Are Christoffel symbols measurable?

[twofish-quant]

I think the source of confusion here came from the different meaning each poster attributes to "observations", in fact this concept is broader and more ill-defined than the more strict concept of measurement of a physical quantity although some physicists use them indistinctly to refer to the latter meaning. When used strictly in the sense of measurement it is clear all of them are scalars in the physical sense as has been explained in this thread.

True. And I'm arguing that there are different meanings in the term "scalar" and that the way that it is used in GR is quite restrictive, and different although clearly related to the way that mathematicians and even other physicists (i.e. people in CFD)
use it. Also the distinction is non trivial since there are some physical quantities that I would argue are "scalar" in the GR sense but "vector" in another. I'm trying to think of something that goes the opposite way, and that is "vector" in the GR sense, but scalar in some other sense.

Also, this logic solves the "paradox of the left-handed glove." If you argue that "scalar" as used in relativity is a very restrictive definition, then the distinction between left and right handed gloves is something that is outside of GR.

 

[twofish-quant]

I don't know about scalar, but one way to make a volume independent of coordinates is to specify the coordinate system. So the volume of Coke in mL is coordinate independent. Of course, this assumes that the people at NIST have done their jobs, and that we have some way of transporting their standards around.

True, but what happens when after specifying a coordinate system, you still end up with something that looks like a vector. Leaving aside social science examples, if you do relativistic fluid dynamics, once you specify the reference frame what you end up is still a "vector."

Velocity fields make things complicated. But color and composition form vector spaces that are independent of the space-time vector spaces. Mathematically you can get into the world of fiber bundles.

 

[twofish-quant]

Isn't the goal of Quantum Gravity to make GR a gauge theory?

The goal of quantum gravity is to unify QM and GR by any means possible. God will tell us the right approach.

 

[atyy]

True, but what happens when after specifying a coordinate system, you still end up with something that looks like a vector. Leaving aside social science examples, if you do relativistic fluid dynamics, once you specify the reference frame what you end up is still a "vector."

Velocity fields make things complicated. But color and composition form vector spaces that are independent of the space-time vector spaces. Mathematically you can get into the world of fiber bundles.

Yes, the point was that in classical GR, in the presence of sufficient matter there isn't a sharp distinction between coordinate dependent and coordinate-independent quantities.

There is the metric which is a tensor field, which is similar to a vector field in that it is a geometric object that eats covectors and spits out "scalars". Its components change with coordinate system, so they are coordinate dependent. But if you use matter to specify a coordinate system, the components then become coordinate independent.

Rovelli gives an example where the metric components are coordinate independent.

 

[juanrga]

Isn't the goal of Quantum Gravity to make GR a gauge theory? Or is this separate issue from the goal of unifying the four forces including gravity but making it part of a larger gauge symmetry? But what perflexed me is how can they make gravitons be indistinguishable from electromagnetic force which is how you do it for example in GUT where and unification produced new physical process that can make quark decay into electrons and neutrinos, hence the search for proton decays.

About Gauge theory of Gravity. I saw this:

http://www.icpress.co.uk/physics/p781.html

It says there are attempts to derive at the gauge theory of gravitation. But in your context how can they do that when "it is not a field theory over flat spacetime. GR is a (geo)metric theory" as you mentioned?

The goal of Quantum Gravity is to describe quantum gravitational phenomena.

Gravity is not a force in GR.

Nobody makes gravitons indistinguishable from electromagnetic force.

People can do all the nonsense that they want including the belief that a covariant derivative can be considered a gauge derivative.

Part of the explanation of why the search for a consistent quantum gravity theory has failed since the 50s is because most of people in the field does not know what are doing.

 

[PAllen]

[two fish posts immediately above clear up some ambiguities for me...]


[1] PAllen posts:

That was at least hinted at elsewhere, and I did not 'get it'...good insight, thanks.


[2]The referenced paper says:



PAllen says:




Although I believe I do understand that components of a vector are themselves vectors...[I had never thought of frequency as a vector component]...I have to think more about this answer...meantime: so what is the referenced paper claiming...Are they wrong, do they have a different definition of scalar, or are they really taking about the 'measurement' ?

There is no discrepancy here. They are just being looser. They said, roughly, it is a single number (vernacular scalar; perhaps, scalar in pre-relativity physics) but it is not a rank0 tensor (= scalar in relativistic theories). I was clarifying what it is in relativity, rather than what it is not. FYI - to see the frequency needs to be treated as a timelike vector component in relativity, just take the 4-momentum of light (E,p) and divide by Planck's constant. Now you have a 4-vector with frequency as its timelike component.

[3] I also did some searching and found this comparison of classical and relativistic scalars which I did not realize [it seems obvious after reading it though] :



http://en.wikipedia.org/wiki/Scalar_(physics)#Scalars_in_relativity_theory

No problem with these ideas, right??

All of this looks fine to me.

 

[PAllen]

But we can stretch this into some absurd conclusions.

I take the a precinct-by-precinct map of the United States containing the election results of the Republican primary in 1980. The candidate votes form a vector and it's s perfectly good vector field. I can also form a vector field containing things like the price of real estate of different types of houses, the probability of default, divorce rates, crime statistics, etc. etc.

