For a Lie group with structure constants $f_{abc}$, its Lie algebra is given by a relation between the generators in the form $$[T_a, T_b]=if_{abc}T_c\tag{1}$$ where the symbol $[T_a, T_b]$, is called the Lie bracket, which as I understand, is a more general concept than that of a commutator bracket.

But how is a Lie bracket different (or more general) from the notion of a commutator bracket, $$[T_a, T_b]=T_aT_b-T_bT_a?$$ Why should the Lie bracket notion enter the discussion instead of the commutator bracket? Is it because the Lie group elements (and therefore, its generators), in general, are not matrices, and without matrices, we cannot define a commutator bracket?

Lie brackets also share many properties of Poisson's bracket. It is tempting to believe that Lie brackets can also reduce to Poisson brackets. Does a Lie group and Lie algebra with Poisson bracket structure ever arise in physics?

My acquaintance with Lie groups is limited and rather elementary. I will not be able to understand a very technical answer.

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    $\begingroup$ A Lie bracket and thus Lie algebra is a more general structure. It is not always the case that commutation is so easily defined, or that sense can be made of the intermediate products (only $T_a T_b$) before the full commutation is "completed". When it makes sense to "look inside" the Lie bracket, it almost always IS a commutator. But this isnt always the case $\endgroup$
    – Craig
    Commented Jul 2, 2023 at 4:37
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Jul 2, 2023 at 4:44
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    $\begingroup$ Example: the cross product defines a Lie bracket on $\mathbb{R}^3$. You could embed this as a matrix product, but no one does. And yet we are not "confused" as to how one combines the objects $\hat{x}$ and $\hat{y}$ to acquire $\hat{z}$. We take it as a definition of the cross product. $\endgroup$
    – Craig
    Commented Jul 2, 2023 at 5:36
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    $\begingroup$ There are Lie groups that cannot be represented as matrices. I assume (although perhaps I’m wrong) that their Lie algebras cannot be represented by matrices. In general, an “algebra” is just a vector space with a bilinear product. A Lie algebra adds the requirement that the bilinear product satisfy the Jacobi identity. What the elements of an algebra are is not relevant to a mathematician; only the algebraic operations that can be done on them are relevant. Abstraction = power. $\endgroup$
    – Ghoster
    Commented Jul 2, 2023 at 5:48
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    $\begingroup$ @Ghoster the previous comment probably means an associative product whose commutator bracket equals the given bracket. $\endgroup$
    – peek-a-boo
    Commented Jul 2, 2023 at 11:46

4 Answers 4


Commutator requires you to define product first and this is not always possible whereas $[T_a , T_b ]$ is a well-defined operation on a Lie algebra even when $T_a T_b$ or $T_b T_a$ separately isn't.

For example, for the group $SO(3)$ or $SU(2)$ elements can be represented as vectors $\vec{v}$ and the Lie bracket is simply given by $$ [ \vec{v} , \vec{w} ] = \vec{v} \times \vec{w} $$ There is no good definition for a "product" of vectors $\vec{v} \vec{w}$ which is sensible whereas the Lie bracket (cross-product) is perfectly well-defined.

