Timeline for Constructions unique up to non-unique isomorphism
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2017 at 22:43 | comment | added | LSpice | @DavidFeldman, do you really think that it's true that it's always a rigidification, and not a basis, that one wants? For example, it's much clearer to me how to think of diagonalising a matrix as choosing a basis than as rigidifying a vector space (unless your model vector space is $(k^{\mathrm{sep}})^n$, regarded as equipped with its projection functions, in which case you're choosing a basis anyway even if you refuse to say so). | |
| Jan 31, 2011 at 5:00 | comment | added | David Feldman | I do think an example lurks here, but you haven't nailed it. You don't want a basis, you want only that for which you thought you wanted a basis, namely rigidification. So for one given field and a given dimension fix a model vector space $M$ of that dimension. Define a rigidification of any given vectors space (over the same field, with the same dimension) as in isomorphism $i:V\rightarrow M$. An isomorphism of rigidified vector spaces is a map $j:M\rightarrow M$ that makes the triangle commute. Now you get $GL(n)$ (and other groups if you force more structure on $M$.) | |
| Jan 30, 2011 at 17:29 | comment | added | Chris Heunen | @Simon: I agree this might only answer the letter of the question and perhaps not its spirit. Neverthelesss, OP asked for as many examples from as many parts of mathematics as possible, and this is one. Moreover, these isomorphisms certainly form interesting groups, namely $SL(n)$ and $U(n)$. Anyway, if one is after insight into the abstract features of such situations, isn't it important to also consider examples that fall outside one's initial intuition? | |
| Jan 30, 2011 at 16:23 | comment | added | Simon Rose | @Chris: We might be splitting hairs, but I think that this misses the spirit of the question. There is some notion of canonicality which defines an algebraic closure, for example, but no such notion that defines a basis of a vector space. | |
| Jan 30, 2011 at 16:14 | comment | added | Chris Heunen | @Simon: but there is an linear isomorphism that maps the former basis to the latter. So indeed a basis is not unique, but unique up to an isomorphism. Finally, this isomorphism itself is not unique, as it can for example be composed with any isomorphism that leaves the latter basis invariant. | |
| Jan 30, 2011 at 16:04 | comment | added | Simon Rose | Wait, how do vector spaces have a unique basis? If $V = \langle x, y\rangle$, then $x + y, y$ is also a basis of $V$, but it is not the same basis. | |
| Jan 30, 2011 at 10:17 | history | answered | Chris Heunen | CC BY-SA 2.5 |