Timeline for Is a vector space naturally isomorphic to its dual?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2019 at 9:50 | comment | added | M. Winter | @Tom This is not the case for $V\to V^{**}$, as $$\forall v\in V:\forall f\in V^* :\phi(v)(f)=f(v)$$ is a totally fine first-order statement that is made true by exactly one isomorphism $\phi:V\to V^{**}$, and so no permutation exists that maps $\phi$ onto any other isomorphism while preserving truth. | |
| Nov 10, 2019 at 9:39 | comment | added | M. Winter | @Tom I have the feeling that your answer does not lie in categroy theory, but in formal logic. What you want is a transitive permutation on the set of all isomorphisms $\phi:V\to V^*$ (with $V$ fixed), so that this permutation preserves the truth of all first-order statements in whatever language used to work with the pair $(V,V^*)$ of vector spaces. Then, any two isomorphisms cannot be distinguished by the means of the language, and so no non-arbitrary choice is possible. | |
| Nov 10, 2019 at 8:17 | comment | added | Tom Ellis | Thank you, yes, I understand this. What I want to know is how does one formally specify the notion of "no additional inputs"? | |
| Nov 9, 2019 at 17:39 | history | answered | Ivan Meir | CC BY-SA 4.0 |