User bs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:50:36Z http://mathoverflow.net/feeds/user/6451 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131196/regularity-of-eigenfunctions-of-schrodinger-operator/131201#131201 Answer by BS for regularity of eigenfunctions of Schrödinger Operator BS 2013-05-20T09:33:35Z 2013-05-20T09:33:35Z <p>If the first (lowest) eigenfunction $f_0$ is smooth, then $V$ is smooth. Indeed, assuming $M$ connected, it is a classical fact that $f_0$ doesn't vanish (it is the first case of Courant's nodal theorem for instance), and obviously $V=\lambda_0 +\Delta f_0/f_0$.</p> <p>With boundary and Neumann condition, the same argument applies, and with Dirichlet condition, $V$ is at least smooth in the interior.</p> http://mathoverflow.net/questions/130062/concrete-examples-of-noncongruence-arithmetic-subgroups-of-sl2-r/130186#130186 Answer by BS for Concrete examples of noncongruence, arithmetic subgroups of SL(2,R) BS 2013-05-09T16:03:39Z 2013-05-09T18:02:13Z <p>If you want to go really concrete, take any connected finite $3$-valent graph in the $2$-sphere with disks as components of the complement. </p> <p>This yields a conjugacy class of genus $0$ finite index torsion free subgroups of $\Gamma(1)=PSL(2,\mathbb{Z})$. In fact the index is $6n$ if the graph has $2n$ vertices (hence $3n$ edges and $n+2$ faces). ADDED: Indeed, since $\Gamma(1)\simeq Z/2*Z/3$, there is a transitive action of $\Gamma(1)$ on oriented edges of the graph, flipping orientation for $Z/2$, and rotating around source vertex for $Z/3$. The conjugacy class of stabilizers subgroups determines the graph.</p> <p>But there are clearly infinitely many such graphs (up to oriented homeomorphism), whereas there are only finitely many congruence subgroups of genus $0$ in $\Gamma(1)$ (about $33$ conjugacy classes of torsion free ones, with maximum index $60$, according to <a href="http://www.uncg.edu/mat/faculty/pauli/congruence/csg0.html" rel="nofollow">this table</a>).</p> http://mathoverflow.net/questions/130171/fourier-transform-of-a-bounded-function/130174#130174 Answer by BS for Fourier transform of a bounded function BS 2013-05-09T14:36:44Z 2013-05-09T14:36:44Z <p>The answer is no : Fourier transform of the signum function is a constant times the distribution $\lim_{\epsilon\to 0} 1_{|x|>\epsilon}/x$, of order $1$, also known as the <a href="http://en.wikipedia.org/wiki/Cauchy_principal_value" rel="nofollow">principal value</a> of $1/x$. Another keyword is <a href="http://en.wikipedia.org/wiki/Hilbert_transform" rel="nofollow">Hilbert transform</a>.</p> http://mathoverflow.net/questions/108033/double-a-manifold-with-boundary/108041#108041 Answer by BS for Double a manifold with boundary BS 2012-09-25T10:29:43Z 2012-09-25T10:29:43Z <p>You can give $D(M)$ a $C^k$ structure, $k=0,\ldots,\infty,\omega$ if $M$ has one. </p> <p>Take a $C^k$ function $f$ on $M$ that is $1$ on the boundary with positive normal derivative, and $0\leq f&lt;1$in the interior. </p> <p>Then you can take $D(M)=\{(x,t)\in M\times\mathbb{R}: f(x)+t^2=1\}$.</p> <p>Do you want $D(M)$ to be a <em>riemannian</em> $C^k$ manifold ? With isometric copies of $(M,g)$ this will not be possible in general for $k>2$, but sometimes it will, for instance in the locally symmetric case.</p> http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation/106821#106821 Answer by BS for Simultaneous diophantine approximation BS 2012-09-10T15:23:08Z 2012-09-10T15:47:06Z <p>This is true, and known as the <a href="http://en.wikipedia.org/wiki/Kronecker%27s_theorem" rel="nofollow">Kronecker Theorem</a> on diophantine approximation.</p> http://mathoverflow.net/questions/106623/vector-space-structure-on-velocity-space-of-manifold/106666#106666 Answer by BS for Vector space structure on velocity space of manifold BS 2012-09-08T12:37:06Z 2012-09-09T09:43:07Z <p>I assume you ask for a "natural" vector space structure, one for which the natural group action of $r$-jets of diffeomorphisms of $M$ fixing $x$ is linear, and that the vector structure is smooth in standard coordinates. </p> <p>Then the answer is negative for all $r>1$ (for $r=1$ it is of course positive, the space is $(T_x M)^k$).</p> <p>For the proof, let $\phi:(\mathbb{R}^n,0)\to(\mathbb{R}^n,0)$ be a germ of diffeomorphism of the form $\phi(x)=x+h(x)$ with $h\neq 0$ homogeneous of degree $r$, and $f:(\mathbb{R}^k,0)\to(\mathbb{R}^n,0)$.</p> <p>Then for $k\lt r$, $$D^k(\phi\circ f)(0)=D^kf(0),$$ and $$D^r(\phi\circ f)(0)=D^rf(0)+D^rh(Df(0),...,Df(0))$$.</p> <p>Hence the action of $J_0^r\phi$ on $J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0$ is tangent to the identity at the jet of the zero map $\mathbf{0}$, but is not the identity.</p> <p>Now assume for contradiction that there is a "natural" vector space structure. Since the $\mathbf{0}$ is the only jet fixed by all $r$-jets of diffeomorphisms fixing $0$, it is the only possible zero of a natural vector space structure.</p> <p>Now, by assumption the action of $J_0^r\phi$ is linearizable by a smooth ($C^1$ is enough) diffeomorphism $$\psi: (J_0^r(\mathbb{R}^k,\mathbb{R}^n)_0,\mathbf{0}) \to (\mathbb{R}^N,0),$$ but the only possible candidate for the linear map $\psi\circ J_0^r\phi\circ\psi^{-1}$ from $(\mathbb{R}^N,0)$ to itself is its derivative at 0, i.e. the identity, a contradiction.