The Tangent Space at the Identity
The geometric idea behind Lie theory is that a curved object, namely a matrix Lie
group \(G\), can, when connected, 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
play a privileged role, and every one of them
has the form
\(\gamma(t) = \exp(tA)\). 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 \(\mathfrak{g}\), that of \(H\) is \(\mathfrak{h}\), and so
on. For named groups, we use the corresponding lowercase name. Thus
\(GL(n, \mathbb{R})\) has Lie algebra \(\mathfrak{gl}(n, \mathbb{R})\) and \(SO(n)\)
has \(\mathfrak{so}(n)\).
The definition offers two characterizations, one via the exponential map and one via
tangent vectors of smooth curves. They 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 asserts that every velocity vector \(A = \gamma'(0)\) of a
smooth curve \(\gamma\) in \(G\) through \(I\) satisfies \(\exp(tA) \in G\) for all
\(t\). That inclusion 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 \(I\). The identification of that tangent space with the
set \(\{A : \exp(tA) \in G \text{ for all } t\}\) is part of the standard argument for
Cartan's theorem, which rests on the inverse function theorem on manifolds. We take
that identification as given and do not prove it on this page.
Throughout this page we take the smooth-curve characterization as our operative
definition. It is the natural one for structural arguments, namely 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. Membership \(X \in \mathfrak{g}\) means
that \(X = \gamma'(0)\) for some smooth curve \(\gamma : \mathbb{R} \to G\) with
\(\gamma(0) = I\). The proof exploits the product structure of \(G\) directly and
lifts 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,
\[
\begin{align*}
C'(0) &= A'(0)\,B(0) + A(0)\,B'(0) \\\\
&= X \cdot I + I \cdot Y \\\\
&= X + Y.
\end{align*}
\]
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 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 \(A \in \mathfrak{g}\), we 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 \(A\). When \(G\) consists of real matrices, the curve takes its values
in \(M_n(\mathbb{R})\), so its velocity \(A\) is real as well. This is the reality
constraint carried by the displays for the real groups below.
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.
One inclusion follows from the fact that \(\exp(tA) \in GL(n, \mathbb{R})\) for
all \(t\) and all \(A \in M_n(\mathbb{R})\), since
\(\det(\exp(tA)) = e^{t\,\mathrm{tr}(A)}\)
is never zero. Conversely, velocity vectors of curves in \(GL(n, \mathbb{R})\)
are real, which gives the reverse inclusion. 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 \(A\) lies in the Lie
algebra of \(SL(n, \mathbb{R})\).
(Only if) Suppose \(A\) lies in the Lie algebra of
\(SL(n, \mathbb{R})\).
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 \(t\). The first equality holds because
transposition is linear and continuous, so it passes through the partial sums of
the defining series, and the third is the
inverse formula \(\exp(B)^{-1} = \exp(-B)\).
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 \(t\). Hence \(A\) belongs to the Lie algebra of
\(O(n)\), and equally to that of \(SO(n)\).
(Only if) Suppose \(A\) belongs to the Lie algebra of \(O(n)\).
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:
\[
\begin{align*}
\left.\frac{d}{dt}\right|_{t=0} \bigl[\gamma(t)^\top \gamma(t)\bigr]
&= \gamma'(0)^\top \gamma(0) + \gamma(0)^\top \gamma'(0) \\\\
&= A^\top \cdot I + I \cdot A \\\\
&= A^\top + A = 0.
\end{align*}
\]
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. Here the diagonal entries are purely imaginary, giving \(n\)
real parameters, while the strictly upper-triangular entries are arbitrary complex
numbers, giving \(2 \cdot \frac{n(n-1)}{2} = n(n-1)\) more, for a total of
\(n + n(n-1) = n^2\).
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:
\[
\begin{align*}
E_1 &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \\\\
E_2 &= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \\\\
E_3 &= \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.
\end{align*}
\]
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.
The
hat map
was introduced for unit vectors in the previous page. The same formula
\(\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)^\top \mapsto \hat{\boldsymbol{\omega}}_\times = \omega_1 E_1 + \omega_2 E_2 + \omega_3 E_3\)
is defined for every \(\boldsymbol{\omega} \in \mathbb{R}^3\), and we use it in that
generality from here on. Carrying the standard basis of \(\mathbb{R}^3\) to the basis
\(\{E_1, E_2, E_3\}\), it
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. The set \(\mathfrak{g}\) 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 \(\mathfrak{g}\):
\[
\begin{align*}
C_s'(0) &= A(s)\,B'(0)\,A(s)^{-1} \\\\
&= A(s)\,Y\,A(s)^{-1} \in \mathfrak{g}.
