What is a space group? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-16T15:11:51Z https://physics.stackexchange.com/feeds/question/363475 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/363475 0 What is a space group? Timothy https://physics.stackexchange.com/users/46757 2017-10-17T20:37:22Z 2017-10-18T01:28:08Z <p>I learned about space groups but never saw a definition of one. For example, the Wikipedia article <a href="https://en.wikipedia.org/wiki/Space_group" rel="nofollow noreferrer">Space group</a> discusses space groups but doesn't say what a space group is.</p> https://physics.stackexchange.com/questions/363475/-/363476#363476 1 Answer by Timothy for What is a space group? Timothy https://physics.stackexchange.com/users/46757 2017-10-17T20:37:22Z 2017-10-18T01:28:08Z <p>A subset of R^3 belongs to a space group if and only if it satisfies the following properties.</p> <ul> <li>All its translational symmetry operations can be generated by 3 linearly independent translation operations.</li> <li>The set of all its translational symmetry operations is a normal subgroup of the group of all its symmetry operations under the operation of composition and the quotient group of that subgroup is finite.</li> </ul> <p>Conversely, for any space group, there exists a subset of R^3 satisfying those properties that belongs to it.</p> <p>It's easy to show that for any subset of R^3, the set of all symmetry operations of that set with the operation of composition is a group.</p> <p>For any 2 subsets of of R^3 that belong to a space group, S and T, they belong to the same space group if there exists a bijective linear transfomation R on R^3 such that the function that assigns to each symmetry operation U of S, RUR^-1 is a bijection from the set of all symmtery operations of S to the set of all symmetry operations of T. RUR^-1 is not necessarily an isometry for all isometries U but will always be an isometry when U is a symmetry operation of S. According to this definition, there are 219 space groups. According to the alternate definition where you instead say there exists a bijective linear transformation R whose matrix representation has a determinant greater than 0, there are 230 space groups.</p> <p>We could define a space group differently by defining a subset of R^3 to belong to a space group if there exists a posative real number r such that for all spheres of radius r, there is at least one and only finitely many symmetry operations that move the origin into that sphere. According to that definition, maybe there are some highly counterintuitive space groups that don't fit the my firse definition of a space group and don't have any translational symmetry operations except for the identity transformation. According to Lee Mosher's answer to my question at <a href="https://math.stackexchange.com/questions/779097/does-there-exist-a-highly-counterintuitive-space-group">https://math.stackexchange.com/questions/779097/does-there-exist-a-highly-counterintuitive-space-group</a>, one of the Bierbach Theorems is that no such space group exists and everything that satisfies the second definition of a space group also satisfies the first definition.</p>