Understanding "infinite" relations in groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:54:23Z http://mathoverflow.net/feeds/question/58555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58555/understanding-infinite-relations-in-groups Understanding "infinite" relations in groups ARupinski 2011-03-15T17:51:57Z 2011-03-19T18:23:43Z <p>Consider the matrices $A = \frac{1}{5}\begin{pmatrix}5&amp;0&amp;0\\ 2&amp;2&amp;1\\ 2&amp;1&amp;2\end{pmatrix}$, $B = \frac{1}{5}\begin{pmatrix}2&amp;2&amp;1\\ 0&amp;5&amp;0\\ 1&amp;2&amp;2\end{pmatrix}$, and $C = \frac{1}{5}\begin{pmatrix}2&amp;1&amp;2\\ 1&amp;2&amp;2\\ 0&amp;0&amp;5\end{pmatrix}$.</p> <p>The group $G\subset GL_3(\mathbb{C})$ they generate is free of rank 3. However, one also has:</p> <p>$A^\infty := \lim_{k\rightarrow\infty}A^k = \begin{pmatrix}1&amp;0&amp;0\\ 1&amp;0&amp;0\\ 1&amp;0&amp;0\\ \end{pmatrix}$, and similarly $B^\infty = \begin{pmatrix}0&amp;1&amp;0\\ 0&amp;1&amp;0\\ 0&amp;1&amp;0\\ \end{pmatrix}$ and $C^\infty = \begin{pmatrix}0&amp;0&amp;1\\ 0&amp;0&amp;1\\ 0&amp;0&amp;1\\ \end{pmatrix}$.</p> <p>This leads to three "infinite" relations among $A$,$B$, and $C:$</p> <p>$A^\infty B = B^\infty A$</p> <p>$B^\infty C = C^\infty B$</p> <p>$C^\infty A = A^\infty C$</p> <p>Of course these are not true group relations since they involve infinite products and furthermore these infinite products are singular, hence do not actually lie in the group generated by $A$, $B$, and $C$, but rather are relations among elements of the "boundary" of this group (in an appropriate sense of the word boundary).</p> <blockquote> <p><b>Question 1:</b> I am wondering if anyone has studied such infinite/boundary relations on groups, and in particular if they have any usefulness in understanding any properties of the original group.</p> <p><b>Question 2:</b> Suppose $H$ is another free group of rank 3 in $GL_3(\mathbb{C})$. $H$ is of course isomorphic to $G$, but may not have any boundary relations. Is there a way to use the infinite relations to define a "stronger" notion of isomorphism which says that $G$ and $H$ are not isomorphic since they do not have the same infinite relations? <b>Edit:</b> Upon further thought, this latter question is more of a question about the particular representation of $G$ chosen, so perhaps it is not so interesting? Or maybe something aside from the usual stuff about isomorphism of representations might be relevant?</p> </blockquote> <p><b>Motivation:</b> Awhile back I was studying harmonic functions on the Sierpinski Gasket; these three matrices arise naturally in the evaluation of such functions. In this context the infinite relations above are equivalent to the statement that such functions are well-defined at each point of the Gasket. The relations above were occasionally useful for proving various facts, but at the time I never bothered to consider them in the context of group relations.</p> <p>More generally they can arise in studying subgroups of $GL_n(R)$ for $R$ a commutative unital ring. In this case the "boundary" of the subgroup may be a subset of the singular elements of $Mat_n(R)$ and one can ask questions about relations between elements of this "boundary".</p> http://mathoverflow.net/questions/58555/understanding-infinite-relations-in-groups/58930#58930 Answer by Greg Kuperberg for Understanding "infinite" relations in groups Greg Kuperberg 2011-03-19T17:19:12Z 2011-03-19T18:23:43Z <p>This answer is partly an expansion of the comments so far. What you are looking for is a type of decoration on groups that (1) lets you evaluate infinite products, and (2) also creates extra points that are the value of some or all of those products. There is one fairly inevitable answer to (1): You should consider topological groups or at least topological spaces, since taking limits in a topology is the main way to extend finite products to infinite products. But for (2), a topology isn't good enough. You get more information from a metric than from a topology, because a metric space has a completion. On the other hand, a metric isn't entirely satisfactory either, among other reasons because it has extra, non-canonical data. A good alternative to a metric is a <a href="http://en.wikipedia.org/wiki/Uniform_space" rel="nofollow">uniformity</a>. A uniform space also has a completion (in which all Cauchy nets converge) and every metric space is canonically a uniform space.</p> <p>In fact, every well-behaved topological group has a canonical associated uniformity, because you can use the neighborhoods of the identity to make uniform neighborhoods. (I am a little hazy on how exactly the topological group should be well-behaved. It should be sufficient for it to be Hausdorff and for the left and right uniformities to agree.) Thus every such topological group has a completion which is also a topological group, and which then has infinite products which are group elements. As Mark Sapir suggests, one such uniformity is the profinite uniformity on a residually finite group.</p> <p>At the other end, I think that a word-hyperbolic group $G$ has a uniformity defined by Gromov whose completion points are called the Gromov boundary. However, multiplication is not uniformly continuous in this construction; rather you only have that the uniformity is left-invariant. Thus, the completion is not a topological group, but it does have a left action of $G$. You can only define right-infinite products, and you can only multiply them by group elements on the left. Still, you can do that much, which still lets you consider many interesting infinite relations.</p> <p>Your example of groups of matrices is modeled by a uniformity of intermediate quality. The uniformity of the additive Lie group structure on $M_3(\mathbb{C})$ is not the same as the uniformity of the multiplicative Lie group structure on $GL_3(\mathbb{C})$, even though the topologies are the same. Multiplication extends to the completion, but inversion does not. (Toy model: The function $1/x$ on $\mathbb{R}^\times$ is not uniformly continuous in the standard metric on $\mathbb{R}$ and obviously does not extend to $0$.) The completion of $GL_3(\mathbb{C})$ is obviously $M_3(\mathbb{C})$, and any subgroup has an inherited uniformity and a completion which is then a semigroup. So an answer to your second question is that a discrete group can have many different semigroup-completed uniformities, some of which come from matrix representations.</p>