From: mschoene@math.rwth-aachen.de (Martin Schoenert) Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: 25 Nov 94 21:16:49 GMT [I post this for Prof. J. Nueb"user] Dear Laura Helen, You are right that it is not too difficult to find out the minimum number of generators of a given crystallographic space group of three dimensional space in each given case, e.g. by considering finite factorgroups. You are also right that the translation subgroup is a normal subgroup, in fact the unique maximal commutative normal subgroup of the space group, i.e. the factorgroup of it acts faithfully on it. You are also right that the minimal number of generators can be bigger for a proper subgroup than for the whole group, as shown indeed by free groups. However this can also happen with space groups. The translation subgroup being free abelian of course has minimal generator number 3 in all cases. But consider the space group R3, (no 146 of IT): This contains a rotation which by transformation cyclically permutes the three generators (basis vectors) of the translation subgroup and hence can be generated by two elements, namely this element of order three and a translation by one of the basis vectors. On the other hand, there are space groups that need more than three generators. For instance the minimal number of generators of Pmmm (IT no 47) is 6. This can be seen by observing that on one hand it can indeed be generated by 6 elements, namely three (basic) translations and three reflections, and that on the other hand at least 6 elements are needed: A group needs at least as many generators as any of its factor groups, but Pmmm has a factor group which is the direct product of 6 cyclic groups of order 2. This you can either infer from the fact that Pmmm has 63 subgroups of index 2, (see footnote on page 33 of IT) or by considering the factorgroup by the normal subroup generated by twice the basic tranlations, which indeed then must be the direct product of six cyclic groups of order 6. This is indeed the kind of argument with which one can settle each individual case without too much difficulty. Let me just remark that our group theoretic program system GAP can be of help in such considerations since on one hand it offers possibilities to calculate with such factor groups and on the other hand with its last release has tables of all space groups (not only in three but also in four dimensions). I do not think however that the minimal number of generators is of particular interest in understanding the space group. Rather the question of how much of its factorgroup by the tranlation subgroup is represented by finite subgroups (or in group theoretic terms some information on the splitting behaviour of it as an extension) might be considered a measure of its 'complication'. Perhaps it is basically a question of that kind that your 'mathy' coworker is looking at. If (s)he does not mind, I would be interested to hear what (s)he is doing more exactly. With kind regards Joachim Neubueser ============================================================================== From: Laura Helen Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: Sun, 27 Nov 94 20:59:43 -0500 mschoene@math.rwth-aachen.de (Martin Schoenert) writes: >[I post this for Prof. J. Nueb"user] >For instance the minimal number of generators of Pmmm (IT >no 47) is 6. This can be seen by observing that on one hand it can >indeed be generated by 6 elements, namely three (basic) translations >and three reflections, and that on the other hand at least 6 elements >are needed: A group needs at least as many generators as any of its >factor groups, but Pmmm has a factor group which is the direct product >of 6 cyclic groups of order 2. This you can either infer from the fact >that Pmmm has 63 subgroups of index 2, (see footnote on page 33 of IT) >or by considering the factorgroup by the normal subroup generated by >twice the basic tranlations, which indeed then must be the direct >product of six cyclic groups of order 6. Maybe a good factor group to look at would be G/G'. It should be easy to get the number of generators since it's abelian. Laura ============================================================================== From: wpt@tampa.berkeley.edu (Bill Thurston) Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: 26 Nov 1994 17:43:58 GMT In article <3b3ulq$oe2@news2.delphi.com>, LAURAHELEN@DELPHI.COM wrote: >wpt@tampa.berkeley.edu (Bill Thurston) writes: > >>However, there are only a finite number (211, I think) of >>distinct space groups >>( a space group = discrete group of isometries of Euclidean 3-space >>with compact quotient <=Bieberbach=> subgroup containing Z^3 of finite >>index which is its own centralizer) > >I'm not sure what the compact quotient subgroup is here -- the lattice >subgroup? This is a way of ruling out sublattices, perhaps? I omitted too many words above, and your parsing is not what I