# Are Lie algebras of non-isomorphic central simple algebras non-isomorphic?

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.

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The last line should read: "For example, a quaternion division algebra does not have a solvable three-dimensional subalgebra, but $M_2(K)$ does. –  B R Dec 15 '11 at 23:09

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.