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    $\begingroup$ You are making the fundamental error of conflating a group or algebra (abstract concept) with its representation (this is where matrix comes in). Lie algebras have NOTHING TO DO with matrices whatsoever. A particular representation can be described in terms of a matrix sometimes. The Lie algebra $so(3)$ has infinitely many representations and matrices of any dimension can be used to describe it. Many other groups do not admit a matrix representation, e.g. Poincare group has no nice matrix structure. You have to treat these two concepts separately. $\endgroup$
    – Prahar
    Commented Jul 2, 2023 at 14:42
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    $\begingroup$ $so(3)$ is a three-dimensional Lie algebra. As a vector space, we can choose a basis of elements $(e_x,e_y,e_z)$ in this Lie algebra. Any element $v$ of this Lie algebra can be expanded in this basis as $v = v_x e_x + v_y e_y + v_z e_z$ which we can then represent as a vector $\vec{v} = (v_x,v_y,v_z)$. The Lie bracket structure of the algebra is then given by the cross-product. That is what I meant in my post. $\endgroup$
    – Prahar
    Commented Jul 2, 2023 at 14:45
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    $\begingroup$ I know that group is an abstract concept. For example, the group $Z_2$ has nothing to do with matrices. It has matrix representations. But for the SO(3), the group elements are not abstract. They are defined to be the set of all $3\times 3$ real orthogonal matrices with unit determinants. I mean to say that unlike $Z_n$ or the permutation group$S_n$, the SO(3) elements are not abstract; they are defined in terms of $3\times 3$ matrices. It's a matrix group. Yes. It is also a representation, called defining representation. $\endgroup$ Commented Jul 2, 2023 at 14:52
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    $\begingroup$ @Solidification - It is often useful to define a group in terms of its so-called "fundamental representation" which is what you have done here. $SO(n)$ or $U(n)$ or $Sp(n)$ have similar definitions. This is especially useful in physics where the representations have actual physical meaning that we can observe. However, at the end of the day, the abstract mathematical definition makes no reference to matrices. The product $T_a T_b$ is NOT well-defined in $SO(3)$. It is only well-defined in a particular matrix representation of $SO(3)$. $\endgroup$
    – Prahar
    Commented Jul 2, 2023 at 14:59
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    $\begingroup$ $so(3)$ is a vector space isomorphic to ${\mathbb R}^3$ along with the Lie bracket $[ e_x , e_y ] = e_z$, $[ e_y,e_z]=e_x$, $[e_z,e_x]=e_y$. That's it. $\endgroup$
    – Prahar
    Commented Jul 2, 2023 at 15:09

Answer to title question is because they’re not the same thing. Let’s provide a review of three definitions, that of algebra over a field, next a Lie algebra, and finally a Poisson algebra. For the sake of concreteness, you may want to assume the field is always $\Bbb{R}$.

Definition 1. (Algebra over a field)

An algebra $A$ over a field $\Bbb{F}$ is by definition a vector space $A$ over the field $\Bbb{F}$, together with a $\Bbb{F}$-bilinear map $A\times A\to A$, typically denoted $\cdot$, and called the algebra product on $A$ or simply product on $A$.

We say $A$ is associative if for each $a,b,c\in A$, we have $(a\cdot b)\cdot c=a\cdot(b\cdot c)$. We say $A$ is commutative if for each $a,b\in A$, we have $a\cdot b=b\cdot a$.

Finally a notational remark: it is common to write simply $ab$ instead of $a\cdot b$, if the meaning of $\cdot$ is clear from the context.

So, in an algebra $A$, you need a notion of addition, scalar multiplication, and a “product”. Examples include the field $\Bbb{F}$, or more non-trivially, the collection of say all continuous real-valued functions on $[0,1]$, $C([0,1])$ (clearly we can add, scalar multiply, and multiply functions… this gives an associative, commutative algebra over $\Bbb{R}$). More generally, continuous functions on a topological space, or smooth functions on a manifold are examples. Another example is $A=M_{n\times n}(\Bbb{F})$, the space of all matrices, with the usual vector space structure and $\cdot$ being matrix multiplication. I should remark that many authors include associativity as part of the definition, and if they intend something more general, they’ll say something like a “non-associative algebra”.

Definition 2. (Lie algebra)

A Lie algebra (in the abstract sense) over a field $\Bbb{F}$ is by definition a vector space $L$ over $\Bbb{F}$ together with the choice of a bilinear map $[\cdot,\cdot]:L\times L\to L$, called the Lie bracket, such that

  • $[\cdot,\cdot]$ is alternating: for all $x\in L$, $[x,x]=0$.
  • $[\cdot,\cdot]$ satisfies the Jacobi identity: for all $x,y,z\in L$, $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$.

