Timeline for Does the linear automorphism group determine the vector space?
Current License: CC BY-SA 3.0
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| Nov 11, 2014 at 21:27 | comment | added | LSpice | Theo, certainly I agree that I could have asked the stronger question that you pose, but I didn't because, as you point out, the answer is pretty obviously 'no'. Although the isomorphism type of $\operatorname{GL}(V)$ does determine that of $V$, per @AntonKlyachko (mathoverflow.net/a/186573/2383), I'm not sure that's the only way that the answer could have played out (though maybe some abstract nonsense says that I am wrong). That is why I asked about equal, not just isomorphic, groups $\operatorname{GL}(V)$. | |
| Nov 11, 2014 at 21:24 | comment | added | LSpice | @ToddTrimble, I'm not sure that I see an appropriate change, but it's obvious that many people found my wording confusing. Could you suggest a specific, improved wording? | |
| Nov 10, 2014 at 1:00 | comment | added | Theo Johnson-Freyd | ... well formed. Indeed, the problem is that there is no functorial construction of $V$ from $\mathrm{GL}(V)$, so that, in particular, bundles of groups isomorphic to $\mathrm{GL}(V)$ do not lift to bundles of vector spaces isomorphic to $V$. (The first example I know of is a bundle whose base space is the suspension of $\mathbb R \mathbb P^2$.) | |
| Nov 10, 2014 at 0:56 | comment | added | Theo Johnson-Freyd | @LSpice Incidentally, I was led astray by your discussion of the underlying set $X$. Normally a "vector space structure" consists of particular maps --- it makes sense, then, to ask whether two given vector space structures on the same set are equal. The Erlangen program, at best, is about actually giving a manifold some structure in this strict sense. Your question, as I now understand, was whether the isomorphism type of the group $\mathrm{GL}(V)$ determines the isomorphism type of the vector space $V$. This is within a family of common questions, but I would argue is not really ... | |
| Nov 9, 2014 at 14:30 | comment | added | Todd Trimble | In view of your last comment under Prasad's answer, shouldn't the opening sentence be edited a little? | |
| Nov 7, 2014 at 18:00 | comment | added | LSpice | I'm confused—$z \mapsto \overline z$ is a linear isomorphism of 'usual' $\mathbb C$ with 'conjugate' $\mathbb C$, so this doesn't seem like a counterexample. (I didn't require that the isomorphism be the identity map of $X$.) | |
| Nov 7, 2014 at 15:48 | history | answered | Theo Johnson-Freyd | CC BY-SA 3.0 |