User andreas thom - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:20:15Z http://mathoverflow.net/feeds/user/8176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60011/trees-in-groups-of-exponential-growth Trees in groups of exponential growth Andreas Thom 2011-03-29T19:16:08Z 2013-05-11T12:58:44Z <blockquote> <p><strong>Question:</strong> Let $G$ be a finitely generated group with exponential growth. Is there a finite generating set $S \subset G$, such that the associated Cayley graph $Cay(G,S)$ contains a binary tree?</p> </blockquote> <p>Some background: </p> <ol> <li><p>The existence of such a tree clearly implies exponential growth. </p></li> <li><p>Kevin Whyte showed in <em>Amenability, Bilipschitz Equivalence, and the Von Neumann Conjecture</em>, Duke Journal of Mathematics 1999, p. 93-112, that such trees exist if $G$ is non-amenable. So the question is only open for amenable groups of exponential growth.</p></li> <li><p>One good reason for such a binary tree to exist is the existence of a free semigroup inside $G$. In fact, if $G$ is solvable, then the existence of such a semigroup is known to be equivalent to exponential growth (and equivalent to being not virtually nilpotent). This is part of some version or extension of the Tits alternative. Grigorchuk constructed an amenable torsion group with exponential growth, which does not contain such a semigroup, but it contains a binary tree.</p></li> </ol> <p><strong>EDIT:</strong> Al Tal pointed out in an answer below that Benjamini and Schramm covered the non-amenable case (this is 2. from above) already in Benjamini and Schramm "<em>Every Graph With A Positive Cheeger Constant Contains A Tree With A Positive Cheeger Constant</em>", GAFA, 1997.</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/46937#46937 Answer by Andreas Thom for Is a map a homotopy equivalence if its suspension is so? Andreas Thom 2010-11-22T12:19:33Z 2013-02-19T18:37:54Z <p>Whitehead's Theorem (it is Corollary 4.33 in Allen Hatcher's <a href="http://www.math.cornell.edu/~hatcher/#ATI" rel="nofollow">book</a>) says that a map between simply connected CW-complexes is a homotopy equivalence if and only if the induced map on homology (with $\mathbb Z$-coefficients) is an isomorphism. If $\Sigma f : \Sigma X \to \Sigma Y$ is a homotopy equivalence, then this is clearly the case, since suspension just shifts dimensions and the spaces are connected, so that there is no problem in dimension $0$.</p> <p>It would be interesting to have an argument which does not use all the machinery that goes into Whitehead's Theorem, since your assumption is rather strong.</p> http://mathoverflow.net/questions/34909/conjugacy-classes-and-reduced-group-c-algebra-of-an-amenable-group Conjugacy classes and reduced group $C^*$-algebra of an amenable group Andreas Thom 2010-08-08T09:30:08Z 2013-02-18T22:16:01Z <p>The reduced $C^*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star_{red} G$. On the other side, for the group ${\mathbb Z}$, separation is clearly possible.</p> <blockquote> <p><strong>Question:</strong> Let $G$ be an amenable group. Does the reduced group $C^\star$-algebra of $G$ support sufficiently many traces to distinguish between conjugacy classes of group elements?</p> </blockquote> <p><strong>EDIT:</strong> The question seems already interesting for $S_{\infty} = \cup_n S_n$. Let's get explicit and pick $g \in S_n$ (for some $n$) and consider the canonical trace $\tau_{g,n}$ which sends every conjugate of $g$ to $1$ and all other elements to zero. (This can be done for any $n' \geq n$ since $S_n \subset S_{n'}$.) The function $\tau_{g,n} \colon {\mathbb C}S_n \to {\mathbb C}$ is a conjugation invariant function and hence, it must be a linear combination of the normalized traces of irreducible representations of $S_n$. </p> <blockquote> <p><strong>Question:</strong> What is the sum of the absolute values of the coefficients that come up in this linear combination of traces? </p> </blockquote> <p>This is (as one can check) the norm of $\tau_{g,n}$, call it $c(g,n)$. So, we see that the compatible family of maps $\tau_{g,n}$ extends from ${\mathbb C}S_{\infty}$ to $C^* S_{\infty}$ if and only if $c(g,n)$ remains bounded.</p> http://mathoverflow.net/questions/34909/conjugacy-classes-and-reduced-group-c-algebra-of-an-amenable-group/122181#122181 Answer by Andreas Thom for Conjugacy classes and reduced group $C^*$-algebra of an amenable group Andreas Thom 2013-02-18T15:24:03Z 2013-02-18T22:16:01Z <p>There is an easy example, namely $SL_3(F)$ where $F$ is the algebraic closure of some finite field. This group does not admit non-trivial characters (a result of Kirillov) and is locally finite, hence amenable. This gives a negative answer to the first question.</p> <p>Sorry, I deleted my previous answer since it contained a mistake.</p> http://mathoverflow.net/questions/121453/mapping-space-between-classifying-spaces/121473#121473 Answer by Andreas Thom for mapping space between classifying spaces Andreas Thom 2013-02-11T14:09:50Z 2013-02-11T14:09:50Z <p>As Neil Strickland points our, this is too difficult to have a good answer in general. In this post, I am only considering discrete groups. A very special case is the following</p> <blockquote> <blockquote> <p><strong>Definition</strong> Let $f : BG \to BK$ be a continuous map. We say that $f$ is a superposition, if for any ${\mathbb Q} K$-module $L$, the induced map on equivariant homology $$H^G_\ast(G,f^\ast(L)) \to H^K_\ast(K,L)$$ is surjective.