I Meaning of each member being a unit vector

AI Thread Summary
The discussion focuses on understanding the concept of unit vectors in the context of directional data and tensor products. Each observed directional data point is treated as a unit vector, indicating that they have a length of one and represent specific directions. The average of these unit vectors is calculated to derive tensor quantities that characterize the data distribution. Clarifications are sought regarding the notation used in the paper, particularly the meaning of indices and the representation of multiple vectors. The conversation also highlights the existence of resources and papers on fabric tensors, which provide further insights into the topic.
AlekM
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Meaning of each member being a unit vector, and how the products of each tensor can be averaged.
Summary: Meaning of each member being a unit vector, and how the products of each tensor can be averaged.

Hello!
I am struggling with understanding the meaning of "each member is a unit vector":
Let n(1), n(2), and n(N) be observed directional data, where each member is a unit vector. The most fundamental quantities are various averages of them. Since we are trying to seek tensor quantities to characterize the data distribution, we first consider the average of their tensor product, or the “moment”, and put
Ni1i2...in = ⟨ni1ni2...nin

I can see that N would represent the number of samples, and the pointy bracket represents an average:
Average of ith component of a = ⟨ai⟩ = (ai1+ai2+aiN)/N

However the notation of the averaged tensors in the text is ni2...nin, which leads me to wonder what i and n represent. It states that each member (of n?) is a unit vector, does that mean that each value of n is a normalized vector representing a direction? Or could a single value of n contain multiple vectors of directional data?

The paper later states that they adopt the summation convention over tensor indices, would this mean that something like i represents the axes that make up each vector (such as x,y,z in Cartesian) and 1 through n represents the axes that make up each vector? In that case it would seem that the directional vectors from each sample are averaged together. I'm not sure if I am on the right track or entirely incorrect.

Thanks in advance for any help,
Alek
 
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I read it as
AlekM said:
... does that mean that each value of n is a normalized vector representing a direction ...
We have ##N## observations and they span the phase space ##n^{(1)},\ldots,n^{(N)}##, i.e. each observed value ##n^{(i)}## is considered a basis vector of length ##1##. From there we can consider various other vectors in this space.

However, without context this is a bit of a guess.
 
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AlekM said:
The paper later states

Is the paper online? Can you give a link to it?
 
I just found that a page earlier direction is represented by unit vector n, so I think this is correct. However my next confusion is what nin represents.

As for the paper it is called "Distribution of Directional Data and Fabric Tensors," and I am referencing the beginning of page 2: https://www.sciencedirect.com/science/article/pii/0020722584900909
 
I can't read the paper since I'm not a member.

It surprised me how many articles are written about "fabric tensors". I hadn't heard of them before.

This thesis https://scholarworks.montana.edu/xm...2264/ShertzerR0811.pdf?sequence=1&isAllowed=y defines them on page 58 of the PDF , page 38 of the document. Perhaps some forum member who is an expert in tensors can explain them.

FabricTensor1.jpg
 
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Wow I'm not sure how I missed that source, it is incredibly useful. Thanks for sharing!

I didn't realize that link wasn't open to everyone, for anyone else reading this thread here is a mirror: http://einsteiniumstudios.com/fabric_tensors.pdf
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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