</p> <p><strong>PS</strong>: the natural structure on jets is a bit complicated. There is a sequence of fiber bundles $$T_k^r M\to T^{r-1}_k M\to...\to T^1_k M\to M,$$ where the rightmost map is a vector bundle, but the others are <em>affine</em> bundles under the vector bundles $P_k^r M \simeq S^r(\mathbb{R}^k)^*\otimes TM$ of jets in $T_k^r M$ which map to constant jets in $T^{r-1}_k M$. </p> <p>To see that the $P_k^r M$ are vector bundles, one may observe that a $r$-jet of diffeomorphism $J_x^r\phi$ of $M$ acts on $(P_k^r M)_x$ only through $J_x^1\phi$.</p> <p>It is remarkable that the choice of an <a href="http://en.wikipedia.org/wiki/Affine_connection" rel="nofollow">affine connection</a> on $M$, which may be viewed as a splitting of $$0\to P_1^2 M\to T_1^2 M\to T^1_1M=TM\to 0,$$ is sufficient to put a vector bundle structure on all jet bundles, by using covariant derivatives, see this wikipedia <a href="http://en.wikipedia.org/wiki/Jet_bundle" rel="nofollow">article</a> and its references for more details, notably on the important <em>Cartan distribution</em> on jet bundles.</p> http://mathoverflow.net/questions/106448/topologies-on-the-field-of-rationals/106454#106454 Answer by BS for Topologies on the field of rationals BS 2012-09-05T17:43:34Z 2012-09-06T14:06:03Z <p>According to <a href="http://www.encyclopediaofmath.org/index.php/Topological_field" rel="nofollow">this link</a>, there are as many field topologies on $\mathbb{Q}$ as there are subsets of $\mathbb{R}$, so I doubt there is a classification. A reference seems to be Wieslaw's book "topological fields" (it doesn't seem to be on google books, unfortunately).</p> <p>PS: note that there is only one non-Hausdorff field (or ring) topology on a field (the two open sets one) since the closure of 0 is an ideal, and the others are completely regular, as any Hausdorff group topology. Also there is only continuum many metrizable topologies on $\mathbb{Q}$, so most of the topologies referred to in the above link are quite pathological, non first countable for instance.</p> <p>PPS: searching "number of field topologies" in MathSciNet returns the following references</p> <p>Podewski, Klaus-Peter The number of field topologies on countable fields. Proc. Amer. Math. Soc. 39 (1973), 33–38. </p> <p>Kiltinen, John O. On the number of field topologies on an infinite field. Proc. Amer. Math. Soc. 40 (1973), 30–36. </p> <p>and they are both freely accessible (thanks AMS!) <a href="http://www.ams.org/journals/proc/1973-039-01/S0002-9939-1973-0311633-9/home.html" rel="nofollow">here</a> and <a href="http://www.ams.org/journals/proc/1973-040-01/S0002-9939-1973-0318118-4/home.html" rel="nofollow">here</a>.</p> <p>Concerning the proof for a countable field $K$, Podewski manages to define a continuum $\mathcal{G}$ of (metrizable) field topologies on $K$ such that the suprema of any two distinct susbsets of $\mathcal{G}$ are distinct field topologies.</p> <p>The details are somewhat complicated, but the idea is quite natural. Suffice it to say that the metrizable field topologies on $K$ are parameterized --- via fundamental sequences of neighbourhoods of $0$ --- by chains $G$ in a partially ordered set $P$ of "conditions" (is it forcing in disguise?), which specify for a finite number of elements of $K$ wether they belong or not to the $n$-th neighborhood in the sequence.</p> <p>Strangely, for uncountable fields $K$ the proof is easier, using valuations instead of chains of conditions, and the fact that the transcendence degree of $K$ over the prime subfield is the cardinality of $K$.</p> http://mathoverflow.net/questions/106216/without-choice-can-every-homomorphism-from-a-profinite-group-to-a-finite-group-b/106232#106232 Answer by BS for Without choice, can every homomorphism from a profinite group to a finite group be continuous? BS 2012-09-03T10:52:44Z 2012-09-03T10:52:44Z <p>Saharon Shelah (elaborating on Solovay) <a href="http://en.wikipedia.org/wiki/Solovay_model#CITEREFShelah1984" rel="nofollow">proved</a> that it is equiconsistent with ZF that there exists a model of ZF+DC (DC=<a href="http://en.wikipedia.org/wiki/Axiom_of_dependent_choice" rel="nofollow">dependent choice,</a> ) in which all sets of reals have the <a href="http://en.wikipedia.org/wiki/Property_of_Baire" rel="nofollow">Baire property</a>. And indeed all subsets of any Polish space have this property.</p> <p>Now if $G$ is a second countable (=metrizable) profinite group, it is a Polish space (indeed a Cantor), and if $H$ is a non-open finite index subgroup, $H$ cannot have the Baire property, by a standard argument (it would have to be meager, which is absurd).</p> <p>Hence it is consistent with ZF(+DC) that there is no discontinuous homomorphism from a <em>second countable</em> profinite group to a finite group.</p> <p>On the other hand, I don't know if one can reduce the general case to the second countable one, e.g. by finding a countable intersection of open normal subgroups inside $H$.