</p> </blockquote> </blockquote> <p>Examples of superpositions include retractions, maps between oriented manifolds of non-zero degree, maps whose homotopy fibre is a finite complex with non-zero Euler characteristic. In the application below, one needs a much weaker condition which might be checked by hand for particular maps. The following result can be proved:</p> <blockquote> <blockquote> <p><strong>Theorem</strong> Let $BK$ be a finite complex. Let $f \colon BG \to BK$ be a continuous superposition. If $\chi(BK) \neq 0$, then the mapping space $map(BG,BK,f)$ (that is the connected component of $f$) is contractible.</p> </blockquote> </blockquote> <p>This is (a special case of) a consequence of results in [Daniel H. Gottlieb. Covering transformations and universal fibrations. Illinois J. Math., 13:432–437, 1969., Daniel H. Gottlieb. Self coincidence numbers and the fundamental group. J. Fixed Point Theory Appl. 2 (2007), no. 1, 73–83.] and was proved in [Thomas Schick and Andreas Thom. On a conjecture of Gottlieb. Algebr. Geom. Topol. 7 (2007), 779–784.]</p> http://mathoverflow.net/questions/121174/pointwise-ergodic-theorem-and-mean-sojourn-time/121179#121179 Answer by Andreas Thom for pointwise ergodic theorem and mean sojourn time Andreas Thom 2013-02-08T12:04:56Z 2013-02-08T16:55:05Z <p>I can show that $$\int_X f(x) \ d \mu= \liminf_{k \to \infty} \frac{1}{|F_k|} \sum_{g \in F_k} f(g.x)$$ for $f \in L^1(X)$ with $f \geq 0$ and almost all $x \in X$.</p> <p>The conclusion is obvious if $f \in L^{\infty}(X)$, since any such $f$ is a uniform limit of step-functions. Now, set $$f_n(x) = \min \lbrace n,f(x)\rbrace.$$</p> <p>The set of points, where point-wise convergence holds for $f_n$ has measure $1$. Hence, taking the countable intersection of these sets, there is a set of measure $1$, where it holds for all $f_n$ at the same time. I denote the mean of $f$ over $F$ at $x$ by $m(F,x,f)$. Now, for all those $x \in X$, we get:</p> <p>$$g(x):=\liminf_{k \to \infty} m(F_k,x,f) \geq \sup_{n} \lim_{k \to \infty} m(F_k,x,f_n) = \sup_{n} \int_X f_n(x) \ d \mu = \int_X f(x)\ d\mu$$</p> <p>However, $\int_X m(F_k,x,f) d\mu = \int_X f(x) d\mu$ for all $k$ and hence $$\int g(x) \ d\mu = \int_X \liminf_{k \to \infty} m(F_k,x,f) \ d\mu \leq \liminf_{k \to \infty} \int_X m(F_k,x,f) \ d\mu = \int_{X} f(x) \ d\mu$$ by Fatou's lemma. It follows that $g(x)=\int_X f(x)\ d\mu$ almost everywhere.</p> http://mathoverflow.net/questions/121083/centraliser-of-the-complex-conjugation-in-the-absolute-galois-group/121085#121085 Answer by Andreas Thom for Centraliser of the complex conjugation in the absolute Galois group Andreas Thom 2013-02-07T15:44:30Z 2013-02-07T15:44:30Z <p>If some element centralizes the complex conjugation, then it must preserve the real numbers as a set. Now, since any automorphism of the real numbers preserves the set of squares, it must preserve the order; and hence be continuous. Since $\mathbb Q$ is fixed, this implies that the real numbers are fixed pointwise. It follows that any element which centralized the complex conjugation must be the identity or the complex conjugation itself.</p> http://mathoverflow.net/questions/119910/how-similar-different-are-dense-subgroups-of-a-compact-group/120097#120097 Answer by Andreas Thom for How similar/different are dense subgroups of a compact group. Andreas Thom 2013-01-28T10:44:29Z 2013-01-28T10:44:29Z <p>Since you are interested in positive results (rather than counterexamples) in the case when the pro-finite completions of two groups agree, let me mention the following result from</p> <p>Martin R. Bridson and Alan W. Reid, <em>Nilpotent completions of groups, Grothendieck pairs, and four problems of Baumslag</em>, <a href="http://people.maths.ox.ac.uk/bridson/papers/BReid12/" rel="nofollow">http://people.maths.ox.ac.uk/bridson/papers/BReid12/</a></p> <blockquote> <p><strong>Corollary 7.6:</strong> Let $p$ be a prime number. Let $\Gamma$ and $\Lambda$ be finitely presented and residually p-finite groups. If $\Gamma$ and $\Lambda$ have isomorphic pro-p-completions, then there first $\ell^2$-Betti number of $\Gamma$ and $\Lambda$ coincides.</p> </blockquote> <p>The techniques in the proof (and hence the result) are more general.</p> http://mathoverflow.net/questions/36693/nelsons-program-to-show-inconsistency-of-zf Nelson's program to show inconsistency of ZF Andreas Thom 2010-08-25T20:50:03Z 2013-01-25T05:53:32Z <p>At the end of the paper <a href="http://arxiv.org/abs/math/0605779v1" rel="nofollow"><em>Division by three</em></a> by Peter G. Doyle and John H. Conway, the authors say:</p> <p><em>Not that we believe there really are any such things as infinite sets, or that the Zermelo-Fraenkel axioms for set theory are necessarily even consistent. Indeed, we’re somewhat doubtful whether large natural numbers (like $80^{5000}$, or even $2^{200}$) exist in any very real sense, and we’re secretly hoping that Nelson will succeed in his program for proving that the usual axioms of arithmetic—and hence also of set theory—are inconsistent. (See Nelson [E. Nelson. Predicative Arithmetic. Princeton University Press, Princeton, 1986.].) All the more reason, then, for us to stick with methods which, because of their concrete, combinatorial nature, are likely to survive the possible collapse of set theory as we know it today.</em></p> <p>Here are my questions:</p> <p>What is the status of Nelson's program? Are there any obstruction to finding a relatively easy proof of the inconsistency of ZF? Is there anybody seriously working on this?</p> http://mathoverflow.net/questions/48182/representability-of-matroids-over-mathbb-r Representability of matroids over $\mathbb R$ Andreas Thom 2010-12-03T15:12:04Z 2013-01-05T13:14:21Z <p>Let $M$ be a matroid, for example viewed as being given by a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that</p> <p>1) $d(\varnothing)=0$, $d(\lbrace x \rbrace)=1$, for all $x \in X$,</p> <p>2) $A \subset B$ implies $d(A) \leq d(B)$, and</p> <p>3) $d(A \cap B) + d(A \cup B) \leq d(A) + d(B)$ for all $A,B \in P(X)$.