr}(A) + O(t^2) = 0\) for all \(t\). The vanishing of the
linear coefficient forces \(\mathrm{tr}(A) = 0\).
Definition: \(\mathfrak{so}(n)\)
The Lie algebra of both \(O(n)\) and
\(SO(n)\)
is
\[
\mathfrak{so}(n) = \{ A \in M_n(\mathbb{R}) : A^\top = -A \},
\]
the space of skew-symmetric (or antisymmetric) real matrices.
The dimension is \(n(n-1)/2\).
Proof:
(If) If \(A^\top = -A\), then
\(\exp(tA)^\top = \exp(tA^\top) = \exp(-tA) = \exp(tA)^{-1}\), so
\(\exp(tA) \in O(n)\) for all \(t\). Moreover, the map
\(t \mapsto \det(\exp(tA)) = e^{t\,\mathrm{tr}(A)} = e^0 = 1\) is constantly 1
(since every skew-symmetric matrix has zero diagonal, hence zero trace), so
\(\exp(tA) \in SO(n)\) for all \(t\).
(Only if) Suppose \(A \in \mathfrak{so}(n)\). Choose a smooth
curve \(\gamma\) in \(O(n)\) with \(\gamma(0) = I\) and \(\gamma'(0) = A\).
The defining equation of \(O(n)\) is
\(\gamma(t)^\top \gamma(t) = I\) for all \(t\). Since transposition acts
entry-wise, it commutes with differentiation:
\(\frac{d}{dt} \gamma(t)^\top = \gamma'(t)^\top\). Differentiating at
\(t = 0\) using the product rule:
\[
\left.\frac{d}{dt}\right|_{t=0} \bigl[\gamma(t)^\top \gamma(t)\bigr]
= \gamma'(0)^\top \gamma(0) + \gamma(0)^\top \gamma'(0)
= A^\top \cdot I + I \cdot A = A^\top + A = 0.
\]
Therefore \(A^\top = -A\).
The proof in fact shows that \(O(n)\) and \(SO(n)\) have the
same Lie algebra. Any smooth curve \(\gamma : \mathbb{R} \to O(n)\) with
\(\gamma(0) = I\) automatically lies in \(SO(n)\): the function
\(t \mapsto \det \gamma(t)\) is continuous and takes only the values \(\pm 1\), so
by connectedness of \(\mathbb{R}\) and \(\det \gamma(0) = 1\), it is constantly
\(1\). Hence the velocity vectors realized by curves through \(I\) are the same
in \(O(n)\) and in \(SO(n)\), giving \(\mathfrak{o}(n) = \mathfrak{so}(n)\). This
reflects a general principle: the Lie algebra captures only the
local structure of a group near the identity, and \(SO(n)\) is
precisely the connected component of \(O(n)\) containing \(I\).
Definition: \(\mathfrak{u}(n)\)
The Lie algebra of
\(U(n)\)
is
\[
\mathfrak{u}(n) = \{ A \in M_n(\mathbb{C}) : A^* = -A \},
\]
the space of skew-Hermitian matrices. The dimension is \(n^2\)
(as a real vector space: the diagonal entries are purely imaginary, giving \(n\)
real parameters, and the strictly upper-triangular entries are arbitrary complex
numbers, giving \(2 \cdot \frac{n(n-1)}{2} = n(n-1)\) real parameters; total
\(n + n(n-1) = n^2\)).
Definition: \(\mathfrak{su}(n)\)
The Lie algebra of
\(SU(n)\)
is
\[
\mathfrak{su}(n) = \{ A \in M_n(\mathbb{C}) : A^* = -A,\; \mathrm{tr}(A) = 0 \},
\]
the space of traceless skew-Hermitian matrices. The dimension is \(n^2 - 1\).
The proofs for \(\mathfrak{u}(n)\) and \(\mathfrak{su}(n)\) are entirely analogous to
those for \(\mathfrak{so}(n)\) and \(\mathfrak{sl}(n)\), replacing the transpose
\(A^\top\) with the conjugate transpose \(A^*\).