</p> http://mathoverflow.net/questions/103287/a-existence-problem-of-p-q-metric/103329#103329 Answer by BS for A Existence Problem of (p,q) metric BS 2012-07-27T16:42:17Z 2012-07-30T16:06:49Z <p>The criterion for existence of a $(p,q)$ metric is (assuming $p+q=dim X$) that the tangent bundle splits as a direct sum of two subbundles of dimensions $p$ and $q$. </p> <p>EDIT : I doubt there is an easy algebraic topology criterion in general, as characteristic classes [EDIT after Lennart Meier's comment: other than Euler class, which cannot help if $p,q>1$] are <em>stable</em> invariants of vector bundles. At least, one has necessary conditions, such as : the total Stiefel-Whitney and Ponttryagin (Chern of complexified bundle) class of $TX$ is the product of two (inhomogeneous) classes of degrees at most $p$ and $q$. Perhaps real $K$-theory (and its operations) gives better necessary conditions (although still not sufficient in general). But it's not easy to compute.</p> http://mathoverflow.net/questions/103365/equivalence-between-statements-of-hodge-conjecture/103385#103385 Answer by BS for Equivalence between statements of Hodge conjecture BS 2012-07-28T14:05:47Z 2012-07-28T14:05:47Z <p>The rotation number in question has to do with the behaviour of a differential form $\omega$ on the manifold $X$ under rotation $e^{i\theta}$ of tangent vectors : a complex $k$-form $\omega$ has rotation number $p-q$ iff $$\omega(e^{i\theta}v_1,\dots,e^{i\theta}v_k)=e^{i(p-q)\theta}\omega(v_1,\dots,v_k)$$ (with $p+q=k$, of course). </p> <p>This should explain the terminology, although it is not much in use today. Such a form is called a $(p,q)$-form, their space is denoted $\Omega^{p,q}(X)$, and the subspace of $H^{p+q}(X,\mathbb{C})$ (de Rham cohomology) of classes having a (closed) $(p,q)$-form representative is denoted $H^{p,q}(X)$. </p> <p>Hodge theory then gives a decomposition $H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X)$. </p> <p>Now a topological cycle $C$ (or its class) of (real) codimension $2p$ has a <em>Poincaré dual</em> de Rham cohomology class $[\omega]$, for a closed $2p$ form $\omega$, such that intersection of cycles of <em>dimension</em> $2p$ with $C$ coincides with integration of $\omega$.</p> <p>"The rotation number of $C$ is zero" means that one can choose $\omega$ of type $(p,p)$.</p> <p>On the other hand, a cohomology class in $H^{2p}(X,\mathbb{Q})$ is said to be Hodge if under the Hodge decomposition $$H^{2p}(X,\mathbb{Q})\otimes\mathbb{C}\simeq \bigoplus_{i=-p}^{p} H^{p-i,p+i}(X)$$ it belongs to $H^{p,p}(X)$.</p> <p>I hope this clarifies the equivalence of the two statements you found.</p> http://mathoverflow.net/questions/102876/is-every-connected-regular-space-having-more-than-one-point-uncountable/102878#102878 Answer by BS for Is every connected regular space having more than one point uncountable? BS 2012-07-22T15:41:40Z 2012-07-22T15:41:40Z <p>Google is your friend : "connected regular space" returns </p> <p><a href="http://topospaces.subwiki.org/wiki/Connected_regular_space" rel="nofollow">http://topospaces.subwiki.org/wiki/Connected_regular_space</a></p> <p>The answer is yes, the proof is by contradiction, using that Lindelöf and regular implies normal (even paracompact) .</p> http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers/102225#102225 Answer by BS for Proving the existence of good covers BS 2012-07-14T12:33:31Z 2012-07-14T12:33:31Z <p>I think you can obtain a good cover of $C^2$ manifold (compact or not) from the charts/atlas definition and a little bit of topology (locally finite atlas and a relatively compact "shrinking" of it).</p> <p>The very simple idea (akin to that in Vitali's answer) is that under a $C^2$ diffeomorphism between open subsets of euclidean $n$-space, the (pre-)image of a sufficiently small ball centered at a point will be convex, as soon as the curvature of its boundary "dominates" the second derivative of the diffeomorphism (or its inverse).</p> <p>In formulas, if $\phi$ is the diffeomorphism, this boils down to the fact that the $C^2$ function $x\mapsto |\phi(x)-\phi(x_0)|^2$ has a positive definite hessian at $x_0$, hence is convex near $x_0$. </p> <p>With a little more care, I think you can still conclude if $\phi$ is only $C^{1+Lip}$. </p> http://mathoverflow.net/questions/101158/constructing-a-simplicial-set-homology-equivalent-to-a-given-cw-complex/101571#101571 Answer by BS for Constructing a simplicial set homology-equivalent to a given CW complex BS 2012-07-07T11:23:21Z 2012-07-07T11:23:21Z <p>If you only want to compute homology of a finite presentation complex $X$ of a group $G$ (as I understand from comments), you can do it by (integer) linear algebra only.</p> <p>Let generators be $x_1,\dots,x_m$, with relations $r_1,\dots,r_n$. </p> <p>Then $H_1(X)=G_{ab}$ is the cokernel of the $m\times n$ integer matrix $A$ of degrees $\deg_{x_i} r_j$ (view as morphism $\mathbb{Z}^n\to \mathbb{Z}^m$), and $H_2(X)$ is the kernel of $A$ (in $\mathbb{Z}^n$).