</p> <p>A matroid is said to be <em>representable</em> over a field $k$, if there exists a collection of vectors $\lbrace \xi_x \in V \mid x \in X \rbrace$ of some $k$-vectorspace $V$, such that</p> <p>$$d(A) = \dim {\rm span}_k \lbrace \xi_x \mid x \in A \rbrace \quad \forall A \in P(X).$$</p> <p>It is well-known by results of Tutte, that representability of $M$ over $GF(2)$ and representability over all fields is characterized by certain finite lists of excluded minors that $M$ should not contain. At the same time Vámos has shown that there is <em>no</em> such finite list of excluded minors which characterizes representability over $\mathbb R$.</p> <blockquote> <p><strong>Question:</strong> What are sufficient conditions for representability of $M$ over $\mathbb R$?</p> </blockquote> <p>By Tutte's result, $M$ is representable over any field if $M$ does not contain $U_{24}$, $F_7$ and $F^\ast_7$ as minors. Here, $U_{24}$ denotes the matroid of four points on a line, $F_7$ is the Fano plane and $F^\ast_7$ its dual. The question is whether there is a general result, that describes a larger class of matroids which are representable over $\mathbb R$.</p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/96657#96657 Answer by Andreas Thom for Is there an i.c.c. nonamenable simple group that is inner amenable? Andreas Thom 2012-05-11T10:54:38Z 2012-12-26T19:52:57Z <p>The group $G:=SL_{\infty}(\mathbb Q) = \cup_n SL_n(\mathbb Q)$ is a concrete example. It is obviously simple and non-amenable. Let $g_n \in SL_{\infty}(\mathbb Q)$ be the matrix which is $$g_n:= 1_n \oplus \left(\begin{matrix} 0 &amp; 1 \newline -1 &amp; 0 \end{matrix}\right) \oplus 1_{\infty}.$$ and let $$m_{n}(A) := \begin{cases} 1 &amp; g_n \in A \newline 0 &amp; g_n \not \in A \end{cases}.$$ be the finitely additive probability measure associated with $g_n$. Now, for any non-principal ultrafilter $\omega \in \beta \mathbb N \setminus \mathbb N$, $$m(A) := \lim_{n \to \omega} m_n(A) \in [0,1]$$ is a conjugation invariant finitely additive probability measure on $G \setminus {e}$. Conjugation invariance follows since the each element in $G$ commutes with $g_n$ for $n$ large enough.</p> http://mathoverflow.net/questions/114955/fuglede-kadison-determinants-in-l-mathbbf-2/114968#114968 Answer by Andreas Thom for Fuglede-Kadison determinants in $L(\mathbb{F}_2)$ Andreas Thom 2012-11-30T11:38:34Z 2012-12-01T18:30:26Z <p>The spectral measures for self-adjoint elements in $\mathbb C F_2$ are very special. In particular, it is known that non of the elements in $\mathbb C F_2$ has a kernel when acting via the left-regular representation on $\ell^2 F_2$. This was shown by Peter Linnell using index-theoretic methods in</p> <p>Linnell, Peter, <em>Division rings and group von Neumann algebras.</em> Forum Math. 5 (1993), no. 6, 561–576. </p> <p>(It also follows from another of Linnell's papers and the fact that $F_2$ is left-orderable.)</p> <p>It is also known that Novikov-Shubin invariants are always positive. This together implies that $\Delta(T) \neq 0$ if $T\neq 0$. </p> <p>The main advantage is that $\sum_{n} \tau(T^n)z^n$ is an algebraic power series for any $T \in \mathbb C F_2$. This implies the result about Novikov-Shubin invariants (and much more). It was proved in</p> <p>Sauer, Roman, <em>Power series over the group ring of a free group and applications to Novikov-Shubin invariants.</em> High-dimensional manifold topology, 449–468, World Sci. Publ., River Edge, NJ, 2003</p> <p>Starting with the computations in this paper (and using the Stieltjes-Inversion formula), you can also make concrete computations for the determinant of specific elements.</p> http://mathoverflow.net/questions/60441/the-multiplicative-order-of-2-modulo-primes The multiplicative order of 2 modulo primes Andreas Thom 2011-04-03T15:10:33Z 2012-11-21T22:50:34Z <p><a href="http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots" rel="nofollow">Artin's Conjecture</a> says that any positive integer, which is not a square, is a primitive root modulo infinitely many primes. Christopher Hooley gave in</p> <ul> <li>Hooley, Christopher (1967). "<em>On Artin's conjecture.</em>" J. Reine Angew. Math. 225, 209-220.</li> </ul> <p>a proof of this conjecture assuming the Generalized Riemann Hypothesis.</p> <p>Roger Heath-Brown showed (not using the GRH) in</p> <ul> <li>Heath-Brown, D.R. (1986). "<em><a href="http://qjmath.oxfordjournals.org/content/37/1/27.extract" rel="nofollow">Artin's conjecture for primitive roots.</a></em>" Quart. J. Math. Oxford Ser. 37(1), 27-38.</li> </ul> <p>that there are at most two primes for which Artin's Conjecture fails. Nevertheless, it seems to be unknown whether any single specific prime number satisfies the conjecture. In particular, it is unknown if 2 is a primitive root modulo infinitely many primes.</p> <blockquote> <p><strong>Question:</strong> What is known about the multiplicative order of 2 modulo primes?</p> </blockquote> <p>More specifically, can one prove interesting statements of the form: For infinitely many primes $p$, the multiplicative order of 2 is larger than some expression in terms of $p$ (which goes to infinity as $p \to \infty$)?</p> <p>I have to say, that I am not an expert on these kind of questions at all. Given the enormous amount of literature on these questions, I tag this as a reference-request.