The following table collects the classical Lie algebras alongside the groups from which
they arise. Compare this with the summary table in
Matrix Lie Groups and the
linear-to-nonlinear correspondence in
The Matrix Exponential:
| Group \(G\) |
Lie Algebra \(\mathfrak{g}\) |
Defining Condition on \(A \in \mathfrak{g}\) |
\(\dim_{\mathbb{R}} \mathfrak{g}\) |
| \(GL(n, \mathbb{R})\) |
\(\mathfrak{gl}(n, \mathbb{R})\) |
(no constraint) |
\(n^2\) |
| \(SL(n, \mathbb{R})\) |
\(\mathfrak{sl}(n, \mathbb{R})\) |
\(\mathrm{tr}(A) = 0\) |
\(n^2 - 1\) |
| \(O(n)\) / \(SO(n)\) |
\(\mathfrak{so}(n)\) |
\(A^\top = -A\) |
\(n(n-1)/2\) |
| \(U(n)\) |
\(\mathfrak{u}(n)\) |
\(A^* = -A\) |
\(n^2\) |
| \(SU(n)\) |
\(\mathfrak{su}(n)\) |
\(A^* = -A,\; \mathrm{tr}(A) = 0\) |
\(n^2 - 1\) |
Explicit Basis for \(\mathfrak{so}(3)\)
The Lie algebra \(\mathfrak{so}(3)\) — the tangent space of the rotation group
\(SO(3)\) at the identity — is a 3-dimensional real vector space. A natural basis
consists of the infinitesimal generators introduced in the previous page:
\[
E_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \quad
E_2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \quad
E_3 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.
\]
Each \(E_i\) is skew-symmetric (\(E_i^\top = -E_i\)), confirming
\(E_i \in \mathfrak{so}(3)\). They are linearly independent and
\(\dim \mathfrak{so}(3) = 3(3-1)/2 = 3\), so \(\{E_1, E_2, E_3\}\) is a basis.
Recall that the
hat map
\(\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)^\top \mapsto \hat{\boldsymbol{\omega}}_\times = \omega_1 E_1 + \omega_2 E_2 + \omega_3 E_3\)
provides an isomorphism \(\mathbb{R}^3 \xrightarrow{\;\sim\;} \mathfrak{so}(3)\) of
vector spaces. We will soon see that this isomorphism also respects additional algebraic
structure: the cross product on \(\mathbb{R}^3\) corresponds to the Lie bracket on
\(\mathfrak{so}(3)\).
The Lie Bracket
The proof of vector-space closure showed that the algebra sum on \(\mathfrak{g}\)
is, in a sense, the linearization of the group product on \(G\) — the velocity
of \(A(t)B(t)\) at \(t=0\) is \(X+Y\). But this is only a partial picture: the
group product is generally non-commutative, while vector addition is always
commutative. The algebra inherits something more from the group than just its
sum, and to capture it we must study the behavior of two smooth paths
\(A(s), B(t)\) when \(s\) and \(t\) vary independently. The non-commutative
content extracted from this two-parameter behavior is the Lie bracket —
a second algebraic operation on \(\mathfrak{g}\) that completes the picture.
Closure of the Lie Algebra under the Commutator
The first question is: if \(A\) and \(B\) belong to the Lie algebra \(\mathfrak{g}\),
does their commutator \([A, B] = AB - BA\) also belong to \(\mathfrak{g}\)? Note that
this is not obvious — \(\mathfrak{g}\) is defined as a subspace of
\(M_n(\mathbb{C})\), and the product of two elements of \(\mathfrak{g}\) need not lie
in \(\mathfrak{g}\) (for instance, the product of two skew-symmetric matrices is
generally not skew-symmetric). The answer is yes, and the proof is
illuminating.
Theorem: Closure under the Commutator
Let \(G\) be a matrix Lie group with Lie algebra \(\mathfrak{g}\). If
\(A, B \in \mathfrak{g}\), then \([A, B] = AB - BA \in \mathfrak{g}\).
Proof:
As with vector space closure, we use the smooth-curve characterization of
\(\mathfrak{g}\). Choose smooth curves \(A(s), B(t)\) in \(G\) with
\(A(0) = B(0) = I\), \(A'(0) = X\), \(B'(0) = Y\). The bracket emerges from
the conjugation of \(B\) by \(A\) — the operation
\(B \mapsto A B A^{-1}\) — which captures non-commutativity directly.
Step 1: Conjugation paths land in \(\mathfrak{g}\).
For each fixed \(s\), define
\[
C_s(t) = A(s)\,B(t)\,A(s)^{-1}.