All of those are perfectly good vector fields.

Now are you trying to tell me that general relativity says something non-trivial about how political scientists can observe election results, or how real estate prices can be calculated? Just because you can represent real estate prices in a vector field, you are telling me that I have to *observe* the price of houses in a component by component way.

Now if you say "Yes, general relativity does restrict the way election results of the Republican primary and real estate prices for different types of houses in the US can be observed, and come up with some convoluted explanation for why, then we can go down that path, and I'll think of something for which that logic is so crazy that you'll have to say "huh?"

Now it's makes more sense to argue that this all happens because of a mix up in terminology. GR and SR state the all measurable quantities must be invariant and scalar *with respect to Lorenz transforms*. The results of the Republican primaries of 1980 are indeed invariant *with respect to Lorenz transforms* and even though a political scientist may represent them as "vectors" within relativity they are "scalars." In other words, GR has nothing non-trivial to say about political science and election results.

In other words, relativity provides some restrictions for how things are measured *with respect to a certain set of transforms*. Arguing that relativity restricts measurement for *all uses of vector spaces* is a bit of a stretch, and if you go down that route I'm sure that I can find something even more ridiculous than the examples I provided. Vector spaces are very useful and widely used in social science and political science, and I could think of some uses for art and literature.

Just thought of something ridiculous. Restaurant and movie reviews. I go on yelp.com or rottentomatoes.com. Restaurant rates form a vector (i.e. atmosphere, decor, service, etc.) You can do movie reviews the same way (quality of plot, amount of action, quality of print, etc.) Are you telling me that GR says that I can't make a measure of the atmosphere of the restaurant and decor, at the same time? I think I can. But wait, you are saying that general relativity says that it's impossible for me to come to non-scalar conclusions about restaurants. If you say yes, then my reaction is "who made Einstein the restaurant review police?" So you are saying that it is *physically impossible* for me to measure restaurant atmosphere and service at the same time?!

If you insist on yes, 1) I'll think of something more ridiculous and 2) I'll introduce you to a group of restaurant reviewers and let you tell them that you as an expert in general relativity have figured out that it is physically impossible to do reviews in a certain way, and if they insist that they can come up with vector conclusions, that Albert Einstein says that its impossible. Regardless of the outcome of 2), it will be worth watching for the entertainment value (Scientists Versus Restaurant Reviewers, the new Food Network reality show).

At some point what I'm trying to get you to do is to say "wait, Lorenz invariance and restaurant reviews are totally separate things! When you are using vector spaces to represent restaurant reviews that's got nothing to do with how vector spaces are used in GR" Which is my point.

Now if you agree with that. Suppose some alien creature creates a chain of restaurants around a black hole...

Also this is a no-lose situation. You might come up with some argument that indeed GR says something non-trivial and non-obvious about restaurant reviews. Like it says a lot of things about foreign exchange rates. (seriously)

Rather than discussing the details above, I will clarify where I am coming from, philosophically. I will specify some beliefs from the most general to the most technically specific:

1) The development of science since 1900 (esp. relativity and QM, but also generalizations outside of science) supports the view that nothing is observable or has 'objective reality' without also specifying the method of observation. An outside of science example is 'popular opinion'. I don't think it exists outside specification of the measurement process, and will be very different depending on how it is measured. Similarly, I don't consider E and B fields (or photon and electron fields) observable or objective; you need to specify characteristics of the measuring device to get an observation.

2) Jumping to physics (possibly extending to other cases), modern physical theories have a variety of internal symmetries. In each such theory, something that changes with these internal symmetries is defined as not observable. One class of mistake in using such theories is failure ensure a prediction is invariant relative to these internal symmetries.

3) The important thing is the achieving the invariance appropriate to the theory - otherwise you have misapplied it. I will concede that I have perhaps overemphasized 'scalar' when the real issue is invariance (and not e.g. covariance), because possibly all invariant quantities can be stretched to be collections of scalars (suitably defined). But the important issue is the invariance; focusing on scalars in GR is the most effective way to make sure you have formulated an observable properly. An example in GR where it is artificial to reduce to scalars to get invariance is: curvature tensor vanishes everywhere. This is an invariant feature of a Riemannian or Semi-Riemannian manifold. Ben gave a few other examples where get an invariant without needing to explicitly produce scalars.

 

[PAllen]

Rather than discussing the details above, I will clarify where I am coming from, philosophically. I will specify some beliefs from the most general to the most technically specific:

1) The development of science since 1900 (esp. relativity and QM, but also generalizations outside of science) supports the view that nothing is observable or has 'objective reality' without also specifying the method of observation. An outside of science example is 'popular opinion'. I don't think it exists outside specification of the measurement process, and will be very different depending on how it is measured. Similarly, I don't consider E and B fields (or photon and electron fields) observable or objective; you need to specify characteristics of the measuring device to get an observation.

2) Jumping to physics (possibly extending to other cases), modern physical theories have a variety of internal symmetries. In each such theory, something that changes with these internal symmetries is defined as not observable. One class of mistake in using such theories is failure ensure a prediction is invariant relative to these internal symmetries.