</p> http://mathoverflow.net/questions/103020/which-polynomials-are-fricke-polynomials/103040#103040 Answer by Andreas Thom for Which polynomials are Fricke polynomials ? Andreas Thom 2012-07-24T19:31:49Z 2012-07-24T19:31:49Z <p>I do not think that there is a complete answer to this question. However, one can give some necessary conditions (which show that any answer must be complicated).</p> <p>One can show that a triple $(x,y,z) \in \mathbb C^3$ comes from traces of matrices in $SU(2)$ (i.e. there exist $u,v \in SU(2)$ such that $(x,y,z)=(t(u),t(v),t(uv))$) if and only if $(x,y,z) \in \mathbb [-2,2]^3$ (this is obviously necessary) and $$x^2 + y^2 + z^2 - xyz \leq 4.$$</p> <p>Hence, a necessary condition on a polynomial $p \in \mathbb Z[X,Y,Z]$ to be a Fricke polynomial is that for all $(x,y,z) \in [-2,2]^3$ we have the implication $$x^2 + y^2 + z^2 - xyz \leq 4 \quad \Rightarrow \quad p(x,y,z) \in [-2,2].$$</p> <p>Indeed, the word evaluated on $u$ and $v$ will give again a matrix in $SU(2)$ and hence its trace is in the interval $[-2,2]$.</p> <p>The semi-algebraic set $$S:=[-2,2]^3 \cap \lbrace (x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 - xyz \leq 4 \rbrace$$ is a spectrahedron (that means it can be defined by a linear matrix inequality) and is called the Elliptope $E_3$.</p> http://mathoverflow.net/questions/97026/kuipers-theorem-via-approximation/101804#101804 Answer by Andreas Thom for Kuiper's theorem via approximation Andreas Thom 2012-07-09T20:01:40Z 2012-07-10T10:47:39Z <p>This is not an answer but too long for a comment.</p> <p>It was shown in</p> <p>Popa, S. and Takesaki,M., <em>The Topological Structure of the Unitary and Automorphism Groups of a Factor</em>, Commun. Math. Phys. 155, 93-101 (1993)</p> <p>that the unitary group $U(R)$ of the hyperfinite $II_1$-factor $R$ is contractible in the strong topology (which is equal to the strong-$*$-topology in this case). </p> <p>Years before, it was shown in</p> <p>Araki, H., Smith, M.-S.B. and Smith, L., <em>On the homotopical significance of the type of von Neumann algebra factors</em>, Commun. Math. Phys. 22, 71-88 (1971)</p> <p>that the first homotopy group of $U(R)$ in the norm toplogy is isomorphic to $(\mathbb R,+)$. Under this isomorphism, the element $\lambda \in [0,1]$ is represented by the loop $$[0,1] \ni t \mapsto \exp(2\pi i\cdot t p) \in U(R),$$ where $p\in R$ is some projection of trace $\lambda$.</p> <p>This shows that the homotopy type depends heavily on the chosen topology and there is no reason (at least in general) that one should find an easy approximation argument.</p> http://mathoverflow.net/questions/101746/commutator-subgroup-group-theory/101798#101798 Answer by Andreas Thom for Commutator Subgroup - Group Theory Andreas Thom 2012-07-09T19:17:33Z 2012-07-09T19:17:33Z <p>The commutator subgroup of the free group $\langle a,b \rangle$ is freely generated by the set $$\lbrace [a^n,b^m] \mid n,m \in \mathbb Z, nm \neq 0 \rbrace.$$</p> http://mathoverflow.net/questions/100331/on-infinite-dimensional-unitary-representations-of-kazhdan-groups/100335#100335 Answer by Andreas Thom for On infinite-dimensional unitary representations of Kazhdan groups Andreas Thom 2012-06-22T09:16:34Z 2012-06-22T09:16:34Z <p>The situation in infinite dimensions is different for Kazhdan groups.</p> <p>If $\Gamma$ contains a non-abelian free group, then the left-regular representation $\lambda \colon \Gamma \to U(\ell^2 \Gamma)$ admits a deformation $\lambda_t$ (for $t \in [0,1]$ say), such that $\lambda_t$ is a unitary representation, $$\sup_{g \in \Gamma} \|\lambda_t(g) - \lambda_s(g)\| \leq |s-t|,$$ $\lambda_0=\lambda$ and $\lambda_t$ not equivalent to $\lambda$ for $t \neq 0$. This is a very strong condition on a deformation; typically one does not require a <em>uniform</em> deformation and usually also using the strong operator topology.</p> <p>An explicit construction of such a deformation for the free group itself goes back to Pytlik-Swarc in</p> <p>T. Pytlik and R. Swarc, <em>An analytic family of uniformly bounded representations of tree groups</em>, Acta Math. 157(3-4):289-309, 1986.</p> <p>The idea is then to induce this deformation to $\Gamma$ and check that the continuity is preserved (this is easy because of the strength of the assumption) and that the resulting representations remain inequivalent.</p> <p>A similar behaviour is conjectured to hold for all non-amenable groups. These ideas have appeared in</p> <p>Marc Burger,Narutaka Ozawa, and Andreas Thom, <em>On Ulam stability</em>, to appear in Israel J. Math., <a href="http://arxiv.org/abs/1010.0565" rel="nofollow">http://arxiv.org/abs/1010.0565</a></p> http://mathoverflow.net/questions/100245/no-injective-groups-with-more-than-one-element/100253#100253 Answer by Andreas Thom for No injective groups with more than one element? Andreas Thom 2012-06-21T15:20:54Z 2012-06-21T15:33:11Z <p>Let $G$ be a non-trivial injective group and $g \in G$ non-trivial. From category theory, $G \times G$ is injective as well. But, now $G \times G$ embeds into a group $H$, where $(g,e)$ and $(e,g)$ are conjugate using an HNN-extension, or just embedding into the group $H:=(G \times G) \rtimes \mathbb Z/2\mathbb Z$. Since $(g,e)$ and $(e,g)$ are not conjugate in $G \times G$, $H$ cannot split back to $G \times G$. This is a contradiction.</p> http://mathoverflow.net/questions/100155/second-homotopy-group-of-cayley-complex/100210#100210 Answer by Andreas Thom for Second homotopy group of Cayley complex Andreas Thom 2012-06-21T06:17:57Z 2012-06-21T10:02:47Z <p>If $\langle X,R \rangle$ is a finite presentation of a group $G$, then there exists an exact sequence of $\mathbb ZG$-modules $$0 \to \pi_2(Z) \to \mathbb Z G^{\oplus R} \to \mathbb Z G^{\oplus X} \to \mathbb Z G \to \mathbb Z \to 0,$$ where $Z$ is the presentation $2$-complex of the presentation above. If one knows in addition that $G$ is of type $FP_3$, then $\pi_2(Z)$ must be finitely generated as a $\mathbb Z G$-module. It is well-known that hyperbolic groups are $FP_{\infty}$, using the Rips complex.