</p> <p>Note that the computation of $H_2$ of the <em>group</em> $G$ from a finite presentation is algorithmically infeasible in general, as shown by C. Gordon in the eighties (see <a href="http://mathoverflow.net/questions/73153/is-there-an-algorithm-for-computing-schur-multiplier/73155#73155" rel="nofollow">this answer</a>). </p> <p>I like to see this problem as akin to that of deciding finite sets of tiles that tile the plane, as it asks for computing all tilings (or at least a "generating" set of tilings) of the 2-sphere by copies of topological disks boundary-decorated with relations (well, maybe introducing some non-reduced relator words). But I don't know if this idea leads to a proof (Gordon's proof is different). Recall that deciding finite sets of tiles that tile the plane is also <a href="http://en.wikipedia.org/wiki/Wang_tile" rel="nofollow">algorithmically infeasible</a>.</p> <p>When possible, this computes $H_2(G)$ of the group as $H_2(X)/h(\pi_2(X))$, where $h$ is Hurewicz homomorphism.</p> http://mathoverflow.net/questions/98992/connected-components-of-the-boundary-of-an-open-subset/100478#100478 Answer by BS for Connected components of the boundary of an open subset BS 2012-06-23T18:03:12Z 2012-06-24T13:53:29Z <p>Let $F=f^{-1}(0)\subset \mathbb{R}^n$. </p> <p>It is a closed subset, which disconnects $\mathbb{R}^n$, and all connected components of the complement are unbounded.</p> <p>The question amounts to see if the connected component of $\infty$ in $\widehat{F}=F\cup\infty \subset S^n$ could be reduced to $\infty$. </p> <p>If it were the case, there would be a decreasing sequence $V_i$ of neighbourhoods of $\infty$ with intersection only this point and $Fr(V_i)$ disjoint from $F$ (in a compact space, <em>quasicomponents</em> are the same as connected components). </p> <p>But then $K_i=F- V_i$ is compact and does not disconnect $\mathbb{R}^n$, otherwise there would be a bounded component of the complement, since it would also disconnect $S^n$ ($n>1$ is used here). But this easily implies that $\mathbb{R}^n-F$ would have a bounded component, contrary to the assumptions.</p> <p>Hence $F$ is the union of a sequence of <em>disjoint</em> non-disconnecting compact sets $L_i=K_i\setminus K_{i-1}$ going to infinity.</p> <p>Now observe that $L_i$ doesn't disconnect any open connected subset $U$ containing it, by <em>excision</em>. More precisely this gives an isomorphism of relative homology groups $$H_1(U_i,U_i-L_i)\simeq H_1(\mathbb{R}^n,\mathbb{R}^n-L_i)\simeq \widetilde{H}_0(\mathbb{R}^n-L_i)=0$$, where the second isomorphism comes from the homology exact sequence of the pair $(\mathbb{R}^n,\mathbb{R}^n-L_i)$. Then the homology exact sequence for $(U_i,U_i-L_i)$ gives $\widetilde{H}_0(U_i-L_i)=0$ for a connected $U_i$, as claimed.</p> <p>It remains to take disjoint connected $U_i$'s containing $L_i$'s, with union $U$ and then $$H_1(U,U-F)\simeq \bigoplus_i H_1(U_i,U_i-L_i) =0,$$ a contradiction since this is also $$H_1(\mathbb{R}^n,\mathbb{R}^n-F)\simeq \widetilde{H}_0(\mathbb{R}^n-F)\neq 0\ .$$</p> http://mathoverflow.net/questions/100451/lagrangian-kleinian-bottles/100458#100458 Answer by BS for Lagrangian Kleinian bottles BS 2012-06-23T13:44:28Z 2012-06-23T18:10:38Z <p>Just googling "Lagrangian klein bottles" returns the following :</p> <p>Stefan Nemirovski, Lagrangian Klein Bottles in R2n Geometric And Functional Analysis Volume 19, Number 3 (2009), 902-909</p> <p>the freely accessible arxiv version is at <a href="http://arxiv.org/abs/0712.1760" rel="nofollow">http://arxiv.org/abs/0712.1760</a> abstract : "It is shown that the n-dimensional Klein bottle admits a Lagrangian embedding into R^{2n} if and only if n is odd." </p> <p>There was a previous result by the same author about the $n=2$ case, but it had a flaw initially, and was completely rewritten. See <a href="http://arxiv.org/abs/math/0106122" rel="nofollow">http://arxiv.org/abs/math/0106122</a></p> <p>There is also another proof of this result by V. Schevchishin <a href="http://front.math.ucdavis.edu/0707.2085" rel="nofollow">http://front.math.ucdavis.edu/0707.2085</a></p> http://mathoverflow.net/questions/100224/which-immersed-plane-curves-bound-an-immersed-disc/100230#100230 Answer by BS for Which immersed plane curves bound an immersed disc? BS 2012-06-21T11:37:07Z 2012-06-21T12:01:44Z <p>The answer was given by Samuel Blank in his Brandeis 1967 phd dissertation, on which Poenaru gave a <a href="http://www.numdam.org/item?id=SB_1966-1968__10__473_0" rel="nofollow">Bourbaki seminar</a>.</p> <p>Then <a href="http://www.sciencedirect.com/science/article/pii/092577219290019O" rel="nofollow">Peter Shor and C. J. Van Wyk</a> gave a polynomial time algorithm to decide if there is an extension.</p> <p>EDIT: Blank's method already gave a polynomial algorithm, but with an exponent too large to make it practical, which is needed for applications (for instance to integrated circuit design). </p> <p>The answer is in term of existence of a chain of "reductions" of a certain kind for the cyclically reduced word in the free group $F_n$ on $n$ generators determined by the immersion, where $n$ is the number of bounded components of the complement of the curve (assumed to have only transverse double points). </p> http://mathoverflow.net/questions/61546/explicit-constructions-of-kg-2/100149#100149 Answer by BS for Explicit constructions of K(G,2)? BS 2012-06-20T16:59:37Z 2012-06-20T16:59:37Z <p>As an explicit construction (not obviously as a CW complex one, but I think it's one, though), I would suggest the <em>Dold-Thom construction</em>. Unfortunately I can't find a link to the precise construction I have in mind (can't reach wikipedia right now!?). </p> <p>Simply put, $K(G,2)$ is obtained as a component of the space $G[S^2]$ of (finite) formal combination of points on $S^2$ with $G$ coefficients, with a topology such that e.g. $ax+by\to (a +b)z$ when $x,y\to z$ in $S^2$ and $a,b$ in $G$. </p> <p>Then as $K(G,2)$ you can take the component where the sum of all coefficients is $0$ ("neutral configurations of particles"). This is an abelian topological group.