3) The important thing is the achieving the invariance appropriate to the theory - otherwise you have misapplied it. I will concede that I have perhaps overemphasized 'scalar' when the real issue is invariance (and not e.g. covariance), because possibly all invariant quantities can be stretched to be collections of scalars (suitably defined). But the important issue is the invariance; focusing on scalars in GR is the most effective way to make sure you have formulated an observable properly. An example in GR where it is artificial to reduce to scalars to get invariance is: curvature tensor vanishes everywhere. This is an invariant feature of a Riemannian or Semi-Riemannian manifold. Ben gave a few other examples where get an invariant without needing to explicitly produce scalars.

Let me add to (3) a few further observations. All measurements, even over time, constitute a finite amount of information. At most, they can be considered a finite set of vector quantities (using a different sense of vectors than GR). In no sense I know of, can any collection of measurements be considered to constitute a vector field. Any vector field you might associate with measurements is an abstraction. Thus, in my view no vector field is directly observable.

A further observation in the GR context is that in context of significant gravity and a region not completely 'local', even supposing you have specified the basis (position and motion -> basis 4 vectors) of each 'vector' observation, there is neither a unique (nor even unique most natural) way to patch these frames (one for each flag, for example) into a coordinate system. Thus any expression you give to a vector field has a significant contribution due to arbitrary convention.

 

[twofish-quant]

All measurements, even over time, constitute a finite amount of information.

I worry about statements like this. I don't have too much problem with the statement that "all measurements that physicists are used to making have characteristic X." But one of the points that I've been making is that there is a big difference between "measurements that physicists are used to making" and "all measurements."

You could argue that "all measurements can be reduced to all measurements that physicists are used to making." That's an extremely strong philosophical statement, and one that I'm inclined to consider to be false. There's one famous counter example called the "marriage meter." There is no set of physical measurements that you could make on me or my wife that could tell you whether or not we are married. By doing some sort of brain scan, you could establish that we *think* we are married. Since there is no such thing as a "marriage meter" then this means that things like marriage/divorce rates aren't physically measurable and you can put these things in vector spaces.

I run into this sort of thing all the time at work. Two of the big, big questions write now is "how do you measure liquidity?" and "how do you measure risk?" Which quickly gets you into some philosophical questions "what is liquidity?" and "what is risk?" The relevance of this to the current discussion is that it seems that whatever liquidity and risk are, they somehow involve rather complicated vector spaces and the same sort of math that you find in GR. (Correlation matrices from hell.)

In no sense I know of, can any collection of measurements be considered to constitute a vector field.

Color. Color requires three components to be specified, but color is independent of those components. You can specific color in terms of RGB, or CMYK or pantone or color temperature, but color is independent of those measurement. Because you need multiple components to specific color, and the existence of color is *independent* of those components, its a vector field, and more than a collection of measurements.

Stock portfolios have similar issues. There are multiple equivalent ways of representing the dynamics of stock portfolio, but the dynamics exists independent of those representations.

Vector spaces and the math associated with it comes in very handy when you have an "underlying reality" that's independent of the measurements. Relativity is one such example, but it's not the only one.

Any vector field you might associate with measurements is an abstraction. Thus, in my view no vector field is directly observable.

But you could argue that scalars are an abstraction. I mean when I measure light intensity, it goes into a meter that goes through my eyes into my brain where who knows what happens. The problem with saying that no vector field is directly observable is that you end up with a very restrictive definition of "observe" under which it's not clear that anything is observable.

There might be a physics reason to do this. In QM, to observe means to "collapse the wavefunction."

Thus any expression you give to a vector field has a significant contribution due to arbitrary convention.

True, but vectors are useful especially in cases where there is a "reality" that is independent of arbitrary convention. GR is one such use case but there are others.

 

[twofish-quant]

The other thing to be careful here is "proof by lack of imagination." In order to prove non-existence, you have to show that something really bad happens if something did exist. I can think of a lot of bad things that would happen if you had physical measurements that were none Lorenz invariant, or if quantum observations didn't reduce to a single number.

However, asserting that something is impossible because one can't think of counterexamples is a bad way of showing that something is impossible.

This is particularly true because vector spaces are really useful, and can represent things that are pure fantasy (i.e. any first person shooter video game has vector space representations of all sorts of imaginary things).

 

[atyy]

I worry about statements like this. I don't have too much problem with the statement that "all measurements that physicists are used to making have characteristic X."

Proof for the modern age: it's stored on digital media.

 

[Ben Niehoff]

The other thing to be careful here is "proof by lack of imagination." In order to prove non-existence, you have to show that something really bad happens if something did exist. I can think of a lot of bad things that would happen if you had physical measurements that were none Lorenz invariant, or if quantum observations didn't reduce to a single number.

However, asserting that something is impossible because one can't think of counterexamples is a bad way of showing that something is impossible.

Throughout this thread, we have been discussing the fact that all measured quantities must be scalars with respect to coordinate transformations of spacetime, a context to which none of your examples has been relevant.

