I ran into this question on math.stackexchange.com "How many 3 dimensional simple Lie algebras are there over the rationals?" The question has been sitting idle for a long time. I thought it was interesting and would like to know myself.

If working over $\mathbb{C}$, we know all 3-dimensional simple Lie algebras are isomorphic to $\mathfrak{sl}_2(\mathbb{C})$. When we move down to the reals, there are 2 non-isomorphic real forms.

The same argument that works over $\mathbb{C}$ works over $\bar{\mathbb{Q}}$ (the field of algebraic numbers). So over $\bar{\mathbb{Q}}$ the only 3-dimensional simple Lie algebra is $\mathfrak{sl}_2(\bar{\mathbb{Q}})$. Running through a similar argument as that which classifies 3-dim simples over $\mathbb{R}$, I get a whole mess of possibilities over $\mathbb{Q}$ (due to the lack of square roots) but have no idea which forms are non-isomorphic or even how to go about proving they are non-isomorphic.

Anybody know what the answer is? Have a good reference?


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    $\begingroup$ The essence of what happens when you go from $\mathbb C$ to $\mathbb R$ is that you have two four-dimensional algebras over $\mathbb R$ that coincide over $\mathbb C$, $M_2(\mathbb R)$ and $\mathbb H$ (the quaternions). For $\mathbb Q$, there are infinitely many (non-isomorphic) quaternion algebras (basically, for every finite set of primes (counting, as number theorists do, infinity as a prime), there exists a unique quaternion algebra over $\mathbb Q$). I don't know an accessible reference, though books on central simple algebras would be a good place to start. $\endgroup$
    – B R
    Oct 18, 2011 at 20:58
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    $\begingroup$ I meant "for every even (finite) set of primes". Also, I think the simplest way to see this is that there is a unique quaternion division algebra over $\mathbb Q_p$, then use quadratic reciprocity to show that you can stitch an even number of them together to form a nontrivial quaternion algebra over $\mathbb Q$. Then the fact that quaternion algebras arising from different sets of primes are not isomorphic is obvious. $\endgroup$
    – B R
    Oct 18, 2011 at 22:50
  • $\begingroup$ Thanks BR! At least that sounds like a good place to start. $\endgroup$
    – Bill Cook
    Oct 18, 2011 at 23:21
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    $\begingroup$ The etiquette in this case is to encourage the OP on math.stackexchange.com to ask the question on mathoverflow.net. $\endgroup$
    – Guntram
    Oct 19, 2011 at 6:01

4 Answers 4


Let me flesh out my comments. First, why there are infinitely many three-dimensional simple Lie algebras over $\mathbb Q$. One of the major steps in proving class field theory is to prove that we have an exact sequence of Brauer groups $$0\rightarrow {\rm Br}(\mathbb Q)\rightarrow\oplus_{p\le\infty}{\rm Br}(\mathbb Q_p)\rightarrow \mathbb Q/\mathbb Z\rightarrow 0$$ where for $p<\infty$, ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$, and of course ${\rm Br}(\mathbb R)\simeq \mathbb Z/(2)$. This says that given a finite set $S$ of primes (regarding $\infty$ as a prime) and division algebras $D_p$ over $\mathbb Q_p$ for all $p\in S$, if the corresponding invariants (in ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$) sum to zero in $\mathbb Q/\mathbb Z$, then there exists a division algebra $D$ over $\mathbb Q$ giving rise to the $D_p$ in the sense that $D\otimes_\mathbb Q\mathbb Q_p\simeq M_{n_p}(D_p)$ (we need to use matrices over a division algebra because the dimensions of $D$ and $D_p$ don't have to match).

Now, for quaternion algebras over $\mathbb Q$, the situation is simpler because (1) a central simple algebra of dimension 4 over $\mathbb Q_p$ is either a quaternion division algebra or $M_2(\mathbb Q_p)$ and (2) the hard algebraic number theory can be done "by hand" (it basically follows from quadratic reciprocity). Note this implies that the invariants for a central simple algebra of dimension 4 is either $0$ or $1/2$, hence the need for an even set of primes to get a division algebra over $\mathbb Q$.

So we have infinitely many quaternion algebras over $\mathbb Q$, all of which split over $\mathbb C$ to be isomorphic to $M_2(\mathbb C)$. To get three-dimensional Lie algebras out of this, you restrict to elements with "reduced trace" equal to zero.

