**In the following, an algebra will always mean a finite-dimensional associative commutative unital algebra (over some field $k$).**


Let $A$ be a $\mathbb{C}$-algebra. I am trying to understand how its group of units $A^{\times}$ sits topologically in $A$ relative to the locus of non-invertible elements $A\setminus A^{\times}$.


In the answer to [this question at MSE](https://math.stackexchange.com/questions/1465479/are-finite-dim-real-comm-algebras-with-path-connected-group-of-units-automatic), Jeremy Rickard gives a nice argument showing that every $\mathbb{R}$-algebra $A$ with connected group of units $A^{\times}$ is necessarily complex.



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> (Q1) Is the converse true, i.e. does every $\mathbb{C}$-algebra have connected group of units?
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If not,



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> (Q2) Can we characterize / classify those $\mathbb{C}$-algebras with connected group of units? Or is that too much to hope for? (Note that in general the classification of finite-dimensional commutative $\mathbb{C}$-algebras is a hard problem, see for example B.Poonen's work for more details on this.)
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Since $A$ *as above* is necessarily complex, it follows that $A^{\times}$ is connected abelian complex Lie group, hence $A^{\times}\simeq\mathbb{C}^n/\Gamma$ for some lattice $\Gamma\subset\mathbb{C}^n$. (In fact, by Remmert-Morimoto $A^{\times}$ is isomorphic to a product of finite number of copies of $\mathbb{C}$, of $\mathbb{C}^{\times}$, and a Cousin group.)



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> (Q3) Conversely, when can a connected abelian complex Lie group be realized as the group of units of some $\mathbb{C}$-algebra $A$?
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Connectedness of units in finite-dimensional commutative complex algebras
For (Q1): A finite dimensional $\mathbb{C}$-algebra $A$ is Artinian, so $A$ is a product of Artin local algebras. The units of a product of algebras is the product of the units, so we may assume $A$ is local, with maximal ideal $\mathfrak{m}$. The complement of a subspace in a complex vector space is connected, so $A^\times=A\backslash\mathfrak{m}$ is connected.


For (Q3): Tom's answer (and I guess mine too) shows that the group of units has the structure of an connected abelian linear algebraic group, so looks like a product of copies of $\mathbb{G}\_m$ ($=\mathbb{C}^\times$) and $\mathbb{G}\_a$ ($=\mathbb{C}$). Assumining $A\neq 0$, then $x\mapsto$ determinant of multiplication by $x$ is a surjective algebraic group homomorphism to $\mathbb{G}\_m$, so there must be at least one factor of $\mathbb{G}\_m$ (I confess I don't know what a Cousin group is, but we won't get them). If $G=\mathbb{G}\_m^a\times \mathbb{G}\_a^b$ with $a\geq 1$, then $G$ is isomorphic to the units of
$$
A:=\mathbb{C}[x\_1,\ldots,x\_b]/(x\_ix\_j)\times\mathbb{C}^{a-1}.
$$