So, you can say that a Lie algebra is an algebra over $\Bbb{F}$, whereby the “product” $[\cdot,\cdot]$ is alternating, and satisfies the Jacobi identity. Note that $[\cdot,\cdot]$ is not associative in general, which is why one typically abstains from calling it a “product”, and simply refers to it as a “bracket operation”. Also, note that the alternating condition implies skew-symmetry: $[x,y]=-[y,x]$, because for any $x,y\in L$, we have \begin{align} 0&=[x+y,x+y]=[x,x]+[x,y]+[y,x]+[y,y]=0+[x,y]+[y,x]+0=[x,y]+[y,x]. \end{align} The converse (i.e skew-symmetry implies alternating) is true for a field of characteristic not equal to $2$ (e.g $\Bbb{R}$ or $\Bbb{C}$, which are the cases of interest in physics).

Perhaps the most common example of a Lie algebra is to start with an associative algebra $(A,\cdot)$, and define the bracket $[\cdot,\cdot]:A\times A\to A$ by setting $[x,y]:=xy-yx$. This is easily checked to be bilinear, alternating, and satisfies the Jacobi identity. We call this the (commutator) Lie bracket on $A$ induced by the product on $A$.

Given this example, it is natural to wonder if the converse holds, namely given a Lie algebra $(L,[\cdot,\cdot])$ over a field $\Bbb{F}$, does there exist a bilinear map $\cdot:L\times L\to L$ which makes $(L,\cdot)$ into an associative algebra such that the given Lie-bracket $[\cdot,\cdot]$ equals the commutator Lie bracket induced by the product $\cdot$. Or more succinctly, are Lie brackets and commutators the same thing. The answer unfortunately is no; see Does every Lie algebra come from commutator of some associative product operation? This is why you need to deal with more abstract stuff than just commutators; in more poetic terms, you’re “cutting out the middle man”. I should remark that while the details of that particular answer may be a little too involved, atleast heuristically you should expect the answer to be no, because in general, on a vector space there is no natural associative product $\cdot$ lying around. So if all you know is $(L,[\cdot,\cdot])$, then where would you even come up with a $\cdot$ from in order to write down the equation $[x,y]=x\cdot y-y\cdot x$?

Lastly, let’s talk about Poisson brackets

Definition 3. (Poisson Algebra)

A Poisson algebra over a field $\Bbb{F}$ is a triple $(P,\cdot,\{\cdot,\cdot\})$ such that $(P,\cdot)$ is an associative algebra over $\Bbb{F}$, $(P,\{\cdot,\cdot\})$ is a Lie-algebra over $\Bbb{F}$ and furthermore, $\{\cdot,\cdot\}$ acts as a derivation: for all $x,y,z\in P$, we have $\{x,y\cdot z\}=\{x,y\}\cdot z+y\cdot\{x,z\}$. We refer to $\{\cdot,\cdot\}$ as the Poisson bracket on $P$.

From the axioms, you can show that $\{\cdot,\cdot\}$ acts as a derivation in the other slot as well: $\{x\cdot y,z\}=x\cdot\{y,z\}+\{x,z\}\cdot y$. The first example one encounters is course starting with some phase space, i.e a symplectic manifold $(M,\omega)$ (which itself the most common examples are cotangent bundles of some configuration manifold $Q$) and looking at all smooth functions $C^{\infty}(M)$ (which is an associative algebra over $\Bbb{R}$ with the usual pointwise product) together with the usual Poisson bracket of functions $\{f,g\}$. On $\Bbb{R}^{2n}$, this gives the familiar formula $\{f,g\}=\frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i}-\frac{\partial g}{\partial q^i}\frac{\partial f}{\partial p_i} $ (or maybe there’s a minus sign).

Another very simple example is to start with an associative algebra $(A,\cdot)$, and define the commutator bracket $[x,y]=xy-yx$ as above. We already know this forms a Lie bracket, but even better, this is forms a Poisson bracket too, because \begin{align} [x,y]z+y[x,z]=(xy-yx)z+y(xz-zx)=xyz-yzx=[x,yz]. \end{align} But, one should keep in mind that there may be other interesting Poisson brackets than simply the commutator bracket.

In light of the definitions, it should now be clear that your statement

Lie brackets also share many properties of Poisson's bracket. It is tempting to believe that Lie brackets can also reduce to Poisson brackets

is wrong. It’s the other way around, a Poisson bracket is in particular a Lie bracket, so it is the Poisson bracket which reduces to a Lie bracket (i.e you forget about the associative algebra structure $(P,\cdot)$, and simply focus on $(P,\{\cdot,\cdot\})$).

