User renato g bettiol - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:38:41Z http://mathoverflow.net/feeds/user/15743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90933/possible-isometries-of-a-positively-curved-s2-times-s2 Possible isometries of a positively curved $S^2\times S^2$ Renato G Bettiol 2012-03-11T21:04:38Z 2013-05-01T16:27:38Z <p>Just to put things in perspective, recall that the Hopf Conjecture asks whether $S^2\times S^2$ admits a metric of positive sectional curvature. By the work of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214443064" rel="nofollow">Hsiang-Kleiner</a>, it is known that, if $S^2\times S^2$ admits such a metric, then its isometry group cannot contain a circle, and is hence finite.</p> <blockquote> <p>Q: If $S^2\times S^2$ admits a metric with $sec>0$, what is known about its isometry group $G$?</p> </blockquote> <hr> <p>The only results I know of are:</p> <p>0) [Edit suggested by Misha] The diagonal antipodal action of $\mathbb Z_2$ on $S^2\times S^2$, i.e., $\pm 1\cdot(x,y)=(\pm x,\pm y)$, cannot be isometric if $S^2\times S^2$ is equipped with a metric of positive curvature. By Weinstein's Thm, an orientation-preserving isometry of an even-dimensional positively curved manifold has a fixed point (and the antipodal map does not). Equivalently, it would induce a positively curved metric on the $2$-fold orientable cover of $\mathbb R P^2\times \mathbb R P^2$, hence on $\mathbb R P^2\times \mathbb R P^2$, but this contradicts Synge's Thm.</p> <p>1) From <a href="http://wwwmath.uni-muenster.de/u/weckerm/sfb/about/publ/wilking3.ps" rel="nofollow">Wilking's thesis</a> (Prop 4.2), any simple subgroup of $G$ is either cyclic or isomorphic to a group in a finite list $F_1,\dots,F_k$ of simple groups. (This is actually true for any finitely generated subgroups of isometries of a manifold with $Ric≥0$).</p> <p>2) From <a href="http://arxiv.org/abs/math/0504504" rel="nofollow">Fang's paper</a> (Thm 1.2), $G$ cannot have a subgroup of sufficiently large odd order (but this lower bound is huge, since it is estimated with Gromov's universal constant for the total Betti number).</p> <hr> <p>Apart from these, are there other known restrictions on what $G$ can be like?</p> http://mathoverflow.net/questions/127972/prescribing-the-lie-derivative-of-the-metric/128010#128010 Answer by Renato G Bettiol for Prescribing the Lie derivative of the metric? Renato G Bettiol 2013-04-18T20:06:24Z 2013-04-18T20:12:19Z <p>If $h\in S^2(TM^*)$ is a symmetric $(0,2)$-tensor such that $\int_M \langle h,\mathcal L_X g\rangle =0$ for all vector fields $X$ on $M$, then $h$ need not be zero. In fact, the above is precisely equivalent to $\delta h=0$, where $$\delta=\nabla^*\big|_{S^2(TM^*)} \colon S^2(TM^*)\to \Omega^1(M)$$ is the <em>divergence operator</em>, i.e., the formal $L^2$-adjoint of the covariant derivative $\nabla \colon \Omega^1(M)$ $\to S^2(TM^*)$.</p> <p>The above claim follows from an infinitesimal version of Ebin's slice theorem for the pull-back action of the diffeomorphism group $\mathcal D:=Diff(M)$ on $S^2(TM^*)$, see <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214429060" rel="nofollow">this paper of Berger and Ebin, JDG '69</a>. The statement I am referring to is the $L^2$-orthogonal decomposition: $$S^2(TM^*)=\ker \delta\oplus Im \ \delta^*.$$ Given any element $g\in S^2(TM^*)$, e.g., a Riemannian metric, the space $\ker\delta$ is the tangent space to the slice at $g$ to the $\mathcal D$-action and $Im\ \delta^*=T_g \mathcal D(g)$ is the tangent space to the orbit of $g\in S^2(TM^*)$. The latter is precisely formed by tensors of the form $\mathcal L_X g$, where $X$ is some field on $M$. In fact, consider $\phi_g\colon\mathcal D\to S^2(TM^*)$, $\phi_g(\eta)=\eta^*(g)$, and let $\eta_t\in\mathcal D$ be a curve of diffeomorphisms, with $\eta_0=id$ and $\dot\eta_0=X$. Then $$\mathrm d\phi_g(id)X=\frac{\mathrm d}{\mathrm dt}\phi_g(\eta_t)\big|_{t=0}=\frac{\mathrm d}{\mathrm dt}\eta_t^*(g)\big|_{t=0}=\mathcal L_X g=2\delta^*(X^b),$$ where $X^b=g(X,\cdot)$ is the $1$-form dual to the field $X$. The last equality follows from $\delta^*(\omega)=\tfrac12\mathcal L_{X_\omega}g$, where $X_\omega$ is the vector field dual to the $1$-form $\omega$. Note the above line also proves that $Im \ \delta^*=T_g\mathcal D(g)$.