Almost but not quite a homomorphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:49:35Z http://mathoverflow.net/feeds/question/14684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism Almost but not quite a homomorphism Vipul Naik 2010-02-08T19:53:20Z 2010-02-09T23:00:22Z <p>I'm interested in general heuristics where, for specific algebraic structures, we introduce new maps that are "almost" homomorphisms (or "almost" isomorphisms) but not quite so. Here are some that I have encountered in group theory (and may also be used in ring theory and commutative/noncommutative algebra):</p> <ol> <li>A "pseudo-homomorphism" (sometimes also called "quasi-homomorphism") which is a set map for groups whose restriction to any abelian subgroup is a homomorphism. In other words, if two elements commute, then the image of the product is the product of the images. General idea: require the composition with certain kinds of injective maps to be homomorphisms.</li> <li>A "1-homomorphism" which is a set map for groups whose restriction to any cyclic subgroup is a homomorphism. General idea: require that the restriction to any subalgebra generated by at most $k$ elements is a homomorphism. Note that for algebras defined using at most 2-ary operations, the only interesting case is $k = 1$.</li> <li>An element map that sends subgroups to subgroups. The induced map on the lattice of subgroups is termed a "projectivity". General idea: Require the map to induce a map on some derivative structure (e.g., the lattice of subalgebras) that looks like it could have come from a homomorphism.</li> </ol> <p>My main interest is from a group theory perspective but I'd also be interested in constructions for other algebraic structures.</p> <p>ADDED LATER: There have been a lot of interesting examples here. My original focus was to look at properties of maps that could be considered, at least in principle, between two arbitrary objects. Preferably something that could be composed to give a new category-of-sorts. But there've been some interesting examples of maps that go to fixed target groups and whose definition uses additional information about the structure of those target groups. These could also be of potential interest, so please feel free to give such examples too.</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14697#14697 Answer by Andrew Stacey for Almost but not quite a homomorphism Andrew Stacey 2010-02-08T20:52:42Z 2010-02-08T20:52:42Z <p>I wouldn't rank this as any sort of answer, but one time when looking at <em>set</em> maps from one group to another I discovered that such things were sometimes called "near rings" and another thing related to them are "composition rings". If you haven't heard those before, it might be worth looking them up. At the very least, your maps will be elements of something like a "near ring" that satisfy some additional property.</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14760#14760 Answer by Boris Bukh for Almost but not quite a homomorphism Boris Bukh 2010-02-09T11:15:01Z 2010-02-09T11:15:01Z <p>Let $G$ and $G'$ be groups. A <em>Freiman homomorphism</em> of order $s$ from a $A\subset G$ to $G'$ is a map $\phi\colon A\to G$ such $$\phi(a_1)\phi(a_2)\cdots\phi(a_s)=\phi(a_1a_2\cdots a_s)$$ for any $s$ (not necessarily distinct) elements $a_1,a_2,\ldots,a_s\in A$. The Freiman homomorphism are the correct notion of equivalence between sets in additive combinatorics. For example, if $A$ and $B$ are Freiman $2$-isomorphic, then $\lvert A+A\rvert=\lvert B+B\rvert$. There are other versions of almost morphisms that arise naturally in combinatorial number theory. For instance, one of the equivalent versions of the polynomial Freiman-Ruzsa conjecture (for $(\mathbb{Z}/2\mathbb{Z})^n$) asserts that if $\phi\colon (\mathbb{Z}/2\mathbb{Z})^n\to (\mathbb{Z}/2\mathbb{Z})^n$ is a map for which $\phi(x+y)-\phi(x)-\phi(y)$ takes only $K$ values, then $\phi$ can be written as $\phi=\phi_0+\psi$ where $\phi_0$ is a genuine linear map, and $\psi$ takes only $K^{O(1)}$ values.</p> <p>Another interesting example of an `almost morphism', this time in non-algebraic setting, are Lipschitz maps between metric spaces. A true morphism of metric spaces is supposed to preserve distances, but there are very few of them. However, Lipschitz embeddings are much more plentiful, and give rise to beautiful geometry, with many applications (notably in theoretical computer science).</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14761#14761 Answer by Yemon Choi for Almost but not quite a homomorphism Yemon Choi 2010-02-09T11:41:07Z 2010-02-09T11:41:07Z <p>Related to Henry Wilton's comments: the following might not quite be what you're looking for, but seems interesting given that quasi-morphisms have been mentioned. I'm doing this from memory so if there's a gap, someone please let me know!