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Are Christoffel symbols measurable?

by waterfall
Tags: christoffel, measurable, symbols
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twofish-quant
#127
Feb23-12, 11:55 PM
P: 6,863
Quote Quote by DaleSpam View Post
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
#128
Feb23-12, 11:56 PM
P: 6,863
Quote Quote by DaleSpam View Post
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
#129
Feb23-12, 11:57 PM
P: 6,863
Quote Quote by JDoolin View Post
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
#130
Feb24-12, 12:03 AM
P: 6,863
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.
DaleSpam
#131
Feb24-12, 06:59 AM
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Quote Quote by twofish-quant View Post
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
#132
Feb24-12, 08:01 AM
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Quote Quote by DaleSpam View Post
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
#133
Feb24-12, 08:15 AM
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P: 706
I'm located in a spot on the earth. By asking me yes-no questions, you can figure out my latitude and longitude.
Quote Quote by JDoolin View Post
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.
Quote Quote by twofish-quant View Post
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 lets 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
#134
Feb24-12, 12:33 PM
P: 177
Quote Quote by DaleSpam View Post
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.

Quote Quote by twofish-quant View Post
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).
DaleSpam
#135
Feb24-12, 05:31 PM
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Quote Quote by JDoolin View Post
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
#136
Feb25-12, 03:48 AM
P: 97
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, wich 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
#137
Feb25-12, 09:59 AM
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Quote Quote by DaleSpam View Post
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?
DaleSpam
#138
Feb25-12, 08:34 PM
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Quote Quote by JDoolin View Post
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
#139
Feb26-12, 02:40 PM
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Quote Quote by DaleSpam View Post
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 12th 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
#140
Feb26-12, 04:34 PM
P: 686
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
#141
Feb26-12, 05:57 PM
P: 177
Quote Quote by JDoolin View Post
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).

Quote Quote by ParticleGrl View Post
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).
DaleSpam
#142
Feb26-12, 06:42 PM
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Quote Quote by ParticleGrl View Post
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.
DaleSpam
#143
Feb26-12, 07:04 PM
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Quote Quote by JDoolin View Post
Okay, that's interesting. The space of the set of functions representable by a 12th 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 [itex]5x^{12}-4x^{11}+8x^{10}+...[/itex] 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 [itex](x^{12},x^{11},x^{10},...)[/itex], 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.

Quote Quote by JDoolin View Post
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)


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