Timeline for Nuances Regarding Naturality
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| when toggle format | what | by | license | comment | |
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| May 2, 2020 at 23:51 | comment | added | LSpice | @HeinrichD, setting aside the issue of whether your $\omega(-)v$ span (which is an issue of finite dimensionality, as you say), how do you even show that the map you specify admits a linear extension to their span without choosing a basis? | |
| Sep 12, 2016 at 9:20 | comment | added | HeinrichD | Thus, finite-dimensionality is rather the statement that this relation is the graph of a function. | |
| Sep 12, 2016 at 9:10 | comment | added | HeinrichD | The natural isomorphism $Hom(V,V) \to V^* \otimes V$ is defined by linear extension of $\omega(-) v \mapsto \omega \otimes v$ for $\omega \in V^*$, $v \in V$. Of course, one has to argue that this works (where the finite-dimensionality of $V$ has to be used), but in any case this is a definition which is clearly free from arbitrary choices. If this isn't convincing enough, here is a definition of the graph of this natural isomorphism: $\{(\sum_i \omega_i(-) v_i : \sum_i \omega_i \otimes v_i) : \omega_i \in V^*, v_i \in V\}$. | |
| Jul 12, 2016 at 22:48 | comment | added | Vladimir Sotirov | Since every vector space having a basis is equivalent to the axiom of choice, one indication that a linear map cannot be defined by a basis-free formula would be if the linear map did not always exist when doing linear algebra in some (boolean) topos other than Sets... which is a fancy way of saying you could in principle use set-theoretic forcing. | |
| Jul 12, 2016 at 18:08 | comment | added | Timothy Chow | Possibly related: mathoverflow.net/questions/131255/… and mathoverflow.net/questions/19644/… and mathoverflow.net/questions/56938/… | |
| Jul 12, 2016 at 14:33 | comment | added | Oscar Cunningham | Maybe this is the same as what Pace Nielsen calls "$\exists$ elimination" in his answer. | |
| Jul 12, 2016 at 14:18 | comment | added | Oscar Cunningham | Here's a possible approach: I think the ur-example of a problem where you have to make an arbitrary choice is the following: "Let $A$ be non-empty and let $f:A\rightarrow B$. Is $B$ non-empty?". Clearly the only way to proceed is to choose an arbitrary $a\in A$ an consider $f(a)\in B$. Any other problem involving an arbitrary choice reduces to this one. For example we have a function from the set of finite bases of $V$ to the set of inverses to $V\rightarrow V^{**}$, but to show that $V\rightarrow V^{**}$ actually has an inverse we have to actually pick one of these bases. | |
| Jul 12, 2016 at 13:57 | comment | added | Oscar Cunningham | @DylanWilson I do like showing off my linear algebra skills! But I think the more general point is that if every example we can think of where it looks like you have to make an arbitrary choice turns out to not in fact need an arbitrary choice, then maybe we should begin to suspect that there aren't any situations where you have to make an arbitrary choice. | |
| Jul 12, 2016 at 13:19 | comment | added | Steven Landsburg | KConrad: "Are you bothered by that state of affairs?", Certainly not, and I hope I haven't given the impression that I'm bothered by anything at all. Mostly, I want to know whether people have thought about making subtler and more precise distinctions than the ones I've suggested in the post. If the answer is yes, I'd like very much to know more about what those people have thought (either in the past or in the present, in response to this question). If the answer is no, I'm not bothered by that a bit! | |
| Jul 12, 2016 at 12:10 | comment | added | Dylan Wilson | It seems like more people want to show off their knowledge of fun facts about vector spaces than want to answer the question... | |
| Jul 12, 2016 at 10:44 | answer | added | Aaron Meyerowitz | timeline score: 4 | |
| Jul 12, 2016 at 6:49 | comment | added | user13113 | Maybe what you want here is to consider the free monoidal closed abelian category generated by a single object $X$ and by $\hom(1,1) = k$, and then observe that $X \to X^{**}$ is not an isomorphism? (I'm not sure if there even exists a nonzero map $X^{**} \to X$) | |
| Jul 12, 2016 at 6:32 | comment | added | Qiaochu Yuan | ...sort of choices are needed to prove this? I don't even know. In some sense I need to make all sorts of choices to prove that something is LCH, e.g. for every point a choice of some neighborhood having some property. It's again unclear to me whether this is an interesting notion of having to make choices. | |
| Jul 12, 2016 at 6:31 | comment | added | Qiaochu Yuan | @Steven: I think part of what makes your question ambiguous is that much depends on the definition of "finite-dimensional" you use; if it's "has a finite basis" then there's a pretty tautological sense in which you can't make any claim about finite-dimensional vector spaces without referring to a choice of basis. I assume that's not interesting to you. Here is a different example where I don't even know what choices, if any, are required: if $A$ is a topological group then there's a natural map from $A$ to its double Pontryagin dual $[[A, S^1], S^1]$ which, if $A$ is LCH, is an iso. What... | |
| Jul 12, 2016 at 5:41 | answer | added | Pace Nielsen | timeline score: 5 | |
| Jul 12, 2016 at 4:40 | comment | added | KConrad | I am suspicious that your goal is possible. As an analogy, writing down homomorphisms out of a quotient group $G/N$ very often involves writing a homomorphism out of $G$ and then checking it is well-defined on cosets mod $N$ or is trivial on elements of $N$, not initially writing a homomorphism directly on $G/N$ unambiguously from the start. Are you bothered by that state of affairs? | |
| Jul 12, 2016 at 1:19 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| Jul 12, 2016 at 0:54 | answer | added | Qiaochu Yuan | timeline score: 10 | |
| Jul 11, 2016 at 23:17 | comment | added | HenrikRüping | At C) The natural map goes in the other direction $V^*\otimes_k V\rightarrow Hom_k(V,V), \quad f\otimes v\mapsto f(-)\cdot v$ and has a base free definition. The problem is that for if $V$ is infinite-dimensional, the image of that map consists of all maps in $Hom_k(V,V)$ with finite-dimensional Range. | |
| Jul 11, 2016 at 22:43 | answer | added | Oscar Cunningham | timeline score: 21 | |
| Jul 11, 2016 at 22:28 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| Jul 11, 2016 at 22:14 | history | asked | Steven Landsburg | CC BY-SA 3.0 |