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

[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?

 

[RockyMarciano]

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

 

[fresh_42]

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.

 

[RockyMarciano]

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.

Thanks for replying.
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).

 

[fresh_42]

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.

 

[RockyMarciano]

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.

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).

 

[WWGD]

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.

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.

 

[fresh_42]

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.

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.

 

[WWGD]

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.

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.

 

[RockyMarciano]

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.

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)

 

[WWGD]

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?

Good point, let me think it through.

 

[fresh_42]

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.

 

[WWGD]

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. View attachment 211282Zariski!
---------
$^*)$ Not intended to be rigorous.

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

 

[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?

 

[fresh_42]

This appears to me a condition for an infimum length, no?

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.

 

[martinbn]

Why do you ask?

 

[RockyMarciano]

Doesn't the infimum of length simply mean, that $\mathbb{R}$ is complete?

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.

 

[fresh_42]

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

No, since I don't understand all of your words, e.g. "countable choice" or here "weaker form". Never heard of them before.

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 ...

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.

 

[martinbn]

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

 

[RockyMarciano]

No, since I don't understand all of your words, e.g. "countable choice" or here "weaker form". Never heard of them before.

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

All can be consulted in wikipedia

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.

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.

 

[RockyMarciano]

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

Distance between 2 points in (pseudo)Riemannian manifolds is defined from the concept of curve arc length as the infimum of the arc lengths of the possible curves uniting the points. I would like to know if attaining this infimum requires some weak form of choice like countable choice.

 

[RockyMarciano]

Perhaps an example might help, would you say the following requires some form of choice or is enpugh with completeness? : "Given a sequence ($A_n$) of non-empty sets, choose $x_n ∈ A_n$ for each $n∈N$".
.

 

[martinbn]

Distance between 2 points in (pseudo)Riemannian manifolds is defined from the concept of curve arc length as the infimum of the arc lengths of the possible curves uniting the points. I would like to know if attaining this infimum requires some weak form of choice like countable choice.

In the beginning it seemed that you were asking about the existence of a metric. Now you say that you have a Riemannian manifold, which comes with the metric, and you want to define distance between two points as the infimum of lengths of curves that join them (in the pseudo-Riemannian case some things are different). But then this has nothing to do with manifolds, topologies and what not. What you are asking, seems to be, whether the result that a bounded from below set of real numbers has an infimum requires the axiom of choice.

 

[martinbn]

Perhaps an example might help, would you say the following requires some form of choice or is enpugh with completeness? : "Given a sequence ($A_n$) of non-empty sets, choose $x_n ∈ A_n$ for each $n∈N$".
.

How is that related to any of the above, for example the completeness of the real numbers or metrics on manifolds?

 

[RockyMarciano]

In the beginning it seemed that you were asking about the existence of a metric. Now you say that you have a Riemannian manifold, which comes with the metric, and you want to define distance between two points as the infimum of lengths of curves that join them (in the pseudo-Riemannian case some things are different). But then this has nothing to do with manifolds, topologies and what not.

I realize that I have had problems making my question precise, I did mention lengths in manifolds in the OP.

What you are asking, seems to be, whether the result that a bounded from below set of real numbers has an infimum requires the axiom of choice.

No, that is not what I'm asking exactly, I am referring to the infimum in relation with the definition of distance in Riemannian geometry.

How is that related to any of the above, for example the completeness of the real numbers or metrics on manifolds?

It is related to my question in this way, I identify the sequence $A_n$ with the sequence of curves(as sets of reals) between the points, and $x_n$ with the arc length in each curve, one of them being the infimum. I think this could be a good way to make it precise what I'm asking.

 

[fresh_42]

It is related to my question in this way, I identify the sequence $A_n$ with the sequence of curves(as sets of reals) between the points, and $x_n$ with the arc length in each curve, one of them being the infimum. I think this could be a good way to make it precise what I'm asking.

But you don't make any choice, you build a limit which defines your choice, not an axiom.

 

[RockyMarciano]

But you don't make any choice, you build a limit which defines your choice, not an axiom.

