Right handed frames and Orientation

[Buri]

My text says, "Note that in an arbitrary n-dimensional vector space, there is no well-defined notion of "right-handed", although there is a well defined notion of orientation."

I don't see why. An n frame (a1,a2,...,an) is called right handed in R^n if det[a1 a2 ... an] > 0, but I guess we'd have to define a determinant function on V (though I don't think this should be a problem). However, I remember that the determinant being the only alternating n-tensor on R^n, so maybe this isn't the case in an arbitrary vector space. So we could possibly get different values depending on the determinant we use since it is no longer unique.

Any clarification?

 

[Buri]

Or is it because > 0 and < 0 aren't defined in an arbitrary vector space?

 

[Buri]

Anyone?

 

[D H]

My text says, "Note that in an arbitrary n-dimensional vector space, there is no well-defined notion of "right-handed", although there is a well defined notion of orientation."

Your text is correct. An orthonormal matrix is a proper rotation if its determinant is one. You are confusing "right" with "proper".

"Right-handed" is special to 3 space because the associative cross product is unique to R³. (There is a non-associative cross product in R⁷.)

 

[Buri]

Your text is correct. An orthonormal matrix is a proper rotation if its determinant is one. You are confusing "right" with "proper".

"Right-handed" is special to 3 space because the associative cross product is unique to R³. (There is a non-associative cross product in R⁷.)

I don't understand. Could you maybe explain it a bit more? I'm just lost..

 

[Buri]

I don't see how I'm confusing "right" with "proper". I didn't even know the definition of proper till you mentioned it lol

And I thought the cross product was only defined for R^3 and besides, the definition is in terms of determinants...

 

[D H]

I don't see how I'm confusing "right" with "proper". I didn't even know the definition of proper till you mentioned it lol

Just because you didn't know that you are confusing "right-handed" (which applies only to R³) with "proper" (which applies generically) does not mean that that is exactly what you are doing.

 

[Buri]

Just because you didn't know that you are confusing "right-handed" (which applies only to R³) with "proper" (which applies generically) does not mean that that is exactly what you are doing.

I never said it did. If you read that correctly, I said 'I don't see HOW I'm confusing "right" with "proper" ' and the second remark, "I didn't even know the definition of proper till you mentioned it lol" was meant to show you how I really don't see how I'm confusing the two since I didn't even know the definition.

 

[Buri]

"Right-handed" is special to 3 space because the associative cross product is unique to R³. (There is a non-associative cross product in R⁷.)

Is that even correct? Because I've been given a definition for right-handed in R^n and I'm assuming its well defined (because otherwise I would think its useless), but I think you're saying its only well defined for R^3.

 

[Buri]

Can anyone else help me out on this?

 

[Buri]

C'mon anyone?

 

[fzero]

Is that even correct? Because I've been given a definition for right-handed in R^n and I'm assuming its well defined (because otherwise I would think its useless), but I think you're saying its only well defined for R^3.

In $$\mathbb{R}^3$$, the cross product of two basis vectors $$e_1\times e_2$$ is orthogonal to both $$e_1$$ and $$e_2$$. Therefore it must be parallel to $$e_3$$. If the basis is orthonormal, then the only possibility is that $$e_1\times e_2 = \pm e_3$$. One of these signs (up to the choice of basis) is called right-handed and the other left-handed.

In $$\mathbb{R}^{n\neq 3}$$, there is no (associative) cross product, so the definition of right and left-handed makes no sense. There's no special way of taking two basis vectors and outputting a 3rd. However, the number

$$\sigma(e) = \text{sgn}\left( \sum_{i_1\cdots i_n} \epsilon_{i_1\cdots i_n} (e_1)_{i_1} \cdots (e_n)_{i_n}\right) = \pm 1.$$

defines an equivalence class that corresponds to the orientation of an ordered basis. If we know that the basis is orthonormal, we don't need to introduce the sign function and we simply have

$$\sigma(e) = \\sum_{i_1\cdots i_n} \epsilon_{i_1\cdots i_n} (e_1)_{i_1} \cdots (e_n)_{i_n}= \pm 1.$$

We would call a basis with $$\sigma(e)=1$$ positively-oriented, not right-handed.

If $$e'$$ is a basis obtained by a linear transformation of $$e$$, $$e' = T(e)$$ then

$$\sigma(e') = \text{sgn}(\det T) \sigma(e).$$


When $$n=3$$, the orientation coincides with the equivalence classes of right or left-handed.

 

[Landau]

My text says, "Note that in an arbitrary n-dimensional vector space, there is no well-defined notion of "right-handed", although there is a well defined notion of orientation."

I don't see why. An n frame (a1,a2,...,an) is called right handed in R^n if det[a1 a2 ... an] > 0, but I guess we'd have to define a determinant function on V (though I don't think this should be a problem).

The determinant is something you can compute for a square matrix / endomorphism. If you have a frame (a1,a2,...,an), how are you going to compute "its determinant"? This only makes sense if you have already chosen some special basis in which you express this frame; then you compute the determinant of the change-of-basis matrix.

Such a choice is precisely an orientation. In R^n you have a standard choice, namely the standard basis (e1,...,en), where e_i has 1 in the i-th slot and else zero. In an abstract vector space, there is no preferred basis. Then, given two frames (a1,a2,...,an) and (b1,b2,...,bn), we say that they 'have the same orientation' if the determinant of the linear map expressing one in the other (the 'change of basis' isomorphism) has positive determinant. This is an equivalence relation with exactly TWO classes. Choosing one such class is choosing an orientation.