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I'm confused as to how orientations and ordered bases relate to the “handedness” of Cartesian coordinates. On page 237 of Tu's An Introduction to Manifolds he's discussing orientations of a vector space and writes, with reference to the following diagram: enter image description here

On $\mathbb{R}^{3}$ an orientation is either right-handed (Figure 21.3) or left-handed (Figure 21.4). The right-handed orientation of $\mathbb{R}^{3}$ is the choice of a Cartesian coordinate system such that if you hold out your right hand with the index finger curling from the vector $e_{1}$ in the $x$-axis to the vector $e_{2}$ in the $y$-axis, then your thumb points in the direction of of the vector $e_{3}$ in the $z$-axis.

On my reading, “the choice of a Cartesian coordinate system” implies that Figure 21.4 should therefore show a left-handed Cartesian coordinate system. But that doesn't make sense as then $e_{2}$ would lie on the $x$-axis and $e_{1}$ would lie on the $y$-axis. On reflection (excuse pun) I'm guessing that an ordered basis defines an orientation on $\mathbb{R}^{3}$, which is then graphically represented by choosing either a right-handed or left-handed Cartesian system. So both Tu's diagrams refer to a right-handed coordinate system. Am I on the right track?

EDIT In other words, how do Tu's diagrams relate to these right and left-handed coordinate systems?

enter image description here

EDIT 2 Part of my confusion stems from this answer by @AndrewD.Hwang here:

It's worth pointing out a final subtlety in $\mathbb{R}^{3}$: If $\left(\mathbf{e}^{i}\right)_{i=1}^{3}$ 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".

Peter4075
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  • The handedness of the coordinate system depends on which finger you assign to each coordinate axis. The axes look the same in both pictures, but they're assigned to different fingers, which is the important part here. – Vercassivelaunos Mar 20 '21 at 09:58
  • @Vercassivelaunos - So are you saying that in Figure 21.4 the $e_{2}$ vector is pointing along the $x$-axis? So now $e_{2}\equiv e_{x}$ and $e_{1}\equiv e_{y}$? – Peter4075 Mar 20 '21 at 10:43
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    Forget about x-, y-, and z-axes. Those are just synonyms for $e_1,e_2,e_3$ in this case. The point is that you have three coordinates, and that you can arbitrarily choose an order among them, like $(e_1,e_2,e_3),~(e_2,e_3,e_1)$, or $(e_3,e_2,e_1)$. Half of these orderings are called left handed, the other half is called right handed, depending on the following: if we draw the axes like in your figure, and assign our index finger to the first coordinate in the ordering, the middle finger to the second and the thumb to three third coordinate, do we need a right or a left hand to draw this? – Vercassivelaunos Mar 20 '21 at 12:37
  • @Vercassivelaunos - Sorry, I still don't see what you mean. Tu labels the $x,y,z$ axes when he talks about the right-handed orientation (see above quote). So why can't I label the $x,y,z$ axes in Figure 21.4? I think I might find it easier to understand what's going on if I can do that. Thank you for your patience. – Peter4075 Mar 20 '21 at 13:43
  • But then you must label the $e_1$-axis with $x$, the $e_2$-axis with $y$ and the $e_3$-axis with $z$. At which point you're just renaming stuff. If that helps you, do it. But then you should forget about $e_1,e_2,e_3$. Don't use multiple names for the same thing. – Vercassivelaunos Mar 20 '21 at 15:37

1 Answers1

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On p. 238 Tu gives a formal definiton of orientation as an equivalence class of orderded bases. In the sense of this definiton any (finite-dimensional) vector space $V \ne \{0\}$ has exactly two orientations.

Note that $V = \{0\}$ is an exception. Formally it has $\emptyset$ as its only base (if we agree that sums over the empty index set are $0$, then each element of $\{0\}$ has a unique representation as a linear combination of basis vectors) and thus it has only one orientation.

Tu's preceding explanations are purely motivational and may in fact be somewhat confusing. What is he doing?

In dimensions $n = 1, 2, 3$ he considers the standard Cartesian coordinate system in $\mathbb R^n$ and assigns two orientations to it. Be aware that such an orientation is an additional component attributed to the coordinate system, it cannot be found in the coordinate system itself. It occurs in the figures in form of rotation arrows which are assigned in two variants to the depicted Cartesian coordinate system.