Right, the choice is of an arc length for each n curve. Sorry if I confused you earlier. So the question is just whether choosing $x_n$ in #22 involves just ZF(and induction) or also some amount of the axiom of choice, what's your opinion?

 

[martinbn]

What do you mean by the choice of an arc length for each curve? If you have a Riemannian manifold and a curve between two points, there is no choice for its length.

 

[RockyMarciano]

Sure for the arc length of a curve there is no choice involved, but this is not what I'm talking about here, I'm talking about the function defined on the collection of curves(as sets of real numbers) between two points that assigns to each of these countably infinite curves simultaneously some element of the set of real numbers of each curve to construct a sequence of which the infimum is defined to be the distance between the points. I'm asking if this choice function requires the axiom of countable choice.

 

[martinbn]

It is still very unclear. What choice is there for the function? Given any curve from the family you compute its length. So the function is the map from those curves to real numbers assigning the each curve its length, no choice involved here. May be you can make it clearer on an example. Take the Euclidean plane and two point, consider all curves (let's say smooth) with those two points as end points. Each of them has a length and you have the function that maps each curve to its length i.e. the domain is the set of these curves the range is positive real numbers. What you have is a set of positive real numbers, does it have an infimum? Where in all this is the choice you are asking about?

 

[RockyMarciano]

It is still very unclear. What choice is there for the function? Given any curve from the family you compute its length. So the function is the map from those curves to real numbers assigning the each curve its length, no choice involved here. May be you can make it clearer on an example. Take the Euclidean plane and two point, consider all curves (let's say smooth) with those two points as end points. Each of them has a length and you have the function that maps each curve to its length i.e. the domain is the set of these curves the range is positive real numbers. What you have is a set of positive real numbers, does it have an infimum? Where in all this is the choice you are asking about?

I'm considering the set X of all curves between the 2 given points, and these curves as sets S of points, and the choice function f(S) as the one that assigns a positive real number to each S in X. I'm using this definition of choice function from Wikipedia:"A choice function (selector, selection) is a mathematical function f that is defined on some collection X of nonempty sets and assigns to each set S in that collection some element f(S) of S".

 

[martinbn]

I'm considering the set X of all curves between the 2 given points, and these curves as sets S of points, and the choice function f(S) as the one that assigns a positive real number to each S in X. I'm using this definition of choice function from Wikipedia:"A choice function (selector, selection) is a mathematical function f that is defined on some collection X of nonempty sets and assigns to each set S in that collection some element f(S) of S".

But in your case $f(S)$ is not an element of $S$, it is a real number, while $S$ consists of points of the manifold.

 

[WWGD]

Perhaps an example might help, would you say the following requires some form of choice or is enpugh with completeness? : "Given a sequence ($A_n$) of non-empty sets, choose $x_n ∈ A_n$ for each $n∈N$".
.

Yes, this is the actual statement, at least as I know it.

 

[RockyMarciano]

But in your case $f(S)$ is not an element of $S$, it is a real number, while $S$ consists of points of the manifold.

It is quite straightforward to identify the points with the real intervals(sets of reals) that the curves map to the points in the manifold, so S consists of real numbers. Or is this not licit?

 

[RockyMarciano]

@martinbn, @fresh_42, @WWGD could you please address #34? I'm not sure if curves can be identified with sets of real numbers in this context.

 

[fresh_42]

@martinbn, @fresh_42, @WWGD could you please address #34? I'm not sure if curves can be identified with sets of real numbers in this context.

You can label them with their lengths. I wouldn't call it an identification as it is a function from $\mathbb{R}^n \longrightarrow \mathbb{R}$ and information is lost. But in the next step, you build a limit to find the shortest of them, which is different from choosing one of them. If you'd chose one, then you will get any number or curve. Only the limit gets you the smallest or shortest. Its existence is guaranteed by completeness of $\mathbb{R}$ not by a process of choice. If we'd switch to $\mathbb{Q}$ we would still have a choice, but no limit.