</p> <p>Thus, going back to your original question, the above result of Ebin and Berger tells you that your symmetric traceless tensor $h$ is <em>geometric</em>, in the sense that it is tangent to the slice of the $\mathcal D$-action on $S^2(TM^*)$, or, equivalently, $L^2$-orthogonal to the tangent space to the $\mathcal D$-orbit. In some sense, this means that it descends'' to an object in the quotient space of tensors modulo diffeomorphisms (where, e.g., the moduli space of Riemannian metrics should live).</p> <p>The above observations also answer your question regarding what symmetric $(0,2)$-tensors are of the form $\mathcal L_Xg$; namely, they are precisely the ones in $Im \ \delta^*$. Hope this helps...</p> http://mathoverflow.net/questions/12827/cone-in-a-metric-space/124710#124710 Answer by Renato G Bettiol for Cone in a metric space Renato G Bettiol 2013-03-16T17:59:45Z 2013-03-16T17:59:45Z <p>There is a natural definition of cone in the context of <em>pointed</em> metric spaces:</p> <blockquote> <p>A pointed metric space $(X,p)$ is a <em>cone</em> if $(\lambda X,p)$ is isometric to $(X,p)$ for all $\lambda>0$.</p> </blockquote> <p>Here, $(\lambda X,p)$ denotes the metric space obtained from $(X,p)$ by multiplying the distance function by $\lambda$. Also <em>isometric</em> should be understood in the pointed category, i.e., $(X,x_0)$ and $(Y,y_0)$ are isometric if there exists a distance preserving map $f\colon X\to Y$ such that $f(x_0)=y_0$. This is, for example, the definition you can find in the <a href="http://www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf" rel="nofollow">book by Burago, Burago &amp; Ivanov, Def 8.2.1</a>. It also coincides with the definition in a Banach space setting.</p> <p>Obviously, you can then say that a subspace $(X,p)$ of your metric space $(\bar X,p)$ is a cone if it is a cone as an abstract metric space...</p> http://mathoverflow.net/questions/123633/characterizing-hessians-among-symmetric-bilinear-tensors Characterizing Hessians among symmetric bilinear tensors Renato G Bettiol 2013-03-05T16:44:27Z 2013-03-07T16:10:49Z <p>I apologize in advance if this is somewhat elementary, but:</p> <blockquote> <p>Let $(M,g)$ be a compact Riemannian manifold. Is there a "characterization" of which symmetric bilinear tensors $B\in Sym^2(M)$ are Hessians of functions $f\colon M\to \mathbb R$? </p> <p>In other words, for which $B\in Sym^2(M)$ there exists $f\colon M\to\mathbb R$ such that $B(X,Y)=Hess \ f(X,Y)$, where $Hess\ f(X,Y)=g(\nabla_X \nabla f,Y)=X(Y(f))-\mathrm df(\nabla_X Y)$? </p> </blockquote> <p>Obviously, since $M$ is assumed compact, any continuous function will attain a min and a max, so a symmetric bilinear form that is a candidate to Hessian cannot be definite. This maybe pretty naive, but I cannot think of any other property that distinguishes Hessians among bilinear symmetric tensors in this context... For example, it doesn't seem to be the case that $B$ has to be parallel, or satisfy any other sort of similar "nice" conditions.</p> <p>Note that even if we impose the extra condition that $B$ is the Hessian of a <em>Morse</em> function, I don't quite see if the topology of $M$ will impose any properties on $B$ via Morse theory, since $B$ does not know if it is at a (future) critical point or not, right?</p> http://mathoverflow.net/questions/122987/proper-subgroups-of-rmsud-that-act-transitively-on-rmcpd-1/123147#123147 Answer by Renato G Bettiol for Proper subgroups of $\rm{SU}(d)$ that act transitively on $\rm{CP}^{d-1}$? Renato G Bettiol 2013-02-27T20:57:14Z 2013-02-28T15:01:11Z <p>Groups acting transitively on $\mathbb C P^n$ (as well as on $\mathbb HP^n$ or on $\mathbb{Ca}P^2$) were classified by <a href="http://www.ams.org/mathscinet-getitem?mr=155935" rel="nofollow">Oniscik</a>; <a href="http://www.mathnet.ru/links/ac6f5c8b1d0835b93941b782e481b961/sm4555.pdf" rel="nofollow">here's the original paper</a> (if you understand Russian). As a consequence of this classification, the <em>only group</em> other than $SU(d)$ that acts transitively (and almost effectively) on $\mathbb C P^{d-1}$ is the example given in Aakumadula's answer: $Sp(d)$, when $d$ is even.