How many smooth vector space structures does $\mathbb R^n$ have?

Question

The title is a bit misleading. I am really asking about how many ways are there to equip $\mathbb R^n$ with a smooth addition $+:\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ so that the (possibly non-standard) addition-map together with the standard scalar multiplication $\mathbb R \times \mathbb R^n \to \mathbb R^n$ give the smooth manifold $\mathbb R^n$ (with its standard smooth structure) the structure of a smooth vector space? A smooth vector space is like an ordinary vector space, only that all involved maps must be smooth.

Background: I was learning a little bit of synthetic differential geometry, and while doing that I seemed to have proven that there is only one such smooth addition map on $\mathbb R^n$ once scalar multiplication is fixed. This seemed a bit extreme, so I thought it is time for a reality check.

Here is my argument: There is a well adapted model of SDG which sends the smooth manifold $\mathbb R$ to the line object $R$ of the model and which sends $\mathbb R^n$ to the product $R^n$. We may also choose a model such that each manifold becomes a microlinear space in it. In particular $R^n$ is microlinear (which is one of the axioms of synthetic differential geometry anyway), and it satisfies the Kock-Lawvere axiom. Such an $R$-vector space is called Euclidean by Lavendhomme. Lavendhomme shows in their book that a map out of any $R$-vector space $V$ into an Euclidean vector space is already linear if it is homogeneous. Now assume $+'$ is any addition map on $\mathbb R^n$ such that we get a smooth vector space together with standard scalar multiplication. Then we use the embedding of manifolds into the well adapted model to get an addition map on $R^n$ such that the resulting structure is a vector space internal to the model. Let us denote this vector space by $V$. Since everything except the addition map is fixed, we have that the identity $V \to R^n$ is homogeneous. Since $R^n$ with its standard addition is an euclidean vector space (in the terminology of Lavendhomme, Basic Concepts of Synthetic Differential Geometry), we see that the identity $V\to R^n$ must be linear. But then we see that the two addition maps must agree, and using that the embedding of manifolds into the model is faithful we see that the abitrary addition $+'$ is in fact the standard one.

What do you think? Did I make a mistake? I often do stupid stuff when I calculate alone, so reality check please!

Answer 1

I don't know enough synthetic differential geometry to check your proof but here is a different proof of the same conclusion. Note first that any smooth vector space structure on $\mathbb{R}^n$ is isomorphic as a Lie group to the addition structure (indeed, any abelian Lie group structure on $\mathbb{R}^n$ is isomorphic to the usual one; this is immediate from the classification of connected abelian Lie groups). Also, any such isomorphism of Lie groups will also preserve the scalar multiplication operation and thus is an isomorphism of smooth vector spaces (for instance, since scalar multiplication by rationals is determined by just the group structure and then scalar multiplication by reals is determined by continuity).

This means that any smooth vector space structure with the ordinary scalar multiplication must be conjugate to the usual addition map by some diffeomorphism $f:\mathbb{R}^n\to\mathbb{R}^n$ which commutes with scalar multiplication. But now differentiability of $f$ at $0$ forces it to be linear. Indeed, for any $x\in\mathbb{R}^n$ and $t\in\mathbb{R}_+$, $f(x)=\frac{f(tx)}{t}$. Taking the limit as $t\to 0$, we get $f(x)=df_0(x)$ so $f$ must coincide with its derivative $df_0$ at $0$, which is a linear map. So $f$ is linear, and so conjugating addition by $f$ does not change the addition operation and our smooth vector space structure is equal to the usual one.