I 3-parameter group of symmetries on the plane

[cianfa72]

TL;DR Summary

About the 3-parameter group of Killing vector fields (symmetries) on the Euclidean plane.

Consider $\mathbb R^2$ as the Euclidean plane. Since it is maximally symmetric it has a 3-parameter group of Killing vector fields (KVFs).

Pick orthogonal cartesian coordinates centered at point P. Then the 3 KVFs are given by: $$K_1=\partial_x, K_2=\partial_y, K_3=-y\partial_x + x \partial_y$$
They are linearly independent in the sense that the null vector field is given only by a linear combination of the above KVFs with zero constant coefficients.

$K_3$ represents rotations about the (fixed) point P with coordinates $(0,0)$.

My question is: can any other set of KVFs that includes the rotations about a different point Q be given as linear combination (with constant coefficients) of the above KVFs ?

Thanks.

 

[Orodruin]

Yes.

 

[cianfa72]

Suppose the point Q has coordinate $(2,3)$ in the given cartesian coordinates. How does one write a set of KVFs that includes the rotations about Q (staring from the KVFs in post#1 and using constant coefficients) ?

 

[Ibix]

Define $x'=x-2$ and $y'=y-3$. The Killing field you want is trivial in these coordinates, as is the transform.

 

[cianfa72]

Ok, so in $(x,y)$ coordinates it is $$K_1^{'}=K_1, K_2^{'}=K_2,K_3^{'}=K_3 + 3K_1 - 2K_2$$ One can extend it to any dimension provided that the space is maximally symmetric.

 

[cianfa72]

Just to check my understanding: elements of $E(2)$ group of Euclidean plane isometries consist of: translations, rotations and composition of translations + rotations. One then can show those elements form a group w.r.t. the composition operator (in dimension 2 it should be abelian/commutative).

BTW, $E(2)$ is not isomorphic to $SO(2)$, yet there is a group homomorphism from the former to the latter.

 

[Orodruin]

BTW, $E(2)$ is not isomorphic to $SO(2)$, yet there is a group homomorphism from the former to the latter.

There is at least one homomorphism (although not necessarily a surjective nor injective one) between any two groups. Consider the map taking any element of the first group to the identity of the other, ie, the trivial homomorphism.

What you likely want to comment is the homomorphism mapping an element of $E(2)$ to the element of $SO(2)$ with the same rotation. More formally, that $E(2)$ is a semidirect product.

 

[cianfa72]

What you likely want to comment is the homomorphism mapping an element of $E(2)$ to the element of $SO(2)$ with the same rotation. More formally, that $E(2)$ is a semidirect product.

You mean the group homomorphism that map a translation + rotation (i.e. an element of $E(2)$) to the corresponding element of $SO(2)$ that represent the "rotation part" of the translation + rotation.

 

[pasmith]

Just to check my understanding: elements of $E(2)$ group of Euclidean plane isometries consist of: translations, rotations and composition of translations + rotations. One then can show those elements form a group w.r.t. the composition operator (in dimension 2 it should be abelian/commutative).

BTW, $E(2)$ is not isomorphic to $SO(2)$, yet there is a group homomorphism from the former to the latter.

E(2) also includes reflections. The rotations and translations generate a subgroup SE(2) of this group.

E(2) is a semidirect product of O(2) and \mathbb{R}^2 under vector addition. It acts on \mathbb{R}^2 as <br /> (A, b) : x \mapsto Ax + b and its composition law is <br /> (A, b)(C, d) = (AC, Ad + b). This is not commutative. SE(2) is similarly a semiriect product of SO(2) and \mathbb{R}^2.

 

[cianfa72]

E(2) is a semidirect product of O(2) and \mathbb{R}^2 under vector addition. It acts on \mathbb{R}^2 as <br /> (A, b) : x \mapsto Ax + b and its composition law is <br /> (A, b)(C, d) = (AC, Ad + b). This is not commutative.

Therefore even in the Euclidean plane the composition of two "translation + rotation" is not commutative.

 

[Orodruin]

Therefore even in the Euclidean plane the composition of two "translation + rotation" is not commutative.

Indeed it is not, since rotations do not commute with translations.