</p> <p>The full classification table of projective spaces written as homogeneous spaces $G/H$ (with the usual assumptions: $G$ connected, action is almost effective etc.) is as follows:</p> <ol> <li>$G=SU(n+1)$, $H=S(U(1)\times U(n))$, $G/H=\mathbb C P^n$, isotropy representation is irreducible;</li> <li>$G=Sp(n+1)$, $H=Sp(n)\times Sp(1)$, $G/H=\mathbb H P^n$, isotropy representation is irreducible;</li> <li>$G=F_4$, $H=Spin(9)$, $G/H=\mathbb{Ca}P^2$, isotropy representation is irreducible;</li> <li>$G=Sp(n+1)$, $H=Sp(n)\times U(1)$, $G/H=\mathbb C P^{2n+1}$, isotropy representation has $2$ irreducible summands.</li> </ol> <p>Just for completeness, the list of homogeneous structures on the only remaining compact rank one symmetric spaces (i.e., spheres) was first obtained by Montgomery, Samelson and Borel and then reobtained in Oniscik's work.</p> http://mathoverflow.net/questions/116548/space-of-sections-of-a-fibre-bundle-with-non-compact-base-space/121414#121414 Answer by Renato G Bettiol for Space of sections of a fibre bundle with non-compact base space Renato G Bettiol 2013-02-10T18:27:16Z 2013-02-10T18:27:16Z <p>In addition to the above references, you may be interested in taking a look at this paper:</p> <p><em>P. Piccione and D. V. Tausk, "On the Banach differential structure for sets of maps on non-compact domains." Nonlinear Anal. 46 (2001), no. 2, Ser. A: Theory Methods, 245–265,</em></p> <p>where they study how to introduce a Banach structure on sets of maps between a possibly non-compact topological space as domain and a smooth manifold as target. Of course the regularity discussed here is less than $C^\infty$, giving you a Banach structure instead of Frechet. This has obvious advantages and disadvantages; but regardless of the differences is possibly worth looking at, given that the non-compactness issues of the domain are dealt with successfully. Needless to say, once you have the desired structure on the set of maps, restricting to the particular case of sections of a bundle is straight-forward.</p> http://mathoverflow.net/questions/121168/group-of-diffeomorphisms-of-a-manifold/121208#121208 Answer by Renato G Bettiol for group of diffeomorphisms of a manifold Renato G Bettiol 2013-02-08T15:44:02Z 2013-02-08T16:27:28Z <p>In addition to the references suggested in the above comments, I'd recommend taking a look at recent work of <a href="http://www.math.toronto.edu/khesin/papers.html" rel="nofollow">B. Khesin</a>, D. Ebin, <a href="http://math.nd.edu/people/faculty/gerard-k-misiolek/" rel="nofollow">G. Misiolek</a>, <a href="http://math.colorado.edu/~prestos/" rel="nofollow">S. Preston</a>, <a href="http://www.mat.univie.ac.at/~michor/" rel="nofollow">P. Michor</a> among others... B. Khesin has a <a href="http://www.math.toronto.edu/khesin/papers/Lecture_notes.pdf" rel="nofollow">nice book</a> (freely available on his webpage) about infinite-dimensional groups, with a whole chapter on Diffeomorphism groups, that is perhaps a good place to start.</p> <p>Just for a glimpse of what goes on beyond "just" topology, exotic structures, etc., the geometry of a few interesting subgroups of the diffeomorphism group $\mathcal D^s(M)$ of a manifold $M$ is also an important object of study. For example, take $\mathcal D_\mu^s(M)$, formed by <em>volume preserving</em> diffeomorphisms. This group is very much related with classical equations of hydrodynamics: you can think of the motion of an incompressible fluid filling a manifold $M$ as a curve in $\mathcal D_\mu^s(M)$. Classic work of Arnold (1966) and Ebin &amp; Marsden (Ann. of Math., 1970), later followed the above mentioned authors, establishes a very beautiful setup for the Euler equations $\partial_t u+\nabla_u u=-\nabla p$, $div\ u=0$ (and something similar can be done for Navier-Stokes), proving that solutions $u(t,x)$ to these PDEs on $M$ are the $1$-parameter families of volume preserving diffeomorphisms that arise as geodesics in $\mathcal D_\mu^s(M)$ for an $L^2$ Riemannian metric in this infinite-dimensional manifold. This approach also works for some other PDEs, like Burgers' equation and KdV, the latter having to do with the diffeomorphism group of the circle. Most of this is discussed in detail in Khesin's book. </p> <p>In this way, weak Riemannian geometry of the diffeomorphism group of $M$ and some of its submanifolds is deeply interconnected with many evolution equations on $M$. Of course, as the OP mentions, low dimension plays a big role in having more answers for now. For example, one can prove global existence of solutions to some of these PDEs of hydrodynamics in 2D using the above setup, but the 3D version is open and worth some big prizes and big money.