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Suppose $\phi:V\rightarrow W$ is a Lie algebra isomorphism between matrix Lie algebras $V$ and $W$. Since $V$ and $W$ are real vector spaces of matrices, we can consider the traces of their elements.

My question is, do we have $\operatorname{tr} \phi(X) = \lambda\cdot\operatorname{tr} X$ for some $\lambda\ne 0$? If not, is there a good counterexample?

I am thinking we can choose a basis $\{E_{1}, \ldots, E_{n}\}$ for $V$, take $F_{j} := \phi(E_{j})$ for all $j$, and then consider structure constants $$ [E_{i}, E_{j}] = c_{ijk}E_{k} $$ where we sum over repeated indices (Einstein summation notation). Since $\phi$ is a Lie algebra isomorphism, we have $$ [F_{i}, F_{j}] = c_{ijk}F_{k}. $$ By taking the traces, and noting that the trace of commutators is zero, we find $$ c_{ijk}e_{k} = 0 \qquad\text{ and }\qquad c_{ijk}f_{k} = 0 $$ where $e_{k} := \operatorname{tr} E_{k}$ and $f_{k} := \operatorname{tr} F_{k}$.

Now my idea is to note that the $c_{ijk}$'s are determined, and our question is simply what is the null space / kernel of the $n^{2}\times n$ matrix $C = (c_{ij, k})$ ($ij$ determine row; $k$ determines column), but it's not clear how to proceed from here, or if this approach is any good to begin with. Unfortunately, I don't know of many non-trivial matrix Lie algebra examples at the moment.

Maximal Ideal
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3 Answers3

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If $V$ and $W$ are abelian Lie algebras, then any linear map $V\to W$ is a Lie algebra homomorphism. So, for instance, if $V$ is some vector space of commuting matrices (say, all the diagonal matrices, or all matrices that are polynomials in some given matrix), you are asking whether every invertible linear map $V\to V$ preserves the trace up to scaling. This is obviously false as long as $V$ contains two linearly independent elements with nonzero trace, since an invertible linear map can (for instance) scale those elements by different factors.

Eric Wofsey
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Consider the endomorphism of $M_1(k)\times M_1(k)\subseteq M_2(k)$ that multiplies the elements in the first factor by $2$ and the ones in the second factor by $3$.

Mariano Suárez-Álvarez
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Aside from the other answers it is important to note that the trace is not really an intrinsic thing on the Lie algebra. A single Lie algebra can viewed as a matrix Lie algebra in a number of ways (effectively we are just picking a faithful representation) and the trace of an element in different representations will vary.

If, for example, the Lie algebra is semisimple we have a lot of control over the possible traces. Indeed they are always zero (see Torsten's comment here).

If however the Lie algebra is abelian we have no control whatsoever and the traces can be anything.

Callum
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