</p> <p>Any example of a finitely presented group which is not of type $FP_3$ gives a counterexample, i.e. $\pi_2$ is not finitely generated. Brady constructed a subgroup of a hyperbolic group with this property in</p> <p>Brady, N. <em>Branched Coverings of Cubical Complexes and Subgroups of Hyperbolic Groups</em> J. London Math. Soc. (1999) 60(2): 461-480.</p> <p>Much earlier, Stallings gave an example where the third homology is not finitely generated as a module over the group ring of a finitely presented group.</p> <p>Stallings, J. <em>A finitely presented group whose 3-dimensional integral homology is not finitely generated</em>. Amer. J. Math. 85 (1963), 541–543.</p> http://mathoverflow.net/questions/100042/additive-integer-valued-functions-on-the-module-category/100095#100095 Answer by Andreas Thom for Additive integer-valued functions on the module category Andreas Thom 2012-06-20T08:19:29Z 2012-06-20T08:34:28Z <p>This has been studied for group rings to some extend. It is a theorem of Wolfgang Lück that a homomorphism $\varphi \colon G_0(\mathbb Z \Gamma) \to \mathbb R$ can be constructed with the property $\varphi([\mathbb Z \Gamma]) = 1$ if $\Gamma$ is amenable. Moreover, such a homomorphism cannot exist if $\Gamma$ contains a non-abelian free group. It is conjectured that the existence is a characterization of amenability. Moreover, if $\Gamma$ is torsionfree and amenable, the conjecture is that the range of $\varphi$ is $\mathbb Z$, this is called Atiyah's conjecture.</p> <p>Sometimes, maps like the one you consider exist on subcategories of the category of f.g. modules. An easy example is the category of f.g. abelian groups $A$, so that $A \otimes_{\mathbb Z} \mathbb Q=0$, i.e. torsion groups. Then, the map $A \mapsto \log |A|$ is additive. </p> <p>There is also a version for f.g. modules over the group ring of an amenable group. It can be shown that assinging to a f.g. module $M$ over $\mathbb Z \Gamma$ ($\Gamma$ is amenable here) the entropy of the natural $\Gamma$-action on the Pontryagin dual of $M$ is additive. This is Yuzvinskii's Additivity Formula as proved by Hanfeng Li in</p> <p>Hanfeng Li, <em>Compact group automorphisms, addition formulas and Fuglede-Kadison determinants</em>, Ann. of Math. (2) 176 (2012), no. 1, 303--347.</p> <p>If $\Gamma$ is finite, then this entropy is essentially the logarithm of the cardinality of $M$. For infinite $\Gamma$, this invariant of $M$ is equal to the so-called $\ell^2$-Torsion of $M$, if it can be defined. For $\Gamma = \mathbb Z^d$, this invariant is related to the Mahler measure and of number theoretic significance.</p> http://mathoverflow.net/questions/99239/non-identity-operator-on-a-separable-hilbert-space/99244#99244 Answer by Andreas Thom for non-Identity operator on a separable Hilbert space Andreas Thom 2012-06-10T15:56:08Z 2012-06-10T15:56:08Z <p>The answer is yes, this is true (assuming that the Hilbert space is complex).</p> <p>If $\langle \xi,A\xi \rangle = \sigma$ for some $\sigma \in \mathbb C$ and all $\xi$, then $B:=A - \bar \sigma 1_H$ has the property that $\langle \xi,B\xi \rangle =0$ for all $\xi \in H$. We need to show $B=0$. Let $\xi \in H$ be arbitrary and consider the vector $\lambda \xi + \mu B\xi$ for some $\lambda,\mu \in \mathbb C$.</p> <p>We get: $$0=\langle \lambda \xi + \mu B \xi, \lambda B\xi + \mu B^2 \xi \rangle = \lambda \bar\mu \langle \xi,B^2 \xi \rangle + \mu \bar\lambda \|B \xi\|^2$$ for all complex $\lambda$ and $\mu$. Taking $\lambda = \mu = 1$, we see $\|B\xi\|^2 = - \langle \xi,B^2 \xi \rangle$. Taking $\lambda=1, \mu=i$, we get $\|B\xi\|^2 = \langle \xi,B^2 \xi \rangle$. This shows $B \xi =0$.</p> http://mathoverflow.net/questions/98594/regarding-cayley-graphs-of-property-t-groups/98727#98727 Answer by Andreas Thom for Regarding Cayley Graphs of Property (T) Groups Andreas Thom 2012-06-03T17:18:55Z 2012-06-03T17:18:55Z <p>If Kazhdan's property (T) is reflected in the structure of the Cayley graph, then not in a very <em>geometric</em> way. </p> <p>Steve Gersten (that is what I read in the book by B. Bekka, P. de la Harpe and A. Valette) was the first who found that Kazhdan's property (T) is not invariant under quasi-isometry. The reason is not complicated. If a central extension $$1 \to \mathbb Z \to \Gamma \to \Lambda$$ is obtained from a bounded cocycle $c \colon \Lambda \times \Lambda \to \mathbb Z$, then $\Gamma$ is quasi-isometric to $\Lambda \times \mathbb Z$. This situation arises for $\Lambda$ a cocompact lattice in a simple real Lie group with infinite fundamental group; such as $SU(2,2)$. Then, $\Gamma$ is the inverse image of $\Lambda$ in the universal covering. In this situation, one actually obtains (for suitable generating sets) a bi-Lipschitz equivalence of Cayley graphs.</p> <p>Now, $\Lambda \times \mathbb Z$ does not have Kazhdan's property (T) since it surjects onto $\mathbb Z$, but $\Gamma$ has Kazhdan's property (T) inheriting it from the universal cover of $SU(2,2)$.</p> http://mathoverflow.net/questions/98241/simultaneous-decomposition-of-three-projectors/98255#98255 Answer by Andreas Thom for Simultaneous decomposition of three projectors Andreas Thom 2012-05-29T07:42:54Z 2012-05-29T08:47:08Z <p>One way to put these questions is to ask for a classification of finite-dimensional <code>$*$</code>-representations of a universal $C^*$-algebra. In the first case it is the universal <code>$C^*$</code>-algebra generated by two projections, i.e. the unital <code>$C^*$</code>-free product of $\mathbb C^2$ with itself. You observed that the maximal dimension of a irreducible <code>$*$</code>-representation of this algebra is two.