</p> <p>It is also worth pointing out that, recently, the infinite-dimensional geometry of diffeomorphism groups has been related to areas other than hydrodynamics, like optimal transport and geometric statistics, see <a href="http://www.math.toronto.edu/khesin/papers/densities_diffeos7.pdf" rel="nofollow">this paper</a>.</p> http://mathoverflow.net/questions/121107/banach-manifold/121115#121115 Answer by Renato G Bettiol for Banach Manifold Renato G Bettiol 2013-02-07T21:05:40Z 2013-02-07T22:13:16Z <blockquote> <blockquote> <p>Yes, $C^k(M,N)$ is a smooth Banach manifold, when $M$ and $N$ are smooth closed manifolds.</p> </blockquote> </blockquote> <p>In fact, you can consider manifolds of maps with more general domains and regularities, this lies in the foundations of "Global Analysis". The classic work in the area is mostly due to <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183528314" rel="nofollow">Eells</a>, <a href="http://vmm.math.uci.edu/PalaisPapers/FoundationsOfGlobalNonlinearAnalysis.pdf" rel="nofollow">Palais</a>, <a href="http://www.intlpress.com/JDG/archive/1967/1-1&amp;2-169.pdf" rel="nofollow">Eliasson</a> and others. I also particularly like this nice short note by <a href="http://www.ime.usp.br/~tausk/texts/ManifoldMaps.pdf" rel="nofollow">Tausk</a>, that considers Banach manifolds of maps from sets to manifolds, with minimum regularity requirements.</p> <p>In the previous links you can find explicit descriptions of how charts are constructed; but, <strong>very roughly</strong> (in the following, I am omitting all regularities for the sake of simplicity), you can regard a neighborhood of a map $f\colon M\to N$ as consisting of maps $g\colon M\to N$ that can be written as $g(p)=\exp_{f(p)} X(p)$, where $X\colon M\to TN$ is a vector field along $f$, and the exponential map is regarded w.r.t. some choice of background metric $h$ on $N$. Of course $X$ has to be <em>small enough</em>, e.g., its norm (that depends of what regularity we're talking about) has to be less than the injectivity radius of $h$ along $f(M)$. The correspondence given by such chart is then $g\leftrightarrow X$, which tells you that the tangent space at $f$ to $Maps(M,N)$ consists of vector fields along $f$ (of the same regularity as the maps considered). Details can be found in any of the links above, but I hope this brief description at least gives you some intuition...</p> <hr> <p><strong>Edit:</strong> Another good reference that I forgot to mention is this book by <a href="http://www.amazon.com/Differential-Topology-Graduate-Texts-Mathematics/dp/0387901485" rel="nofollow">Hirsch</a>, see Chap 2.</p> http://mathoverflow.net/questions/121122/how-submanifolds-evolve-under-ricci-flow How submanifolds evolve under Ricci flow? Renato G Bettiol 2013-02-07T21:48:51Z 2013-02-07T21:48:51Z <p>This may be very naive, since I just started trying to learn Ricci flow; but I couldn't really find any answer after looking for a while in all the textbooks and lecture notes I found online...</p> <blockquote> <blockquote> <p>If $(M,g_t)$ is a solution of the Ricci flow (normalized or not, I don't care), and $i\colon N\hookrightarrow (M,g_0)$ is a submanifold (with the induced metric), what is known about what happens to $(N,i^*g_t)$ in terms of its intrinsic/extrinsic geometry?</p> </blockquote> </blockquote> <p>This is somewhat vague, so, to be more precise: under what conditions a totally geodesic (resp. minimal) submanifold remains totally geodesic (resp. minimal)? What evolution equation is satisfied by the second fundamental form $B^t_{\xi^t}(X,Y)=g_t(\nabla^t_X Y,\xi^t)$ of $N\subset (M,g_t)$, or shape operator, in the codimension $1$ case? Note that here almost everything depends on $t$: the connection $\nabla^t$, the normal field $\xi^t$ and obviously the metric $g_t$. I tried to take the $t$ derivative using formulas for each of the objects (e.g., the ones found in <a href="http://homepages.warwick.ac.uk/~maseq/RFnotes.html" rel="nofollow">Topping's notes</a>), but it got incredibly messy very fast -- and there was nothing I could really read off the formulas. I then did some examples, but the only ones I could do all the computations for were somewhat trivial.</p> <p>I would be interested in <em>any</em> intuition/results related to the above, it could be for hypersurfaces (instead of general submanifolds), only in low dimensions, etc...