However, this goes beyond merely spacetime scalars. Any measurement of a vector quantity (in the vector space sense, not in the computer science sense) must come by choosing a basis and projecting out components, thereby measuring a set of scalars. If you think about it for a moment, you will see that this statement is a trivial tautology. What I'm really saying here is that every measurement is a comparison.

After all, that is what we really mean, isn't it? When we say something is "2 meters", all we're saying is it's twice as long as a certain metal bar in France. (Or, in modern SI, it's twice the distance light travels during so many oscillations of Cesium 133.)

Color. Color requires three components to be specified, but color is independent of those components. You can specific color in terms of RGB, or CMYK or pantone or color temperature, but color is independent of those measurement. Because you need multiple components to specific color, and the existence of color is *independent* of those components, its a vector field, and more than a collection of measurements.

RGB, HSB, CMYK, and temperature are all coordinates on color space. They emphatically do not obey the linear transformation law and axioms of a vector basis. In fact, I'm not convinced color space is a vector space at all, if we mean the color space that is relevant to "perceived color". (Of course, we can define color spaces that are vector spaces, but these might have nothing to do with actual color perception).

You could probably geometrize the idea of color space if you like, and make it a manifold, possibly with notions of parallel transport. Who knows, maybe there's a useful way to model some psycho-physical process using a color space bundle over spacetime.

At any rate, "color" is measured by first comparing incoming light against certain bands of frequency; i.e., taking an inner product in frequency space against certain basis vectors in order to form a collection of scalars. These scalars can then be used as a coordinate system on color space. This method works for RGB, CMYK, and color temperature type coordinates. Mapping physical observables to HSB coordinates is more complicated, and will require projecting the incoming spectrum onto several bands and doing some analysis with the results.

Since frequency space is infinite-dimensional, there is no real reason for color space to be 3-dimensional; the ultimate reason is that we have 3 kinds of color receptors in our eyes (which project the spectrum onto 3 bands). Other animals have 2-dimensional, or sometimes 4-dimensional color spaces.

 

[Ben Niehoff]

But we can stretch this into some absurd conclusions.

I take the a precinct-by-precinct map of the United States containing the election results of the Republican primary in 1980. The candidate votes form a vector and it's s perfectly good vector field. I can also form a vector field containing things like the price of real estate of different types of houses, the probability of default, divorce rates, crime statistics, etc. etc.

All of those are perfectly good vector fields.

No they are not. You can't just take a collection of data and call it a vector. This isn't computer science, it's geometry. Vector spaces have to obey certain axioms; you have to show how those axioms are physically reasonable if you want to say some physical quantity is modeled by a vector space.

Just thought of something ridiculous. Restaurant and movie reviews. I go on yelp.com or rottentomatoes.com. Restaurant rates form a vector (i.e. atmosphere, decor, service, etc.)

Nope, that's a collection of data, not a vector.

You can do movie reviews the same way (quality of plot, amount of action, quality of print, etc.)

Same problem.

So you are saying that it is *physically impossible* for me to measure restaurant atmosphere and service at the same time?!

Nobody said you couldn't measure more than one scalar at a time. This isn't quantum mechanics.

In fact, I think I specifically mentioned that a collection of scalars can be used to construct a tensor quantity in the basis used to measure all the scalars. So, e.g., I can use a set of (scalar) measurements to deduce all 6 components of E and B in whatever basis I choose. The point is what constitutes a measurement.

 

[twofish-quant]

Vector spaces have to obey certain axioms; you have to show how those axioms are physically reasonable if you want to say some physical quantity is modeled by a vector space.

And what axioms are violated by having election results represented by a vector? It seems to me to be a perfectly good vector space. All you have to do is to define addition and multiplication operations and you are done.

Axioms are axioms. I don't understand how "physically reasonable axioms" is a grammatical statement. Is "addition" physically reasonable? Now once you've defined addition and multiplication, you can then use them to make statements that are physically true or false. But the fact that you can make a statement that "the scalar multiplication of "red" by 2 gets us outside the set of physically valid colors" means that the axioms are defined.

And it's also a *useful* vector space. Once you've defined the operations, then you can define a "norm" which then describes how "close" two elections are. You can then do matrix transformations from one set of coordinates to another.

I define a C++ class RGB color, I define the operations of 2*Color and Color A + Color B. Once I've defined those operations, it's a vector space. I can even start do to tensor algebra.

 

[twofish-quant]

What I'm really saying here is that every measurement is a comparison.

Which is an extremely strong statement that I'm not sure is true once you go outside the range of physical measurements that physicists are used to. Just to give a trivial example, how do you measure "bank liquidity"? If you state "physical measurement" then I don't object.

RGB, HSB, CMYK, and temperature are all coordinates on color space. They emphatically do not obey the linear transformation law and axioms of a vector basis.

All you need for a vector space to exist is for the transformations to be defined, and I can clearly define a set of vector addition and scalar multiplication operations for RGB numbers. Now whether I get something *physically* meaningful if I perform those operations is another issue.

RGB numbers are physically bound within a range, but if I measure x, y, z coordinates of the earth, there are some values which are invalid.

Also there *isn't* much of a difference between the mathematical concept of a vector space and the computer science one. All mathematics requires is that you have a defined addition and multiplication that has eight axioms. Once you have a collection of data for which you do that, then you have a vector space.