 

[mathwonk]

Thanks, I enjoyed this. It took me a little work, even though it is visually plausible that an element of SE(2) has a unique rotational part, to check that indeed a rotation R conjugates translation by v into translation by (R^-1)(v), and that thus mapping an oriented isometry to its rotational "part", is a homomorphism. (the "rotational part of f", being f followed by translation through -f(0). I.e. to change f into a rotation, force it to have a fixed point.). In particular, the translation subgroup of SE(2) is a normal subgroup, which when modded out, leaves the rotation subgroup as quotient. (When I taught isometries of the plane, we did not have available abstract group theory, and did not use coordinates or matrices.)

 

[Orodruin]

In particular, the translation subgroup of SO(2)

I assume you mean SE(2)? SO(2) doesn’t have a translational part. Just a single rotational parameter.

 

[mathwonk]

yes thank you! edited.

 

[mathwonk]

This is really interesting, and I did not realize how little I knew about it. Forgive me experts, since this is all well known to you.
I stated things a bit imprecisely above. Actually an element of SE(2) does NOT have a unique rotational part. In fact if A is any element of SE(2), and p is any point of the plane, and T is the translation from A(p) back to p, then the composition TA leaves p fixed, and is orientation preserving, hence R = TA is a rotation about the point p.
Thus we can write A as the composition A = T^-1.R, of a translation and a rotation about p, for any point p. Thus many rotations, one for each point of the plane, can be regarded as the rotational part of A. However, no matter what point we choose to base the rotational part of A at, all will have the same angle, so it is the angle of the rotation, not the fixed point, that is unique to A.

Thus in the abstract affine plane, without coordinates, there is no unique rotational part of an isometry. But if we introduce a base point, such as the origin 0 of a coordinate system, as previous posters have done, then we can choose the rotation to be based at 0, and obtain a unique element of SO(2) as the rotational part of A.

I also erred slightly in stating that the quotient group SE(2)/translations, is equal to the "rotation subgroup" of SE(2), indeed the rotations in SE(2) do not form a subgroup. (The composition of two rotations with complementary angles, based at different points, has rotational part with angle zero, hence is a translation.)
Rather it is SO(2), the rotations based at 0, that form a subgroup, and the quotient group of SE(2) by translations is not equal to, but rather naturally isomorphic to, that subgroup. In fact for any point p, the rotations based at p also form a "rotation subgroup" of SE(2). I.e. the rotations about 0 form a non- normal subgroup, and these other rotation subgroups are the conjugates of that one. Since these conjugate subgroups are all isomorphic, the quotient group SE(2)/translations, is naturally isomorphic to any one of them.

I.e. the elements of the quotient group are cosets of the subgroup of translations, and each coset except the trivial one consists entirely of rotations. Thus the group SE(2) is decomposed into infinitely many disjoint cosets, one for each angle. The trivial coset consists of all translations. Each non trivial coset contains all rotations through a given ≠0 angle (mod 2π), one based at each point of the plane. So in a sense almost all elements of SE(2) are rotations. In the trivial coset, the trivial translation can be regarded also as a rotation, about any point, through angle zero.

If we choose any point p of the plane as base point, then each coset contains a unique rotation based at p, and we have a homomorphism from SE(2), hence an isomorphism from the quotient group, onto the subgroup of rotations based at p. I.e. each element of SE(2) determines a unique coset of the translations, and a unique rotation based at p, in that coset.

 

[mathwonk]

Well, what a lot of nonsense I have been saying. An element of SE(2) is either a rotation or a translation, according to whether or not it has a fixed point. Hence if it is a translation it has no rotational part, and if it is a rotation, it does not does not make sense to ask what is its rotational "part", since it is itself a rotation. So contrary to what I said, both the angle and the base point of the rotation are intrinsic to the element A of SE(2), (unless A is a translation).
I was trying to think of the homomorphism from SE(2) to SO(2) as picking off the rotational part of an element A. Rather, if A is a rotation based at p, this map asks for the rotation based at 0 that is conjugate to A via a translation.
Obviously I am still struggling, but at the moment it seems a bit unnatural to think of SE(2) as the semi direct product of SO(2) and the translations, since it seems unnatural to think of a rotation about p as a product of a translation and a rotation about 0. I guess I am saying that putting coordinates on the plane seems somewhat unnatural, but it makes computations more convenient.

Pardon me for sharing my confusion.

