Does the topology determine the vector space structure of a topological vector space?

Question

Let $(V, \cdot, +)$ be a topological vector space over $\mathbb{R}$ or $\mathbb{C}$ with topology $\mathcal{T}$ and let $0 \in V$ be the zero vector. Is then the linear structure $(\cdot, +)$ on the pointed topological space $(V,0)$ uniquely determined by the topology $\mathcal{T}$ and the distinguished zero element $0$?

If not in general, does this hold, if one specifies further requirements on $\mathcal{T}$, i.e. separation axioms?

Answer 1

I don't think so.

Consider $\mathbb{R}$ with the trivial topology. Note that any $\mathbb{R}$-vector space structure on $\mathbb{R}$ makes it into a topological vector space.

Now you can put different vector space structures on $\mathbb{R}$ such that $0$ is the zero vector.

For instance, take a bijection of $\mathbb{R^2}$ onto $\mathbb{R}$ sending $(0,0)$ ont $0$ and consider the induced vector space structure on the codomain obtained from $\mathbb{R}^2$.

Answer 2

No, the topology does not determine the vector space structure uniquely.

Consider any homeomorphism $f : \mathbb R \to \mathbb R$ that fixes $0$; for example, $f(x)=x^3$.

Using $f$, we can define a vector space structure on $\mathbb R$, by using $f$ to "transport" the usual vector space structure: $$x \oplus y = f^{-1}(f(x) + f(y)) $$ $$r \odot x = f^{-1}(r \cdot f(x)) $$ So in our example with $f(x)=x^3$ we have $1 \oplus 2 = \sqrt[3]{9}$.

The same trick works in any dimension. The point is that a topological vector space can have many nonlinear self-homeomorphisms that fix $0$, and each of those can be used to transport the structure to a distinct vector space structure.