Let $A$ be a finite-dimensional unital associative algebra over $\mathbb{R}$. Then its group of units $A^{\times}$ is a Lie group. I am interested in the case when $A$ is also commutative and $A^{\times}$ is path-connected. Then $A^{\times}$ is a path-connected abelian Lie group, hence $A^{\times}\cong\mathbb{R}^m\times\mathbb{T}^n$ for some $m,n\in\mathbb{N}_0$. Now, since the irreducible representations of $(\mathbb{R},+)$ are of the form $\exp(kt)$ for some $k\in\mathbb{C}$ and $(-1)$ is a unit in $A$, a representation $\rho$ of $A$ must realize $A^{\times}$ in a strictly complex way.

Is it true that a finite-dimensional real commutative associative unital algebra with path-connected group of units must be automatically complex?

The missing part is about what is going on with the non-units, and to that effect I've (unsuccessfully) tried to construct counterexamples, hence my question above. Intuitively, the obstruction is that the complex numbers have no action on the dual numbers and the split-complex numbers (as subalgebras of $\mathrm{Mat}_2(\mathbb{R})$ and looking at matrices of 2x2 blocks), but I feel like I'm missing some rigorous key observation to properly finish the argument.


1 Answer 1


The natural surjection $A\to A/\text{rad}A$ induces a surjection $A^\times\to (A/\text{rad}A)^\times$, and so $(A/\text{rad}A)^\times$ is path connected.

$A/\text{rad}A$ is a finite-dimensional commutative semisimple $\mathbb{R}$-algebra, and so is a finite product of copies of $\mathbb{R}$ and $\mathbb{C}$, and since its group of units is path connected it must be a product of copies of $\mathbb{C}$.

Therefore we can pick $x\in A$ so that $x^2=-1$ modulo $\text{rad}A$, and so, in $A$, $x^2=-1+r$ for some $r\in\text{rad}A$. Since $A$ is finite dimensional, $r$ is nilpotent.

So $1-r$ has an inverse $1+s=1+r+r^2-\dots$, which in turn has a square root $1+t=1+s/2-s^2/8+s^3/16-\dots$.

So if $x'=x(1+t)$ then $x'^2=-1$ and so the subalgebra of $A$ generated by $x'$ is isomorphic to $\mathbb{C}$, and so $A$ has the structure of a $\mathbb{C}$-algebra.

  • $\begingroup$ Thanks for your answer! That settles my question nicely. $\endgroup$
    – M.G.
    Oct 5, 2015 at 18:38

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