 

[Orodruin]

Well, what a lot of nonsense I have been saying. An element of SE(2) is either a rotation or a translation, according to whether or not it has a fixed point. Hence if it is a translation it has no rotational part, and if it is a rotation, it does not does not make sense to ask what is its rotational "part", since it is itself a rotation. So contrary to what I said, both the angle and the base point of the rotation are intrinsic to the element A of SE(2), (unless A is a translation).
I was trying to think of the homomorphism from SE(2) to SO(2) as picking off the rotational part of an element A. Rather, if A is a rotation based at p, this map asks for the rotation based at 0 that is conjugate to A via a translation.
Obviously I am still struggling, but at the moment it seems a bit unnatural to think of SE(2) as the semi direct product of SO(2) and the translations, since it seems unnatural to think of a rotation about p as a product of a translation and a rotation about 0. I guess I am saying that putting coordinates on the plane seems somewhat unnatural, but it makes computations more convenient.

Pardon me for sharing my confusion.

It is clear that SE(2) is a semidirect product. Translations $T$ clearly form a normal subgroup and you can pick rotations $R$ about any point as the other subgroup. It is naturally isomorphic to $SO(2)$. It should be evident that $TR = SE(2)$ and that $T \cap R = \{e\}$.

 

[mathwonk]

Yes, it is clear that SE(2) is indeed the semi direct product of the translation subgroup T with any rotation subgroup R based at any point. I am just saying that is to me a somewhat unnatural way of viewing the group SE(2). Forgive me if my confusions are obvious to you. I.e. it is to me not so clear how to read off properties of elements of SE(2) in terms of this representation.
For example, suppose we have a base point 0, and represent SE(2) as the semi direct product of the normal translation subgroup T and the rotation subgroup R based at 0. Then, as you say, TR = SE(2), i.e. every element of SE(2) is uniquely represented as a product tr for t in T and r in R. But what element is it? E.g. if r is a non trivial rotation, and t is a non trivial translation, then the product tr is a rotation about some other point, but what point is it?

I.e. a rotation is only well understood when both its angle w (which is given by r), and also its base point is known. By way of contrast, when we view a rotation in SE(2) as a conjugate of a rotation in R, i.e. as trt^-1, then this rotation obviously has base point at t(0).

This may well be clear to you, but I found it a bit less obvious to compute the base point of tr. If we let q = t(0) = the image of 0 under the translation t, and w = the angle of the rotation r, and if we let r* be rotation about 0 through an angle of π- (w/2), with the same orientation as the rotation r, then it seems the fixed point of tr lies on the ray <0,r*(q)>, from 0 to r*(q), but where is it on there?

By similar triangles, if |s| denotes length of a segment s = <a,b>, then it seems we should have |<0,t(0)>| / |<r(r*(q)),r*(q)>| = |<0,p>| / |<0,r*(q)>|. Hence to find the point p, we could multiply the vector <0,r*(q)> by the scalar |<0,t(0)>| / |<r(r*(q)),r*(q)>|, to obtain the vector <0,p>. But this is a bit clunky.

Is there a more obvious way to find the fixed point of tr?
.................
Well, I guess if we use coordinates, and the explicit action as recorded in post #9, where the product bA = (A,b), of a translation by b and rotation by matrix A, act on a vector x by x-->Ax+b, we can. set Ax+b = x, and get Ax-x = -b, or x = (A-Id)^-1(b), where A-Id is invertible when A is a non trivial rotation.

This does solve for the coordinates of the fixed point x, of the rotation bA, but does not clearly show me what it means geometrically. In fact since our group SE(2) contains only rotations and translations, it is not clear to me what the difference matrix A-Id means geometrically. ... Oh I see, it just uses the other side of the parallelogram I was using to do the calculation by similar triangles! (Instead of taking q = t(0), you just let q be a variable, and complete the parallelogram 0, q, r(q), z, so that the vector <0,z> = -b.)

So this does illustrate the computational power of the representation of SE(2) as a semi direct product, and one can even recover the geometry.

How clever. Thank you again!

 

[Orodruin]

But what element is it? E.g. if r is a non trivial rotation, and t is a non trivial translation, then the product tr is a rotation about some other point, but what point is it?

I.e. a rotation is only well understood when both its angle w (which is given by r), and also its base point is known.

Well, yes and no. That is just one way of writing down a rotation though. You could just as well uniquely charachterize it by giving the rotation angle and the subsequent translation. Which you already have. It will have a fixed point, but what that fixed point is easily computable.

Besides, this classification by considering any transformation without a fixed point a pure translation is limited to SE(2). For $SE(N)$ with $N \geq 3$ you lose this property (because you can start having translations in the direction of the rotational axis

 

[mathwonk]

Thank you for this, Orodruin. Pardon my slowness, but moving up only to SE(3), I see your point. Now if we define a "glide" as a non trivial rotation followed by a non trivial translation parallel to the axis of the rotation, then it seems all elements of SE(3) are either translations, rotations, or glides.