Second, we wonder if we have found all three-dimensional simple Lie algebras over $\mathbb Q$. This is somewhat outside of my comfort zone. We switch to talking about simple algebraic groups of dimension three. We are interested in classifying forms of $SL_2(\mathbb Q)$. These are classified by the (non-abelian) Galois cohomology group $H^1(G(\bar{\mathbb Q}/\mathbb Q),{\rm Aut}_\bar{\mathbb Q}(SL_2(\mathbb Q))$. These can further be split into two classes, inner forms and outer forms, depending on whether the corresponding automorphism is inner or outer. Inner forms correspond to quaternion algebras. Outer forms correspond to certain unitary groups. This is from Platonov and Rapinchuk's "Algebraic Groups and Number Theory", section 2.3.4 (propositions 2.17 and 2.18). So it seems like we might be missing a bit, but the simple nature of our situation may mean that the outer forms are isomorphic to the inner forms, like $SU(1,1)\simeq SL_2(\mathbb R)$ (I have no clue).

As Vladimir notes in the comments to this answer, I am assuming that if two quaternion algebras are non-isomorphic, then their Lie algebras are non-isomorphic. This is a legitimate worry, as if $K$ is a quadratic extension of $\mathbb Q$, then ${\rm Lie}(K)\simeq {\rm Lie}(\mathbb Q^2)$. What happens for quaternion algebras? First, note that for a field $k$ and a quaternion division algebra $D$ over $k$, ${\rm Lie}\big(M_2(k)\big)$ is not isomorphic to ${\rm Lie}(D)$, since, for example, ${\rm Lie}\big(M_2(k)\big)$ has a solvable three-dimensional subalgebra and ${\rm Lie}(D)$ does not (seeing this by explicitly calculating the brackets of basis elements). Second, for two quaternion algebras $D_1$ and $D_2$, if a quadratic extension $K$ splits $D_1$ but not $D_2$, then $D_1\otimes_\mathbb Q K\simeq M_2(K)$, but $D_2\otimes_\mathbb Q K$ remains a division algebra. Since ${\rm Lie}(D_1\otimes_\mathbb Q K)\not\simeq {\rm Lie}(D_2\otimes_\mathbb Q K)$, we have ${\rm Lie}(D_1)\not\simeq {\rm Lie}(D_2)$. Finally, since quaternion algebras over $\mathbb Q$ are determined by the primes where they split, we can always find a quadratic extension $K$ so that exactly one of the $D_i\otimes_\mathbb Q K$ splits.

This argument only partly extends to general division algebras, since there are non-isomorphic division algebras with the same splitting field.

After I typed the above, I happened to see that Chapter X of Jacobson's "Lie Algebras" is devoted to classifying simple Lie algebras over arbitrary fields, which he also does in this paper. In particular, he proves that for central simple algebras $A$ and $B$ over a field $k$,

An isomorphism between ${\rm Lie}(A)$ and ${\rm Lie}(B)$ extends uniquely to either an isomorphism or the negative of an anti-isomorphism between $A$ and $B$. If they are quaternion algebras, it is always an isomorphism.

  • $\begingroup$ I am a bit confused - why those infinitely many Lie algebras that you constructed are non-isomorphic? Somehow, it is not entirely obvious to me, since ``adjoining the unit element'' will, in a sense, carry most information about the quadratic form that defines the quaternion algebra, and so the "infinitely many" might be lost when you pass to the Lie algebra. $\endgroup$ Dec 10, 2011 at 6:40
  • $\begingroup$ Vladimir, I'm not sure I understand your objection (I could be missing something). Are you saying it might be that giving the Lie algebra structure to non-isomorphic division algebras could make them isomorphic as Lie algebras? I actually hadn't thought of that as a potential problem, so I'll think on it. Do you have a reason to suspect that could happen? $\endgroup$
    – B R
    Dec 10, 2011 at 6:59
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    $\begingroup$ You are quite welcome! I agree that it is unclear (though intuitive! and true!) that non-isomorphic algebras have non-isomorphic associated Lie algebras. I'm glad you were able to supply a proof for this case. $\endgroup$
    – B R
    Dec 10, 2011 at 8:07
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    $\begingroup$ (And actually, I did mean ${\mathfrak sl}_1$, since $GL_1(D)\simeq D^\times$. E.g., ${\mathfrak gl}_2(D)$ would be $2\times 2$ matrices with entries in $D$, so four-dimensional over $D$, instead of over $\mathbb Q$!) $\endgroup$
    – B R
    Dec 10, 2011 at 8:08
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    $\begingroup$ Dear BR, There are no outer forms of $SL_2$, since the Dynkin diagram (a single point!) has trivial automorphism group. (So here we see a genuine distinction between the $SL_2$ case and the $GL_2$ case, since in the latter case there are outer forms --- e.g. $U(2)$ is an outer form of $GL_2$ over $\mathbb R$; this distinction is somewhat elided in your answer.) Thus the (necessarily inner) forms of $SL_2$ are classified by $H^1(G_{\mathbb Q}, PSL_2) = H^2(G_{\mathbb Q},\mu_2)$ which is the $2$-torsion in the Brauer group, and so is precisely the classification you've given! Regards, Matthew $\endgroup$
    – Emerton
    Dec 21, 2011 at 8:48