 

[WWGD]

I'm considering the set X of all curves between the 2 given points, and these curves as sets S of points, and the choice function f(S) as the one that assigns a positive real number to each S in X. I'm using this definition of choice function from Wikipedia:"A choice function (selector, selection) is a mathematical function f that is defined on some collection X of nonempty sets and assigns to each set S in that collection some element f(S) of S".

You see, the thing is that, as Fresh mentioned, the (a?) problem with choice is that the selection of elements is " uncontrolled". In your case, there is, in a sense, a unique and well-defined method for choosing the length of the curve, if I understood your layout correctly --please correct me if I did not) so this does not, in this regard create a problem; the functional value is given by the inf. of all lengths, and we are not making AC choices regarding the curves, so I really don't see AC being an issue here.

 

[RockyMarciano]

You can label them with their lengths. I wouldn't call it an identification as it is a function from $\mathbb{R}^n \longrightarrow \mathbb{R}$ and information is lost. But in the next step, you build a limit to find the shortest of them, which is different from choosing one of them. If you'd chose one, then you will get any number or curve. Only the limit gets you the smallest or shortest. Its existence is guaranteed by completeness of $\mathbb{R}$ not by a process of choice. If we'd switch to $\mathbb{Q}$ we would still have a choice, but no limit.

You see, the thing is that, as Fresh mentioned, the (a?) problem with choice is that the selection of elements is " uncontrolled". In your case, there is, in a sense, a unique and well-defined method for choosing the length of the curve, if I understood your layout correctly --please correct me if I did not) so this does not, in this regard create a problem; the functional value is given by the inf. of all lengths, and we are not making AC choices regarding the curves, so I really don't see AC being an issue here.

Thanks. I'm myself convinced that there is no choice involved in the existence of the inf of the lengths. Sorry about creating this confusion in several of my posts. I am concerned just with the picking of the lengths from the countably infinite curves. This seems to hinge according to the above linked definition of choice function on whether one can consider each curve as a set S of real numbers, out of which a specific positive real number is chosen as a length by a choice function, and this is my actual question, can each of the curves between two points in the manifold be considered as a set of real numbers in this contex?.

 

[fresh_42]

This seems to hinge according to the above linked definition of choice function on whether one can consider each curve as a set S of real numbers, out of which a specific positive real number is chosen as a length by a choice function, ...

If you regard it as a choice, then even an ordinary function $f(x)=x$ is a choice and the detour on manifolds isn't needed at all. In your sense every assignment of values is a choice. You can think of it like this, but it again has nothing to do with the axiom of choice and it will cause exactly the same trouble everywhere as here: the rest of us call it function or limit, not choice.

... and this is my actual question, can each of the curves between two points in the manifold be considered as a set of real numbers in this contex?.

No. This is as if you considered every article in a supermarket as the price tag it's labelled with. You can do this if you're calculating the value of its inventory, but not if you're looking for (the cheapest) chocolate. Even if we only labelled chocolate, then the prizes would still be just one aspect of it. It is a function, not an identity. So if curves can be "considered as a set of real numbers", depends on what you want to do with it. In some contexts it will be sufficient and in many others it's not.

 

[WWGD]

Thanks. I'm myself convinced that there is no choice involved in the existence of the inf of the lengths. Sorry about creating this confusion in several of my posts. I am concerned just with the picking of the lengths from the countably infinite curves. This seems to hinge according to the above linked definition of choice function on whether one can consider each curve as a set S of real numbers, out of which a specific positive real number is chosen as a length by a choice function, and this is my actual question, can each of the curves between two points in the manifold be considered as a set of real numbers in this contex?.