intended. A 3-D crystallographic group is a group acting on 3-dimensional Euclidean space (ordinary space) in such a way that the quotient space is compact, or equivalently, that there is a bounded fundamental domain for the action. Bieberbach proved a bunch of facts about groups like these, in arbitrary dimension. Among other things, an abstract group is a crystallographic group if and only if it contains a subgroup isomorphic to Z^3 of finite index which is its own centralizer. This would correspond to the maximum group of translations. The word "lattice" is used inconsistently in different related mathematical fields, so I'd prefer not to call the group of translations the lattice subgroup, but I guess that's also what it's called. However, the condition of being its own centralizer is to rule out things like the product of Z^3 with a finite group; this could act as a discrete group only in a higher-dimensional Euclidean space, and the quotient space wouldn't be compact. >> One good way to think about >>a group G such as this is by studying the quotient space or quotient orbifold >>E^3/G. I've actually gone through and worked out the classification from >>this point of view, with John Conway, and figured out what all the >>quotient pictures "look like". About 90% of them are fibered over >>two-dimensional orbifolds; the complication is that they sometimes fiber >>in 2 or 3 different-looking ways. The remainder are related to >>the figure eight knot assigned 3, up to operations of quotienting. > >Does this mean that by looking at the topology of the orbifold, that you >could know a priori that one spacegroup has the same minimal no. of >generators as another? Yes, you would often know this information by looking at the orbifolds. However, I'm wondering if the "minimum number of generators" is the real question, or whether there's something more. >The space groups we're concerned with don't include reflections, since >they're space groups for non-mirror-symmetric protein molecules. There are also some space groups that don't contain reflections but reverse orientation; I suppose that these are also not what your coworker is looking at. They're the hardest ones to analyze, so this would simplify things a lot. >> A technique >>to get a lower bound is to look at a finite quotient group, and find >>the inimum number of generators for the quotient. These groups (as well >>crystallographic groups in any dimension, >>as all "genuine" 3-orbifold groups) are residually finite, meaning that >>they have many finite quotient groups, almost certainly enough quotients >>to establish the correct minimum number of generators. > >I suggested to my co-worker to find a generating set of minimal size for >the factor group: space group /{unit cell translations}, then to see if >that generating set could be used to generate the whole space group, >perhaps by adding translations. Does this seem a reasonable way to do it? > >Is it possible to show that a generating set of minimal size for the space >group induces a generating set of minimal size for the factor group? In general, a minimum generating set for the space group would not yield a minimum generating set for the quotient. For the factor group you suggest above, this would often not be the case, e.g. if the entire group consists of translations, but in lots of other cases as well. I think, though, if you factor out by the subgroup generated by 3 times any translation in the space group, it would be likely to have the right minimum number of generators. >Do you -- or anyone -- know of references on this problem? Maybe it's >all been done already. I don't know the literature very well. >Thanks! > Laura >not a "space groupie" Bill Thurston wpt@math.berkeley.edu ============================================================================== From: steph@escher.mbi.ucla.edu (Stephanie Wukovitz) Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: 27 Nov 1994 06:27:48 GMT Hi, I'm the one with the space group problem...Laura just mentioned that she'd posted it here (I believe I tried posting this last summer but perhaps the distribution wasn't set right or something because I didn't get any replies. Ah well...) In article <3b7s4u$ovj@agate.berkeley.edu>, wpt@tampa.berkeley.edu (Bill Thurston) writes: |> LAURAHELEN@DELPHI.COM wrote: |> >wpt@tampa.berkeley.edu (Bill Thurston) writes: |> Yes, you would often know this information by looking at the orbifolds. |> However, I'm wondering if the "minimum number of generators" is the real |> question, or whether there's something more. I have a silly question...what's an orbifold? Is there a good reference for this? I have taken a year of graduate level algebra and graded it twice, but I don't know that word (perhaps I've lost some memory while drifting from pure math into the realm of the crystallographers). I really *do* want to know minimum number of generators. I'd like to know