3
$\begingroup$

Hi, I'm looking for a motivation for the names genus and spinor genus of a lattice (and spinor norm of an isometry).

Is there any relation between the genus of a lattice and the genus of an algebraic curve ?

Is the spinor norm used in number theory related to physics ? I mean, if I have an isometry in a quadratic or skew-hermitian space and I calculate its spinor norm... Is there a physical interpretation for that ?

Thanks a lot !!

Improve this question
$\endgroup$
2
  • $\begingroup$ The term spinor just means an element of a certain representation of an orthogonal group (namely a spin representation) (or maybe they have the spin group itself lying around). The group is usually considered as the underlying symmetry of the physics that's going on. Similarly, when a physicist says the word vector, they usually implicitly mean an element of the standard representation of O(3) (or a bigger O(n)) that transforms under the quotient SO(3). This is why they have the term pseudovector: these flip under reflections. $\endgroup$
    – Rob Harron
    Oct 13, 2011 at 14:46
  • $\begingroup$ The genus of a lattice developed from the genus of quadratic forms. See Frei, On the development of the genus of quadratic forms, Ann. Sci. Math. Quebec 3, 5-62 (1979). $\endgroup$
    – Franz Lemmermeyer
    Oct 14, 2011 at 18:19

3 Answers 3

Reset to default
5
$\begingroup$

The word "genus" means "kind," more or less. So no, there is no relation between the genus of a lattice and the genus of an algebraic curve; in both cases the word "genus" just reflects that we like to bundle together sets of objects which have important features in common.

Improve this answer
$\endgroup$
4
$\begingroup$

Given a quadratic space $(V, q)$, and an orthogonal transformation $\sigma \in O(V,q)$, the spinor norm of $\sigma$ is nonzero if and only if $\sigma$ is not in the image of the canonical map $Pin(V,q) \to O(V,q)$. If $\sigma \in SO(V,q)$, we can say the same using the map $Spin(V,q) \to SO(V,q)$. It is a refined way to measure the failure of an element to come from the spin group, and you can view it as a boundary map in Galois cohomology (see, e.g., Wikipedia).

I don't know of a deep connection to physics, but it does come up in the following sense: if $(V,q)$ is an indefinite real space of dimension at least 3, like $\mathbb{R}^{3,1}$, then the pin group only has 2 connected components, while the orthogonal group has 4 - we can reflect in vectors of either positive or negative norm to get P or T type discrete symmetries. The pin group only maps to the subgroup of $O(3,1)$ generated by reflections in positive norm vectors, as these are precisely the transformations with positive spinor norm. The spin group is connected, and maps to the connected group $SO_0(3,1)$, while $SO(3,1)$ contains PT symmetries that reverse orientation of both space and time.

The spinor genus is a version of this that uses additional valuations - two lattices have the same spinor genus if for all completions you can transform the respective base changes by orthogonal transformations with trivial spinor genus.

Improve this answer
$\endgroup$
3
$\begingroup$

Adding to the previous answer: Actually in its original form due to Eichler and Kneser one uses not $Pin$ but the spin group, which is the two fold simply connected cover of the special orthogonal group and has its name from its use in quantum theory.

The term genus was introduced by Gauss (genus in latin, Geschlecht in german). As already remarked, this is just a notion of grouping things together.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.