 

[JDoolin]

I'm not sure if this makes any difference to your argument, but in this "election-result-space" do you mean that the space is the discrete set of electoral districts, or do you mean the space is a continuous geographical/political map of the region?

Also, the "election-results" are vector-like, in the sense that they are multi-valued arrays, but they are non-vector-like in the sense that they do not have any direction. i.e. they don't point from one district toward another.

 

[IsometricPion]

Which is an extremely strong statement that I'm not sure is true once you go outside the range of physical measurements that physicists are used to. Just to give a trivial example, how do you measure "bank liquidity"?

Suppose one has a collection of numbers arranged into categories. If these numbers are to represent quantitative data, then for each category there must be a unit reference against which all data in that category are compared (if two numbers do not share the same unit reference they cannot be immediately compared numerically). Therefore, in order to obtain any data representable as numbers or collections of categories thereof one must compare the quantity(ies) of interest to a reference value (or set thereof in the case of multiple categories). Thus, once a tuple of numbers and corresponding units are given the values of the quantities measured have been specified and any other description of the same values must yield the same (tuple of) numbers when converted into those units (and arranged by category into the same order in the tuple).

The measurement of a tuple of values either entails many different comparisions to unit references, and thus many measurements, or the the comparison of a smaller number of measurements to tuples of unit references. In any case, each comparison must yield precisely one number. It is the combination of this number and the associated unit tuple that represents the measured value. No matter what mappings are done on the collection of numbers, if the appropriate inverse mappings are done such that the necessary unit tuples map back to themselves the measured values must be represented by the same numbers (this places restrictions on what is considered a valid mapping and/or the types of valid measurements (in the context of this thread I prefer to think of it as the former rather than the latter)).

It is in this sense that all measurable quantities are collections of single numbers obtained from single measurements.

If the units involve space or time references, then they must pick out a set of vectors in space-time. Any such set of vectors can be used to construct a tensor of appropriate rank such that after any transformation, the evaluation of the resulting tensor on the images of the aforementioned set of vectors yields the same number as obtained originally (since one requires covarient tensors and vectors to transform in such a way that this is true).

Sorry if the vagueness (or triviality) of the above was excessive, I was going for an abstract approach but may have overreached the bounds of my knowledge and/or conventional nomenclature.

 

[twofish-quant]

Therefore, in order to obtain any data representable as numbers or collections of categories thereof one must compare the quantity(ies) of interest to a reference value (or set thereof in the case of multiple categories).

No. I just thought of a counter example. Twenty questions.

I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude. Am I on land? Yes. Do I see taxicabs? Yes. Are they green? No. Are they yellow? No. Do I see water? Yes.

With each question, you can eliminate parts of the vector space. The fact that I see tax cabs and they are not yellow, means that I'm not in Manhattan. Now if you can ask enough questions, you can figure out my location and convert to GPS coordinates.

Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).

One other way of thinking about it is that you can specify points in a vector space as the interaction of subsets of that vector space, which allows you to specify a point in that space without reference to basis vectors at all.

 

[twofish-quant]

I'm not sure if this makes any difference to your argument, but in this "election-result-space" do you mean that the space is the discrete set of electoral districts, or do you mean the space is a continuous geographical/political map of the region?

I suppose it depends on the data. It's pretty easy to fit anything into a vector space. One thing about vector spaces is that scalars don't have to be real and neither do vectors. Election *results* certainly form a vector space (since you can add and scale vote totals). It's not obvious to me how to represent discrete electoral districts in a vector space, but that's just due to my lack of imagination.

Also, the "election-results" are vector-like, in the sense that they are multi-valued arrays, but they are non-vector-like in the sense that they do not have any direction. i.e. they don't point from one district toward another.

Again this depends on the structure of the data, but my point is that if you consider anything other than relativistic vectors to be "bags of unconnected data" you lose the structural information about the data. If you treat everything as "bags of data" you lose type information, which is a bad thing.

One thing that got me started thinking along these lines is the fact that you can call functions in C++ "covariant" and "contravariant". So what does tensor calculus have to do with C++. Well, that got me into the world of category theory...

 

[Dale]

Election *results* certainly form a vector space (since you can add and scale vote totals).

They certainly do not. One of the requirements of a vector space is that there must be an operation where multiplication of a vector by a real number* leads to another vector. If you multiply an arbitrary election result by any negative number or by any irrational number you will get negative or fractional votes, neither of which are members of the space of possible election results.

You cannot scale election results by arbitrary real numbers, nor even by arbitrary integers.

*Vectors can be generalized to multiplication over other fields besides the real numbers, but the conclusion remains. There is an additive identity element, but no additive inverse in the space of election results.

 

[IsometricPion]

Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).

I don't think any of my assertions would be altered if I replace "unit reference" with something like "way of mapping a number to a measured value".

 

[JDoolin]

They certainly do not. One of the requirements of a vector space is that there must be an operation where multiplication of a vector by a real number* leads to another vector. If you multiply an arbitrary election result by any negative number or by any irrational number you will get negative or fractional votes, neither of which are members of the space of possible election results.

