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Sep 25, 2019 at 21:40 comment added Kimball One idea is to invert the compact version of the map from PGL(2) to SO(2,1)
Sep 25, 2019 at 20:32 history became hot network question
Sep 25, 2019 at 15:22 answer added wlad timeline score: 3
Sep 25, 2019 at 14:43 answer added David E Speyer timeline score: 10
Sep 25, 2019 at 14:41 comment added wlad @DavidESpeyer is that clear or not?
Sep 25, 2019 at 14:30 review Close votes
Sep 30, 2019 at 3:05
Sep 25, 2019 at 14:19 comment added wlad But it's possible that the question is too vague to get an answer
Sep 25, 2019 at 14:18 comment added wlad My axiomatic approach is based on observing that if we want to represent affine transformations using matrices, then we need to treat the matrices as ratios. It then makes sense that if we want to represent the group of rotations, we might also use ratios. My approach effectively pre-supposes what the transformation group should be, whether they be affine transformations or 3D rotations or translations, and then builds a Clifford algebra where almost every element has a well-specified geometric meaning
Sep 25, 2019 at 14:14 comment added wlad @DavidESpeyer I guess my problem with that idea is: What does it mean to add a reflection to a scalar? The advantage of starting with a transformation group and building a "projective coordinate system" for it is that you know what each point in the coordinate system represents. It always represents some transformation (or a degenerate one if it's a zero divisor). You don't formally add scalars to vectors without any understanding of what that might mean
Sep 25, 2019 at 13:48 comment added wlad @DavidESpeyer so you're endowing the reflections with a vector space structure. That is an interesting way to think about it
Sep 25, 2019 at 13:14 comment added David E Speyer When $n=3$, we have $R \cong \mathbb{H} \oplus z \mathbb{H}$, where $z$ is a central element corresponding to negation in $O(3)$. The unit vectors in $\mathbb{H}$ then give rotations and the unit vectors in $z \mathbb{H}$ give reflections.
Sep 25, 2019 at 13:14 comment added David E Speyer I don't understand your objection to Clifford algebras. Suppose I want an $\mathbb{R}$-algebra $R$ which (1) has an element $s_{\vec{v}}$ for each vector $\vec{v} \in \mathbb{R}^n$ (2) the addition and scalar multiplication of $R$ and $\mathbb{R}^n$ are compatible, in the sense that $\vec{v} \mapsto s_{\vec{v}}$ is a linear map and (3) for a unit vector $\vec{v}$, we have $s_{\vec{v}}^2 = 1$, which is what you would expect if $s_{\vec{v}}$ is a reflection. Then you have just defined the Clifford algebra. (continued)
Sep 25, 2019 at 13:14 comment added GreginGre Maybe this link will help: en.wikipedia.org/wiki/Quaternions_and_spatial_rotation
Sep 25, 2019 at 12:39 comment added Andrej Bauer One way to attack such a question is to identify the structure of interest as the unique (up to isomorphism) one having a certain universal property. Then different proofs of its existence will give different constructions of it. What is the universal property of quaternions, expressed so that it relates to rotations?
Sep 25, 2019 at 12:37 comment added wlad @AndrejBauer Yes
Sep 25, 2019 at 12:36 comment added Andrej Bauer Ok, so for instance you don't particularly care that they form a complete meric space? You're shooting for the algebraic structure, and in particular the relationship with rotations, is that it?
Sep 25, 2019 at 12:32 comment added wlad In applications like computer graphics, quaternions are often criticised for being unintuitive. These criticisms could be answered by a "principled" construction of them
Sep 25, 2019 at 12:30 comment added wlad For a construction of the quaternions, I just want to construct them as an algebra where the conceptual connection with 3D rotations is clear, and which isn't pulled out of a hat
Sep 25, 2019 at 12:29 comment added wlad In which case, the axiom states that every open set contain a point that corresponds to a valid rotation. Maybe "corresponds to" is vague
Sep 25, 2019 at 12:28 comment added wlad @AndrejBauer For axiom 2, I assume the usual Euclidean topology.
Sep 25, 2019 at 12:26 comment added Andrej Bauer Isn't it going to be quite important what structure (algebraic and topological) you want on quaternions?
Sep 25, 2019 at 12:18 history asked wlad CC BY-SA 4.0