\]
Since \(A(s) \in G\) and \(B(t) \in G\) and \(G\) is closed under products
and inverses, \(C_s(t) \in G\) for every \(t\). Furthermore \(C_s\) is smooth
in \(t\) and \(C_s(0) = A(s) \cdot I \cdot A(s)^{-1} = I\). So \(C_s\) is a
smooth curve in \(G\) through \(I\), and its velocity at \(t = 0\) lies in
\(\mathfrak{g}\):
\[
C_s'(0) = A(s)\,B'(0)\,A(s)^{-1} = A(s)\,Y\,A(s)^{-1} \in \mathfrak{g}.
\]
Step 2: The conjugation curve in \(\mathfrak{g}\) is itself smooth.
Define \(D : \mathbb{R} \to M_n(\mathbb{C})\) by
\[
D(s) = A(s)\,Y\,A(s)^{-1}.
\]
Since \(A(s)\) is smooth in \(s\) and matrix inversion is smooth on
\(GL(n, \mathbb{C})\), \(D\) is smooth. By Step 1,
\(D(s) \in \mathfrak{g}\) for every \(s\). Now \(\mathfrak{g}\) is a real
vector subspace of the finite-dimensional space \(M_n(\mathbb{C})\) (viewed
as a real vector space of dimension \(2n^2\)), hence itself finite-dimensional
and therefore closed in \(M_n(\mathbb{C})\). A smooth curve lying entirely in
a closed subspace has its derivative also in that subspace, so
\[
D'(0) \in \mathfrak{g}.
\]
Step 3: The derivative is the commutator. Differentiate
\(D(s) = A(s)\,Y\,A(s)^{-1}\) using the product rule and the identity
\(\frac{d}{ds} A(s)^{-1}\big|_{s=0} = -A'(0) = -X\) (which follows from
differentiating \(A(s)\,A(s)^{-1} = I\) at \(s = 0\)):
\[
D'(0) = A'(0)\,Y\,A(0)^{-1} + A(0)\,Y \cdot \left(-A'(0)\right)
= X\,Y \cdot I + I \cdot Y \cdot (-X) = XY - YX = [X, Y].
\]
Combining with Step 2, \([X, Y] \in \mathfrak{g}\).
The proof reflects the structural relationship between group and algebra:
where the vector sum \(X + Y\) arises from multiplying two paths
in \(G\), the commutator \([X, Y]\) arises from conjugating one path
by another. Conjugation \(B \mapsto A B A^{-1}\) is the canonical operation
that detects non-commutativity in any group: it is trivial precisely when
\(A\) and \(B\) commute. At the level of \(\mathfrak{g}\), the same operation
becomes the bracket — making \([X, Y]\) the algebraic shadow of group
non-commutativity.
A useful specialization of the proof is obtained by choosing the smooth curves
to be one-parameter subgroups: take \(A(s) = \exp(sX)\) and \(B(t) = \exp(tY)\).
Then the conjugation curve becomes
\[
D(s) = \exp(sX)\,Y\,\exp(-sX),
\]
and the Step 3 calculation gives the same conclusion in a particularly clean form:
\[
[X, Y] = \left.\frac{d}{ds}\right|_{s=0} \exp(sX)\,Y\,\exp(-sX).
\]
This identity expresses the bracket as the infinitesimal generator of conjugation
by the one-parameter subgroup \(\exp(sX)\), and it will be the operative form when
we construct the adjoint representations in
The Lie Correspondence.
Definition and Properties
Having established that the commutator preserves the Lie algebra, we now formalize it as
an algebraic operation.
Definition: Lie Bracket (Matrix Case)
Let \(\mathfrak{g}\) be the Lie algebra of a matrix Lie group. The
Lie bracket is the operation
\([\,\cdot\,,\,\cdot\,] : \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}\)
defined by
\[
[X, Y] = XY - YX.
\]
Theorem: Properties of the Lie Bracket
For all \(X, Y, Z \in \mathfrak{g}\) and \(a, b \in \mathbb{R}\):
(a) Bilinearity:
\[
\begin{align*}
[aX + bY,\, Z] &= a[X, Z] + b[Y, Z], \\\\
[Z,\, aX + bY] &= a[Z, X] + b[Z, Y].
\end{align*}
\]
(b) Antisymmetry:
\[
[X, Y] = -[Y, X].
\]
In particular, \([X, X] = 0\) for all \(X \in \mathfrak{g}\).
(c) Jacobi identity:
\[
[X,\, [Y, Z]] + [Y,\, [Z, X]] + [Z,\, [X, Y]] = 0.
