Comments to my answer to this MO question, which is isomorphic to this MSE question, point out that I was tacitly assuming that the associated Lie algebras to non-isomorphic quaternion algebras over $\mathbb Q$ remain non-isomorphic. This should have a simple, decisive answer, but I'm unable to really get anywhere.

Specifically, let $A_1$ and $A_2$ be central simple algebras over a field $k$ (of characteristic zero, but not algebraically closed), so $A_i\simeq M_{n_i}(D_i)$, where $D_i$ is a division algebra. Put a Lie algebra structure on the $A_i$ in the standard way by $[x,y]=xy-yx$. Write ${\rm Lie}(A_i)$ for $A_i$ considered as a Lie algebra.

The question is:

If ${\rm Lie}(A_1)\simeq{\rm Lie}(A_2)$, is $A_1\simeq A_2$?

If the $A_i$ are not central (meaning the center strictly contains $k$), the answer is no. For example, if $K$ is a quadratic extension of $k$, then $K$ and $k^2$ will be isomorphic as Lie algebras (since they are isomorphic as vector spaces and both have trivial bracket operation). [Edit: Of course, $k^2$ is not simple. To salvage this example, replace $k^2$ with a second quadratic extension $K'$.]

Added later: Coming back to this, assume $A_1$ is split by an extension $K$ of $k$, and $A_2$ is not. Then $A_1\otimes_kK\simeq M_n(K)$, while $A_2\otimes_kK\simeq M_m(D)$, where $D$ is a non-trivial division algebra over $K$ and $n=md$. Then, we just need to show that ${\rm Lie}\big(M_n(K)\big)\not\simeq{\rm Lie}\big(M_m(D)\big)$. This would answer my question in the affirmative, under the assumption that the $A_i$ are split by different extensions.

In this case, the question becomes:

Is there a simple proof that ${\rm Lie}\big(M_n(K)\big)\not\simeq {\rm Lie}\big(M_m(D)\big)$?

I still can't get this to work. One strategy is to show that $M_n(K)$ has a subalgebra that $M_m(D)$ can't have (e.g., by dimension considerations). (Note that the Lie subalgebras are in 1-1 correspondence with the "normal" subalgebras.) For example, a quaternion division algebra does not have a three-dimensional subalgebra, but $M_2(K)$ does.

  • $\begingroup$ The last line should read: "For example, a quaternion division algebra does not have a solvable three-dimensional subalgebra, but $M_2(K)$ does. $\endgroup$
    – B R
    Dec 15, 2011 at 23:09

1 Answer 1


This turns out to be in some old work of Jacobson's. See, e.g., Theorem 10 in Chapter X of his "Lie Algebras" (p. 305) which can be summarized as

An isomorphism between ${\rm Lie}(A_1)$ and ${\rm Lie}(A_2)$ extends uniquely to either an isomorphism or the negative of an anti-isomorphism between $A_1$ and $A_2$. When the $A_i$ are quaternion algebras, it is always an isomorphism.

I'm pretty sure that central simple algebras are generally not isomorphic to their opposite rings, so the answer to the specific question is no (at least for algebras of dimension greater than four). A satisfying no, though.

Much more can be said about classifying simple algebras over general fields. See the above book or an earlier paper of Jacobson, available online (with subscription, unfortunately).

  • 2
    $\begingroup$ See here or here. $\endgroup$ Dec 21, 2011 at 11:20
  • $\begingroup$ Thanks for those pointers to division algebras not isomorphic to their opposite rings. $\endgroup$
    – B R
    Dec 21, 2011 at 17:34

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