</p> <blockquote> <p>Let $E$ be a Hilbert space, $B(E)$ the algebra of all bounded linear operators on $E$. (Even the case $E={\mathbb R}^n$ is of interest.) Fix a small $\epsilon>0$. Then there exists $\delta>0$ with the following property:</p> <p>Let $G$ be an abelian group, and let $f:G \to B(E)$ be a <em>bounded</em> function (i.e. $\sup_{x\in G} \| f(x) \| &lt; \infty$) which satisfies $$\sup_{x,y}\| f(x)f(y) - f(xy) \| \leq \delta.$$ Then there is some representation $\rho: G \to B(E)$ such that $\sup_x \| f(x)- \rho(x)\| \leq \epsilon$.</p> </blockquote> <p>So, less formally, bounded "almost representations" of abelian groups are "near to" genuine representations.</p> <p>I imagine this could be proved by an averaging argument: the way I learned of this result is as a special case of a more general one, in which the word "abelian" is replaced by the word "amenable", and the word "Hilbert" is replaced by "nice reflexive Banach". That in turn is a special case of a general result on almost multiplicative maps between Banach algebras satisfying certain conditions (due to B. E. Johnson).</p> <p>Anyway, sorry this has wandered off track. The point was to say that there are contexts where things which are close to being group homomorphisms $H\to K$, might under a small perturbation be genuine homomorphisms when restricted to a specified abelian subgroup of $H$. However, in general this can't be done so as to work simultaneously for all abelian subgroups of $H$.</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14788#14788 Answer by Gerhard Paseman for Almost but not quite a homomorphism Gerhard Paseman 2010-02-09T17:34:31Z 2010-02-09T17:34:31Z <p>This may be further afield: maps which preserve or promote some relation, while "encoding" the basic operation. I dimly recall groupoid self-maps called cryptomorphisms, which had some property like f(a*b)= p(f(a))*q(f(b)) for p and q some permutations on the set. Also, in classifying Latin squares, there is some notion of isotopy and some other relations that allow a quasigroup table to be related to others through certain maps.</p> <p>There are similar examples where one is concerned with preserving some property like essential arity (e.g. binary operations which depend on both variables), and will insist on maps between structures that share or preserve such a property. Unfortunately all I remember at the moment is some variant of Mazurkiewicz, who along with similarly named people (speaking as an ignorant American) used variants of morphism to help study groupoids and other structures, partly to understand their spectra (number of algebras/operations/essential operations on an underlying set of n elements, n varying over finite numbers).</p> <p>As an example of something I haven't seen but can imagine: an inequality map, where you care that f(A*B) &lt;> f(C*D) if A*B &lt;> C*D .</p> <p>I invite others to add to this post as details occur and memories sharpen.</p> <p>Gerhard "Ask Me About System Design" Paseman, 2010.02.09</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14789#14789 Answer by jef for Almost but not quite a homomorphism jef 2010-02-09T17:39:19Z 2010-02-09T17:39:19Z <p>There is a construction of the real numbers directly from the integers using quasihomomorphisms of Z. <a href="http://www.maths.mq.edu.au/~street/EffR.pdf" rel="nofollow">http://www.maths.mq.edu.au/~street/EffR.pdf</a></p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14793#14793 Answer by MTS for Almost but not quite a homomorphism MTS 2010-02-09T17:54:38Z 2010-02-09T17:54:38Z <p>In operator algebras, there is something called an asymptotic morphism. If $A$ and $B$ are $C^*$-algebras, an asymptotic morphism is a family of maps $T_h : A \to B$ for $0 &lt; h \le 1$, such that for any $a,b \in A$ and $\lambda \in \mathbb{C}$, $$T_h(a + \lambda b) - T_h(a) - \lambda T_h(b) \to 0$$ $$T_h(a^*) - T_h(a)^* \to 0 \$$ $$T_h(ab) - T_h(a)T_h(b) \to 0$$ as $h \to 0$.</p> <p>These are used to construct maps in K-theory which don't come from $*$-homomorphisms.</p> http://mathoverflow.net/questions/14684/almost-but-not-quite-a-homomorphism/14827#14827 Answer by SB for Almost but not quite a homomorphism SB 2010-02-09T22:48:33Z 2010-02-09T23:00:22Z <p>You could also do the obvious generalization of quasimorphisms:</p> <p>Consider groups with a metric and look at maps maps $\phi: G \to H$ such that $d_H(\phi(g_1)\phi(g_2),\phi(g_1g_2))$ is bounded in some sense.</p> <p>For instance Lipschitz maps $\phi: (G,d_G) \to (H,d_H)$ such that $d_H(\phi(g_1)\phi(g_2),\phi(g_1g_2))$ is uniformly bounded, form a category.</p> <p>Homogeneous quasimorphisms (maps $G \to \mathbb{R}$) are actual homomorphisms on abelian subgroups and are invariant under conjugation. By adjusting the above setup, perhaps one could get such features as well.</p>