B One-to-many relations in group theory

[redtree]

TL;DR Summary

Mappings from one group to another

I apologize for the simple question, but it has been bothering me. One can write a relationship between groups, such as for example between Spin$(n)$ and SO$(n)$ as follows:

\begin{equation}
1 \rightarrow \{-1,+1 \} \rightarrow \text{Spin}(n) \rightarrow \text{SO}(n) \rightarrow 1
\end{equation}
when $n \neq 2$

In this context, Spin$(n)$ is the double covering of SO$(n)$, which, as far as I understand, means there is a 2-to-1 mapping from Spin$(n)$ to SO$(n)$ with neighborhood isomorphism between the groups.

How would one write the inverse relation, i.e., the many-to-one relation between groups. In the case of SO$(n)$ and Spin$(n)$, how would one write the the 1-to-2 relation from SO$(n)$ to Spin$(n)$ where neighborhood isomorphism is preserved?

 

[redtree]

Just to be clear. I understand that a function or mapping can be one-to-one or many-to-one but cannot be one-to-many. However, a binary relation can be one-to-many, such that a binary relation theoretically could connect a base space to a covering space.

 

[redtree]

How would one write the inverse relation, i.e., the many-to-one relation between groups. In the case of SO$(n)$ and Spin$(n)$, how would one write the the 1-to-2 relation from SO$(n)$ to Spin$(n)$ where neighborhood isomorphism is preserved?

I switched terms. It should read as follows:

How would one write the inverse relation, i.e., the one-to-many relation between groups. In the case of SO$(n)$ and Spin$(n)$, how would one write the the 1-to-2 relation from SO$(n)$ to Spin$(n)$ where neighborhood isomorphism is preserved?

 

[Paul Colby]

Why not map each element onto the set of possible values?

 

[mathwonk]

as paul colby says, this reduces to finding, for each transformation in SO(n), the 2 elements of Spin that map to that element. Spoiler alert: I know nothing about this stuff, not even what the clifford algebra is, BUT I scanned the wikipedia article on this topic, and one can get some insight from it. I.e. the clifford algebra, or rather the spin group built from it, has as elements, certain formal products of elements of a vector space V, and the target group SO(n) consists of orthogonal transformations of that space. The key remark to me, in that wiki article, says that in the special case of products with only one factor x of V, i.e. elements of degree one, the target transformation is the reflection in the hyperplane orthogonal to the vector x. Since a hyperplane has two unit length orthogonal vectors, this gives the two preimages of that reflection. Now since more general elements are obtained by taking products of more factors, and the map is a homomorphism, presumably an orhogonal transformation which is a composition of reflections, has as preimage the product of their orthogonal unit vectors. Presumably the equivalence relation on the spin group makes those products of vectors equivalent except for minus, so there are two preimages. Moreover, perhaps every orthogonal transformation in SO(n) is a composition of reflectiopns, not necessarily orthogonal, i.e. consists of orthognal reflections and rotations, and possibly the preimages are products of the unit orthogonal vectors of these hyperp[lanes.

remember, I did not actually learn this stuff before positing this answer, but at least the part quoted from wikipedia is probably right, that the inverse image of a reflection is the two orthogonal unit vectors of the hyperplane fixed by the reflection.