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Can anyone explain what Iwahori order is? All I know is that it is mentioned here.

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Questions will get better answers, and more upvotes, if you provide a little more background and/or motivation. –  Scott Morrison Nov 8 '09 at 6:29
    
My background info is that someone used it in the answer I linked to and a Google search doesn't give me anything useful –  Casebash Nov 8 '09 at 6:49
    
It would help even more if you could indicate what level (roughly) your mathematical work/studies/interests are. Otherwise people might waste time being overly technical, or pointing to more sophisticated variants that you have no interest in or use for... –  Yemon Choi Nov 8 '09 at 8:21
    
I am completing my second honors level maths subject at Sydney University (roughly equivalent to graduate level in the US) –  Casebash Nov 8 '09 at 8:32
    
Thanks. Well, it seems that Iwahori order and Iwahori subgroups are an order of magnitude harder, or more technical, than your original question about the Jacobson radical - though since this is outside my own "safe zone", I might be wrong here. –  Yemon Choi Nov 8 '09 at 8:57
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3 Answers 3

up vote 2 down vote accepted

To translate Rob's answer above, you take a ring $R$ and ideal $I$, and take the ring of consisting of matrices with coefficients in $R$ such that the image in $R/I$ is upper-triangular. This is a subring of all matrices.

The name "Iwahori order" is borrowed from the name "Iwahori subgroup" which refers to the invertible elements in this ring (usually in the case where R is a valuation ring, and I the valuation ideal).

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I normally see this in terms of an "Iwahori subgroup" (ie the the $\operatorname{GL}(2,A)$ version, versus the $M_2(A)$ version of the question you reference). For a general reductive group over a valuation ring $A$, you can consider the set of group elements that are in the Borel subgroup modulo the valuation ideal $I$. If you pick a larger parabolic subgroup (instead of the Borel) you can analogously define "parahoric" subgroups. These are generalizations of the standard $\Gamma_0(N)$ congruence subgroup of $\operatorname{SL}(2,\mathbb{Z})$. If instead of $\operatorname{GL}(n,A)$, you took $M_n(A)$, you would have a subring of matrices that are upper-triangular modulo $I$.

The Iwahori subgroups have nice properties related to buildings, and are useful as "levels" in the theory of admissible representations of reductive groups over local fields. Is there something specific you're wondering about?

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That used a large number of words I have never even heard of –  Casebash Nov 8 '09 at 6:19
    
That's mathematics! (at least on some days...) –  Scott Morrison Nov 8 '09 at 6:27
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I can give a concrete description of David Savitt's example (following your link). Let $A = \mathbb{C}[\![ t]\!]$, the ring of formal Taylor series with complex coefficients. It has a maximal ideal $\mathfrak{m} = t\mathbb{C}[\![t]\!]$, whose elements are those power series with vanishing constant term. $M$ is then the ring of 2 by 2 matrices whose entries are formal Taylor series with complex coefficients. The Iwahori order is the subring of matrices whose lower left entry lies in the maximal ideal, i.e., the constant term of the lower left entry is zero.

If you reduce the entries modulo $t$, $M$ maps onto the ring of 2 by 2 matrices with complex entries, and the Iwahori order maps onto the subring of upper triangular matrices. Reduction modulo $t$ yields identical kernels for both rings, so you can think of an Iwahori as a pullback of rings. If you consider the corresponding groups of invertible matrices, you get $GL_2(A)$, and the Iwahori subgroup, and reduction modulo $t$ yields $GL_2(\mathbb{C})$ and a Borel (i.e., maximal solvable) subgroup.

This construction generalizes straightforwardly to $n$ by $n$ matrices, where you demand that the constant terms vanish on the lower triangular part of the matrix. The general definition of Iwahori subgroup does not need the general linear group - for any reductive algebraic group, it is defined by the property that reduction modulo the maximal ideal yields a Borel subgroup of the constant group.

In reference to Rob H.'s answer, a Bruhat-Tits building can be viewed as a combinatorial skeleton of a symmetric space. If you are working over a finite field $F$ of order $q$, you can make a (poly-) simplicial complex whose vertices are elements of the quotient $G\big(F(\!(t)\!)\big)\big/G\big(F[\![t]\!]\big)$, and whose higher dimensional simplices are elements of parahoric quotients. For $G = SL_2$, the only parahoric group is the Iwahori (so $SL_2\big(F(\!(t)\!)\big)\big/I$ defines the edge set), and you get an infinite $(q+1)$-regular tree. Subgroups of $G$ (like the torus) let you define additional structures in the building, like chambers, apartments, galleries, etc. The geometry of the building is quite useful for studying the representation theory of $G$ over local fields.

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Lots of ideas here. I'll have to come back to this once I have a better grasp of some of the concepts mentioned here –  Casebash Nov 8 '09 at 21:11
    
Since @AndrewStacey has done some MathJaxification: do you prefer $\operatorname{SL}_2$ or $SL_2$? –  Yemon Choi Jul 10 '13 at 17:00
    
I have a slight preference for the latter, but I don't think either choice will pose a significant obstruction to understanding. –  S. Carnahan Jul 17 '13 at 4:43
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