\]
Proofs:
(a) Bilinearity follows directly from the linearity of matrix
multiplication in each factor:
\[
\begin{align*}
[aX + bY, Z] &= (aX + bY)Z - Z(aX + bY) \\\\
&= a(XZ - ZX) + b(YZ - ZY) = a[X, Z] + b[Y, Z].
\end{align*}
\]
The second identity is proved identically.
(b) Immediate:
\[
[X, Y] = XY - YX = -(YX - XY) = -[Y, X].
\]
Setting \(Y = X\) gives \([X, X] = -[X, X]\), hence \([X, X] = 0\).
(c) We expand each term. Writing \([X, [Y, Z]] = X(YZ - ZY) - (YZ - ZY)X = XYZ - XZY - YZX + ZYX\) and cyclically permuting:
\[
\begin{align*}
[X, [Y, Z]] &= XYZ - XZY - YZX + ZYX, \\\\
[Y, [Z, X]] &= YZX - YXZ - ZXY + XZY, \\\\
[Z, [X, Y]] &= ZXY - ZYX - XYZ + YXZ.
\end{align*}
\]
Adding these three expressions, every term cancels:
\(XYZ\) appears once with \(+\) (first line) and once with \(-\) (third line);
similarly for all other terms. The sum is zero.
The Jacobi identity is the Lie-algebraic analogue of associativity. While the Lie bracket
is not associative — in general,
\([X, [Y, Z]] \neq [[X, Y], Z]\) — the Jacobi identity provides a weaker constraint
that governs how brackets interact. It can be rewritten as
\[
[X, [Y, Z]] = [[X, Y], Z] + [Y, [X, Z]],
\]
which states that the operation \(\mathrm{ad}(X) : Y \mapsto [X, Y]\) is a
derivation with respect to the bracket: it satisfies a Leibniz-type
rule. This perspective will become central when we study the adjoint representations in
The Lie Correspondence.
The Abstract Definition
The three properties above — bilinearity, antisymmetry, and the Jacobi identity —
characterize Lie algebras in full generality.
Definition: Lie Algebra (Abstract)
A Lie algebra over a field \(\mathbb{F}\) is a vector space
\(\mathfrak{g}\) over \(\mathbb{F}\) equipped with a bilinear operation
\([\,\cdot\,,\,\cdot\,] : \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}\)
(the Lie bracket) satisfying:
- Antisymmetry: \([X, Y] = -[Y, X]\) for all \(X, Y \in \mathfrak{g}\).
- Jacobi identity: \([X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0\) for all \(X, Y, Z \in \mathfrak{g}\).
Every Lie algebra of a matrix Lie group is a Lie algebra in this abstract sense, with
\(\mathbb{F} = \mathbb{R}\) and the bracket \([X, Y] = XY - YX\). The converse — that
every finite-dimensional (abstract) Lie algebra is isomorphic to a matrix Lie algebra —
is a deep result known as Ado's theorem. Its proof is far beyond our
scope, but the theorem assures us that, at the level of Lie algebras, the matrix
setting loses no generality. (At the level of Lie groups, the situation is
subtler: there exist Lie groups — such as the universal cover
\(\widetilde{SL(2, \mathbb{R})}\) — that cannot be realized as matrix subgroups of
any \(GL(n)\). Such examples lie beyond this curriculum, but the distinction between
"every Lie algebra is matrix" and "every Lie group is matrix" is worth noting.)
Structure Constants and the Algebra of Rotations
A Lie algebra is determined, relative to a chosen basis, by its
structure constants — the coefficients that express each bracket of
basis elements as a linear combination of basis elements. We compute these for
\(\mathfrak{so}(3)\) and discover a striking connection to the cross product.
Structure Constants
Let \(\mathfrak{g}\) be a finite-dimensional Lie algebra with basis
\(\{E_1, \dots, E_d\}\). Since the bracket \([E_i, E_j]\) belongs to \(\mathfrak{g}\),
it can be written as a linear combination of basis elements:
\[
[E_i, E_j] = \sum_{k=1}^{d} c_{ij}^{\,k}\, E_k.
\]
The scalars \(c_{ij}^{\,k}\) are called the structure constants of
\(\mathfrak{g}\) with respect to the basis \(\{E_i\}\). By bilinearity, the bracket
of any two elements is determined by these constants: if
\(X = \sum_i x_i E_i\) and \(Y = \sum_j y_j E_j\), then
\[
[X, Y] = \sum_{i,j} x_i y_j\, [E_i, E_j]
= \sum_{i,j,k} x_i y_j\, c_{ij}^{\,k}\, E_k.