</p> <p>In the second case, you are considering the unital $C^*$-free product of $\mathbb C^3$ with $\mathbb C^2$. The two projections $\Pi_1$ and $\Pi_2$ correspond to $(1,0,0)$ and $(1,1,0)$ in $\mathbb C^3$. This <code>$C^*$</code>-algebra is isomorphic to the universal group $C^*$-algebra of $PSL_2(\mathbb Z)$, being the free product of $\mathbb Z_3$ and $\mathbb Z_2$. It is well-known that the unitary representation theory of $PSL_2(\mathbb Z)$ (which is equivalent to the $*$-representation theory of its universal group <code>$C^*$</code>-algebra) is <em>wild</em> and there is no bound on the dimension of irreducible finite-dimensional representations. </p> http://mathoverflow.net/questions/96782/vector-balancing-problem/97726#97726 Answer by Andreas Thom for vector balancing problem Andreas Thom 2012-05-23T08:06:47Z 2012-05-23T08:47:38Z <p>This is mainly a comment. My first guess would be that if it is true, then a random subset of density $1/2$ works. A result in this direction is Lemma 3.2 in </p> <p>M. Rudelson, <em>Contact points of convex bodies</em>, Israel Journal of Mathematics, 1997, Volume 101, Number 1, Pages 93-124.</p> <p>It says:</p> <blockquote> <p><strong>Lemma</strong> :Let $x_1,...,x_k$ be vectors in $\mathbb R^n$, $\varepsilon_1,...,\varepsilon_k$ be independent Bernoulli variables, taking values $1,-1$ with probability $1/2$. Then $${\mathbb E} \left\| \sum_{i=1}^k \varepsilon_i |x_i\rangle {\langle x_i} | \right\|\leq C \log(n) \sqrt{\log(k)} \max_i \|x_i\| \left\| \sum_{i=1}^k |x_i\rangle \langle x_i| \right\|^{1/2}$$ for some absolute constant C.</p> </blockquote> <p>In your case, this says that a random set of density $1/2$ solves the easier problem where $\delta$ may depend on $n$ and $k$. At the same time it proves much more, since there is even concentration around $1/2$. More precisely, if $\sum_{i=1}^k |x_i\rangle \langle x_i| =1$ and $\|x_i\|&lt; \delta$, then</p> <p>$${\mathbb E} \left\|\frac12 - \sum_{i=1}^k \eta_i |x_i\rangle {\langle x_i} | \right\|\leq C/2 \log(n) \sqrt{\log(k)} \cdot \delta,$$ where $\eta_i$ are independent Bernoulli with values in ${0,1}$. </p> <p>Since $n \leq k \delta^2$ (looking at the trace), an easy calculation shows that $\delta$ only depends on $k$. At the same time, it seems to me that the problem is getting easier if $k$ is larger, but I cannot substantiate this claim. Remark 3.3 in the same paper shows that the inequality cannot be improved to become independent of $n$ and $k$. This somehow shows that choosing a random subset is too naive; at least when one studies the expected value of the norm as in the inequality above.</p> http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits/96256#96256 Answer by Andreas Thom for Are semi-direct products categorical limits? Andreas Thom 2012-05-07T19:35:33Z 2012-05-07T19:35:33Z <p>There is (another ?) description of the crossed product in categorical terms. </p> <p>Let ${\rm Mor}(Gp)$ be the category whose objects are homomorphisms of groups and morphisms are commutative diagrams. Let $C$ be the category of "groups acting on groups" whose objects are pairs of groups $(H,G)$ together with a homomorphism $H \to {\rm Aut}(G)$. Morphisms in this category are equivariant homomorphisms.</p> <p>Now, there is a natural forgetful functor $T \colon {\rm Mor}(Gp) \to C$ which sends $H \to G$ to the pair $(H,G)$ with the homomorphism $H \to {\rm Aut}(G)$ given by conjugation. Now, almost by definition, the crossed product is the left-adjoint of this forgetful functor. Indeed, the left adjoint is easily seen to map $(H,G)$ with $H \to {\rm Aut}(G)$ to the inclusion $H \to G \rtimes H$.</p> <p>Being a left-adjoint, the "crossed product" maps colimits to colimits.</p> http://mathoverflow.net/questions/96225/extension-of-conjugacy-problem Extension of conjugacy problem Andreas Thom 2012-05-07T15:09:52Z 2012-05-07T19:26:23Z <p>Let $F = \langle a,b \rangle$ be a non-abelian free group.</p> <blockquote> <p><strong>Question:</strong> Is there an algorithm that takes as input $x,y,z \in F$ and answers the question whether $x$ is a product of conjugates of $y$ and $z$, i.e. whether there exists $g,h \in F$ with $$x = gyg^{-1} \cdot hzh^{-1} ?$$</p> </blockquote> <p>It is obvious that there exists an algorithm that enumerates all products of conjugates of $y$ and $z$. What is missing is a certificate that $x$ cannot be written in this form. Sometimes (like in the case of the word problem or the membership problem for finitely generated subgroups), this part is done by looking at the finite quotients and finding one finite quotient so that $x$ is not in the product of the conjugacy classes. Hence, one starting point would be to ask if the product of conjugacy classes is closed in the pro-finite topology. This was conjectured by Stallings to hold even in the pro-p topology and disproved (in the pro-p case) by Howie in</p> <p>The p-adic Topology on a Free Group: A Counterexample, Math. Z. 187,25-27 (1984)</p> <p>It might hold in the pro-finite topology, but this seems to be difficult. Anyhow, I am just looking for an algorithm, maybe this can be done without looking at the finite quotients.</p> http://mathoverflow.net/questions/93677/amenable-groups-of-deficiency-1 Amenable groups of deficiency $1$ Andreas Thom 2012-04-10T16:16:08Z 2012-04-21T15:19:28Z <p>Let $G=\langle X;R\rangle$ be a finitely presented group. The rank of $G$ is defined to be the size of smallest generating set of $G$. The deficiency ${\rm def}(G)$ of $G$ is defined to be the maximum of $|X| - |R|$ over all finite presentations $G = \langle X;R \rangle$.