I was recently thinking about what it means to put structure on a set. It seems to me that, in my area (representation theory), the two main ways of imposing structure on a set $X$ are:
- distinguishing certain permutations of $X$ as structure preserving;
and
- distinguishing certain test functions (here I think of $X \to \mathbb C$, but that's just because it fits my example) are structure respecting.
For example, given a manifold, we can look at the diffeomorphisms among its permutations, or the smooth functions among its test functions; and, given a vector space, we can look at the linear automorphisms among its permutations, or the functionals among its test functions. Under reasonable hypotheses, the test-functions perspective (more 'analysis'-flavoured, maybe) determines the structure; but does the group perspective (more 'geometry' / Erlangen-flavoured)?
A colleague pointed out that, for manifolds, the diffeomorphism group does determine the manifold, in a very strong sense: http://www.ams.org/mathscinet-getitem?mr=693972. Given that success, I was moved to ask: does the linear automorphism group determine the vector space?
Here's one way of making that informal question precise. Suppose that $V_1$ and $V_2$ are vector spaces (over $\mathbb C$, say) with the same underlying set $X$, and that the set of permutations of $X$ that are linear automorphisms for $V_1$ is the same as the analogous set for $V_2$. Then are $V_1$ and $V_2$ isomorphic?
Another colleague, and The Masked Avenger, both thought of the axiom of choice when I asked this question; but I'm not sure I see it. It's just a curiosity, so I have no particular investment in whether answers assume, negate, or avoid choice.
EDIT: Since I think it may look like I am making some implicit assumptions, I clarify that I do not mean to assume that the vector spaces are finite dimensional, or that the putative isomorphism from $V_1$ to $V_2$ must be the identity as a set map of $X$. Thus, for example, Theo Johnson-Freyd's example (https://mathoverflow.net/a/186494/2383) of letting $X$ be $\mathbb C$, and equipping it with both its usual and its 'conjugate' $\mathbb C$-vector space structure, is perfectly OK.
Best Answer
The dimension of $V$ is the least non-negative integer $n$ such that there exist $v_1,\dotsc, v_n$ in $V$ such that there exists a unique $g\in G:=GL(V)$ that fixes each of $v_1,\dotsc,v_n$. So the isomorphism class of $V$ is determined by the group action of $G$ on $V$.