\]
The structure constants encode the Lie algebra completely (relative to the chosen basis).
From antisymmetry, the structure constants satisfy
\(c_{ij}^{\,k} = -c_{ji}^{\,k}\), and the Jacobi identity imposes further
quadratic relations among them.
The Bracket of \(\mathfrak{so}(3)\)
We now compute the Lie bracket for the basis \(\{E_1, E_2, E_3\}\) of
\(\mathfrak{so}(3)\) introduced in the previous section.
Computation:
We compute \([E_1, E_2] = E_1 E_2 - E_2 E_1\) by direct matrix multiplication:
\[
E_1 E_2 =
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}
= \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\]
\[
E_2 E_1 =
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}
= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\]
\[
[E_1, E_2] = E_1 E_2 - E_2 E_1
= \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = E_3.
\]
By analogous (or cyclic) computations:
\[
[E_1, E_2] = E_3, \qquad [E_2, E_3] = E_1, \qquad [E_3, E_1] = E_2.
\]
Together with the antisymmetry relations
\([E_2, E_1] = -E_3\), \([E_3, E_2] = -E_1\), \([E_1, E_3] = -E_2\), and the
vanishing brackets \([E_i, E_i] = 0\), these determine the full bracket on
\(\mathfrak{so}(3)\).
The structure constants of \(\mathfrak{so}(3)\) with respect to
\(\{E_1, E_2, E_3\}\) are therefore
\[
c_{ij}^{\,k} = \varepsilon_{ijk},
\]
where \(\varepsilon_{ijk}\) is the Levi-Civita symbol:
\(\varepsilon_{123} = \varepsilon_{231} = \varepsilon_{312} = +1\),
\(\varepsilon_{213} = \varepsilon_{132} = \varepsilon_{321} = -1\), and
\(\varepsilon_{ijk} = 0\) whenever two indices coincide.
The Cross Product Isomorphism
The bracket relations \([E_1, E_2] = E_3\), \([E_2, E_3] = E_1\),
\([E_3, E_1] = E_2\) are identical to the relations defining the
cross product on \(\mathbb{R}^3\):
\(\mathbf{e}_1 \times \mathbf{e}_2 = \mathbf{e}_3\),
\(\mathbf{e}_2 \times \mathbf{e}_3 = \mathbf{e}_1\),
\(\mathbf{e}_3 \times \mathbf{e}_1 = \mathbf{e}_2\). This is not a coincidence.
The hat map \(\boldsymbol{\omega} \mapsto \hat{\boldsymbol{\omega}}_\times\) from
\(\mathbb{R}^3\) to \(\mathfrak{so}(3)\) is an isomorphism of Lie algebras:
\[
\widehat{\boldsymbol{\omega}_1 \times \boldsymbol{\omega}_2}
= [\hat{\boldsymbol{\omega}}_{1,\times},\, \hat{\boldsymbol{\omega}}_{2,\times}]
\]
for all \(\boldsymbol{\omega}_1, \boldsymbol{\omega}_2 \in \mathbb{R}^3\). In
words: the cross product on \(\mathbb{R}^3\) is the Lie bracket on
\(\mathfrak{so}(3)\), transferred via the hat map.
This isomorphism \((\mathbb{R}^3, \times) \cong (\mathfrak{so}(3), [\,\cdot\,,\,\cdot\,])\)
is an exceptional phenomenon: it relies on the fact that both spaces are 3-dimensional
and that the Levi-Civita symbol is totally antisymmetric. There is no analogous
cross-product isomorphism for \(\mathfrak{so}(n)\) when \(n \neq 3\) (since
\(\dim \mathfrak{so}(n) = n(n-1)/2 \neq n\) for \(n \neq 3\)).
Angular Velocity and the Equation of Rotation
The cross product isomorphism gives physical meaning to the Lie algebra
\(\mathfrak{so}(3)\). In mechanics, the angular velocity of a
rotating body is a vector
\(\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)^\top \in \mathbb{R}^3\).