</p> http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120065#120065 Answer by Renato G Bettiol for Geometric picture of scalar curvature Renato G Bettiol 2013-01-27T22:57:33Z 2013-01-27T22:57:33Z <p>Have you taken a look at <a href="http://en.wikipedia.org/wiki/Scalar_curvature" rel="nofollow">wikipedia page for Scalar curvature</a>? [BTW, always a great resource!] There you can find the standard geometric interpretation of Scalar curvature, as measuring the volume distortion on balls of small radius, compared to Euclidean balls of such radius. Analogously, Ricci curvature in a direction measures volume distortion of small rods'' along that direction (and then these interpretations make sense together in terms of scalar curvature being an average of Ricci in all directions). I think that's possibly the most geometric picture you can get (apart from interpretations from mathematical general relativity).</p> <p>As for what "scalar curvature forgets, but Ricci still sees" you can think in terms of topological obstructions for these curvatures to have a certain sign; e.g., manifolds with positive Ricci curvature (bounded from below) must be compact (you even have an estimate for their diameter) and have finite fundamental group. Instead, manifolds with positive scalar curvature can be much wilder (in particular, they might not have finite fundamental group). Also, you can look at the problem of prescribing Ricci vs. prescribing scalar curvature, one is clearly way more flexible than the other, see <a href="http://mathoverflow.net/questions/30035/some-questions-about-scalar-curvature" rel="nofollow">this post</a> regarding Kazdan-Warner's stuff for scalar curvature and compare with obstructions to positive (and non-negative Ricci curvature). Although endowing a given manifold with a metric with positive/nonnegative ricci may be impossible, it certainly always has tons of metrics with negative ricci curvature (see <a href="http://www.jstor.org/discover/10.2307/2118552?uid=3739256&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=21101716283427" rel="nofollow">this paper</a>), in that such metrics are actually $C^0$-dense in the space of all metrics.</p> http://mathoverflow.net/questions/116294/hamiltonian-polar-action-with-lagrangian-section Hamiltonian polar action with Lagrangian section Renato G Bettiol 2012-12-13T16:11:23Z 2012-12-17T01:53:13Z <blockquote> <p>I am looking for examples of Hamiltonian polar isometric actions of a compact Lie group on a Kahler-Einstein (or perhaps just Kahler) manifold, that admits a Lagrangian section.</p> </blockquote> <p>Recall that an isometric action on $M$ is <em>polar</em> if there exists a submanifold $\Sigma\subset M$, called <em>section</em>, that meets all orbits orthogonally. Any section is automatically a totally geodesic submanifold, and there may be many different sections on $M$. The examples I am looking for should have a section $\Sigma\subset M$ that is also a Lagrangian submanifold, i.e., $\omega|_\Sigma=0$ and $\dim M=2\dim \Sigma$.</p> <blockquote> <p>So far, the only example I can see is the standard torus action of $T^n$ on $\mathbb C P^n$, with the Fubini metric.</p> </blockquote> <p>This action is Hamiltonian and isometric. It is also polar (in fact, by a Theorem of Podesta-Thorbergsson, if a torus $T^n$ acts on a compact Kahler manifold $M^{2n}$ of complex dimension $n$ and positive Euler characteristic, then this action is polar). Finally, the usual totally real embedding of $\mathbb R P^n$ into $\mathbb C P^n$ gives a Lagrangian section for the action.</p> <hr> <p>With such strong hypothesis I would imagine that such actions are perhaps even classified, but I could not find anything in the literature about Lagrangian sections. Most classification results have to do with <em>coisotropic</em> actions, where it is required that the <em>orbits</em> satisfy $\omega|_{G(x)}=0$; while I am interested in an "orthogonal" version of that, i.e., I want the <em>section</em> to be <em>coisotropic</em> (even more, <em>Lagrangian</em>).