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Can any one explain, the geometrical meaning of orientation in a vector space?

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closed as off-topic by Mark Viola, SchrodingersCat, user99914, Ivo Terek, Harish Chandra Rajpoot Nov 30 '15 at 0:27

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    $\begingroup$ This is a fairly subtle question. The usual definition relies on the determinant of a linear operator, and the fact that an invertible operator has non-zero determinant (or the equivalent). Have you encountered a cryptic formal definition that you're trying to understand? Are you looking for more than "a way of distinguishing space from its mirror image"? $\endgroup$ – Andrew D. Hwang Nov 29 '15 at 18:10
  • $\begingroup$ Yes, I need geometric meaning of that definition. $\endgroup$ – Selvakumar A Nov 29 '15 at 19:01
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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\ve}{\Vec{e}}\newcommand{\vv}{\Vec{v}}\newcommand{\vw}{\Vec{w}}$To ensure we're on the same page algebraically, here are some preliminaries:

Let $V$ be a finite-dimensional real vector space. If $(\vv_{i})_{i=1}^{n}$ and $(\vw_{i})_{i=1}^{n}$ are ordered bases of $V$, there exists a unique invertible linear operator $T_{\vv}^{\vw}:V \to V$ satisfying $$ T_{\vv}^{\vw}(\vv_{i}) = \vw_{i}\quad\text{for all $i = 1, 2, \dots, n$,} $$ and this operator has a well-defined determinant $\det T$, a non-zero real number. (The usual definition is to express $T$ as a matrix with respect to an arbitrary basis of $V$ and show the determinant of the resulting matrix does not depend on the choice of basis.)

Definition: We say $(\vv_{i})_{i=1}^{n}$ and $(\vw_{i})_{i=1}^{n}$ are consistently oriented if $\det T_{\vv}^{\vw} > 0$; otherwise $(\vv_{i})_{i=1}^{n}$ and $(\vw_{i})_{i=1}^{n}$ are oppositely oriented.

Lemma: "Consistently oriented" is an equivalence relation on the set of ordered bases of $V$.

Fix an orientation of $V$, i.e., an ordered basis $B = (\vv_{i})_{i=1}^{n}$. Let's try to understand the geometry carried by the orientation.

Imagine fixing the first $(n - 1)$ elements of $B$. The hyperplane $H_{n}$ spanned by $(\vv_{i})_{i=1}^{n-1}$ may be viewed as a "mirror" that separates $V$. Precisely, if $\vv$ is an arbitrary vector in $V$, one of three things happens:

  • Replacing $\vv_{n}$ by $\vv$ gives a consistently ordered basis.

  • $\vv \in H_{n}$, so $\{\vv_{1}, \dots, \vv_{n-1}, \vv\}$ is not a basis of $V$.

  • Replacing $\vv_{n}$ by $\vv$ gives an oppositely ordered basis.

You can "interpolate" between these three contingencies by allowing $\vv$ to move continuously along a path from $\vv_{n}$ to $-\vv_{n}$. At some point(s), this path must "cross the hyperplane $H_{n}$" or "travel through the mirror", whereupon the orientation changes, i.e., the resulting basis is oppositely-oriented.

There's nothing special about fixing the $n$th vector, or even about fixing all the vectors but one. You can allow each $\vv_{i}$ to vary continuously: As long as the resulting set remains a basis "at each instant", that basis is consistently oriented.

Note that changing the ordering of the basis elements generally changes the orientation. For example, swapping the labels of two basis elements changes the orientation. Generally, a re-ordering defines a consistently-oriented basis if and only if the permutation of indices is an even permutation, i.e., can be written as a composition of evenly many transpositions.


It's worth pointing out a final subtlety in $\Reals^{3}$: If $(\ve_{i})_{i=1}^{3}$ denotes the standard basis, the (algebraic) ordering of the vectors fixes an orientation, but we must still pick a conventional geometric representation, i.e., must choose to depict the standard basis as "right-handed" or "left-handed".

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