I don't know if you can find the complete list written down anywhere (and I don't know how "nice" such a list would turn out to be), but let me point out that there is a lowbrow approach to this problem that you might be able to work through to a satisfactory conclusion.

It turns out that there is a one-to-one correspondence between 3-dimensional simple Lie algebras over an arbitrary field $k$ and equivalence classes of invertible symmetric 3x3 matrices with entries in $k$, under the equivalence relation $$ A \sim B \iff \exists \text{ } \rho \in k^\times \text{ and } N \in \operatorname{GL}_3(k) \text{ such that } A=\rho N^t B N. $$ Moreover, if $\text{char}\: k \neq 2$, each equivalence class contains a diagonal matrix of the form $\text{diag}\{\alpha,\beta,1\}$.

For the details, see p.13 of Jacobson's Lie Algebras. Incidentally, the last chapter of Jacobson has some theoretical material concerning the classification of simple Lie algebras over arbitrary fields of characteristic zero, but I don't know how helpful it'll be in answering your specific question.


Here is my own favorite way of understanding this problem: Let $\bigl(L,[,]\bigr)$ be a $3$-dimensional Lie algebra over a field $K$ (assumed of characteristic $0$ for my comfort, though this probably is overkill). The bracket $[,]:L\times L \to L$, which is bilinear and skew-symmetric, can be regarded as an element $\beta \in \text{Hom}\bigl(\Lambda^2L,L\bigr)\simeq L\otimes \Lambda^2 (L^\ast)$. Since the dimension of $L$ is $3$, we have that $\Lambda^2(L^\ast)$ is naturally isomorphic to $L\otimes \Lambda^3(L^\ast)$. It follows that we can regard $\beta$ as an element of $L\otimes L\otimes \Lambda^3(L^\ast)$. Since there is a canonical decomposition $L\otimes L\simeq \Lambda^2(L)\oplus S^2(L)$, there is a canonical decomposition $\beta = \lambda + \sigma$ with $\lambda\in \Lambda^2(L)\otimes \Lambda^3(L^\ast)\simeq L^\ast$ and $\sigma\in S^2(L)\otimes \Lambda^3(L^\ast)$.

Now, there is a canonical bilinear pairing $$ \langle,\rangle: L^\ast\times\bigl(S^2(L)\otimes \Lambda^3(L^\ast)\bigr)\longrightarrow L\otimes \Lambda^3(L^\ast)\simeq \Lambda^2(L^\ast), $$ and the Jacobi identity is easily seen to be just $\langle\lambda,\sigma\rangle = 0$.

If $\lambda\not=0$, then, because $\lambda$ must be fixed by any automorphism of $L$ (in particular, the inner automorphisms), it follows that $N = \text{ker}(\lambda)\subset L$ is an ideal of codimension $1$ in $L$, so $L$ is not simple. One could follow this thread further to classify the Lie algebras in this case, but let me pass on to the simple case.

Thus, assume that $\lambda = 0$, so that $\beta = \sigma \in S^2(L)\otimes \Lambda^3(L^\ast)$. Now, there is a natural (i.e., $\text{GL}(L)$-equivariant) cubic polynomial mapping $$ \text{det}: S^2(L)\otimes \Lambda^3(L^\ast) \longrightarrow \Lambda^3(L^\ast) $$ (again, this exploits the assumption that $L$ has dimension $3$). It is easy to see that, if $\text{det}(\sigma) = 0$, then $\sigma$ lies in $S^2(N)\otimes \Lambda^3(L^\ast)$ for some nontrivial proper subspace $N\subset L$ and that this $N$ will be an ideal in $L$, so, again, $L$ is not simple.