</p> http://mathoverflow.net/questions/115269/lagrangian-submanifolds-with-parallel-mean-curvature-vector/115643#115643 Answer by Renato G Bettiol for Lagrangian submanifolds with parallel mean curvature vector Renato G Bettiol 2012-12-06T21:12:33Z 2012-12-06T21:12:33Z <p>I was recently reading this nice survey by <a href="http://www.sci.osaka-cu.ac.jp/~ohnita/paper/RIMS02c%28Kokyuroku%29.pdf" rel="nofollow">Amarzaya-Ohnita</a>, and they have an entire section on the subject you are interested (see Section 3). They claim that totally real submanifolds with parallel mean curvature vector in $\mathbb C^n$ and $\mathbb C P^n$ have been classified by Naitoh and Takeuchi (references 8, 9, 10 &amp; 11 in their bibliography). Maybe you should take a look at those papers too.</p> <p>As a side note, just googling "Lagrangian submanifold parallel mean curvature vector" you get lots of interesting links...</p> http://mathoverflow.net/questions/112394/volume-of-exceptional-group-orbits/114350#114350 Answer by Renato G Bettiol for volume of exceptional group orbits Renato G Bettiol 2012-11-24T17:10:27Z 2012-12-05T23:24:54Z <blockquote> <p>To my understanding, Proposition 1 in this paper of <a href="http://arxiv.org/pdf/math/0207130v3.pdf" rel="nofollow">Pacini, TAMS 2003</a> gives exactly the proof that you ask for in the Riemannian case; namely, that the volume of orbits: $$vol\colon M\to \mathbb R, \quad vol(x)=\int_{G(x)} i^*(vol_M),$$ where $i\colon G(x)\hookrightarrow M$ is the immersion of the $G$-orbit through $x$ and $vol_M$ is the volume form of $M$, is a continuous function on $M$, vanishing exactly at singular orbits.</p> </blockquote> <p>More precisely, he proves that:</p> <ol> <li><p>the volume function on regular (i.e., principal or exceptional) orbits $vol\colon M^{reg}\to\mathbb R$ is a smooth function;</p></li> <li><p>it has a continuous extension $vol\colon M\to\mathbb R$ that is zero on the singular points $M^{sing}=M\setminus M^{reg}$;</p></li> <li><p>$vol^2\colon M\to\mathbb R$ is smooth.</p></li> </ol> <p>Note that Pacini defines the volume of an orbit not by using the volume of the image, but rather by integrating the pull-back of the volume form. These are the same thing only if the immersion is $1$-to-$1$ (e.g., for principal orbits $G/P$). For an exceptional orbit $G/K$, the immersion is $k$-to-$1$, where $k$ is the number of sheets on the covering by a principal orbit $G/P\to G/K$, so the volume of the image is multiplied by $k$. This correction factor is precisely the cardinality of the fiber, $k=|P/K|$, as pointed out by the OP.</p> <p>Regarding the second question, adapting the proof to the nondegenerate semi-Riemannian case should be straightforward.</p> <blockquote> <p><strong>Edit:</strong> I recently realized that also the classic paper by <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214429775" rel="nofollow">Hsiang-Lawson, JDG 1971</a> (see first few lines of page 7) cites the continuity of the volume function above in $M$ (being zero on singular points) and smoothness in the set of regular points. Although they do not provide an explicit proof, they say it is straightforward from the Slice Theorem. There are also many nice examples following that.</p> </blockquote> http://mathoverflow.net/questions/114825/homogeneity-of-the-hopf-fibration-s7-to-s15-to-s8 "Homogeneity" of the Hopf fibration $S^7\to S^{15}\to S^8$ Renato G Bettiol 2012-11-28T23:20:02Z 2012-11-29T02:16:52Z <p>My question has to do with an apparent contradiction I get regarding the Hopf fibration $S^7\to S^{15}\to S^8$. Namely, the two following statements cannot be true at the same time (but I do not see any problem with any of them):</p> <ol> <li><p>The Hopf fibration $S^7\to S^{15}\to S^8$ is not homogeneous, i.e., there is no isometric group action on the round sphere $S^{15}$ whose orbits are the Hopf fibers. This is claimed by <a href="http://www.uam.es/personal_pdi/ciencias/lguijarr/papers/holonomy.pdf" rel="nofollow">Guijarro-Walschap</a>, Corollary 3.2; and was also previously observed, e.g., by Gromoll-Grove.