## October 21, 2019

### Screw Theory

#### Posted by John Baez

Screw theory’ was invented by a guy named Ball. There should be a joke in there somewhere. But what is screw theory?

For starters, any rigid motion of 3d Euclidean space has a screw axis: a line mapped to itself. We translate along this axis, and rotate about it.

Screw theory is about the Euclidean group: the group of rigid motions of Euclidean space. A screw is an element of the Lie algebra of this group. It’s a 6d vector built from a pair of 3d vectors: an infinitesimal translation and an infinitesimal rotation.

An object moving through space and rotating has a velocity and an angular velocity. These combine to form a screw.

When you push on this object you exert a force and a torque on it. These also combine to form a screw.

Screw theory was developed in the 1800s, and its terminology is charmingly mechanical. The screw combining velocity and angular velocity was called the twist, while the screw combining force and torque was called the wrench.

A lot of good mathematicians worked on screw theory: for example Poinsot, Chasles, Plücker, Klein and Clifford. The fact that every element of the Euclidean group has at least one screw axis is called Chasles’ theorem.

These mathematicians defined a dot product and cross product of screws, and did a lot of other things:

What’s going on here? Let me translate it into the language of modern math!

The Euclidean group is the semidirect product of the 3d rotation group $\mathrm{SO}(3)$ and the translation group $\mathbb{R}^3$. Thus, we can write it as $\mathrm{SO}(3)\ltimes \mathbb{R}^3$ It’s isomorphic to $\mathrm{SO}(3)\ltimes \mathfrak{so}(3)$, where we use the fact that any Lie group acts on its Lie algebra — which we can treat as a vector space, and thus an abelian Lie group. In fact any Lie group $G$ acts on its Lie algebra $\mathfrak{g}$ and gives a Lie group $G\ltimes \mathfrak{g}$. This is isomorphic to the tangent bundle $\mathrm{T}G$. So the tangent bundle of a Lie group is again a Lie group!

Putting all this together, the Euclidean group is isomorphic to the tangent bundle $\mathrm{TSO}(3)$. A screw is an element of the Lie algebra of this! A less fancy way to say it: screws live in $\mathfrak{so}(3)\ltimes \mathbb{R}^3$.

The cross product of screws is the Lie bracket in $\mathfrak{so}(3)\ltimes \mathbb{R}^3$. The dot product is the obvious invariant inner product on this Lie algebra.

But in the 1800s, people preferred quaternions! So they had a different story.

We can think of $\mathfrak{so}(3)$ as the imaginary quaternions $\{a i+b j+c k\}$. Thus, we can think of $\mathfrak{so}(3)\ltimes \mathbb{R}^3$ as the imaginary quaternions tensored with the dual numbers $\mathbb{R}[\epsilon]/\langle \epsilon^2\rangle$. (Later, Grothenedieck and Lawvere thought of the dual numbers as the algebra of functions on an ‘infinitesimal arrow’.)

Using these ideas, Clifford thought of screws as sitting inside the algebra of quaternions tensored with the algebra of dual numbers. He called this 8-dimensional algebra the dual quaternions:

In other words, the dual quaternions are the algebra generated by $i,j,k$ obeying the usual quaternion relations together with an element $\epsilon$ commuting with $i,j,k$ and squaring to $0$.

In the dual quaternions, the infinitesimal rotations are guys like $a i+b j+c k$, while the infinitesimal translations are guys like $\epsilon(a i+b j+c k)$. Together these form the ‘screws’.

The screws are closed under commutators! They form the Lie algebra of the Euclidean group.

The dual quaternions can also be seen as the Clifford algebra of a real vector space with a quadratic form of signature $+\!+0$, if we use the convention that the Clifford algebra on a vector space $V$ with quadratic form $Q$ is generated by $v \in V$ with relations $v^2 = -Q(v)$. The three generators are $i, j$ and $\epsilon k$. But this is annoying asymmetrical! A better description is that the dual quaternions are the even part of the Clifford algebra of a vector space with quadratic form of signature $+\!+\!\!+0$:

I really like this, because the vector space with quadratic form of signature $+\!+\!\!+0$ can be seen as the dual of Minkowski spacetime in the $c \to \infty$ limit.

Dual quaternions are still used in engineering, especially robotics. There will be even be a workshop about applications of dual quaternions to robotics at ICAR 2019, the International Conference on Advanced Robotics. Dan Piponi writes:

They’re also used in movie visual effects to simulate rigid-body dynamics. E.g. if you want to simulate a body thrown off a building without putting a stuntperson at risk.

I like the idea of the dual quaternions as an ‘infinitesimal thickening’ of the quaternions, and the Euclidean group $TSO(3)$ as an ‘infinitesimal thickening’ of the rotation group. Rogier Brussee writes:

In algebraic geometry what you would do is consider the affine algebraic group $SO(V,g)\ltimes V$ which is defined by polynomial equations inside $\mathrm{End}(V)\times V$ i.e. by an algebra $H = k[X_{i j}, T_i]/I$. The group structure is equivalent to a commutative but not cocommutative Hopf algebra structure $\Delta \colon A \to A\otimes A$ and $S\colon A \to A$.

Let $\mathfrak{m}$ be the maximal ideal of the identity. Then we can consider the Hopf algebra $A/\mathfrak{m}^2$.

