1063: Orientation of Finite-Dimensional Real Vectors Space

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definition of orientation of finite-dimensional real vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of basis of module.
• The reader knows a definition of quotient set.

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Target Context

• The reader will have a definition of orientation of finite-dimensional real vectors space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V$: $\in \{\text{ the }d\text{ -dimensional }\mathbb{R}\text{ vectors spaces }\}$
$S$: $=\{\text{ the ordered bases for }V\}$
$\sim$: $\in \{\text{ the equivalence relations for }S\}$, $\mathrm{\forall }{B}_1=({b_1}_1,...,{b_1}_d),B_2=({b_2}_1,...,{b_2}_d)\in S(B_1\sim B_2⟺\text{ for }M\text{ such that }({b_2}_1,...,{b_2}_d{)}^t=({b_1}_1,...,{b_1}_d{)}^{t}M\text{ , }0<detM)$
$S/sim$: $=\text{ the quotient set }$
$\ast o$: $\in S/sim$
//

Conditions:
//

$d$ above is supposed to be positive.

When $d=0$, any orientation is defined to be $1$ or $-1$.

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2: Note

Let us see that $\sim$ is indeed an equivalence relation.

1st, for each $(B_1,B_2)$ pair, $M$ is uniquely determined: ${b_2}_j={b_1}_l{M}_j^l$, which uniquely determines $(M_j^1,...,M_j^d)$, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

$detM$ is not $0$, because otherwise, $B_2$ would not be linearly independent. So, $0<detM$ or $detM<0$.

$B_1\sim B_1$, because $M=I$ and $0<detI$.

When $B_1\sim B_2$, $B_2\sim B_1$, because from $({b_2}_1,...,{b_2}_d{)}^t=({b_1}_1,...,{b_1}_d{)}^{t}M$, $({b_1}_1,...,{b_1}_d{)}^t=({b_2}_1,...,{b_2}_d{)}^t{M}^{-1}$, but $det{M}^{-1}=1/detM$, which is positive.

When $B_1\sim B_2$ and $B_2\sim B_3$, $B_1\sim B_3$, because from $({b_2}_1,...,{b_2}_d{)}^t=({b_1}_1,...,{b_1}_d{)}^t{M}_1$ and $({b_3}_1,...,{b_3}_d{)}^t=({b_2}_1,...,{b_2}_d{)}^t{M}_2$, $({b_3}_1,...,{b_3}_d{)}^t=({b_1}_1,...,{b_1}_d{)}^t{M}_1{M}_2$, but $det(M_1{M}_2)=det{M}_{1}det{M}_2$, which is positive.

So, $S/sim$ is well-defined.

Let us see that $S/sim$ consists of $2$ elements.

Let us take any $B_1\in S$ and the $[B_1]\in S/sim$.

$\{[B_1]\}\ne S/sim$, because we can take a matrix, $M$, such that $detM<0$, and then, $B_2:=({b_2}_1,...,{b_2}_d{)}^t=({b_1}_1,...,{b_1}_d{)}^{t}M$ is an ordered basis, but as $detM<0$, $B_2\notin [B_1]$.

So, there is the $[B_2]\in S/sim$ such that $[B_1]\ne [B_2]$.

Then, for each $B^{\prime }=(b_1^{\prime },...,b_d^{\prime })\in S$, $(b_1^{\prime },...,b_d^{\prime }{)}^t=({b_1}_1,...,{b_1}_d{)}^t{M}^{\prime }$, but if $0<det{M}^{\prime }$, $B^{\prime }\in [B_1]$; if $det{M}^{\prime }<0$, $(b_1^{\prime },...,b_d^{\prime }{)}^t=({b_1}_1,...,{b_1}_d{)}^t{M}^{\prime }=({b_2}_1,...,{b_2}_d{)}^t{M}^{-1}{M}^{\prime }$, but $det(M^{-1}{M}^{\prime })=det(M^{-1})det{M}^{\prime }$, which is positive, so, $B^{\prime }\in [B_2]$.

So, $S/sim=\{[B_1],[B_2]\}$.

Let us see the relation between orientation and so-called "right-screw" or "counterclockwise" for ${\mathbb{R}}^2$.

For each orientation, $[(b_1,b_2)]$, rotate $b_1$ toward $b_2$ in the shorter-direction, then the direction the right-screw proceeds is the right-screw direction: the orientation determines the right-screw direction.

For example, for the usual picture of ${\mathbb{R}}^2$, $[(\hat{x},\hat{y})]$ determines the upward right-screw direction, but $[(\hat{y},\hat{x})]$ determines the downward right-screw direction.

On the other hand, any choice of right-screw direction determines the orientation.

For example, for the usual picture of ${\mathbb{R}}^2$, the upward right-screw direction determines $[(\hat{x},\hat{y})]$.

In order to begin to talk about "counterclockwise", we need to have chosen a right-screw direction: any rotation is counterclockwise by looking on a face and is clockwise by looking on the other face.

When the rotation is counterclockwise by looking on the face from which the right-screw is coming up, the rotation is called "counterclockwise".

As any orientation determines the right-screw direction, the orientation determines the counterclockwise rotation direction.

For the usual picture of ${\mathbb{R}}^2$, as $[(\hat{x},\hat{y})]$ is usually implicitly chosen as the orientation, "counterclockwise" is determined.

On the other hand, any choice of counterclockwise rotation direction determines the orientation.

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References

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