definition of orientation of finite-dimensional real vectors space
Topics
About: vectors space
The table of contents of this article
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.
Target Context
- The reader will have a definition of orientation of finite-dimensional real vectors space.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( V\): \(\in \{\text{ the } d \text{ -dimensional } \mathbb{R} \text{ vectors spaces }\}\)
\( S\): \(= \{\text{ the ordered bases for } V\}\)
\( \sim\): \(\in \{\text{ the equivalence relations for } S\}\), \(\forall B_1 = ({b_1}_1, ..., {b_1}_d), B_2 = ({b_2}_1, ..., {b_2}_d) \in S (B_1 \sim B_2 \iff \text{ for } M \text{ such that } ({b_2}_1, ..., {b_2}_d)^t = ({b_1}_1, ..., {b_1}_d)^t M \text{ , } 0 \lt det M)\)
\( S / sim\): \(= \text{ the quotient set }\)
\(*o\): \(\in S / sim\)
//
Conditions:
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\(d\) above is supposed to be positive.
When \(d = 0\), any orientation is defined to be \(1\) or \(-1\).
2: Note
Let us see that \(\sim\) is indeed an equivalence relation.
1st, for each \((B_1, B_2)\) pair, \(M\) is uniquely determined: \({b_2}_j = {b_1}_l M^l_j\), which uniquely determines \((M^1_j, ..., M^d_j)\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(det M\) is not \(0\), because otherwise, \(B_2\) would not be linearly independent. So, \(0 \lt det M\) or \(det M \lt 0\).
\(B_1 \sim B_1\), because \(M = I\) and \(0 \lt det I\).
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 \(det M^{-1} = 1 / det M\), 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 \([B_1] \in S / sim\).
\(\{[B_1]\} \neq S / sim\), because we can take a matrix, \(M\), such that \(det M \lt 0\), 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 \(det M \lt 0\), \(B_2 \notin [B_1]\).
So, there is the \([B_2] \in S / sim\) such that \([B_1] \neq [B_2]\).
Then, for each \(B' = (b'_1, ..., b'_d) \in S\), \((b'_1, ..., b'_d)^t = ({b_1}_1, ..., {b_1}_d)^t M'\), but if \(0 \lt det M'\), \(B' \in [B_1]\); if \(det M' \lt 0\), \((b'_1, ..., b'_d)^t = ({b_1}_1, ..., {b_1}_d)^t M' = ({b_2}_1, ..., {b_2}_d)^t M^{-1} M'\), but \(det (M^{-1} M') = det (M^{-1}) det M'\), which is positive, so, \(B' \in [B_2]\).
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.