Thus, assume that $\text{det}(\sigma)\in\Lambda^3(L^\ast)$ is not zero. In particular, $L$ is endowed with a canonical volume form, $\text{det}(\sigma)$. Since $\text{GL}(L)$ acts transitively on the nonzero elements of $\Lambda^3(L^\ast)$, it follows that classifying the $3$-dimensional Lie algebra structures on $L$ up to the action of $\text{GL}(L)$ with $\lambda=0$ and $\text{det}(\sigma)\not=0$ is equivalent to fixing a (nonzero) volume form $\Upsilon\in \Lambda^3(L^\ast)$ on $L$ and classifying, up to the action of $\text{SL}(L)$, the elements $s\in S^2(L)$ that satisfy $\text{det}(s\otimes \Upsilon) = \Upsilon$.

It is not hard to see that for such $s$, the corresponding Lie algebra is indeed simple, so this reduces the classification of the $3$-dimensional simple Lie algebras over $K$ to the classification of the unimodular quadratic forms in $3$ variables over $K$ up to the action of $\text{SL}(3,K)$, a classical problem.


Disclaimer: This answer is mostly an extended comment coming from my attempt to understand the answer of BR. However, the time I invested in it made me think that someone else would find it useful. Essentially, it's a (somewhat fruitful) union of ideas contained in answers of BR and Faisal.

So, first of all, Jacobson indeed classifies all Lie algebras of dimension three over a field $k$ in his book; it is done pretty much by hand, and, as Faisal noted, Jacobson shows that if $\mathop{\mathrm{char}}(k)\ne 2$, every three-dimensional simple Lie algebra is isomorphic to the one with the bracket $[e_1,e_2]=e_3$, $[e_2,e_3]=\alpha e_1$, $[e_3,e_1]=\beta e_2$. And, in general, there is a bijection with the simple Lie algebras of dimension three over $k$ and $GL_3(k)$-orbits on the projectivisation of the space of symmetric bilinear forms on $k^3$.

There is also an answer of BR who suggests to build simple Lie algebras of dimension three as Lie subalgebras of elements of trace zero in quaternionic algebras, similarly to how we can obtain $sl_2$ and $so_3$. I actually like this answer a lot, though in the first place it made me really worry if those algebras are pairwise non-isomorphic. You know, it's that silly reasoning that made me worry: when we define the usual quaternions from the usual $so(3)$, we adjoint the unit and say something like $gh=-(g,h)\cdot1+[g,h]$. So if one is as slow as I am, it is tempting to think that the quaternionic structure is all hidden in the unit element. However, it is not, and thinking about that formula for a bit more, I came up with a simple and surely well known idea that I enjoyed enough to type this reply. The scalar product used above is, up to a constant, the Killing form! The form $(g,h)=\mathop{\mathrm{tr}}_L(\mathop{\mathrm{ad}}_g\mathop{\mathrm{ad}}_h)$ is something that both captures the relevant information about the quaternionic algebra we have in mind, and is totally intrinsic for our Lie algebra.

So in brief, for me the slogan is: there are as many three-dimensional simple Lie algebras as there are quadratic forms up to coordinate change and proportionality; the bijection is described via the Killing form.

  • $\begingroup$ Hi, the part where you construct the quaternions from $so(3)$ ("we adjoin the unit and say something like gh=−(g,h)⋅1+[g,h]") is of interest to me. Is there a good reference for this construction? $\endgroup$ Nov 27, 2013 at 13:47
  • $\begingroup$ If you recall that the Lie algebra $so(3)$ is the same as 3d vectors under cross product, this would be found in virtually any textbook discussing quaternions, I assume! $\endgroup$ Nov 27, 2013 at 14:19
  • $\begingroup$ Ah, thanks. I thought there was a more general principle (having to deal with Lie algebras inside skew fields), but this looks kind of specific to the standard (Hamiltonian) quaternions. $\endgroup$ Dec 10, 2013 at 22:09
  • $\begingroup$ There are some generalisations of this construction, related to Freudenthal-style constructions of exceptional Lie algebras, but that is also something very well known. $\endgroup$ Dec 12, 2013 at 12:21

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