</p> <p>I think this is due to Dusa McDuff, Configuration spaces of positive and negative particles, Topology 14 (1975), 91-107, elaborating on a slighltly different construction by Dold and Thom. The original construction was an infinite symmetric product with of $S^2$, meaning the topological monoid of combinations with nonnegative coefficients, and a base point $*$ identified to $0$.</p> http://mathoverflow.net/questions/99758/convexity-of-a-specific-semialgebraic-set/100018#100018 Answer by BS for Convexity of a specific semialgebraic set BS 2012-06-19T15:55:33Z 2012-06-19T17:20:27Z <p>Your set is indeed a convex cone.</p> <p>Since it is a cone, it suffices to show that the $m=1$ section is convex. </p> <p>But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.</p> <p>Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.</p> <p>EDIT : in fact any set of $(x,y)$ defined by an inequality $S(y)-A(x)^*A(x) \succ 0$, with $S(y)$ $n\times n$ symmetric and linear in $y$, and $A(x)$ $n\times d$ and linear in $x$, is convex and moreover <em>defined by a Linear Matrix Inequality</em>.</p> <p>Indeed this is equivalent to </p> <p>$$\left(\begin{matrix} S(y) &amp; A(x)^* \\ A(x) &amp; I \end{matrix}\right) \succ 0$$</p> <p>as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.</p> http://mathoverflow.net/questions/93926/is-rsu-4-cong-rso-6/93963#93963 Answer by BS for Is $R(su_{4})\cong R(so_{6})$? BS 2012-04-13T15:05:40Z 2012-04-13T15:05:40Z <p>It is a standard fact that $Spin(6)$ and $SU(4)$ (hence $so_{6}$ and $su_{4}$) are isomorphic. An easy way to see this is to observe that $SU(4)$ is simply connected and acts on the exterior square of $\mathbb{C}^4$, preserving a (complex) quadratic form (coming form exterior squaring and $det=1$) and a real structure (a conjugate-linear involution, hodge star followed by conjugation) for which the quadratic form is definite. Then the equality of dimensions (and easy calculation of the kernel) shows that $SU(4)\to SO(6)$ is a double covering.</p> <p>From a higher standpoint, you may also observe that the Dynkin diagrams for $D_3$ and $A_3$ are the same, hence the maximal compact subgroups of the corresponding simply connected complex Lie groups are isomorphic.</p> http://mathoverflow.net/questions/92136/kazhdans-property-t-for-kahler-surfaces/92167#92167 Answer by BS for Kazhdan's property T for Kahler surfaces BS 2012-03-25T13:13:41Z 2012-03-25T13:13:41Z <p>According to this <a href="http://library.msri.org/books/Book28/files/arapura.pdf" rel="nofollow">survey</a> by Donu Arapura, Toledo proved that many arithmetic lattices in higher rank algebraic $\mathbb{Q}$-groups (with hermitian symmetric space) are fundamental groups of smooth <em>projective</em> surfaces.</p> <p>In particular $Sp(2n,\mathbb{Z})$ for $n>2$, is such a group, and has property (T).</p> <p>Note that once you get a group as fundamental group of a smooth projective variety you obtain a smooth projective surface with the same fundamental group by intersecting with some generic hyperplanes.</p> http://mathoverflow.net/questions/91924/why-is-this-a-lattice/91932#91932 Answer by BS for Why is this a lattice? BS 2012-03-22T16:58:42Z 2012-03-22T16:58:42Z <p>As to the second question, this is pretty elementary : use the "fractional part" in $\mathbf{Z}[1/2]$ to put any element of $\mathbf{R}\times\mathbf{Q}_2$ in $\mathbf{R}\times\mathbf{Z}_2$ (by subtraction), and the "remaining" $\mathbf{Z}$ to put the $\mathbf{R}$ component in $[0,1]$. This way you see that the quotient is a so-called <em>solenoid</em>, the compact quotient $[0,1]\times\mathbf{Z}_2/(1,x)\sim (0,x+1)$ (this is a compact <em>connected</em> topological group).</p> <p>Informally, elements of $\mathbf{R}$ have dyadic expansion that is infinite to the right and finite to the left, and $2$-adic numbers have the opposite situation. Then $\mathbf{Z}[1/2]$ is their "intersection". The (diagonal) quotient just "blends" them.</p> <p>As to the first question, you might first use that $\text{SL}_3(\mathbf{Z}[1/2])$ is dense in $\text{SL}_3(\mathbf{Q}_2)$ to put the $2$-adic component of any element of $\text{SL}_3(\mathbf{R})\times\text{SL}_3(\mathbf{Q}_2)$ in the <em>open</em> subgroup $\text{SL}_3(\mathbf{Z}_2)$, then $\text{SL}_3(\mathbf{Z})$ to put the $\text{SL}_3(\mathbf{R})$ component in a finite volume fundamental domain $D$. Then $D\times\text{SL}_3(\mathbf{Z}_2)$ is a finite volume fundamental domain.</p> http://mathoverflow.net/questions/91257/2-transitive-and-3-transitive-lie-groups/91269#91269 Answer by BS for 2-transitive and 3-transitive Lie groups BS 2012-03-15T11:12:04Z 2012-03-15T11:12:04Z <p>I found the following by googling :</p> <p>Kramer, Linus. Two-transitive Lie groups. J. Reine Angew. Math. 563 (2003), 83-113.