3-parameter group of symmetries on the plane

[cianfa72]

Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?

 

[Orodruin]

This is what makes parallel parking possible.

 

[Orodruin]

Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?

Yes.

 

[mathwonk]

yes. if T is the subgroup of translations, then T is a normal subgroup, so if r is any rotation, the subgroup r^-1Tr is again equal to T. so there is a translation t', with r^-1tr = t', so in fact tr = rt', i.e. you can even take r' = r.

 

[Orodruin]

you can even take r' = r.

In fact, if R is the subgroup of rotations about a specific point, you must take r’ = r.

 

[mathwonk]

since each element (presumably) has a unique such expression in RT, hence t' is also unique.

 

[mathwonk]

cianfa72, For the orthogonal matrix and rigid motion stuff, you may see LADW sections 5.7 and 6.5.

 

[cianfa72]

so if r is any rotation, the subgroup r^-1Tr is again equal to T.

You mean the set $\{ r^{-1}tr, t \in T \}$ is a subgroup and is equal to T itself.

 

[Orodruin]

You mean the set $\{ r^{-1}tr, t \in T \}$ is a subgroup and is equal to T itself.

Yes. This is the definition of a normal subgroup. That this holds regardless of what $r$ is.

 

[mathwonk]

this might be a reasonable intro to groups:
https://link.springer.com/book/10.1...6CyjlqTIMnbL_oHnPIfBJqDXwsy0aAk_gEALw_wcB#toc

 

[pasmith]

Another point: since SE(n) is not commutative then if we pick a=tr do exist t' and r' such that a=r't' ?

Yes: instead of doing $Ax + b$ you do $A(x + A^{-1}b)$.

 

[mathwonk]

without using normal subgroups, if M is any isometry, and p is any point, then the translation t taking M^-1(p) to p, makes the composition Mt^-1 have a fixed point at p, so Mt^-1 =r is a rotation fixing p, and M = rt.

In pasmith's post, p = 0 the origin of coordinates, M(x) = Ax+b, so M^-1(0) = -A^-1(b), and the translation t by A^-1(b) takes this point back to 0, so Mt^-1 is a rotation fixing 0.
In fact Mt^-1 (x) = A(x-A^-1(b))+b = Ax, so Mt^-1 = A, as Orodruin also predicted.

Hence M = rt, i.e. M(x) = A.(x+A^-1(b)).

So this post of pasmith calculates for you, when given M = t'r', exactly what r and t will give you M = rt, where r and r' are both presumed to fix the point 0. In particular, r = r', and t is translation from 0 to r^-1(t'(0)) = t(0).

Of course once we know r = r', we can solve t'r' = t'r = rt, for t, hence r^-1t'r = t, and find t(0) = r^-1(t'(0)) = A^-1(b).

i.e. just solve A.x+b = A.(x+c), for c, by applying A^-1 to both sides and setting x=0, (or vice versa).

 

[mathwonk]

Clearly, the vector notation is very powerful here, since matrix multiplication is linear, i.e. preserves vector addition.

Thus if A is a rotation matrix and b a vector then the composition rt of the corresponding rotation and translation, takes x to A(x+b) = Ax + Ab, which is visibly a composition t'r, where t' is translation by Ab.

I.e. on the left, we first translate by b, then rotate by A, and on the right we first rotate by A then translate by Ab. This compatibility is not at all as obvious to me just using geometry and the abstract group theoretic approach.

This "linearity" thus proves that the translation subgroup is invariant under conjugation by rotations, a corollary of normality. I.e. to show r^-1tr is again a translation, just note that A^-1.(Ax+b) = x + A^-1(b), is a translation.

So for ease of computation and understanding, I am finally coming around to the very helpful representation of SE(n) as a semi direct product of SO(n) with the subgroup of translations.

 

[mathwonk]

apologies again to the experts for these details, which are for my benefit.

I want to understand geometrically what seems to be the key point, i.e. why the translation subgroup is normal in the group of oriented isometries, SE(n). We saw this is a corollary of the fact that the elements having a fixed point are “linear”, which can in fact be treated abstractly.

I.e. let A be an isometry with a fixed point 0, and for any point x, let t_x be the translation taking 0 to x. Then I claim for all points p,q, that (A.(t_p)) (q) = ((t_A(p)).A) (q), i.e. for all p, that A.(t_p) = (t_A(p)).A. Equivalently, for all p, that A.(t_p).A^-1 = t_A(p).

This will prove the conjugate by A, of a translate, is again a translate, which is the key step in proving the subgroup of translates is normal.

The essential point is that an oriented isometry fixing 0, takes a parallelogram based at 0, to another parallelogram based at 0. I.e. the unique parallelogram with sides 0p and 0q, has 4th vertex at (t_p)(t_q)(0) = (t_p)(q) = (t_q)(p) = (t_q)(t_p)(0). This is the geometric version of linearity.

I.e. the unique parallelogram with vertices 0 = A(0), A(p), and A(q), has 4th vertex equal to (t_A(p))(A(q)), and since this parallelogram is the image under A of the earlier one, this 4th vertex is the image of (t_p)(q) under A.

I.e. for all p, q, (t_A(p))(A(q)) = A((t_p)(q)).
Thus for all p, the maps (t_A(p))(A) = A.(t_p), are equal.

Hence, for all p, the conjugate A.(t_p).A^-1 of the translate t_p, by A, equals the translate (t_A(p)).

Then write any isometry M as a product M = t’A, of a translate t’ and an isometry A with a fixed point, as we have seen is possible. For any translate t, we claim the conjugate MtM^-1 is a translate.

Here, MtM^-1 = (t’A)t(A^-1t’^-1) = t’(AtA^-1)t'^-1.

But now we know that AtA^-1 is a translate, say t’’, so we have
MtM^-1 = t’.t’’.t’^-1, a composition of translates, hence a translate.

Thus the subgroup of translates is a normal subgroup of the group SE(n) of isometries of n space.

So although the subgroup of translations does not depend on a choice of base point 0, it helps to choose one for computations, since that represents each translation t by a unique point t(0) of the underlying space. Then the translations, and the isometries fixing 0, have nice algebraic representatives as vectors and matrices, which can be combined to represent all elements of SE(n).

 

[cianfa72]

Sorry, I was thinking again about post#34 e #35. I read also LADW as you advised.

A is by hypothesis an orthogonal matrix, however (I - A) is not. As said there, if we pick a basis that includes the vectors from the null space of (I-A) as the first m vectors (m is the dimension of kernel/null space), then the first m columns of the "new" (I-A) representation (similar matrix) are null.

What about the last (n-m) columns ? Is the orthocomplement of (I-A) null space actually invariant under the (I-A) linear transformation ?

 

[mathwonk]

have you tried to answer this yourself? Let N be the null space of I-A, i.e. the null space of A-I. Then N is the subspace where A restricts to the identity I. In particular, A leaves this space invariant, i.e. A maps N into N. Then since A is orthogonal, if H is the ortho-complement of N, then A also leaves H invariant, i.e. for all v in H, Av is also in H. What about (I-A)v? well (I-A)v = v - Av, and both v and Av are in H, hence also their difference.

you see how easy this is? I am tempted to think you did not spend much effort trying to solve it yourself, since you are obviously smart, as you routinely catch my careless statements. Before asking such questions, do you practice spending at least 30 minutes trying to solve it as an exercise? You will soon find that you can do many of them yourself.
In fact it is clear you are already doing this to some extent, since you have suggested several corrections to my typos. keep up the good work.

 

[cianfa72]

Yes, that makes sense. Thank you