You cannot scale election results by arbitrary real numbers, nor even by arbitrary integers.

*Vectors can be generalized to multiplication over other fields besides the real numbers, but the conclusion remains. There is an additive identity element, but no additive inverse in the space of election results.

So, hypothetically, what if they changed the election rules so instead of voting yes/no, each voter would turn an analog dial to determine how much they liked each candidate, yielding some real number between zero and one?

 

[Dale]

So, hypothetically, what if they changed the election rules so instead of voting yes/no, each voter would turn an analog dial to determine how much they liked each candidate, yielding some real number between zero and one?

It still wouldn't be a vector space because no candidate can have a negative result so the axioms of a vector space are not satisfied. I.e. if A is a non-null election result then there is no election result B such that A+B=0 where 0 is the null election vector.

 

[JDoolin]

It still wouldn't be a vector space because no candidate can have a negative result so the axioms of a vector space are not satisfied. I.e. if A is a non-null election result then there is no election result B such that A+B=0 where 0 is the null election vector.

Hmmmm. And that motivates my next question, what if the dials are allowed to turn anywhere from -1 to 1?

 

[Dale]

Hmmmm. And that motivates my next question, what if the dials are allowed to turn anywhere from -1 to 1?

That would overcome my previous objection. However, there are other problems, for instance you can always take an election result A representing 100% turnout with everyone voting the maximum allowed for one candidate. Then A+A would not be a valid election result since it would represent 200% turnout.

 

[JDoolin]

No. I just thought of a counter example. Twenty questions.

I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude. Am I on land? Yes. Do I see taxicabs? Yes. Are they green? No. Are they yellow? No. Do I see water? Yes.

With each question, you can eliminate parts of the vector space. The fact that I see tax cabs and they are not yellow, means that I'm not in Manhattan. Now if you can ask enough questions, you can figure out my location and convert to GPS coordinates.

Note that you've figured out my GPS coordinates without actually measuring my latitude and longitude or doing any reference comparisons at all. You can show that no reference comparisons were done, because you can play this game without knowing anything about latitude and longitude at all, and it's the same game that you can play with things that are *not* vector spaces (i.e. words in a dictionary).

One other way of thinking about it is that you can specify points in a vector space as the interaction of subsets of that vector space, which allows you to specify a point in that space without reference to basis vectors at all.

In your twenty questions, aren't you likely to eventually ask something about a specific object? Not just the color of the taxi-cabs in the region in general, but you need to ask about a specific street-corner, or a specific building?

Once you pick a specific landmark, you now have a reference comparison--it's not latitude or longitude, but it is a reference comparison.

 

[twofish-quant]

They certainly do not. One of the requirements of a vector space is that there must be an operation where multiplication of a vector by a real number* leads to another vector.

No. You can use any field and it's a vector space. One way you can get a vector space out of probability is to use the trivial field [0, 1] as your scalar field, and modulo 1 everything for your vector field.

If you multiply an arbitrary election result by any negative number or by any irrational number you will get negative or fractional votes, neither of which are members of the space of possible election results.

So restrict your scalars and vectors to integers. Still a vector space.

*Vectors can be generalized to multiplication over other fields besides the real numbers, but the conclusion remains. There is an additive identity element, but no additive inverse in the space of election results.

You can invent one that is meaningful by doing component by component subtraction. You can talk about the difference in votes between two districts, and the difference between a district and itself is zero.

Again, this is not esoteric math.

 

[twofish-quant]

That would overcome my previous objection. However, there are other problems, for instance you can always take an election result A representing 100% turnout with everyone voting the maximum allowed for one candidate. Then A+A would not be a valid election result since it would represent 200% turnout.

Then modulo everything 1.

 

[twofish-quant]

In your twenty questions, aren't you likely to eventually ask something about a specific object? Not just the color of the taxi-cabs in the region in general, but you need to ask about a specific street-corner, or a specific building?

We can set up the rules to exclude those questions.

 

[twofish-quant]

Also the rules of vector spaces are that the mathematical operations are *defined*, but necessarily that they always lead to physically possible results. I can count the number of banana trees with integers. The fact that it's possible to talk about a trillion banana trees when in fact there are not that many trees in the world doesn't invalidate the use of numbers to describe banana trees. Also, I haven't seen a "negative banana tree" but it's possible to define an additive inverse anyway.

Also for election results, it's possible to do factor analysis and all sorts of pretty complicated linear algebra with those results. The fact that it is *possible* to do those mathematical operations is what renders it a vector space. What's really cool is that once you've defined some basic operations, then you end up getting the mathematics of vector spaces.

And people *do* use these things in "real life". You assign each election district a number indicated for example "rural-ness" and do coorelations, and that tells you how to plan out your next campaign.

 

[Dale]

You can invent one that is meaningful by doing component by component subtraction. You can talk about the difference in votes between two districts, and the difference between a district and itself is zero.

This doesn't work. You cannot do subtraction if the negative vectors are not members of the vector space. I.e. the difference in votes between two districts is the sum of the votes of one district plus -1 times the votes of another district. Since -1 times the votes of another district is not a valid election result it is not a valid vector and you cannot sum it.

