Are Christoffel symbols measurable?

DaleSpam

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

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

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?

DaleSpam

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

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

DaleSpam

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

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

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.

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

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

Then modulo everything 1.

twofish-quant

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.

DaleSpam

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

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

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

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.

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

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, 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