# A Need (or not) for invoking axiom of choice in a certain case

1. Sep 13, 2017

### RockyMarciano

I was wondering about the following scenario, we have a certain differentiable manifold with the standard topology not induced by any previous metric structure on the manifold. There is no natural way to identify a vector with its dual(no canonical isomorphism between them),

If we had to define the length of a curve and we were not allowed to use an induced metric from a prespecified metric structure, would it be necessary to make a choice of inner product for each of the uncountable points in the curve and therefore would we have to appeal to the axiom of choice?

2. Sep 16, 2017

### RockyMarciano

I meant "induced topology" instead of "induced metric" in the second paragraph. Sorry about that.

3. Sep 16, 2017

### Staff: Mentor

A differentiable manifold is a Hausdorff space with countable basis (definition) and every differentiable manifold admits a Riemannian metric (theorem). Therefore your question can be translated to: Does the proof of this theorem require the axiom of choice? As far as I have seen on a quick glimpse this is not the case as the main tool is the countable basis. However, AC is sometimes well hidden somewhere in a result which is used by propositions which themselves are used to prove the theorem. Personally I do not think that AC is needed somewhere, as all required properties are already contained in the definition of a differentiable manifold. So neither an embedding nor the AC itself should be required at some stage of the proof. Differential geometry is about neighborhoods, not about points.

4. Sep 16, 2017

### RockyMarciano

I completely agree with everything you wrote. My question is trying to complicate a bit the picture by imposing some constraint that might seem to be artificial but that has some important examples in mathermatical physics. So the requirement is that even though our differential manifold admits a Riemannian metric, let's say that it has a pseudoriemannian metric and therefore the topology in the manifold is not the one that the pseudo-Riemannian metric tensor would induce, but the standard topology instead. And we want to determine the length( of a timelike curve in this manifold(let's say for concreteness that it is Minkowski 4-space).

Now in the usual case in Riemannian geometry with a differentiable manifold that has a defined riemannian metric that induces the standard topology I can easily see how integrating infinitesimal lengths at each point to obtain a length doesn't require the AC as the metric tensor is naturally built up on the differentiable and topologic structuures, but when the metric tensor doesn't agree with the manifold topology I don't know if this could require choice at each poit of the curve even if just to decide the signature convention for timelike vectors at each point given that it being purely conventional there is no canonical choice but still at each tangent space it is needed in order to integrate them to a curve length(or proper time).

5. Sep 16, 2017

### Staff: Mentor

The difficulty with the axiom of choice is, that you cannot control the way individual points are picked. Thus you don't have any control of their behavior. Esp. you can neither make any assumptions on their neighborhood nor on any equations like linearity, or inequalities like the triangle inequality, which doesn't make it helpful in the context of geometries. Things might change in the infinite dimensional case, where you need it to guarantee the existence of a basis of the tangent space. However, a metric on a finite dimensional manifold is an equation (bilinear form) about tangent vectors, or connections to compare different ones. All of which are given by an equation, i.e. a closed set. These are algebraic, resp. topological properties and the axiom of choice doesn't really apply on them. Or put it another way: the axiom of choice is about well-orderings. So either this well-order has something to do with the metric, in which case you don't need it anymore, or it has not, in which case it isn't helpful.

I once listened to a talk about the proof of the Banach-Tarski paradoxon and I remember, that the closing statement was: "This shows less the absurdity of the axiom of choice than it demonstrates our difficulties with what a point is." I think there is much truth to it. And translated to the case you mentioned, it means, that manifolds are at least continuous objects, and all their properties are of topological nature where AC has no role in it, i.e. open sets, not points.

6. Sep 17, 2017 at 5:52 AM

### RockyMarciano

Thanks, I see. So I would have to go for something weaker than the full AC, right? For instance the axiom of countable dependent choice(DC).

I'm not completely sure if DC is assumed either in the differentiable manifold definition(here I tend to think that countable choice(CC), which is weaker than DC, would be enough) or in the (pseudo)Riemannian manifold definition(here I'm pretty sure non-compact metric spaces need DC).

