Talk:Commensurability (mathematics)

Is the usage of the rather strange math expression ${\mathcal {O}}type$ in the text intentional? Terry 20:55, 26 Mar 2005 (UTC)

Link to stacks

There was a link to Stack (category theory). As the latter has been split, the link has been updated accordingly.

However, the context in which the term appears in the article is hardly specific enough or even understandable as it stands. It should be either expanded or dropped. Stca74 10:10, 10 September 2007 (UTC)[reply]

commensurator

What is a commensurator? 82.18.24.252 (talk) 08:43, 1 January 2010 (UTC)[reply]

I know a meaning in group theory, which a google search indicates is rather ubiquitous: the commensurator of a subgroup H of a group G is the set of all elements g in G, such that the intersection of H and its conjugate g^-1Hg has finite index in both groups. Incidently, the commensurator is itself a subgroup of G.

However, I suspect a more general and more concrete meaning to be older. Commensurability in the classical sense always indicates the existence of a common "unit", such that the two entities are integer multiples of that unit. It would be consistent to call succh a common unit a commensurator. (In the case of two commensurable subgroups of a given group, a "commensurator" in this sense would be their intersection, or any other common subgroup with finite indices in both.) JoergenB (talk) 18:43, 11 January 2010 (UTC)[reply]

Classical sense

IMHO, the article initial paragraph now limits itself too much to a special case, expressed in modern terminology. The theory of commensurable and incommensurable ratios predates the modern concept of real number, and does not depend of it. On the contrary, it may be used in order to lay a logically sound foundation for the concept of real numbers themselves. JoergenB (talk) 18:43, 11 January 2010 (UTC)[reply]

Citation needed

For

"Sometimes in fact this relation is called commensurate, and to be commensurable requires only to be conjugate to a commensurate subgroup." —Preceding unsigned comment added by 84.146.253.136 (talk) 10:05, 4 January 2011 (UTC)[reply]