Let us focus to $n = 3$. Tu considers the usual $x$-$y$-$z$-coordinate system with standard unit vectors $e_1$ on the $x$-axis, $e_2$ on the $y$-axis and $e_3$ on the $z$-axis. Drawing the $e_i$ as arrows gives a direction to each of the axes (the positive direction). From the viewer's perspective the graphical representation of Cartesian coordinates in 21.3 and 21.4 shows the $x$-axis pointing forward, the $y$-axis to the right and the $z$-axis upwards. This is an arbitrary convention, you could do it as you want. As an example take the figure in your edit. Anyway, if you decided for a particular graphical representation, you can glue the coordinate system to your right hand or your left hand according to the following convention:

  1. Arrange index finger (I), middle finger (M) and thumb (T) so that they are perpendicular to each other. Then the fingers of your right and of your left hand form two ordered $3$-frames, the right-handed frame $(I_R, M_R, T_R)$ and the left-handed frame $(I_L, M_L, T_L)$. No spatial movement of your hands will transform the right-handed into the left-handed frame - they are mirror images. These two frames are the intuitive prototypes of an orientation of $\mathbb R^3$.

  2. Gluing to the right hand ("right-handed orientation"): Point with $I_R$ in direction $e_1$, with $M_R$ in direction $e_2$ and with $T_T$ in direction $e_3$.

  3. Gluing to the left hand ("left-handed orientation"): Point with $I_L$ in direction $e_2$, with $M_L$ in direction $e_1$ and with $T_L$ in direction $e_3$.

This attributes a handedness or orientation to the Cartesian coordinate system. Formally this is described via the two bijections $$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_2 & e_1 & e_3 \end{pmatrix}$$

In my opinion the graphical representation in your edit is more intuitive than Tu's.

enter image description here

In both cases you use assign $e_1$ to $I$, $e_2$ to $M$ and $e_3$ to $T$ which gives you two variants of a Cartesian coordinate system which are optically distinguishable. Here you have $$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & e_2 & e_3 \end{pmatrix}$$

However, if you take the graphical representation on the right side as the standard representation of a Cartesian coordinate system (this agrees with Tu, just imagine to look at this coordinate system with your eye facing the arrow tip of $x$), then you will admit that the graphical representation on the left side is non-standard. And in fact, if you use your left hand in the standard representation, then you get $$\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & -e_2 & e_3 \end{pmatrix}$$ Working with equivalence classes of ordered bases, you can easily see that the latter representation of $\omega_L$ coincides with Tu's variant.

Paul Frost
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  • So if you had a vector $\mathbf{v}=\left(3,4,5\right)$ in Fig. 21.3, would that vector in Fig. 21.4 have components $\left(4,3,5\right)$? And referring to the figure in my edit, that same vector would have components $\left(3,4,5\right)$ for the right-handed orientation and $\left(4,3,5\right)$ for the left-handed orientation? – Peter4075 Mar 25 '21 at 18:43
  • Components of a given vector have nothing to with the orientation of the coordinate system. Thus the vector $\mathbf v$ has always the same coordinates $(3,4,5)$ whatever the orientation may be. However, in the figure in your edit, $(3,4,5)$ in the graphical representation on the right and $(4,3,5)$ in the graphical representation on the left look identical if you forget the labels of the axes. A simpler example is $e_1$. In the left diagram it looks like $e_2$ in the right diagram. But the optical difference is only due to the different representations of the same coordinate system. – Paul Frost Mar 25 '21 at 23:32
  • Say in Fig. 21.3, $\mathbf{v}=3e_{1}+4e_{2}+5e_{3}=\left(3,4,5\right)$. In Fig. 21.4 isn't the same vector (ie the same thing in space) – this time using the ordered basis $\left(e_{2},e_{1},e_{3}\right)$ – written as $\mathbf{v}=4e_{2}+3e_{1}+5e_{3}=\left(4,3,5\right)$,ie with different components? – Peter4075 Mar 27 '21 at 16:09
  • Certainly $\mathbf v = 3e_1 +4e_2 + 5e_3 = 4e_2 + 3e_1 + 5e_3$, the components $x_i$ belonging to $e_i$ are the same in both sums. You can write this in the symbolic form $\mathbf v = \begin{pmatrix} e_1 & e_2 & e_3 \ 3 & 4 & 5 \end{pmatrix}$ or $\mathbf v = \begin{pmatrix} e_2 & e_1 & e_3 \ 4 & 3 & 5 \end{pmatrix}$, but it is still the same vector although the rows of both matrices look different. – Paul Frost Mar 27 '21 at 17:03
  • Personally, I find the notion of orientation more intuitively obvious in the two $x,y,z$ labelled diagrams in the figure in my edit. Namely, you can't go from one diagram to the other except via a reflection. I find the opposite orientations of Tu's two figures harder to grasp. The easiest way for me to see what's going on is to say (going from 21.4 to 21.5) that we want to transform $e_{1}$ to $e_{2}$, $e_{2}$ to $e_{1}$ and $e_{3}$ to $e_{3}$. The only way to do this is, again, via a reflection. It's taken me a week to see this. – Peter4075 Mar 27 '21 at 20:02
  • I agree. As I said "In my opinion the graphical representation in your edit is more intuitive than Tu's." – Paul Frost Mar 27 '21 at 22:50
  • This is indeed subtle stuff. I got there eventually. Many thanks for your help. – Peter4075 Mar 28 '21 at 07:26