the minimal number of unique contacts that molecules must make if they are in a crystal with a specified space group and the crystal is to be connected (i.e. there is a path of contacts from every molecule to every other molecule through the contacts). This corresponds to the minimal number of generators for the group. |> There are also some space groups that don't contain reflections but |> reverse orientation; I suppose that these are also not what your coworker |> is looking at. They're the hardest ones to analyze, so this would |> simplify things a lot. We're only concerned with the *biological* space groups, so that rules out mirrors, inversions (what you call reverse orientations, I believe) and glide planes. That gives us 65 space groups to worry about, although I guess it would be nice to know the min gen set for all spgrps. |> >I suggested to my co-worker to find a generating set of minimal size for |> >the factor group: space group /{unit cell translations}, then to see if |> >that generating set could be used to generate the whole space group, |> >perhaps by adding translations. Does this seem a reasonable way to do it? Simply adding translations wouldn't do it. Screw axes generate translations, for instance...also, you can sometimes compose a rotation and a translation in one direction to get translations in other directions (such as a 3-fold rotation and unit translation in P432). |> In general, a minimum generating set for the space group would not yield |> a minimum generating set for the quotient. For the factor group you |> suggest above, this would often not be the case, e.g. if the entire |> group consists of translations, but in lots of other cases as well. |> I think, though, if you factor out by the subgroup generated by |> 3 times any translation in the space group, it would be likely to have |> the right minimum number of generators. Hm. Wouldn't there be a problem with P4_3? The screw axis really does generate one of the translations, but it would not appear to do so if you factored the group this way. Or maybe I'm missing something. Of course every time you have a 4_3 screw axis, there is an enantiomeric space group (with a 4_1 axis) and min gen set is the same for both of these, so maybe there's no problem here? What do you think? Factoring by the subgroup generated by 12 times the translations would do it (I think), but that would be a HUGE group in many cases, not something I'd particularly like to compute with. *Is* there a well-known fast algorithm for finding the min gen set of a finite group? I can of course think of a way to do it, but not a nice way. |> >Thanks! |> > Laura |> >not a "space groupie" Thanks! I wish I had known about this thread sooner... -Stephanie space grouped out ============================================================================== From: LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: 27 Nov 1994 07:55:49 -0000 >>I suggested to my co-worker to find a generating set of minimal size for >>the factor group: space group /{unit cell translations}, then to see if >>that generating set could be used to generate the whole space group, >>perhaps by adding translations. Does this seem a reasonable way to do it? >> >>Is it possible to show that a generating set of minimal size for the space >>group induces a generating set of minimal size for the factor group? What I meant here was the question: if you find a generating set of minimal size for a factor group, if you "lift" it -- I think that's the word -- to the whole space group, can you in this way find a generating set of minimal size for the space group? That is, if I take representatives from cosets in the generating set for the factor group -- maybe more than one representative from a coset, can I be sure of being able to create a generating set of minimal size for the space group. Well, all this may be pretty much settled. Stephanie (who doesn't check her mail too often :) ) found *upper* bounds for the number of generators. And *lower* bounds can be found by looking at finite factor groups. So if those are in agreement, it's settled. Laura ============================================================================== From: steph@escher.mbi.ucla.edu (Stephanie Wukovitz) Newsgroups: sci.math Subject: Re: generators of crystallographic space groups Date: 28 Nov 1994 03:17:24 GMT Hi again...please ignore my comments about the 4_3 screw axes...Laura pointed out to me what was wrong with the picture I drew, and boy was it embarrassing. Now that I have no more objections to factoring by 3{unit translations}, here's another question: Why can't one factor by 2{unit translations}? Or is that OK? I can't think offhand of any examples of problems, but of course that doesn't mean there aren't any. Thanks for all of the help so far... -- Stephanie W. Wukovitz steph@escher.mbi.ucla.edu "...and, here we are...victims of mathematics." -Londo, _Babylon 5_