More precisely, it seems an element is a rotation if and only if it has a fixed point, hence infinitely many fixed points. It is a glide if and only if it has no fixed points, and exactly one invariant line (a line mapped into itself); and a (non trivial) translation if and only if it has no fixed points and at least 2, hence infinitely many, invariant lines.

Again, SE(3) seems to be the semi direct product of the rotations about 0, and the translations. This time if t is a non trivial translation, and r a non trivial rotation about 0, then tr is apparently a rotation if and only if the direction of the translation t is perpendicular to the axis of r, and a glide otherwise.

If A is the matrix of a non trivial rotation, and b a non zero translation vector, then x-->Ax+b is a rotation if and only if b lies in the column space of (A-Id).

 

[Orodruin]

Again, SE(3) seems to be the semi direct product of the rotations about 0, and the translations.

Yes, but there is no requirement to pick 0. Any point will do (as it should, no zero exists in Euclidean space until you pick one).

 

[cianfa72]

If A is the matrix of a non trivial rotation, and b a non zero translation vector, then x-->Ax+b is a rotation if and only if b lies in the column space of (A-Id).

You mean that if and only if $b$ is in the column space of $(A -I)$, i.e. $b = (A - I)c$ for some $c$ then $$Ax + b = Ax + (A -I)c = A(x + c) - c$$
i.e. the vector $-c$ is left invariant, hence x-->Ax + b is a "pure" rotation about the point $-c$.

 

[mathwonk]

yes cianfa72, if b is in that column space then so is -b, so there exists x with (A-I)x = -b, so Ax-x = -b, so Ax+b = x.


Thank you Orodruin; with your guidance, I have come up with a version for all dimensions, generalizing the results from dim 3: (I have not written down all the proofs, so errors are certainly possible.)

(Oriented) Isometries of Euclidean space:

Every orientation preserving isometry of Euclidean space is an orthogonal product of plane rotations, and a translation parallel to the intersection of the axes of the rotations.

More precisely:

Rotations
If p is a point of a Euclidean 2-plane ∏, a rotation centered at p is an orientation preserving isometry of E that fixes p.

If a Euclidean space E is a product of ≥1 orthogonal 2-planes ∏j and another orthogonal space C, the product of a non trivial rotation in each ∏j, with the identity map on C, is called a (non-trivial) rotation of E, with axis (fixed set) C. (Every matrix in SO(n) defines such a map of R^n.)

The fixed set of a rotation is always non empty; the trivial rotation fixes all of E. A product of non trivial rotations, (possible only on an even dimensional space E), fixes only a point.

Theorem:
An isometry of E is a rotation if and only if it has a fixed point.

Translations
A translation is an isometry T such that for any two points p,q, the points p,q,T(p),T(q), all lie in the same plane, and the oriented segments pT(p), and qT(q), are opposite sides of a parallelogram. A translation is entirely determined by any point p and its image T(p). In particular a translation with a fixed point is the identity.

An isometry A is a translation if and only if there is a point p such that, if T is the translation from p to A(p), then T^-1A is the identity map of E, if and only if E (of positive dimension), is the union of the invariant lines of A.

General (oriented) isometries:
Theorem:
Every orientation preserving isometry A of E is uniquely a product A = TR of a (possibly trivial) rotation R and a (possibly trivial) translation T that leaves the axis of R invariant. The axis of R is called the axis, or center, of A. T is the (intrinsic) “translation part” of A, and R is the (intrinsic) “rotation part” of A.

Given any isometry A of E, the axis of A is the unique maximal subspace C on which A restricts to a (possibly trivial) translation T. C is always non empty. If A has a fixed point, its axis C is the union of all fixed points of A. If A has no fixed points, its axis C is the union of all invariant lines of A.

Given A, and hence C and T, the product T^-1A is the identity on C, hence equals a rotation R. Then A = TR is the unique decomposition of A into a rotation R, and a translation T leaving the axis of R invariant.

Next: how to represent such isometries with matrices, and translation vectors, after choosing a base point 0.

Briefly, then an isometry A is uniquely a product tr, where r is a rotation with axis containing 0, and t is a translation not necessarily leaving the axis of r invariant. If we then decompose t into two translations, t = t1.t0, where t0 is orthogonal to the axis of r, and t1 is parallel to that axis, then the product t0.r is again a rotation R, with axis C parallel to the axis of r, and t1 = T is then a translation parallel to the axis C of R = t0r. hence we have decomposed A into a product A = tr = t1.(t0.r) = TR, as above.

