Does there exist an $\mathbb{R}$-algebra in which exponentials are especially well-behaved?

Question

Question. Does there exist a finite-dimensional commutative $\mathbb{R}$-algebra $X$ equipped with a homeomorphism $$\exp : X \rightarrow X \setminus \{0\}$$ satisfying $\exp(0) = 1$ and $\exp(x+y) = \exp(x)\exp(y)$?

1. Note that $\mathbb{R}$ doesn't work because the range of the real exponential is $(0,\infty)$.

2. Note that $\mathbb{C}$ doesn't work because the complex exponential is not a bijection.

3. Obviously, my guess is that getting better behaviour than $\mathbb{C}$ offers is fundamentally impossible. I can't actually prove that this is true, so I thought I'd ask.

Addendum. This probably makes little difference, but anyway I changed $\exp$ from being a continuous bijection to being a homeomorphism. If you can solve the original version of the question, in which it was just required to be continuous bijection, that's also fine, and you'll still get the green checkmark.

Answer 1

This is impossible even if you relax "homeomorphism" to "bijection". Indeed, consider $x=\exp^{-1}(-1)$. Then $\exp(2x)=\exp(x)\exp(x)=1$, so by injectivity $2x=0$ which implies $x=0$, which is a contradiction. (More abstractly, the group $(X,+)$ is torsion-free, so it cannot be isomorphic to $(X\setminus\{0\},\cdot)$ which has the torsion element $-1$.)

Also, if you relax "homeomorphism" to "surjection" then this already implies that $X$ can only be $\mathbb{R}$ or $\mathbb{C}$ (up to isomorphism). Indeed, if $\exp$ is surjective, then $(X\setminus\{0\},\cdot)$ must be a group because $(X,+)$ is, which then means $X$ is not just a commutative $\mathbb{R}$-algebra but a field. The only finite field extensions of $\mathbb{R}$ are $\mathbb{R}$ and $\mathbb{C}$.

Answer 2

No for the obvious reason: $\mathbb R^n$ is not homeomorphic to $\mathbb R^n\setminus\{0\}$, since the latter is homotopic to $S^{n-1}$ which is not contractible.

From the algebraic point of view, note that $\exp(x)\exp(-x)=\exp(0)=1$, hence $\exp(x)$ is invertible. Therefore if $\exp$ is surjective, then $X$ is a field. So $X$ can only be $\mathbb R$ or $\mathbb C$. We can classify all Lie group homomorphisms from $(X, +)$ to $(X^{*}, \times)$ in both cases.

If homeomorphism is dropped, we can always define $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$ in any Banach algebra (over $\mathbb R$ or $\mathbb C$). And we can always make a Banach algebra out of a finite-dimensional algebra and the choice of the norm wouldn't change the convergence since there is a unique Hausdorff topology on any finite-dimensional real vector space compatible with the vector space operations.