Let $\mathfrak{g}$ be a rational, nilpotent Lie algebra. Then its adjoint representation will consist of elements which are nilpotent matrices over rationals. But this representation generally is not faithful. My question is does a rational nilpotent Lie algebra always admit a faithful representation such that in this representation there is a choice of basis elements which are all nilpotent rational matrices? Would it possible to explicitly construct these matrices using simply the structure coefficients of the basis (as in the case of adjoint representation)? The reason for this question is for instance this is the case for the Heisenberg algebra which has a representation made of nilpotent matrices, which looks like adjoint representation (up to plus minus signs and one element being non-zero).

  • $\begingroup$ You are right that the adjoint representation is not faithful - nilpotent Lie algebras always have centers. I am not sure if rational Lie algebras have faithful representations. However, if you tensor the Lie algebra over $\mathbb{C}$, then you get a complex Lie algebra, which always has a faithful representation (Ado's Theorem.) $\endgroup$ – Siddharth Venkatesh Apr 28 '14 at 3:42
  • $\begingroup$ Ado's theorem holds for every field (in characteristic $p$ the result is due to Iwasawa). $\endgroup$ – Dietrich Burde Jan 12 '17 at 14:45

The answer is yes, and this was proved first by Birkhoff for nilpotent Lie algebras. A modern proof can be found in Terence Tao's What's new - Ado's Theorem, Theorem $3$. Actually Ado proved the Theorem for all fields of characteristic zero, and Iwasawa gave a proof for fields of characteristic $p>0$.
There is an "explicit construction" of a finite-dimensional faithful linear representation, using quotients of the universal enveloping algebra. For details see here.


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