Maths question, with surprise at the end.

[minkwe]

Hello mathematical analysts, I have a question for your idle brains

Given two unit vectors vectors and a function , how much can you tell about the nature of this function if you are told only two pieces of information:

1. is piece-wise smooth.
2. is invariant under a basis transformation.

[Guest]

Hello mathematical analysts, I have a question for your idle brains :)

Given two unit vectors vectors and a function , how much can you tell about the nature of this function if you are told only two pieces of information:

1. is piece-wise smooth.
2. is invariant under a basis transformation.

Hi Michel

Can you be a bit more precise?

Are x and y unit vectors in R^2 or R^3?

By basis transformation, do you mean a rotation about the origin?

[minkwe]

Hi,
I was hoping to be as general as possible. But for the sake of being precise, let us say R^3. By basis transformation, I include rotation or translation of just the basis vectors, ie, they could be pointing in different directions, or the origin could be translated to an arbitrary position in R^3.

[Joy Christian]

It seems to me all one can say about such a function is that it is piecewise smooth and invariant under basis transformations in .

You have specified the domain of the function, which is (because and are unit vectors), but left the co-domain wide open:



Until you specify what is, there is not much restriction on the nature of . In other words, it could be almost anything you like.

[Guest]

Hi,
I was hoping to be as general as possible. But for the sake of being precise, let us say R^3. By basis transformation, I include rotation or translation of just the basis vectors, ie, they could be pointing in different directions, or the origin could be translated to an arbitrary position in R^3.

So the vectors x, y are not restricted to unit length? The function f maps R^3 x R^3 to R (I suppose) and is invariant under affine transformations of R^3.

Not difficult to see that the function must be constant.

[minkwe]

Joy,

Hint 1: if the is invariant under basis transformation, then f must also be a function of a joint property of the two unit vectors that is invariant under a basis transformation and what properties of two unit vectors do we know about that have that property. Geometric product?

You can probably guess where I'm going with this now.

[Joy Christian]

Joy,

Hint 1: if the is invariant under basis transformation, then f must also be a function of a joint property of the two unit vectors that is invariant under a basis transformation and what properties of two unit vectors do we know about that have that property. Geometric product?

You can probably guess where I'm going with this now.

Yes.

[minkwe]

What if I added that instead of just being piecewise-smooth, the function must be "differentiable at every point". Because, I want a small perturbation of the directions of the unit vectors, to cause only a small perturbation of the result of the function?

Can we say anything about periodicity, if we know that must be a function of the geometric product of the vectors?

[minkwe]

http://arxiv.org/pdf/1303.4574v3.pdf
http://arxiv.org/pdf/1504.04944.pdf
http://arxiv.org/pdf/1506.03373.pdf