Here r is given by a matrix in SO(n), and t is given as translation by a vector in R^n, which makes computations doable. But note that even though the expression A = tr is unique, this t is not usually the (intrinsic) translation part of A, and r is not usually the (intrinsic) rotation part of A.

When A = tr, still r does contain a lot of information about R, e.g. r is a product of the same number of non trivial plane rotations as R, and its axis is parallel to that of R and of the same dimension. Moreover r=Id if and only if t0 = Id, if and only if R=Id, if and only if A = t=T; i.e. A = tr is a translation iff r = Id. Moreover, if t = Id, then A = r = R is a rotation. However it is possible for A = tr, to be a rotation even if t ≠ Id. Indeed this happens iff t is a translation orthogonal to the axis of r.

 

[cianfa72]

Every orientation preserving isometry A of E is uniquely a product A = TR of a (possibly trivial) rotation R and a (possibly trivial) translation T that leaves the axis of R invariant. The axis of R is called the axis, or center, of A. T is the (intrinsic) “translation part” of A, and R is the (intrinsic) “rotation part” of A

The product A=TR is the action of a "rotation" R followed from a Translation T. From the point of view of matrix representation of SE(n), R is an orthogonal matrix in SO(n). Which is the matrix representation of T ?

 

[mathwonk]

no matrix is needed for T, just adding a vector b as in post #9.

 

[cianfa72]

no matrix is needed for T, just adding a vector b as in post #9.

So when you write A=TR what you have in mind is actually the group composition law (i.e. the element T of the group composed with the element R). In the specific case of SE(2), T is an element from a normal subgroup and R from the rotation subgroup about any point in $\mathbb E^2$. Such a rotation subgroup is isomorphic to $SO(2)$.

It is clear that SE(2) is a semidirect product. Translations $T$ clearly form a normal subgroup and you can pick rotations $R$ about any point as the other subgroup. It is naturally isomorphic to $SO(2)$. It should be evident that $TR = SE(2)$ and that $T \cap R = \{e\}$.

Here $T$, instead, is a normal subgroup of $SE(2)$ and is not a generic element of it. The same holds for $R$.

Therefore $TR$ is actually the expression of the semidirect product of subgroups $T$ and $R$ and turns out to be equal to $SE(2)$.

 

[mathwonk]

correct, in both cases.

and when I said:..." it is possible for A = tr, to be a rotation even if t ≠ Id. Indeed this happens iff t is a translation orthogonal to the axis of r."

moreover, when this happens, i.e. when t is translation by a vector t(0) which is orthogonal to axis (r) = null space of (I-r), then the matrix (I-r) is invertible when restricted to the orthogonal complement of axis(r), and the axis of the rotation tr is the translate of the axis of r by the vector (I-r)^-1(t(0)), where (I-r)^-1 denotes the inverse of the restriction of (I-r) to the ortho-complement of axis(r). (here t(0) = b is the vector giving the translation t.)

Note here the advantage of matrix methods. I.e. the matrix I-r does not represent an element of the group SE(n), and it is hard for me to see clearly by just geometry and group theory methods how to show that tr is a rotation when t(0) is orthogonal to axis(r). I had trouble even for n=2, and had to use ideas of continuity. No doubt knowledgable people know how to do this though.

 

[cianfa72]

and when I said:..." it is possible for A = tr, to be a rotation even if t ≠ Id. Indeed this happens iff t is a translation orthogonal to the axis of r."

moreover, when this happens, i.e. when t is translation by a vector t(0) which is orthogonal to axis (r) = null space of (I-r), then the matrix (I-r) is invertible when restricted to the orthogonal complement of axis(r), and the axis of the rotation tr is the translate of the axis of r by the vector (I-r)^-1(t(0)), where (I-r)^-1 denotes the inverse of the restriction of (I-r) to the ortho-complement of axis(r). (here t(0) = b is the vector giving the translation t.)

Sorry, here r is both a rotation and its matrix representation. (I - r) is a matrix and your claim is that when t(0) is orthogonal to the null space of (I - r) then A=tr is a "pure" rotation about some axis (and vice-versa).