\end{align*}
\]
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 \(s = 0\)):
\[
\begin{align*}
D'(0) &= A'(0)\,Y\,A(0)^{-1} + A(0)\,Y \cdot \left(-A'(0)\right) \\\\
&= X\,Y \cdot I + I \cdot Y \cdot (-X) \\\\
&= XY - YX = [X, Y].
\end{align*}
\]
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, and
\([X, Y]\) is 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. We take \(A(s) = \exp(sX)\) and \(B(t) = \exp(tY)\), which
lie in \(G\) by the exponential characterization of \(\mathfrak{g}\). 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, we find that every term cancels. The product
\(XYZ\) appears once with \(+\) (first line) and once with \(-\) (third line), and
the same holds for every other term. The sum is zero.
The Jacobi identity is the Lie-algebraic analogue of associativity. The bracket itself
is not associative, since in general \([X, [Y, Z]] \neq [[X, Y], Z]\), but
the 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, meaning that it satisfies a
Leibniz-type rule. This perspective will become central when we study the adjoint
representations in
The Lie Correspondence.
The Abstract Definition
Bilinearity, antisymmetry, and the Jacobi identity characterize Lie algebras in full
generality.
Definition: Lie Algebra (Abstract)
A Lie algebra over \(\mathbb{F}\) (either \(\mathbb{R}\) or
\(\mathbb{C}\)) 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 \([X, Y] = XY - YX\). A deep result in the
opposite direction, Ado's theorem, states that every
finite-dimensional abstract Lie algebra is isomorphic to a Lie algebra of matrices,
that is, to a subspace of some \(M_n(\mathbb{F})\) closed under the commutator. The
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 in
terms of the basis itself. We compute these for \(\mathfrak{so}(3)\) and discover that
the bracket reproduces the cross product.
Structure Constants
Let \(\mathfrak{g}\) be a finite-dimensional Lie algebra with basis
\(\{E_1, \ldots, E_d\}\). 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
\[
\begin{align*}
[X, Y] &= \sum_{i,j} x_i y_j\, [E_i, E_j] \\\\
&= \sum_{i,j,k} x_i y_j\, c_{ij}^{\,k}\, E_k.
\end{align*}
\]
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:
\[
\begin{align*}
E_1 E_2 &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \\\\
&= \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\end{align*}
\]
\[
\begin{align*}
E_2 E_1 &= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \\\\
&= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\end{align*}
\]
\[
\begin{align*}
[E_1, E_2] &= E_1 E_2 - E_2 E_1 \\\\
&= \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = E_3.
\end{align*}
\]
By analogous (or cyclic) computations:
\[
\begin{align*}
[E_1, E_2] &= E_3, \\\\
[E_2, E_3] &= E_1, \\\\
[E_3, E_1] &= E_2.
\end{align*}
\]
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 conventions
related by \(\boldsymbol{\omega} = R\,\boldsymbol{\omega}^b\).
The equation 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, are the
coefficients of the cross product, which also appears in the equations of motion
of a spinning rigid body and so in gyroscopic precession. The relation
\([E_1, E_2] = E_3\) says that the failure of rotations about the \(x\)- and
\(y\)-axes to commute is itself an infinitesimal rotation about the \(z\)-axis,
with the sign fixed by the right-hand rule.
Abelian and Non-Abelian Lie Algebras
A Lie algebra \(\mathfrak{g}\) is called abelian if \([X, Y] = 0\)
for all \(X, Y \in \mathfrak{g}\). Suppose \(\mathfrak{g}\) comes from a matrix Lie
group \(G\), and call \(G\) locally commutative when it is commutative in a
neighborhood of \(I\). Such a group has abelian Lie algebra, because \(\exp(sX)\) and
\(\exp(tY)\) then commute for small \(s, t\) and the bracket is the derivative of
conjugation. The reverse implication also holds, but it needs the fact that a
neighborhood of \(I\) consists of exponentials, which belongs to the Lie
correspondence.
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 every bracket takes the form \([cJ, dJ] = cd[J, J] = 0\), 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. The Lie bracket provides the precise infinitesimal
measure of that failure.
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?
The Lie group-Lie algebra correspondence, developed in
The Lie Correspondence, answers
this question. 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 \(G\).
This is 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, since it marks exactly the gap between
local and global. The example of \(\mathfrak{su}(2)\) and \(\mathfrak{so}(3)\)
exhibits this gap concretely. These are isomorphic Lie algebras whose groups \(SU(2)\)
and \(SO(3)\) have different global topology and are connected by a 2:1 covering map.
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.