</p> <p>also available as arxiv:math/0106108</p> <p>It completely classifies locally compact sigma-compact groups $G$ acting effectively and 2-transitively on a non totally disconnected space $X$ (hausdorff I presume) : then $G$ is a Lie group, $X$ is a <em>connected</em> manifold, and the examples are listed sperately, according to wether $X$ is compact or not. </p> <p>It remains to extract the 3-transitive cases, which should not be too hard.</p> http://mathoverflow.net/questions/90805/example-of-special-lagrangian-submanifold/90808#90808 Answer by BS for example of special lagrangian submanifold BS 2012-03-10T11:09:40Z 2012-03-10T16:59:20Z <p>On the contrary, R. Bryant has shown that any closed oriented real analytic 3-dimensional riemannian manifold is the real locus of an antiholomorphic, isometric involution of a Calabi-Yau 3-fold (see <a href="http://arxiv.org/abs/math/9912246" rel="nofollow">http://arxiv.org/abs/math/9912246</a>). </p> http://mathoverflow.net/questions/89950/does-the-length-spectrum-determine-the-volume/89965#89965 Answer by BS for Does the length spectrum determine the volume? BS 2012-03-01T15:18:58Z 2012-03-01T15:18:58Z <p>This seems like a difficult question, even for closed hyperbolic manifolds. </p> <p>Indeed Marcos Salvai, in </p> <p>"On the Laplace and complex length spectra of locally symmetric spaces of negative curvature." Math. Nachr. 239/240 (2002), plus erratum on his web page</p> <p>proved that the complex length spectrum (with multiplicities) <em>and the volume</em> of a closed (oriented) hyperbolic manifold (real, complex, quaternionic or octonionic) determine the laplace spectrum (with multiplicities, even on forms) but cannot dispense with the volume. Hence even the complex length spectrum (length spectrum and holonomies) is not shown to determine volume. In the erratum, he seems to be confident that this can be repaired, but this is not published.</p> <p>However, there is a recent <a href="http://front.math.ucdavis.edu/1108.2977" rel="nofollow">preprint</a> by Dubi Kelmer, which shows among other things, that the length spectrum determines the laplace spectrum (hence the volume) for compact hyperbolic manifolds (real, complex, quaternionic or octonionic). The methods seem strongly Lie group representation-theoretic, not generalizing to non locally symmetric (or homogeneous) manifolds.</p> <p>Interestingly, he leaves open the question wether the laplace spectrum determines the multiplicities in the length spectrum (it is known that the length <em>set</em> is determined), whereas Salvai proves that the laplace-beltrami spectrum on forms determines the complex length spectrum, by following the proof by Gordon and Mao Math. Res. Lett. 1 (1994), no. 6, that it determines the length spectrum.</p> <p>Complicated situation...</p> http://mathoverflow.net/questions/89882/length-spectrum-of-spheres/89891#89891 Answer by BS for Length spectrum of spheres BS 2012-02-29T18:19:37Z 2012-02-29T20:27:15Z <p>As to the question of closed geodesics all of the same length, there are <a href="http://en.wikipedia.org/wiki/Zoll_surface" rel="nofollow">Zoll metrics</a>, so the length spectrum doesn't characterize even the round metric. See e.g. Besse's "Manifolds all of whose geodesics are closed" Springer 1978, or recent work by LeBrun and Mason on twistor methods applied to Zoll metrics.</p> <p>On the other hand (and extremally opposite case), for a suitably generic metric, there is the Duistermaat-Guillemin trace formula which implies that the length and Laplace spectra determine each other. But there are isospectral non isometric surfaces. I don't know if there are spherical examples (this seems plausible via Sunada -- or rather Buser -- construction), and even if Zoll metrics are isospectral (which I doubt).</p> <p>Hope this helps.</p> <p>EDIT : in fact Berger proved that the round 2-sphere is determined by its laplace spectrum inside smooth metrics, so Zoll's aren't all isospectral.</p> http://mathoverflow.net/questions/89664/finite-abelian-yet-fugitive-orthogonal-subgroups/89676#89676 Answer by BS for Finite, abelian, yet "fugitive" orthogonal subgroups BS 2012-02-27T16:02:27Z 2012-02-27T16:14:04Z <p>These results belong to what is called <em>duality in finite abelian groups</em>, a theory that has been generalized by <a href="http://en.wikipedia.org/wiki/Pontryagin_duality" rel="nofollow">Pontryagin</a> and others in the 30's to locally compact abelian groups. </p> <p>Another keyword here is "Discrete Fourier Transform", although it is mainly used for cyclic groups (of order $2^N$ for FFT) or $GF(2)$ vector spaces (analysis of boolean functions).