This construction works for all algebraic groups, and gives an algebraic group $\mathrm{Spec}(A/\mathfrak{m}^2)$ with only one point and a non reduced structure (i.e. one that has nilpotent ‘functions’). Apparently if $V$ is 3-dimensional and we take the algebraic group $\mathrm{SO}(V,g)\ltimes V$ the algebra of ‘functions’ splits over $k[\epsilon]/\langle\epsilon^2\rangle$!

I think there’s a bit more to be done with screws using modern mathematical ideas. Not exactly earth-shaking… but it would be sad not to carry on the noble 19th-century work connecting abstractions like dual quaternions to very visceral things like force and torque.

Posted at October 21, 2019 1:21 AM UTC

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### Thickening

I’m trying to understand this “infinitesimal thickening” concept. Is it that we regard G as a principal (Z/2)-bundle over G/(Z/2), then take the associated Spec(dual numbers)-bundle?

Posted by: Allen Knutson on October 21, 2019 3:29 AM | Permalink | Reply to this

### Re: Thickening

I was just trying to sound cool: Urs Schreiber talks about the dual numbers as the algebra of functions on the infinitesimally thickened point, since it has first-order nilpotent fuzz, and if we think of it as an affine scheme a map from it to a variety picks out a tangent vector in that variety. More generally:

The site of definition for the Cahiers topos is the category of spaces that are products of an $\mathbb{R}^n$ with the dual of a Weil algebra. So these are infinitesmally thickened Cartesian spaces.

Here I think ‘Weil algebra’ is being used to mean local Artin algebra, a class of algebras that includes the dual numbers.

I don’t know if it’s really sensible to think of the tangent bundle of a manifold as an infinitesimally thickened version of that manifold… I guess only if we treat the algebra of functions on the fibers as nilpotent? Maybe it’s a red herring when it comes to screw theory. But there’s something interesting about how the dual numbers are showing up in this game.

Posted by: John Baez on October 21, 2019 4:51 AM | Permalink | Reply to this

### Re: Screw Theory

The dual quaternions are equal to the Clifford algebra “- - 0” not ++0.

Posted by: laptopman on October 21, 2019 7:17 AM | Permalink | Reply to this

### Re: Screw Theory

Rather, it’s the Clifford algebra for “- - 0”. The one for “++0” contains a non-trivial square root of 1, so it can’t be equivalent to the Dual Quaternions – rather, it’s equivalent to the tensor product of the Dual Numbers with the Split Quaternions. This ends up being a relativistic counterpart to the Dual Quaternions. (The Split Quaternions are more commonly known as the 2x2 real matrices).

Posted by: laptopman on October 21, 2019 9:17 AM | Permalink | Reply to this

### Re: Screw Theory

I said “using the conventions of my last diary entry”, which is not very useful here… but my conventions are that the Clifford algebra for a vector space $V$ with a quadratic form $Q$ has generators $v \in V$ obeying

$v^2 = -Q(v)$

This minus sign means we actually agree. I like this minus sign because I want the quaternions to be the even part of the Clifford algebra associated to the signature +++. In other words, I think square roots of -1 are more fundamentally connected to Euclidean geometry than square roots of +1. But this is ultimately just a convention.

I’ll change my post to make this clearer.

Posted by: John Baez on October 22, 2019 5:20 PM | Permalink | Reply to this

### dot product

This is cool. I’ll point out that the dot product of screws isn’t actually an inner product, since it takes values in the dual numbers. I remember reading somewhere that $\mathfrak{so}(3) \ltimes \mathbb{R}^3$ doesn’t admit an invariant inner product.

But I suppose the dot product of screws can be viewed as a “thickening” of the invariant inner product on $\mathfrak{so}(3)$.

Posted by: Rajan Mehta on October 25, 2019 7:22 PM | Permalink | Reply to this

### Re: dot product

Hmm, maybe I was mixed up. In physics, elements of $\mathfrak{so}(3) \ltimes \mathbb{R}^3$ describe angular momentum and linear momentum, which are the generators of rotations and translations respectively. I was thinking that the magnitude of angular momentum and linear momentum are separately conserved by both rotations and translations, so we could sum the squares of these to get an invariant norm on $\mathfrak{so}(3) \ltimes \mathbb{R}^3$. But now I’m thinking the magnitude of angular momentum is not preserved by translations. This is confusing to me since mathematically it’s $\mathfrak{so}(3)$ that’s acting on $\mathbb{R}^3$, not the other way around!

Posted by: John Baez on October 25, 2019 7:56 PM | Permalink | Reply to this

### Re: dot product

I think the problem is that while the infinitesimal translations don’t act on the infinitesimal rotations, if we apply a group element that involves a translation to an infinitesimal rotation, the infinitesimal rotation will act on that translation to produce an infinitesimal translation. As a result, the Lie algebra element that we thought was just angular now has a linear component via a coordinate change. But because it isn’t purely linear, we call it all angular momentum anyway.

Posted by: Layra on October 26, 2019 6:54 AM | Permalink | Reply to this

### HASSEN

I kindly would like to ask the group if someone has experience how to find the reciprocal screw when the limb is constrained by one DoF while having six joints screws. Example: PRPS (prismatic-revolute-prismatic-spherical) have two linear actuators and six joint screws in general. How one can obtain the active wrenches in this case ? or How to invert the Jacobian ? I usually invert Jacobian analytically when screws are less than six. But, screws are more than six including the constraint for the example I introduced or other similar cases. So, I am asking if someone has experience in this group. Thanks in advance.
Posted by: HASSEN on February 26, 2021 8:18 AM | Permalink | Reply to this

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