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## Are Christoffel symbols measurable?

 Quote by DaleSpam 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?

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 Quote by JDoolin 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.

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 Quote by DaleSpam 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.

 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.

 Quote by JDoolin 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 by ParticleGrl 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).

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 Quote by ParticleGrl 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.

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 Quote by JDoolin 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 $5x^{12}-4x^{11}+8x^{10}+...$ 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.

 Quote by JDoolin 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)