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commensurable

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commensurable
commensurable, a. (n.) (kəˈmɛnsjʊərəb(ə)l, -ʃər-) [ad. L. commensūrābil-is (Boeth.) having a common measure, f. com- together + mensūrābilis that can be measured, f. mensūrā-re to measure (see -ble), f. mensūra measure, f. mens- ppl. stem of metīri to measure, mete. Also in French (Oresme 14th c.), ... Oxford English Dictionary
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Commensurability (group theory)
The group SL(2,Z) is also commensurable with these free groups. Any two surface groups of genus at least 2 are commensurable with each other. By the relation between covering spaces and the fundamental group, commensurable spaces have commensurable fundamental groups. wikipedia.org
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commeasurable
commeasurable, a. (kəˈmɛʒ(j)ʊərəb(ə)l, -ʒər-) [f. com- + measurable.] = commensurable.1670 Walton Life Donne 42 A commeasurable grief took as full a possession of him as joy had done. 1814 Southey Roderick xviii, Their gather'd multitudes..With more than commeasurable strength Haste to prevent the d... Oxford English Dictionary
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Commensurability
Two concepts or things are commensurable if they are measurable or comparable by a common standard. wikipedia.org
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Are commensurable subgroups contained in a finite overgroup? Let $\Gamma$ and $\Lambda$ be commensurable discrete lattices in a semi-simple Lie group $G$, i.e. the intersection $\Gamma\cap\Lambda$ has finite index in ...
Let me assume that you implicitly mean $G$ to be semisimple and the discrete subgroups to be (arithmetic) lattices. Here's an example where they are not contained in a finite index overgroup. Namely $G=\mathrm{SL}_2(\mathbf{R})$ (you can do something similar in $G=\mathrm{SL}_d(\mathbf{R})$ for $d\g...
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Apotome (mathematics)
In the historical study of mathematics, an apotome is a line segment formed from a longer line segment by breaking it into two parts, one of which is commensurable In this definition, two line segments are said to be "commensurable only in power" when the ratio of their lengths is an irrational number but the ratio wikipedia.org
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bimedial
bimedial, a. (and n.) (baɪˈmiːdɪəl) [f. bi- prefix2 + medial, f. L. medi-us middle.] † a. Algebra. (See quot. 1557.) Obs. b. Geom. The sum of two medial lines; a medial line being the geometric mean between two incommensurable lines, which have commensurable squares.1557 Recorde Whetst. P p iv, The ... Oxford English Dictionary
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Bianchi group
It is well known that every non-cocompact arithmetic Kleinian group is weakly commensurable with a Bianchi group. wikipedia.org
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For the sum to be periodic, $f_1$ and $f_2$ must be commensurable; For the sum to be periodic, $f_1$ and $f_2$ must be commensurable; that is, there most be a number $f_0$ contained in each an integral number of times...
DON'T turn the $f_i$ into decimals. You have $$f_1={2\over2\pi},\qquad f_2={5\over2}$$ You want to know whether there are integers $n_1$ and $n_2$ such that $f_1/n_1=f_2/n_2=f_0$. That would make $f_2/f_1=n_2/n_1$, a rational number. So: is $f_2/f_1$ a rational number?
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171 (number)
This generates 171 moonshine groups within associated with that are principal moduli for different genus zero congruence groups commensurable with the wikipedia.org
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Proving that commensurability is transitive We have that two groups $\Gamma$ and $\Gamma'$ are _commensurable_ if there exist finite index subgroups $G \leq \Gamma$ and $G' \leq \Gamma'$ such that $G \cong G'$. We den...
An idea: $$G\cong G'\implies \,\forall\,H'\le \Gamma'\;\exists\, H\le \Gamma\;\;s.t.\;\;G\cap H\cong G'\cap H'$$ because isomorphic groups have isomorphic subgroups. But then in fact $$G'\cap H'\cong K'\cap H''\;,\;\;\text{for some finite index}\;\ K'\le\Gamma'\;\text{(same argument as above)} $$ an...
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Arithmetic hyperbolic 3-manifold
An arithmetic Kleinian group is any subgroup of which is commensurable to a group derived from a quaternion algebra. They are not cocompact, and any arithmetic Kleinian group which is not commensurable to a conjugate of a Bianchi group is cocompact. wikipedia.org
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Commensurability of Subgroups Let $G$ be some group, and let $S_1$ and $S_2$ be some subgroups. Then $S_1$ and $S_2$ are said to be commensurable iff $|S_1 : S_1 \cap S_2|$ and $|S_2 : S_1 \cap S_2|$ are finite. I am...
To simplify notation, let me write $S_{12}=S_1\cap S_2$, $S_{123}=S_1\cap S_2\cap S_3$, etc. Define $f:S_{12}/S_{123}\to S_2/S_{23}$ by $f(aS_{123})=aS_{23}$. To prove this is well-defined, we must show that if $aS_{123}=bS_{123}$ for $a,b\in S_{12}$ then $aS_{23}=bS_{23}$. But if $aS_{123}=bS_{123}...
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Trace field of a representation
The invariant trace field of Fuchsian groups is stable under taking commensurable groups. In the case where is arithmetic then it is commensurable to the arithmetic group defined by . wikipedia.org
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Free groups of rank $\geq 2$ are commensurable I am reading Bowditch's notes on Geometric Group Theory and I am trying to understand a proof. He proves that if $F_p$, $F_q$ are free groups of rank $p,q\geq 2$, then th...
To describe the covering map $K_{mn} \mapsto K_n$, first map the main circle $R / (mnZ) \subset K_{mn}$ around the main circle $R / nZ \subset K_n$ by a covering map of degree $m$ (taking the point $x + mnZ$ to the point $x + nZ$ for each $x \in R$). Next, the loop of $K_{mn}$ attached to the main c...
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