</p> <p>It is also a chapter in the representation theory of finite groups, founded by Frobenius in the end of 19th century, namely the case of finite abelian groups, where the irreducible characters are all homomorphisms to $S^1$. This case was already known to Gauss and Dirichlet. </p> <p>By the way, the orthogonal you define is more naturally seen as a subgroup of the <em>dual group</em> $\hat{G}=Hom(G,S^1)$, which in (multiplicative) duality with $G$ via the evaluation $G\times \hat{G}\to S^1$. It is isomorphic to $G$, but not naturally so. </p> <p>I would bet that almost any book on either Pontryagin duality or representation theory of finite groups has the results you want to cite, but I have only been able to find <a href="http://www.encyclopediaofmath.org/index.php/Pontryagin_duality" rel="nofollow">this link</a> to an online encyclopedy, which easily implies your claims, since 3 and 4 follow quickly from 1 and 2. </p> <p>Hope this helps.</p> http://mathoverflow.net/questions/89168/stiefel-whitney-classes-of-a-projective-space-bundle/89175#89175 Answer by BS for Stiefel-Whitney classes of a projective space bundle BS 2012-02-22T10:15:35Z 2012-02-23T11:48:10Z <p>The situation seems more complicated.</p> <p>In fact, let $V$ be the vector bundle $\gamma_1 \oplus \mathbb{R}^{2k-1}$ over $S^1$. </p> <p>Then $TP(V)$ is isomorphic (via the choice of a connection) to $q^*TS^1 \oplus q^*V / L$, where $L\subset q^*V$ is the tautological bundle.</p> <p>Hence the total Stiefel-Whitney class of $TP(V)$ is $q^* w(V)\cup w(L)^{-1}$ in the algebra $H^*(TP(V),\mathbb{F}_2)$.</p> <p>This <s>algebra</s> is isomorphic to $\mathbb{F}_2[x,y]/(x^2,y^{2k})$, [<b>EDIT</b>: as a module over $H^*(S^1)$] since the $\mathbb{F}_2$-cohomology spectral sequence of $P(V)\to S^1$ necessarily has zero differentials on the $E_2$ page. Here $x=q^*(w_1(\gamma_1))$. Note that $\pi_1(S^1)$ acts trivially on $H^*(P^{2k-1},\mathbb{F}_2) \simeq \mathbb{F}_2[y]/(y^{2k})$. [<b>EDIT</b> : $y\in H^1(P(V ))$ is a class that restricts to the generator of $H^1$ of any fiber. <em>But this doesn't characterize it</em> : one may add $x$ to it. Hence the algebra structure must be determined by other means. See the comments].</p> <p>But $w_1(L)\in H^1(P(V),\mathbb{F}_2)\simeq Hom(\pi_1(P(V)),\mathbb{F}_2)$ is easily checked to be $x+y$ : first note that $\pi_1(V)\simeq \mathbb{Z}\times \mathbb{Z}/2$, then that $L$ is non trivial along the section of $q$ given by $P(\gamma_1)$. Hence the $x$ summand. The $y$ summand comes from restriction to a fibre. [<b>EDIT</b> : here I may precise a choice of $y$. It is Poincaré dual to the "hyperplane section" $S^1\times P(\mathbb{R}^{2k-1})$ in $P(V)$. But this doesn't determine the multiplicative structure.]</p> <p>[<b>EDIT</b> : The following calculation was wrong, due to a wrong algebra structure. See the comments for calculations with the correct one, given by $x^2=0$ and $(x+y)y^{2k-1}=0$.] </p> http://mathoverflow.net/questions/88259/nonvanishing-of-jacobians-implies-global-injectivity/88352#88352 Answer by BS for Nonvanishing of Jacobians implies global injectivity? BS 2012-02-13T15:44:49Z 2012-02-13T15:44:49Z <p>You might be interested by C. Soulé, M.Kaufman, R.Thomas <a href="http://www.ihes.fr/~soule/documents/publications.html" rel="nofollow">results</a> (search "multistationarity"). This might seem unrelated to your question but it is in fact related.</p> <p>Briefly, they study various <em>necessary</em> conditions for a differential equation $dx/dt=F(x)$, $x\in\mathbb{R}^n$ to have <em>several</em> non degenerate stationary points $F(x)=0$. </p> <p>The conditions depend on a signed "interaction graph" $G(x)$ deduced from the signs in the Jacobian matrix of $F$ at $x$. </p> <p>By taking the contrapositive, applied to $F-c$ or various other simple transform , you obtain <em>sufficient</em> conditions for $F$ to be injective (assuming non-vanishing jacobian determinant).</p> <p>Hope this helps.</p> http://mathoverflow.net/questions/87524/which-vector-bundle-are-the-christoffel-symbols-sections-of/87530#87530 Answer by BS for Which vector bundle are the Christoffel symbols sections of? BS 2012-02-04T13:42:24Z 2012-02-04T13:42:24Z <p>Connections on a vector bundle $E\to M$ are sections of an <em>affine</em> bundle associated to $E$.</p> <p>Namely there is a vector bundle $J^1E$ of $1$-<a href="http://en.wikipedia.org/wiki/Jet_%28mathematics%29" rel="nofollow">jets</a> of sections of $E$, and an exact sequence of bundles $$0\to T^*M\otimes E\to J^1E \overset{p}\to E\to 0$$ where the map $p$ is the evaluation ("$0$-jet").</p> <p>Then a connection is a section of the affine bundle of sections (sic) of $p$, namely the $s\in Hom(E,J^1E)$ such that $p\circ s=id_E$. The associated vector bundle is $Hom(E,T^*M\otimes E)\simeq T^*M\otimes End(E)$, where one can view the Christoffel symbols (if $E=TM$) as living : once (local) a trivialisation is chosen there is an associated "trivial" connexion, and any other connection differs from it by a section of this vector bundle.</p> http://mathoverflow.net/questions/87491/definition-of-hessian-with-respect-to-connection/87518#87518 Answer by BS for definition of Hessian with respect to connection BS 2012-02-04T10:50:33Z 2012-02-04T10:50:33Z <p>There is no real need for a connection in this situation. </p> <p>The underlying fact is that if $s$ is a section of a vector bundle $E$ over $X$, and $s$ vanishes at a point $x_0$, then it has an intrinsic derivative $Ds(x_0):T_{x_0}X\to E_{x_0}$, defined by $Ds(x_0)\cdot v=\lim_{t\to 0} s(x_t)/t$, where $t\mapsto x_t$, $t\in]-1,1[$ is any curve in $X$ with velocity $v$ at $t=0$.