7. Sep 17, 2017 at 2:41 PM

### WWGD

The proof I remember to show the existence of a globally-defined metric does not require any infinite construction. Essentially, the local diffeomorphisms allow us to pullback ( pushforward) the needed properties of $\mathbb R^n$ onto the manifold locally. Then, using paracompactness ( 2nd Countable+ something else) we can show the existence of partitions of unity subordinate to the cover by open sets, so that we can glue the locally-defined inner-products into a single globally-defined inner-product.

8. Sep 17, 2017 at 2:46 PM

### Staff: Mentor

That's the proof I've found, too. I didn't read the details, but I thought for the partition of unity the countable basis would have been needed to avoid unpleasant sums.

9. Sep 17, 2017 at 2:48 PM

### WWGD

Precisely, no need to deal with convergence. Differential Geometers are somewhat-spoiled in dealing with manifolds, which are overall very well-behaved. It seems like Algebraic Geometry or Geometric Measure Theory would be nastier -- no wonder it is called "Ze-Risky" topology in Alg Geo. ;). Good point on the " points v open sets" , though.

10. Sep 17, 2017 at 3:11 PM

### RockyMarciano

The example of determining a length in the OP doesn't just require existence, wouldn't it be needed to attain an infimum? Isn't some countable choice needed for this?

Edit:(Actually to be precise I should be referring to the distance between two points in the manifold, rather than just the length of a curve)

Last edited: Sep 17, 2017 at 3:41 PM
11. Sep 17, 2017 at 3:13 PM

### WWGD

Good point, let me think it through.

12. Sep 17, 2017 at 3:15 PM

### Staff: Mentor

I love the risky topology. Maybe because the pun doesn't work here but more likely because the Zariski topology offers such a pretty short imagination: equations ergo closed, inequality ergo open. And everything is dense!$^*)$

There is another pun possible: May I introduce the honorable Prof. Dr.

Zariski!
---------
$^*)$ Not intended to be rigorous.

13. Sep 17, 2017 at 3:17 PM

### WWGD

But it is "anti-Haudorff", aka strongly-connected in most cases; no two points can be separated. Open sets are just way too big

14. Sep 18, 2017 at 8:59 AM

### RockyMarciano

A quick PF search of "countable choice" got this http://dml.cz/dmlcz/118951 showing several uses of various degrees of choice. The sentence it ends with: "Let us sum things up: Topology with “choice” may be as unreal as a soap-bubble dream, but topology without “choice” is as horrible as nightmare", seems to be in contrast with the previous comments about the null role of choice in topology.

Anyway, specifically for what concerns the OP it says that: the assertion "in a (pseudo)metric space X, a point x is an accumulation point of a subset A iff there exists a sequence in A \ {x} that converges to x" is equivalent to the axiom of countable choice. This appears to me a condition for an infimum length, no?

15. Sep 18, 2017 at 10:07 AM

### Staff: Mentor

Doesn't the infimum of length simply mean, that $\mathbb{R}$ is complete? I don't see the connection. Of course we could debate the justification of complete scalar fields, but this is a different topic and again independent of AC. We don't chose the limits in existence, we require they're already there.

16. Sep 18, 2017 at 10:21 AM

### martinbn

17. Sep 18, 2017 at 11:07 AM

### RockyMarciano

I don't see how defining distance in an arbitrary (pseudo)metric space with the infimum of a sequence of lengths o the curves that connect two points, which seems to require the assertion equivalent to countable choice in my previous post, could be equivalent to every cauchy sequence converges in the space, I don't think defining a distance requires completeness.

I'm not even sure if you are making any distinction between full AC and weaker forms in this thread.

18. Sep 18, 2017 at 11:22 AM

### Staff: Mentor

No, since I don't understand all of your words, e.g. "countable choice" or here "weaker form". Never heard of them before.
Me neither, since for the infimum to exist you need the limit to exist, which is completeness in my understanding and has nothing to do with any kind of choice.

19. Sep 18, 2017 at 11:36 AM

### martinbn

I am a bit lost. Infimum of what, taken over what?

20. Sep 18, 2017 at 11:41 AM

### RockyMarciano

Some weaker forms of choice:
Axiom of countable dependent choice
Axiom of countable choice
Ultrafilter lemma

All can be consulted in wikipedia

Ok, hopefully somebody might step in and clarify.
Added: You're probably right that it requires completeness but I don't think it is the same as completeness.

Last edited: Sep 18, 2017 at 11:55 AM