Isometries of R2 and R3

**Bernhard** · January 8th 2012, 08:48 PM

I am reading Papantonopoulou: Algebra Ch 14 Symmetries. I am seeking to fully understand Theorem 14.21 (see attached Papantonopoulou pp 459 -463)

On page 462, Papantonopoulou defines translations, rotations and reflections for R2 and R3 (see attached). Note that the rotations are defined as about the origin and the reflections are about the X-axis or $e_1$ axis.

Then on Page 463 he states Theorem 14.21 as follows:

14.21 Theorem An isometry S of $\mathbb{R}^2$ or $\mathbb{R}^3$ can be uniquely expressed as $S = t_b \circ \rho_{\phi} \circ r^i$ where i = 0 or 1

The proof (see attached) proceeds as follows:

Proof: To prove existence, let b = S(0). Then $t_{-b} \circ S$ fixes the origin and so $t_{-b} \circ S = T_A$ for some A $\in$ O(n) ... ... etc ... ( see attached for definition of $T_A$ )

I need some help with this first line of the proof!

Restating the proof explicitly in $\mathbb{R}^2$ we have S(0,0) = ( $b_1 , b_2$ )

i.e. $S(0,0) = t_b \circ \rho_{\phi} \circ r^i (0,0)$ = ( $b_1 , b_2$ )

Thus $t_{-b}\circ S(0,0) = t_{-b} \circ t_b \circ \rho_{\phi} \circ r^i (0,0) = \rho_{\phi} \circ r^i (0,0) = (0,0)$

But ... why does $\rho_{\phi} \circ r^i (0,0) = (0,0)$ ???

Presumably this is because Papantonopoulou defines rotations as about the origin (0,0) and reflections are defined to be about the X-axis or $e_1$ axis. (see definitions of rotations and reflections at bottom of page 462 - attachment)

A further question is this ... what is Papantonopoulou actually trying to do when he starts his proof with "To prove existence, let b = S(0)"? What is his strategy here? What is he tryinh to do - I am somewhat lost regarding the overall point of this!

I would be really appreciative of some help here - or at least confirmation of my reasoning.

Peter

**Deveno** · January 9th 2012, 03:01 AM

the point is, that $\rho_{\phi} \circ r^i$ is a linear map (in the sense of vector spaces), and linear maps always preserve the identity element of the vector space, as an abelian group (linear maps are abelian group homomorphisms that preserve scalar multiplication, as well).

it seems that the linear algebra might be throwing you off, a bit. it is accepted as so basic a fact of linear algebra (that the 0-vector of a vector space is in the kernel of a linear map), that most authors just state it without proof.

**Bernhard** · January 9th 2012, 12:50 PM

I think it is that I just keep thinking geometrically and can see that a rotation about a point Q - not at the origin (see Figure 2 attached) - would not, of course, fix the origin, but would be a good way to describe an isometry that maps $L_1$ onto $L_2$ .

I guess I just have to think of rotations as defined to be about the origin.

How would you mathematically describe an isometry that consisted of a rotation about Q - or would that not be an isometry?

Peter

**Deveno** · January 10th 2012, 06:02 AM

a rotation about a point Q (not the origin), can be considered the following way:

translate by -Q, rotate about the origin, and translate by Q.

being a composition of isometries, this is still an isometry.

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