Take for example $GL(2,\mathbb R)$, the group of $2\times2$ invertible matrices with real entries. By considering small variations from the identity, it is clear that one needs four parameters to parametrize this group, and hence we will need four "infinitesimal" generators. If I am thinking about this correctly, we could take as generators the matrices

$$ J_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad J_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad J_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \quad J_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. \tag{1}$$

But we could equally well take some inspiration from the Pauli matrices and use

$$ L_1 = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad L_2 = \frac{1}{2} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad L_3 = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad L_4 = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \tag{2}$$

Both of these sets are linearly independent, so they seem like they should be fine, but they produce different Lie bracket (commutation) relations. Specifically, for the first set we have the (somewhat messy) relations

$$[J_1, J_2] = J_2, \quad [J_1, J_3] = -J_3, \quad [J_1, J_4] = 0, \quad [J_2, J_3] = J_1 - J_4, \quad [J_2, J_4] = J_2 \quad [J_3, J_4] = -J_3,\tag{3}$$ while for the second set we have $$[L_1, L_2] = L_3, \quad [L_1, L_3] = L_2, \quad [L_2, L_3] = L_1, \quad [L_i, L_4] = 0.\tag{4}$$

My question

When speaking about certain Lie algebras, I often read/hear people refer to the Lie bracket structure, say $[S_\alpha, S_\beta] = i\hbar\epsilon_{\alpha\beta\gamma}S_\gamma$, as the Lie algebra of the group (representation) in question. But, if my argument above is correct, it seems that this structure is not unique, but indeed depends on the generators chosen. So I can see two possible resolutions:

  1. One (or both) of my suggested sets of generators is wrong. If this is the case, could you tell me why?

  2. The Lie bracket structure for a given group is not unique, and those that say so are being sloppy with language somehow. Some of the more mathematically oriented sources seem to imply that the Lie algebra is actually a sort of tangent space about the identity transformation, with the generators as basis vectors. Then maybe it is this space that is unique, while the bracket structure is basis dependent?

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    $\begingroup$ Structure constants are component of a tensor....That tensor is unique for a given Lie algebra. $\endgroup$ Commented Jun 4, 2021 at 21:34
  • 3
    $\begingroup$ The Lie algebra is unique up to change-of-basis isomorphisms. These change the components of the tensor, but not the tensor itself. This is just like changing the basis for a vector: the numerical components change but the vector is unchanged. This is what @ Walter Moretti is saying. $\endgroup$
    – mike stone
    Commented Jun 4, 2021 at 21:38

2 Answers 2


Like most mathematical objects, when we ask whether two Lie algebras are "the same" we should really be asking whether these is an isomorphism of Lie algebras between them. Since a Lie algebra is a vector space with extra structure (the Lie bracket), such an isomorphism is an isomorphism of vector spaces $f : L_1 \to L_2$ that also has the property that $$ [f(v), f(w)]_2 = f([v,w]_1)$$ for all $v,w\in L_1$. So when someone writes down a definition of a Lie algebra on $n$ generators $e_i$ with some bracket $[e_i,e_j] = f_{ijk}e_k$ and someone else a definition on $n$ generators $e'_i$ with $[e'_i, e'_j] = f'_{ijk}e'_k$ these two definitions describe the "same" Lie algebra if there is an isomorphism of Lie algebras between them.

As you have explicitly demonstrated by choosing different bases for the same Lie algebra, two such descriptions can be isomorphic even if their structure constants $f$ and $f'$ look different. This should be familiar - $f$ and $f'$ are basis-dependent expressions of a structure on the vector space like matrices for inner products: If you have an orthonormal basis, the inner product matrix is just the identity matrix, but if your basis isn't orthonormal, it's something else. That doesn't mean the inner product is somehow not uniquely given by taking one basis and saying it's a certain matrix in that basis, nor does it mean any other choice of basis is somehow less "valid" than another.

  • $\begingroup$ Thank you for a very quick and excellent answer. This clears it up very well! $\endgroup$
    – ummg
    Commented Jun 4, 2021 at 21:53
  • $\begingroup$ Do you have any advice as to how to find a Lie algebra isomorphism, like $f$ in your answer, given two sets of generators? Say, if I want to show explicitly that there is such an isomorphism for the sets (1) and (2) in my question. I tried to deduce one, but it seems to be a little bit more difficult than i expected. $\endgroup$
    – ummg
    Commented Jun 4, 2021 at 23:37
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    $\begingroup$ @ummg since you have them explicitly as matrices, just express one set of matrices in the form of the other, i.e. $J_1 = L_3 + L_4$ and likewise for the other $J_i$ in terms of $L_i$. The isomorphism is the map that sends $J_1 \mapsto L_3 + L_4$ and likewise for the other $J_i$. $\endgroup$
    – ACuriousMind
    Commented Jun 4, 2021 at 23:49
  • $\begingroup$ Ah, I see. Thank you! $\endgroup$
    – ummg
    Commented Jun 5, 2021 at 0:09

The Lie algebra of a group is unique, but only up to the choice of basis. Most common algebras have a canonical basis one often works with (there's a complete classification of the finite representations of compact Lie groups, so there's only so many options). So, no, the algebra is not literally unique in the sense of the coefficients of the algebra being uniquely defined.

However, saying that something unique up to choice in basis is indeed unique is a common point of language. After all, if you are working in special relativity, you would say that the metric is the Minkowski metric. Whether the components of that metric are very literally $\text{diag}(-1,1,1,1)$ depends on your coordinates...working in polar coordinates obviously the components of the metric would change, but you would still call it the Minkowski metric.

  • $\begingroup$ Thank you for your answer as well. I would have accepted it if it weren't for the fact that an other answer provided some useful additional information. $\endgroup$
    – ummg
    Commented Jun 4, 2021 at 21:54

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