But you see, we really don't need a choice function, since we have exactly one value and this value is selected, not by a random choice, but by a well-defined method. And, yes, a curve can be identified with a collection of Real numbers, e.g., through a parametrization; a point p in the manifold can be identified with t with C(t)=p , but then each curve is a continuum, but you are not really selecting any of the values t here; you are choosing from a collection of lengths, specifically the inf of all lengths. EDIT The process of finding an inf does not necessarily require AC that I can tell; e.g., inf(a,b)=b. And the collection of all curves is uncountable, e.g., for every pair of points (x,y) ; $x, y \in \mathbb R - \mathbb Q$ you can draw a curve passing through them.So it comes down to deciding if in this case, finding the inf requires AC, but I don't see how; you are selecting _one_ number from a collection of values ( the lengths of all possible curves between two points), and I don't see any choice being made.

 

[RockyMarciano]

I think I understand your point now. I only still have a doubt concerning how this applies to the Minkowskian case I mentioned at the beginning of the thread. I can see now how in the Riemannian case I'm not selecting the lengths as there is well-defined method to select lengths from curves and this is a given so there is no choice function here. But the integral giving the lengths in the Minkowskian case is not exactly like in the riemannian case, it includes a sign choice that is dependant on the previous choices to be consistently a timelike or spacelike curve, would this be an example of dependent choice?

 

[fresh_42]

This is a choice about the geometry or topology you want to perform your calculations in. The same way as if you chose real or complex analysis. It will change the set-up, possibilities and outcome, but it is a choice among different systems, not a choice within a given system.

 

[RockyMarciano]

This is a choice about the geometry or topology you want to perform your calculations in. The same way as if you chose real or complex analysis. It will change the set-up, possibilities and outcome, but it is a choice among different systems, not a choice within a given system.

Not sure what you mean by system here. Using timelike curves and congruences is a decision dictated within the spacetime notion "system", very specifically and well defined mathematically.

 

[martinbn]

I think I understand your point now. I only still have a doubt concerning how this applies to the Minkowskian case I mentioned at the beginning of the thread. I can see now how in the Riemannian case I'm not selecting the lengths as there is well-defined method to select lengths from curves and this is a given so there is no choice function here. But the integral giving the lengths in the Minkowskian case is not exactly like in the riemannian case, it includes a sign choice that is dependant on the previous choices to be consistently a timelike or spacelike curve, would this be an example of dependent choice?

No, it would not be such an example. If you want to study the geometry forget the axiom of choice. It only confuses you.

 

[fresh_42]

Not sure what you mean by system here. Using timelike curves and congruences is a decision dictated within the spacetime notion "system", very specifically and well defined mathematically.

Yes, but this is the same framework we discussed all over the thread. In your last post you've asked about the choice between Minkowski and Euclid, which is a choice of systems. Once you made this choice, we are back at post #1.

 

[RockyMarciano]

In your last post you've asked about the choice between Minkowski and Euclid, which is a choice of systems. Once you made this choice, we are back at post #1.

I didn't ask about any choice between Minkowski and Euclid, I asked about the integral of the metric in Minkowski incorporating a ± sign choice that must be consistently maintained at every point of the integrated path.

 

[martinbn]

What would be the situation where the sign is not only plus or only minus along the curve?

 

[RockyMarciano]

What would be the situation where the sign is not only plus or only minus along the curve?

You mean mathematically? I'd say only with dependent choice(where previous choices of plus or minus condition later ones) one can mathematically guarantee that such situation doesn't arise.

 

[martinbn]

No, I mean what would be a specific situation, and why would you be interested in it?

 

[RockyMarciano]

No, I mean what would be a specific situation, and why would you be interested in it?

As you can see below such situation(last sentence of quoted paragraph) is not considered, the reason is physical.
I'm not interested in it, rather I'm interested on how it is mathematically granted this situation doesn't arise, and I thought it was done by using dependent choice, but you claim it isn't, right?
https://en.wikipedia.org/wiki/Arc_length#Generalization_to_.28pseudo-.29Riemannian_manifolds

"The sign in the square root is chosen once for a given curve, to ensure that the square root is a real number. The positive sign is chosen for spacelike curves; in a pseudo-Riemannian manifold, the negative sign may be chosen for timelike curves . Thus the length of a curve in a non-negative real number. Usually no curves are considered which are partly spacelike and partly timelike."