</p> <p>In your case $E$ is $T^*X$ and $s=df$. </p> http://mathoverflow.net/questions/130400/triangle-area-on-surfaces-of-constant-curvature/130428#130428 Comment by BS BS 2013-05-16T09:46:56Z 2013-05-16T09:46:56Z What about the scissors congruence definition af area of simple polygons ? http://mathoverflow.net/questions/130171/fourier-transform-of-a-bounded-function/130174#130174 Comment by BS BS 2013-05-09T16:24:32Z 2013-05-09T16:24:32Z No, because modifying the signum function, say on [-1,1], to render it continuous (or even smooth), would just add an $L^1$ function to it, hence a $C_0$ function to its Fourier transform. On the other hand continuous [positive definite][1] functions on $\mathbb{R}$ are exactly the Fourier transforms of finite positive measures, by Bochner's theorem. [1]: <a href="http://en.wikipedia.org/wiki/Positive-definite_function" rel="nofollow">en.wikipedia.org/wiki/Positive-definite_function</a> http://mathoverflow.net/questions/118801/why-the-group-gl6-v-has-an-open-orbit Comment by BS BS 2013-01-15T11:05:04Z 2013-01-15T11:05:04Z The proof is given in section 2, pages 3 to 6, first the complex case, then in the real case. http://mathoverflow.net/questions/118801/why-the-group-gl6-v-has-an-open-orbit Comment by BS BS 2013-01-13T12:59:32Z 2013-01-13T12:59:32Z This is very well explained in Hitchin's article freely available at <a href="http://arxiv.org/abs/math/0010054v1" rel="nofollow">arxiv.org/abs/math/0010054v1</a> http://mathoverflow.net/questions/116692/orthogonality-wrt-ext-tor-in-commutative-noetherian-rings Comment by BS BS 2012-12-18T11:36:07Z 2012-12-18T11:36:07Z Could you explain what is $E(R/p)$ ? http://mathoverflow.net/questions/108528/choosing-a-kahler-metric-which-restricts-the-norms-of-some-forms Comment by BS BS 2012-10-06T08:56:06Z 2012-10-06T08:56:06Z The hypothesis is still not satisfied if the $f_i$ converge to, say, the indicator function of a compact set (which is possible, by convolutions). Then the $\bar{\partial f_i}$ are <i>locally</i> unbounded. http://mathoverflow.net/questions/108508/injective-maps-on-cohomology-and-kahler-manifolds/108510#108510 Comment by BS BS 2012-10-01T11:34:13Z 2012-10-01T11:34:13Z In fact $Y$ is not required to be symplectic. It suffices that $f$ is surjective and $X$ has a closed $2$ form which is symplectic on a generic fiber (so that dimensions have equal parity). When $f$ is a submersion, it is called a symplectic fibration, but there are important examples with critical points, the so-called (symplectic) Lefschetz fibrations. http://mathoverflow.net/questions/108528/choosing-a-kahler-metric-which-restricts-the-norms-of-some-forms Comment by BS BS 2012-10-01T10:45:49Z 2012-10-01T10:45:49Z You should add the hypotheses on the $\alpha_i$ that you have in mind, otherwise the answer is obviously &quot;no&quot;: take the integer multiples of a non zero $\bar{\partial}f$. For instance, boundedness on every compact subset (in some/any hermitian metric) forbids this kind of stupid example, and is obviously necessary. http://mathoverflow.net/questions/108456/how-to-understand-ext-mathcalo-y-mathcalo-z-for-subvarieties-y-z-su Comment by BS BS 2012-09-30T10:15:35Z 2012-09-30T10:15:35Z A minor remark : the exact sequences you've written are classified by $Ext(B,A)$ (extensions of $B$-quotient by $A$-kernel) not $Ext(A,B)$. http://mathoverflow.net/questions/108421/a-closed-immersed-submanifold Comment by BS BS 2012-09-30T10:06:56Z 2012-09-30T10:06:56Z I think that <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> would be more suitable for your question. Please read <a href="http://mathoverflow.net/faq" rel="nofollow">mathoverflow.net/faq</a> , and you will learn that MO is for research level math questions. http://mathoverflow.net/questions/108402/decomposition-of-matrices-in-semisimple-and-nilpotent-parts Comment by BS BS 2012-09-30T10:00:35Z 2012-09-30T10:00:35Z There is no uniqueness if you don't require that the two matrices in the decomposition commute. http://mathoverflow.net/questions/108188/tensorial-decomposition-of-bh/108229#108229 Comment by BS BS 2012-09-27T10:42:57Z 2012-09-27T10:42:57Z I think that the tensor norm should be a $C^*$-cross norm not a general cross norm http://mathoverflow.net/questions/13089/why-do-so-many-textbooks-have-so-much-technical-detail-and-so-little-enlightenmen/13098#13098 Comment by BS BS 2012-09-23T09:41:25Z 2012-09-23T09:41:25Z And Bourbaki's volume also have historical notes, not so frequent in textbooks. http://mathoverflow.net/questions/13089/why-do-so-many-textbooks-have-so-much-technical-detail-and-so-little-enlightenmen/13098#13098 Comment by BS BS 2012-09-23T09:37:32Z 2012-09-23T09:37:32Z Don't miss that the real flesh of Bourbaki <b>is in the exercises</b>. There are real gems there. The main text is the skeleton. http://mathoverflow.net/questions/106888/when-does-a-planar-ternary-ring-uniquely-coordinitise-a-projective-plane/106961#106961 Comment by BS BS 2012-09-12T08:57:15Z 2012-09-12T08:57:15Z This isn't the question asked, which is to characterize projective planes which have only one isomorphism class of ternary ring, or equivalently whose automorphisms act transitively on quadrangles.