</p> <p>The deficiency of an amenable group can be at most $1$. One (maybe the only?) way to see this is to note that there is a Morse inequality for $\ell^2$-homology $$1-{\rm def}(G) \geq b_0^{(2)}(G) - b_1^{(2)}(G) + b_2^{(2)}(G),$$ where $b_i^{(2)}(G)$ denotes the $i$-th $\ell^2$-Betti number of $G$. Cheeger and Gromov showed that an amenable group satisfies $b_i^{(2)}(G)=0$ for $i \geq 1$. This implies in particular that ${\rm def}(G) \leq 1$ for $G$ amenable.</p> <p>Now, apart from $\mathbb Z$ and Baumslag-Solitar groups $BS(1,n) = \langle a,b; a^nba^{-1}b^{-1} \rangle$, I do not know of any amenable groups which realize ${\rm def}(G) =1$. In particular, I do not know any examples of rank $\geq 3$.</p> <blockquote> <p><strong>Question:</strong> Does every amenable group of deficiency $1$ have rank $\leq 2$.</p> </blockquote> <p>It can be shown that any amenable group with ${\rm def}(G)=1$ must have cohomological dimension $\leq 2$, which puts severe restrictions on $G$. What else is known about amenable groups with deficiency $1$?</p> http://mathoverflow.net/questions/66206/fourier-transforms-of-characteristic-functions Fourier transforms of characteristic functions Andreas Thom 2011-05-27T17:54:49Z 2012-04-20T14:14:28Z <p>I am wondering how badly summable the Fourier transform of the characteristic function of a measurable subset of $S^1$ can be.</p> <blockquote> <p><strong>Question:</strong> Let $\alpha \colon \mathbb N \to [1,\infty)$ be a monotone increasing function with $\lim_{n \to \infty} \alpha(n) = \infty$. Is there a measurable subset $E \subset S^1$, such that $$\sum_{n \in \mathbb Z} | \widehat \chi_E(n)|^2 \cdot \alpha(|n|) = \infty \ ?$$ Here, $\widehat \chi(n)$ are the usual moments $$\widehat \chi_E(n):= \int_E z^n \ dz.$$</p> </blockquote> <p>The only example I know is the Fourier transform of the characteristic function of an interval, which grows like $1/n$. On the other hand, one can easily see that the growth cannot be better than $1/n$ (something like $1/n^{1 + \varepsilon}$), since $\ell^1 \mathbb Z \subset C(S^1)$.</p> <p>More concretely: </p> <blockquote> <p><strong>Question:</strong> Can anyone compute the growth of the Fourier transform of the characteristic function of something like a Cantor set of non-zero measure?</p> </blockquote> <p>Again, more abstractly:</p> <blockquote> <p><strong>Question:</strong> What can be said about the growth of the Fourier transform of the characteristic function of a generic subset of $S^1$?</p> </blockquote> http://mathoverflow.net/questions/93677/amenable-groups-of-deficiency-1/93939#93939 Answer by Andreas Thom for Amenable groups of deficiency $1$ Andreas Thom 2012-04-13T07:50:30Z 2012-04-13T19:02:31Z <p>Inspired by John Wilson's result which was mentioned by Mark in his answer (which he has deleted by now), I can answer my own question now for elementary amenable groups.</p> <p>Let us first argue that any amenable $G$ with ${\rm def}(G)=1$ has cohomological dimension $\leq 2$. This follows from an argument that I learned from [1]. Consider the presentation $2$-complex $X$ (with one 0-cell) of a presentation which realizes the deficiency. The cellular chain complex of the universal covering looks like $$0 \to {\mathbb Z}G^n \stackrel{d}\to {\mathbb Z}G^{n+1} \to {\mathbb Z}G \to 0.$$</p> <p>The kernel of $d$ is $\pi_2(X)$ and the only thing to show is that $\pi_2(X)=0$. Then $X$ is aspherical and ${\rm cd}(G)\leq 2$. Now, the $\ell^2$-homology of $X$ is computed by $$0 \to ({\ell^2}G)^n \stackrel{d}\to ({\ell^2}G)^{n+1} \to {\ell^2}G \to 0.$$ Now, the zeroeth and first $\ell^2$-Betti number of $X$ agrees with the $\ell^2$-Betti number of $G$ and has to vanish by Cheeger-Gromov since $G$ is amenable. This implies that $d$ must be injective on $(\ell^2 G)^n$. Hence, it is injective on $(\mathbb Z G)^n$. We conclude that $\pi_2(X)=0$.</p> <p>Let us come to the main part of the argument. Jonathan Hillman has defined the Hirsch length $h(G)$ for any elementary amenable group $G$ (Theorem 1 in [3]) and shown that it is bounded above by the cohomological dimension of $G$, see Lemma 2 in [3]. Now, Theorem 2 of [3] implies $G/T$ is solvable where $T$ is the maximal locally finite normal subgroup of $G$. However, since the cohomological dimension is finite, $T$ is trivial and we conclude that $G$ itself is solvable. Now, Theorem 5 from [2] says that every solvable group of cohomological dimension $\leq 2$ must be solvable Baumslag-Solitar, $\mathbb Z^2$ or $\mathbb Z$.</p> <p>[1] Jon Berrick and Jonathan Hillman "The Whitehead conjecture and $L^2$-Betti numbers", Guido's Book of Conjectures, Monographies de L'Enseignement Mathématique, L'Enseignement Mathé matique, (2008), 35–37. </p> <p>[2] Dion Gildenhuys, "Classification of soluble groups of cohomological dimension two" Math. Z. 166, 1 (1979), 21-25.</p> <p>[3] Jonathan Hillman, "Elementary amenable groups and 4-manifolds with Euler characteristic 0" J. Austral. Math. Soc. (Series A) 50 (1991), 160-170.</p> http://mathoverflow.net/questions/93668/generalization-of-finitely-generated-finitely-presented-modules/93674#93674 Answer by Andreas Thom for Generalization of finitely generated, finitely presented modules? Andreas Thom 2012-04-10T15:48:52Z 2012-04-10T15:48:52Z <p>All these notions have been defined and studied long time ago. Serre called a module type $FL_n$ if it is finitely $n$-presented in your terminology. Type $FL_\infty$ and type $FL$ is used for finitely $\infty$-presented and finitely $\omega$-presented. They are studied a lot for group rings ($R$ does not need to be commutative) and show up in the definition of $G$-theory for general rings. Modules of type $FP_\infty$ are sometimes also called pseudo-coherent, a name/definition that goes back to SGA 6, I.2.9, see for Example 7.1.4 in Chuck Weibel's book on Algebraic K-theory.