I am not sure about modulo, but I doubt that a modulo arithmetic results in vectors which satisfy all of the axioms of a vector space. I would have to see a proof.

In any case, as you reach further and further to get valid vectors from election results you also get further and further from a mathematical structure that has any natural relationship to election results. The reason that e.g. momentum is a vector is that the properties of momentum have a natural relationship with the properties of vectors. I.e. the result of adding two objects' momentum vectors corresponds naturally to the momentum of the system of the two objects.

What election result does the sum of two election results modulo 1 represent? Does such a result have any natural relationship to the two original election results?

 

[JDoolin]

That would overcome my previous objection. However, there are other problems, for instance you can always take an election result A representing 100% turnout with everyone voting the maximum allowed for one candidate. Then A+A would not be a valid election result since it would represent 200% turnout.

What if we don't change it to percentages, but simply leave it in units of votes?

You can ask questions like, What's the total number of votes in those three counties? Or How many more votes did the candidates receive in this county than that county?

What do you have then?

 

[JDoolin]

I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude.

In your twenty questions, aren't you likely to eventually ask something about a specific object? Not just the color of the taxi-cabs in the region in general, but you need to ask about a specific street-corner, or a specific building?

Once you pick a specific landmark, you now have a reference comparison--it's not latitude or longitude, but it is a reference comparison.

We can set up the rules to exclude those questions.

So let's say I'm in the ocean, outside of the sight of land. What kinds of questions would you ask me to determine my latitude and longitude?

You can ask me what species of fish are in the water
You can ask me, when I call out on the radio, what language is being answered in...

And let's say by asking some general questions like these, you manage to figure out that I must be in the Pacific Ocean, somewhere southwest of Hawaii.

The thing is, we set up the rules so that you can't ask about any specific landmarks, but the whole goal of the game is to circumvent those rules, and to find a specific landmark, which I am next to. If you have figured out that there is a unique location on Earth where you have trout and starfish and they speak Hawaiian on the radio, etc, then that unique intersection of properties actually is, in itself, a landmark.

 

[IsometricPion]

I am not sure about modulo, but I doubt that a modulo arithmetic results in vectors which satisfy all of the axioms of a vector space. I would have to see a proof.

The definition (wikipedia Vector space) is clearly satisfied for vectors composed of elements of a field and scalars from the same field. It remains to show that {0,1} with modular arithmatic form a field. Modular artihmatic implies closure under addition and multiplication, distributivity, commutativity, associativity, the existence of additive and multiplicative identities, and additive inverses. The only remaining operation to be checked (and one which is not, in general, satisfied by modular arithmatic) is the existence of a multiplicative inverse for all non-zero elements (i.e. the existence of a element for each element such that the product of the two yields 1). The only element to be checked is 1, since 1*1=1, 1 has a multiplicative inverse. Thus, {0,1} forms a field and can be used to construct a vector space. It turns out that for the integers modulo primes modular arithmatic yields fields (in this case the prime is 2).

If one tries a similar construction for the integers or non-prime mods, then one obtains an R-module (wikipedia Module, where R denotes a ring), which is not a vector space.

With each question, you can eliminate parts of the vector space. The fact that I see tax cabs and they are not yellow, means that I'm not in Manhattan. Now if you can ask enough questions, you can figure out my location and convert to GPS coordinates.

That's the thing, one can think of the answers to the questions (that give location information) as mappings from the vector space to itself plus the value {false}. Since the mapping does not require coordinates for its definition, it must be possible to formulate it in a coordinate invarient manner. In particular, it should be possible to construct a covarient tensor (the codomain of which will not equal its domain if location information is provided by the answer) to represent any such answer. The result of allowing this tensor to act on sufficently many elements of the pre-image of its codomain will be a number (which by construction is a rank-0 tensor).

 

[Dale]

What if we don't change it to percentages, but simply leave it in units of votes?

That is what I was assuming above.

I am tired of this game. It does not seem to me that "election results are a vector" is a very natural concept. You can continue to do little tweaks and may eventually come up with something that is mathematically a vector, but as you do it seems that you are getting further away from a useful representation of elections.

In general, to say of some real-world concept that "X are vectors" requires the following:
1) There needs to be a bijection of the different X to different mathematical objects
2) There needs to be a correspondence between real-world operations on X and operations on the mathematical objects
3) Those mathematical objects and operations need to satisfy all of the axioms of vector spaces (http://en.wikipedia.org/wiki/Vector_space)

I don't see a way of doing that for election results. Even step 1) seems questionable to me. If you really want to do it then please go ahead, but just make sure that you are careful because it is a really unclear fit.

 

[Ilmrak]

To answer the original question, we have to agree on what is measurable.

To me a measure is some set of elements each of which is invariant with respect to everything. In this sense they are "scalars".
Strictly speaking, only scalars are physical quantities because results of experiments are not arbitrary: as someone said before, the result of a measure of a given quantity done with a specific instrument in a specific condition cannot possibly change.

A couple examples.

