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This is a continuation of a previous question: Connectedness of groups of units in finite-dimensional commutative algebras.

Let $k$ be an algebraically closed field of characteristic $0$. Which algebraic groups / Lie groups (for $k=\mathbb{C}$) can occur as the groups of units of finite-dimensional noncommutative associative $k$-algebras $A$?

Even though a classification of finite-dimensional commutative $k$-algebras seems (at the moment) out of reach, it turned out that their groups of units are easy to understand (see the linked question). So I am hoping the answer to this new question might also be feasible despite the fact that a classification of finite-dimensional noncommutative $k$-algebras is hopeless.

Obviously, $U(A)$ has to contain a copy of $\mathbb{G}_m$, which automatically excludes all compact Lie groups. Moreover, the following remark of Tom Goodwillie applies:

"Even without assuming commutativity, the set of non-units is the vanishing set of a not-identically-zero complex polynomial function, the function sending $x$ to the determinant of $y\mapsto xy$. This has complex codimension $\ge 1$, so its complement is connected." (answer at Oct 28 '15 at 18:41 in op. cit.)

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    $\begingroup$ If you factor the algebra by its radical you get a product of matrix algebras so the group of units has a connected unipotent normal subgroup such that the quotient is a product of general linear groups (unipotent since the radical is nilpotent, connected because it is a coset of the radical which is a linear space). Moreover, this quotient splits by the Wedderburn-Malcev theorem. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 7, 2017 at 21:36
  • $\begingroup$ @BenjaminSteinberg: Thanks! If I understand you correctly, that means that the group of units is then recovered as a semidirect product of $\mathrm{GL}$-s and a connected unipotent group (and the latter is a closed subgroup of upper-triangular matrices). So, a natural question is then if any connected unipotent group can be realized (in the way you described above)? $\endgroup$
    – M.G.
    Commented Oct 8, 2017 at 9:52
  • $\begingroup$ Im not sure if you can get any. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 8, 2017 at 10:48

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