</p> <p>A good starting point might be K.S. Brown's book "Cohomology of groups", Chapter VIII is about finiteness conditions.</p> http://mathoverflow.net/questions/125853/unbounded-representations-of-groups Comment by Andreas Thom Andreas Thom 2013-03-28T21:26:13Z 2013-03-28T21:26:13Z wccanard, the question is whether there exists <i>some</i> representation with <i>some</i> proper cocycle. http://mathoverflow.net/questions/125853/unbounded-representations-of-groups Comment by Andreas Thom Andreas Thom 2013-03-28T21:14:53Z 2013-03-28T21:14:53Z What do you mean precisely by an unbounded representation? One fixed dense domain for all group elements? http://mathoverflow.net/questions/122982/sum-of-two-self-adjoint-unbounded-operators Comment by Andreas Thom Andreas Thom 2013-02-26T14:48:33Z 2013-02-26T14:48:33Z Usually, $T+S$ is not even densely defined. http://mathoverflow.net/questions/122432/famous-mathematicians-with-severe-mental-health-issues Comment by Andreas Thom Andreas Thom 2013-02-20T18:21:29Z 2013-02-20T18:21:29Z The difference might lie in the difference between &quot;having mental health issues&quot; and &quot;receiving a PhD late in life&quot;. http://mathoverflow.net/questions/122353/not-measure-equivalent-icc-groups-g-and-h-but-lg-cong-lh Comment by Andreas Thom Andreas Thom 2013-02-19T22:07:44Z 2013-02-19T22:07:44Z The way you stated it, you can take any group $G$ which admits a finite index subgroup $H$ and so that $L(G)$ has some restrictions on the fundamental group. I believe such groups have been constructed. If you ask only for stable equivalence of the von Neumann algebras, I do not think that there are examples. http://mathoverflow.net/questions/121255/ring-theoretic-version-of-a-matrix-problem Comment by Andreas Thom Andreas Thom 2013-02-09T12:29:05Z 2013-02-09T12:29:05Z ... they study <b>linear combinations</b> (not just sums) of orthogonal matrices. http://mathoverflow.net/questions/120802/are-all-endomorphisms-of-c-just-power-maps Comment by Andreas Thom Andreas Thom 2013-02-09T10:44:56Z 2013-02-09T10:44:56Z All continuous endomorphisms are of the form $z\mapsto z^n\bar z^m$, for some $n,m \in \mathbb Z$. http://mathoverflow.net/questions/121083/centraliser-of-the-complex-conjugation-in-the-absolute-galois-group/121085#121085 Comment by Andreas Thom Andreas Thom 2013-02-07T21:54:38Z 2013-02-07T21:54:38Z What really makes a difference here is that $\mathbb Q$ is dense in the real algebraic numbers (as well as the real numbers) in the order-sense. If this is not the case (i.e. if the real closed field is more general and contains infinitesimal elements etc.), one has to find indeed different arguments and use Artin-Schreier theory. http://mathoverflow.net/questions/121083/centraliser-of-the-complex-conjugation-in-the-absolute-galois-group/121085#121085 Comment by Andreas Thom Andreas Thom 2013-02-07T21:46:30Z 2013-02-07T21:46:30Z You do not need to extend (if you do not want to), it is still just the same argument. If the real algebraic numbers have only one ordering, then any automorphism must preserve it; and then any automorphism must fix any real algebraic number, since any real number is determined by the set of rational numbers below it (no theory needed for this). It is true that I was secretly thinking about the complex numbers (in fact I misread the question), but it does not make any difference. http://mathoverflow.net/questions/121083/centraliser-of-the-complex-conjugation-in-the-absolute-galois-group/121085#121085 Comment by Andreas Thom Andreas Thom 2013-02-07T19:22:36Z 2013-02-07T19:22:36Z The argument is just the same. An element is non-negative in the ordering if and only if it is a square - this is obviously invariant under automorphisms. No sophisticated theory is needed at this point. http://mathoverflow.net/questions/108045/a-product-on-the-square-roots-of-unit-matrix Comment by Andreas Thom Andreas Thom 2012-09-25T12:25:15Z 2012-09-25T12:25:15Z Note that $A+B$ need not be invertible. http://mathoverflow.net/questions/100331/on-infinite-dimensional-unitary-representations-of-kazhdan-groups/100335#100335 Comment by Andreas Thom Andreas Thom 2012-08-11T08:39:39Z 2012-08-11T08:39:39Z Alain, you are right. What I wrote is relevant for another notion of rigidity; where one is interested in uniform estimates. http://mathoverflow.net/questions/103020/which-polynomials-are-fricke-polynomials/103040#103040 Comment by Andreas Thom Andreas Thom 2012-07-25T08:22:31Z 2012-07-25T08:22:31Z I think your new (or initial) question is really interesting. Why don't you ask this as a separate question? http://mathoverflow.net/questions/103020/which-polynomials-are-fricke-polynomials/103038#103038 Comment by Andreas Thom Andreas Thom 2012-07-24T18:58:05Z 2012-07-24T18:58:05Z Fricke showed (I think) that every polynomial function on $SL_2 \times SL_2$ which is conjugation invariant, is a polynomial in $X,Y$ and $Z$ as above. This applies to traces of words. The question which such polynomials are coming from words in the free group is more complicated. I do not think that this subset is structured in any obvious way. http://mathoverflow.net/questions/102328/ultraweak-closure-inside-a-closed-ball Comment by Andreas Thom Andreas Thom 2012-07-16T05:14:39Z 2012-07-16T05:14:39Z The equality does not hold in general. Take $S$ to be unit sphere in $H$. Then $\bar S=B_1$, whereas $S \cap B_r = \varnothing$ for $r&lt;1$. Maybe you want $S$ convex?