</p></li> <li><p>The Hopf fibration $S^7\to S^{15}\to S^8$ is of the form $K/H \to G/H \stackrel{\pi}{\to} G/K$, where $\pi(gH)=gK$ and $H&lt; K &lt; G$ are the groups $Spin(7)$, $Spin(8)$ and $Spin(9)$, respectively. The inclusion of $K$ in $G$ is the usual one; however, the inclusion of $H$ in $K$ is the usual one followed by a <strong>nontrivial triality automorphism</strong> of $Spin(8)$, see, e.g., <a href="http://arxiv.org/pdf/0806.3824.pdf" rel="nofollow">Section 4 of this paper</a> or Besse's book "Einstein manifolds", p. 258, 9.84 Example 4. Such an automorphism is <em>outer</em>, and is not the restriction of any automorphism of $Spin(9)$. Now, consider the $K$-action on $G/H$ given by $k\cdot gH:=gk^{-1} H$. Its orbits are clearly of the form $gKH\subset G/H$, and I claim these are exactly the fibers of $G/H\to G/K$, i.e., the Hopf fibers. Indeed, if both $aH$ and $bH$ are mapped to $gK$ under the projection $G/H\to G/K$, then $aK=bK$, i.e., $b^{-1}a\in K$, which means $a\in bK$; so the subset of $G/H$ that get mapped to $gK$ is exactly $(gK)H$; and these were the $K$-orbits. The above construction (for abstract Lie groups $H, K, G$; not specifically for the Hopf fibration as above) shows up in many places, e.g., in <a href="http://www.math.upenn.edu/~wziller/papers/survey_noneg_curvature_Final.pdf" rel="nofollow">Ziller's survey</a>, p. 16.</p></li> </ol> <p><strong>Edit:</strong> Shortly after posting this question, I received an email from Ziller in which he answers the question. Statement (1) is correct, and the problem is in statement (2), as suspected. In fact, the claim that the $K$-orbits are always fibers of $G/H\to G/K$ is false in general, unless $H$ is a <strong>normal subgroup</strong> of $K$. This is because the action by multiplication (by the inverse) on the right on cosets defined above is only well-defined (i.e., independent of choice of coset representative) if $H$ is normal in $K$, as was also pointed out in Emerton's comment. The rest of the claims in (2) are correct. As a side note, for the other Hopf fibrations $S^1\to S^{2n+1}\to \mathbb CP^n$ and $S^3\to S^{4n+3}\to \mathbb H P^n$ the corresponding subgroup $H$ is normal in $K$ -- after all the fiber $K/H$ is a group -- and the $K$-action is hence well-defined, so statement (2) holds in full for these cases. In general, however, the $K$-orbits are not fibers of $G/H\to G/K$, as the $S^7\to S^{15}\to S^8$ example illustrates.</p> <hr> <p>I am not sure how useful these comments are, but here is some more information about a $Spin(8)$ action on $S^{15}$. The representation $\rho_8\oplus\Delta^\pm_8$ of $Spin(8)$ in $R^{16}$ gives a cohomogeneity one action on the unit sphere $S^{15}$, and the orbit space is the interval $[0,\pi/2]$. This action has two singular orbits, whose isotropy is $Spin(7)$, and the principal isotropy is $G_2$. The inclusion of the singular isotropies is the usual one followed by a nontrivial triality automorphism that is $\pm$, according to the choice $\Delta^\pm_8$ of spinorial representation. The principal orbits of this $Spin(8)$ action have dimension $14$, and are of course not Hopf fibers. The singular orbits, however, give a pair of antipodal $S^7$'s inside $S^{15}$ and these are Hopf fibers. [Note this action is different from the action $k\cdot gH=gk^{-1}H$ described in (2) above.] From what I have heard, the only subactions of the transitive $Spin(9)$ action on $S^{15}$ that preserve a fixed Hopf fiber (and hence its antipodal fiber as well) are actions by some $Spin(8)\subset Spin(9)$ conjugate to the one I have just described. The $Spin(9)$ action on $S^{15}$ is $g_1\cdot g_2H=g_1g_2H$, so the action described in (2) is not a restriction of it; however the statements in (1) seem to not specify any particular action on $S^{15}$, i.e., to my understanding, they show no group can act isometrically and have orbits that are precisely the Hopf fibers.</p> http://mathoverflow.net/questions/100349/constant-scalar-curvature-metrics-in-a-conformal-class/100416#100416 Answer by Renato G Bettiol for Constant scalar curvature metrics in a conformal class Renato G Bettiol 2012-06-23T00:36:53Z 2012-11-22T17:14:08Z <p>Yes, as the OP mentions, in many cases the solution to the Yamabe problem <strong>may not be unique</strong> (we obviously ignore the effect of rescaling a metric). Perhaps a quick review of some non-uniqueness results could be of interest. Recall that the existence of solution to the Yamabe problem was proved by finding a <em>minimizer</em> of the Hilbert-Einstein functional on unit volume metrics in the conformal class $[g]$, however any critical points give a solution. For example, in the conformal class of an Einstein metric (except the round sphere), the solution is unique. On the one hand,</p> <ul> <li><a href="http://www.math.sunysb.edu/~anderson/yamabe.pdf" rel="nofollow">Anderson</a> proved that on generic conformal classes, the solution is unique.