Via the hat map, this corresponds to the skew-symmetric matrix
\(\hat{\boldsymbol{\omega}}_\times \in \mathfrak{so}(3)\). The kinematic equation
of a rotating rigid body is the ODE on \(SO(3)\):
\[
\frac{dR}{dt} = \hat{\boldsymbol{\omega}}_\times\, R,
\]
where \(R(t) \in SO(3)\) describes the orientation of the body at time \(t\)
and \(\boldsymbol{\omega}\) is the angular velocity expressed in the
spatial frame (the body-frame convention instead writes
\(\dot{R} = R\,\hat{\boldsymbol{\omega}}^b_\times\), with the two related by
\(\boldsymbol{\omega} = R\,\boldsymbol{\omega}^b\)).
This is an ODE on the Lie group \(SO(3)\), driven by a time-varying element of
the Lie algebra \(\mathfrak{so}(3)\). When \(\boldsymbol{\omega}\) is constant,
the solution is
\(R(t) = \exp(t\,\hat{\boldsymbol{\omega}}_\times)\,R(0)\) — a one-parameter
subgroup applied to the initial orientation.
The structure constants of \(\mathfrak{so}(3)\) — the Levi-Civita symbol — encode
the physics of gyroscopic precession. The relation \([E_1, E_2] = E_3\) means that
combining infinitesimal rotations about the \(x\)- and \(y\)-axes produces an
infinitesimal rotation about the \(z\)-axis. This is the mathematical content of
the right-hand rule.
Abelian and Non-Abelian Lie Algebras
A Lie algebra \(\mathfrak{g}\) is called abelian if
\([X, Y] = 0\) for all \(X, Y \in \mathfrak{g}\). This occurs precisely when the
corresponding Lie group is locally commutative (i.e., commutative in a
neighborhood of the identity).
Examples:
(a) \(\mathfrak{gl}(1, \mathbb{R}) = \mathbb{R}\) is abelian:
the bracket of two real numbers is \([a, b] = ab - ba = 0\). The group
\(GL(1, \mathbb{R}) = \mathbb{R} \setminus \{0\}\) is commutative (multiplication
of real numbers is commutative).
(b) \(\mathfrak{so}(2) \cong \mathbb{R}\) is abelian: it is
1-dimensional, so any bracket \([cJ, dJ] = cd[J, J] = 0\) where
\(J = \bigl(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\bigr)\).
The group \(SO(2) \cong S^1\) is commutative (rotations of the plane commute).
(c) \(\mathfrak{so}(3)\) is non-abelian:
\([E_1, E_2] = E_3 \neq 0\). This reflects the non-commutativity of 3D rotations,
which we first encountered in the context of the
dihedral group
\(D_n\) and its non-commuting generators. The passage from \(D_n\) (discrete,
non-abelian) to \(SO(3)\) (continuous, non-abelian) preserves the essential
failure of commutativity — and the Lie bracket provides the precise infinitesimal
measure of it.
Looking Ahead
We have defined the Lie algebra \(\mathfrak{g}\) of a matrix Lie group \(G\) as the
tangent space at the identity, computed the classical Lie algebras, and discovered that
the commutator \([X, Y] = XY - YX\) equips \(\mathfrak{g}\) with the structure of an
abstract Lie algebra. The worked example of \(\mathfrak{so}(3)\) revealed that the
Lie bracket encodes rotational physics — the structure constants are the Levi-Civita
symbol, and the bracket is the cross product.
A fundamental question remains: to what extent does the Lie algebra determine the
Lie group? We have seen that different groups can share the same Lie algebra
(\(O(n)\) and \(SO(n)\) both have Lie algebra \(\mathfrak{so}(n)\)). How much group
information is captured by the algebra, and how much is lost?
In the next page, we answer this
question through the Lie group-Lie algebra correspondence. The
Baker-Campbell-Hausdorff formula will show that the group multiplication
near the identity is entirely encoded by the Lie bracket, establishing that
\(\mathfrak{g}\) determines the local structure of \(G\) — the precise sense
in which a flat tangent space "almost completely" captures a curved group, as we
promised at the outset. The qualifier almost is not idle: it is exactly the
gap between local and global. The dramatic example of \(\mathfrak{su}(2)\) and
\(\mathfrak{so}(3)\) — isomorphic Lie algebras whose groups \(SU(2)\) and \(SO(3)\)
have different global topology and are connected by a 2:1 covering map — exhibits
this gap concretely. Finally, the
adjoint representations \(\mathrm{Ad}\) and \(\mathrm{ad}\) will show
how the group acts on its own Lie algebra, bridging the theory toward
Representation Theory.