# Connectedness of units in finite-dimensional commutative complex algebras

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, Jeremy Rickard gives a nice argument showing that every $\mathbb{R}$-algebra $A$ with connected group of units $A^{\times}$ is necessarily complex.

(Q1) Is the converse true, i.e. does every $\mathbb{C}$-algebra have connected group of units?

If not,

(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.)

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.)

(Q3) Conversely, when can a connected abelian complex Lie group be realized as the group of units of some $\mathbb{C}$-algebra $A$?

• Just to clarify: is the topology you are putting on A the one it inherits by choosing a ${\mathbb C}$-vector-space basis of $A$ and using it to identify A with some ${\mathbb C}^n$? Oct 28 '15 at 18:30
• @YemonChoi: yes, $A$ is topologized as any finite-dimensional $\mathbb{C}$-vector space.
– M.G.
Oct 28 '15 at 18:39

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 $$A:=\mathbb{C}[x_1,\ldots,x_b]/(x_ix_j)\times\mathbb{C}^{a-1}.$$

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.
• If I am not mistaken: one needs to know that if $L: A \to End(A)$ is the embedding in your answer, then $L_x$ is bijective if and only if $x$ is a unit in $A$; but this is easily done, of course. Oct 28 '15 at 18:52