- Let us take one man jumping off a cliff. The acceleration acting on his center of mass is not measurable, because of the fact we can perform a change of coordinates such that it goes to zero (free fall frame).
Nonetheless in every possible frame that man is going to die in a horrible way. What is measurable is the set of components his acceleration has with respect to the ground in a given frame in some chosen units. This numbers are not going to change with the frame, because they are defined in a specific coordinate system (like the mass, which is the zeroth component of the four-momentum in the rest frame). So those numbers are "scalars".
Similarly the colour of a laser is not a physical quantity, but colour of a laser in a given frame it is.

- Try now to do something similar with components of the electromagnetic four-potential. This time it's components in a given frame are no well defined because of the gauge invariance. To perform a measure of the four-potential we have to choose not only the frame and the units but also the gauge.

What we usually call physical quantities are simply the set of all possible measure we can get of a object (note that this set is, by definition, invariant).
The four-potential is a Lorents vector: if we know the result of a measure of its component in a given frame, gauge and units, then we can calculate every possible result of any other experiment in another frame, gauge and units by performing a Lorentz, gauge and units transformation.

Please tell me if you agree ^^

Ilm

P.s. sorry for my bad english

 

[JDoolin]

That is what I was assuming above.

I am tired of this game.

I'm actually kind of enjoying the game, but let me see if I understand what the game is... I'm trying to come up with an example of a non-directional vector space. You're trying to avoid coming up with an example of a non-directional vector space.

If you want to end the game, answer these questions:

Are there any examples of non-directional vector spaces? If so, give one. If not, why not?

 

[Dale]

Are there any examples of non-directional vector spaces? If so, give one.

Sure, there are many non-directional vector spaces. One easy example is the space of, say, 12 th order polynomial functions.

 

[JDoolin]

Sure, there are many non-directional vector spaces. One easy example is the space of, say, 12 th order polynomial functions.

Okay, that's interesting. The space of the set of functions representable by a 12^th order polynomial is a 12 dimensional, non-directional vector space. Am I correct that the space of the 12th order polynomial functions is different from the 12th order polynomial functions themselves? The space of the polynomial functions is just a list of the 12 coefficients.

I'm still a little confused about how an election, as we've set up above, with each voter being allowed to vote using a real number between -1 and 1, for each of, say, 12 candidates, doesn't also create a 12 dimensional, non-directional vector space, though.

The 12th order polynomial functions are essentially a list of 12 real coefficients associated with 12 powers of x, whereas the election results are 12 real numbers associated with 12 voting-results.

 

[ParticleGrl]

It does not seem to me that "election results are a vector" is a very natural concept.

I'm surprised no one has pointed this out- political scientists do regressions on election results that rely on the ability to do linear analysis. So the fact that you can fit election results into some sort of vector space is useful enough that people actually do it.

 

[IsometricPion]

Am I correct that the space of the 12th order polynomial functions is different from the 12th order polynomial functions themselves? The space of the polynomial functions is just a list of the 12 coefficients.

No, the vector space has as elements the polynomials themselves. Vectors formed by the thirteen coefficients would also constitue a vector space, but a different one (which is isomorphic to the previous one).

political scientists do regressions on election results that rely on the ability to do linear analysis.

They rely on the ability to form vectors from the results (this operation can be done with any two sets of numbers with an equal number of elements). This does not mean that election results make sense as a vector space (since the vectors formed do not generally fulfill the properties required of a vector space if one requires all the vectors correspond to possible election results).

 

[Dale]

I'm surprised no one has pointed this out- political scientists do regressions on election results that rely on the ability to do linear analysis. So the fact that you can fit election results into some sort of vector space is useful enough that people actually do it.

Linear analysis uses a linear model, so you could consider the best-fit model to be a specific member of the vector space of all possible models of the given form. That doesn't imply that the data itself is a sampling of a vector space.

 

[Dale]

Okay, that's interesting. The space of the set of functions representable by a 12^th order polynomial is a 12 dimensional, non-directional vector space. Am I correct that the space of the 12th order polynomial functions is different from the 12th order polynomial functions themselves? The space of the polynomial functions is just a list of the 12 coefficients.

As IsometricPion mentioned the vector space is the set of all 12 th order polynomials, e.g. the polynomial 5x^{12}-4x^{11}+8x^{10}+... is a vector in the space, but list of numbers (5,-4,8,...) is not.

You can consider many different sets of basis vectors for this space, the most obvious being (x^{12},x^{11},x^{10},...), and in terms of that basis you can write the vector as (5,-4,8,...) for notational convenience, but the 13 numbers are no longer just an arbitrary list of numbers, but are instead coordinates in a specified basis.

You could consider different basis vectors, such as the Legendre polynomials. Then you could write the same vector as a different list of numbers corresponding to coordinates in a different basis. Although the list of numbers would be different, the vector is the same, since the vector is the polynomial and not the list of numbers.

I'm still a little confused about how an election, ...

I am still not interested in this game. It seems clear to me that they are not, and I gave several good reasons above. I am not going to critique your ideas one by one. If you think that they form a vector space then please
1) Define the bijection between different election results and different vectors
2) Demonstrate the correspondence between real-world operations on elections and operations on the vectors
3) Prove that the vectors and operations satisfy all of the axioms of vector spaces (http://en.wikipedia.org/wiki/Vector_space)