If you want to read up more, at the group and manifold level, see also Poisson manifolds and Poisson-Lie groups.

Hopefully it’s now clear which is more special than which.


An algebra $\mathcal A=(V, \star)$ consists of a vector space $V$ equipped with a bilinear product operation $\star:V\times V\rightarrow V$. Examples of algebras include:

  • The vector space of formal polynomials in $x$ equipped with polynomial multiplication $$p_1=x \qquad p_2 = 1+2x^2 \qquad p_1\star p_2 = x + 2x^3$$
  • The vector space of $2\times 2$ matrices equipped with matrix multiplication $$m_1 = \pmatrix{1&2\\0&1} \qquad m_2 = \pmatrix{0&0\\1&0} \qquad m_1\star m_2 = \pmatrix{2&0\\1&0}$$
  • The vector space $\mathbb R^3$ equipped with the standard cross product $$v_1 = \pmatrix{1\\0\\1} \qquad v_2 = \pmatrix{1\\1\\0} \qquad v_1\star v_2 = \pmatrix{-1\\1 \\ 1}$$

If $\star$ happens to be alternating (so $x\star x = 0$) and obeys the Jacobi identity $$x\star(y\star z) + y\star (z\star x) + z\star(x\star y) = 0$$ then the resulting algebra has some very special and interesting properties. Such a product is traditionally written $$x\star y \equiv [\![ x,y ]\!]$$ and is called a Lie bracket; the algebra itself is called a Lie algebra.

Historically, Lie algebras arose during the study of Lie groups - which is presumably where you, as a physicist, are encountering them for the first time. Given a Lie group $G$ - which can be understood in particular as a smooth manifold with a group structure - the tangent space to $G$ at the identity element $\mathrm{id}_G$ is a vector space (as all tangent spaces are) which can be equipped with a special product operation which is inherited from the group composition operation on $G$. No matter what $G$ is and what the properties of its composition operation are, this inherited operation on the tangent space to $\mathrm{id}_G$ obeys the conditions above and is therefore a Lie bracket, and the tangent space to $\mathrm{id}_G$ therefore constitutes a Lie algebra. Each Lie group gives rise to a unique Lie algebra, and every Lie algebra can be associated to at least one Lie group (though in general, the latter association is one-to-many).

In physics, we are often interested in Lie groups which are meant to act as linear transformations on some vector space $\mathcal V$. If we can find a representation of $G$ in terms of such linear transformations, then we can automatically find a representation of the corresponding Lie algebra $\mathfrak g$. In this representation, the Lie bracket takes the form of a commutator.

Why should the Lie bracket notion enter the discussion instead of the commutator bracket? Is it because the Lie group elements (and therefore, its generators), in general, are not matrices, and without matrices, we cannot define a commutator bracket?

It comes from a desire to be more general with the discussion. Equating Lie brackets with matrix commutators is like equating vectors with lists of numbers - it's not an unreasonable way to encounter the subject, but it is a limited view which should be generalized sooner rather than later.

Lie brackets also share many properties of Poisson's bracket. It is tempting to believe that Lie brackets can also reduce to Poisson brackets. Does a Lie group and Lie algebra with Poisson bracket structure ever arise in physics?

This is not a coincidence - the Poisson bracket is an example of a Lie bracket. The corresponding Lie algebra is the vector space of smooth functions on the phase space, equipped with the Poisson bracket as the product operation. In fact, this is a special kind of Lie algebra called a Poisson algebra, which has extra properties above and beyond those of an ordinary Lie algebra. But that's a subject for another day.


In the context of generators of groups of transformations, the Lie Bracket is given by the Lie Derivative of one generator with respect to the other, in other words, the Lie Bracket tells you how much one generator's flow affects another generator. Now since your operators are Lie-Algebra Valued, you can show that this relative Lie Derivative simply reduces to the commutator (using that they are derivations this is not that difficult), thus you get that the Lie Bracket is given by the commutator bracket.


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