 

[mathwonk]

right. we are trying to find a fixed point of tr, to prove it is a rotation. if (tr)(x) is represented as Ax+b, then we want Ax+b = x, or (A-I)x = -b, or (I-A)x = b. So we need b to be in the column space of (I-A) = column space of (A-I) = axis of r [oops, the axis of r is actually the null space of (I-A).].
But since A is orthogonal, the column space of (A-I) is orthogonal to the null space, so we need b to be orthogonal to the axis of r. Given that, our fixed point will be an x with (I-A)x = b, and if we restrict to the orthocomplement of the axis of r, then (I-A) is invertible, so we take x = (I-A)^-1(b), where b = t(0).

Then having found one solution x of (I-A)x = b, we get the other solutions, i.e. the other fixed points of tr, i.e. the full axis of tr, by adding x to all elements of the null space of (I-A), i.e. by translating the axis of r, by x = (I-A)^-1(b) = (1-r)^-1(t(0)).

 

[cianfa72]

Therefore to avoid confusion one would stick to use lowercase letters for elements of group (t and r) and uppercase for subgroups T and R respectively.

right. we are trying to find a fixed point of tr, to prove it is a rotation. if (tr)(x) is represented as Ax+b, then we want Ax+b = x, or (A-I)x = -b, or (I-A)x = b. So we need b to be in the column space of (I-A) = column space of (A-I) = axis of r. But since A is orthogonal, the column space of (A-I) is orthogonal to the null space, so we need b to be orthogonal to the axis of r.

Sorry, you said that we want b in the column space of (A-I) and that the latter is the axis of r. Then b can't be orthogonal to the axis of r. I believe the point is that the null space of (A-I) is actually the axis of r, hence, since A is orthogonal, b turns out to be orthogonal to the (A-I) null space (i.e. orthogonal to the axis of r). In this case (A=tr is a "pure" rotation about a point) the action of tr basically amounts to "translating back by $(I-A)^{-1}b$", rotate according A (the matrix representation of r) and translate again by $(I-A)^{-1}b$.

BTW in dimension 2 for a not null angle rotation r the kernel/null space of (A-I) is the zero vector.

Given that, our fixed point will be an x with (I-A)x = b, and if we restrict to the orthocomplement of the axis of r, then (I-A) is invertible, so we take x = (I-A)^-1(b), where b = t(0).

Ok, here t(0) is the vector resulting from the action of translation t on the point "labelled" 0 in the Euclidean space (having fixed a cartesian coordinate system in it). What does it mean that restricting to a vector subspace (here the orthocomplement of axis of r) the matrix (I-A) turns out to be invertible when actually it is singular ?

 

[Vanadium 50]

since rotations do not commute with translations.

This is what makes parallel parking possible.

 

[mathwonk]

cianfa72, you are right I made a brief misstatement about axis r, later stated correctly, I added a correction, thank you.

As to I-A being singular, it is only singular because it has a null space, so if we restrict it to the ortho-complement of that null space, it becomes invertible as a map on that smaller space, which actually equals the column space. I wrote (I-A)^-1 also for the inverse of that restriction, when perhaps to avoid confusion I could have called it ((I-A)res)^-1.

 

[cianfa72]

As to I-A being singular, it is only singular because it has a null space, so if we restrict it to the ortho-complement of that null space, it becomes invertible as a map on that smaller space, which actually equals the column space.

Ok yes, but how does one actually write down such an inverse matrix (restricted to (I-A) column space) ?

 

[mathwonk]

you find a basis of the null space, then augment it by a basis of the orthocomplement (i.e. the column space), then in that basis the matrix is a block matrix with a zero block in the upper left corner, and the desired invertible restricted block matrix in the lower right hand corner. Then you invert that lower right hand block matrix.

you might enjoy reading a good book on linear algebra, like that of Sergei Treil, at Brown, his LADW (linear algebra done wrong.)
https://www.math.brown.edu/streil/papers/LADW/LADW.html

 

[cianfa72]

you find a basis of the null space, then augment it by a basis of the orthocomplement (i.e. the column space), then in that basis the matrix is a block matrix with a zero block in the upper left corner, and the desired invertible restricted block matrix in the lower right hand corner. Then you invert that lower right hand block matrix.

Ok, so you mean rewrite the (I-A) matrix in that basis. Of course said m the dimension of the kernel of (I-A) then the first m column of the "new" matrix are null, and the remaining (n-m) basis vectors (the picked orthocomplement basis) map to linear combination of themselves.

Therefore the target inverse will have a zero block in the upper left corner and the inverse of the (invertible) lower right hand block at the lower right corner.

 

[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 ?