</li> </ul> <p>On the other hand, there are many non-uniqueness results, e.g.,</p> <ul> <li><p>Ambrosetti and Malchiodi (JFA, 1999) and Berti and Malchiodi (JFA, 2001) proved non-uniqueness results for conformal classes of deformations of the round metric on spheres;</p></li> <li><p>Pollack (CAG, 1993) proved existence of arbitrarily small $C^0$ perturbations of any given metric (on any closed manifold!), with arbitrarily large number of solutions in its conformal class;</p></li> <li><p>Schoen (1991) proved that taking the product $S^1\times S^n$ with the product of round metrics, and making the radius $r$ of $S^1$ go to $+\infty$, the number of solutions also goes to $+\infty$ as $r\to+\infty$;</p></li> <li><p>More recently, <a href="http://arxiv.org/abs/1012.1497" rel="nofollow">de Lima, Piccione and Zedda</a> obtained a sort of generalization, proving existence of infinitely many <em>bifurcation points</em> for many $1$-parameter family of product metrics, obtained by scaling one of the factors. In particular, this yields existence of a countable set of metrics in this family in which conformal class there exist <em>at least three</em> distinct solutions to the Yamabe problem;</p></li> <li><p>Even more recently, in a <a href="http://arxiv.org/abs/1107.5335" rel="nofollow">joint paper with Piccione</a>, we obtained a similar result (infinitely many bifurcations) for families of homogeneous metrics on spheres. Recall that essentially all homogeneous metrics on a sphere are given by scaling the round metric by some factor in the direction of the Hopf fibration, e.g., take $S^3\to S^{4n+3}\to HP^n$ and shrink the round metric on $S^{4n+3}$ in the vertical directions by a factor of $t^2$, obtaining a metric $g_t$. Then as $t\to 0$ (i.e., as the fibers collapse), there are infinitely many values of $t$ for which the Yamabe problem in $[g_t]$ has <em>at least 3 solutions</em>. In a preprint to appear soon (which I'd be happy to discuss privately, if you're interested) we extend this result to any homogeneous fibration $K/H\to G/H\to G/K$ where $K/H$ has positive scalar curvature.</p></li> </ul> <p>In most cases above, where explicit non-uniqueness of solutions to the Yamabe problem is proven, it is easy to give a qualitative description of the values of the (constant) scalar curvatures of the various solutions. For general results, however, the best available are the type of compactness results of Brendle, Khuri, Marques, Schoen etc mentioned in macbeth's answer.</p> http://mathoverflow.net/questions/108673/open-problems-about-cmc-hypersurfaces-with-symmetries Open problems about CMC hypersurfaces with symmetries? Renato G Bettiol 2012-10-02T23:18:47Z 2012-10-04T02:55:22Z <p>Recently, <a href="http://arxiv.org/pdf/1204.5007v3.pdf" rel="nofollow">Andrews and Li</a> announced a complete classification of CMC ($H=const.$) tori in $S^3$, confirming a conjecture of Pinkall and Sterling. Their main result is that any such torus is rotationally symmetric, and this reduces the problem to a previous classification due to <a href="http://arxiv.org/pdf/0903.1321.pdf" rel="nofollow">Perdomo</a>, also studied by <a href="http://www.cims.nyu.edu/~rhynd/symmetryS3.pdf" rel="nofollow">Hynd, Park and McCuan</a>. [As a side note, some of the techniques used are similar to <a href="http://arxiv.org/pdf/1203.6597v3.pdf" rel="nofollow">Brendle's solution of the Lawson Conjecture</a>, that states that the only minimal ($H=0$) torus in $S^3$, up to congruences, is the Clifford torus $S^1(\tfrac{1}{\sqrt2})\times S^1(\tfrac{1}{\sqrt2})$.] It is also known that there are CMC tori in $S^3$ that are not congruent to the family $S^1(r)\times S^1(\sqrt{1-r^2})$ of Clifford tori, so, in a certain sense, their only (continuous) symmetries are encoded in an isometric circle action, as confirmed by Andrews' and Li's result.</p> <p>Given this context, my question is:</p> <blockquote> <p>Are there open questions, conjectures, etc., regarding existence or non-existence of CMC hypersurfaces with symmetries in highly symmetric compact manifolds, analogously to the above case of CMC tori in $S^3$? Actually, I'd be particularly interested in questions regarding <em>non-minimal</em> CMC embeddings ($H=const\neq 0$).</p> </blockquote> <p>To be (slightly) more precise, I am interested in knowing if there are questions of the form "Let $(M,g)$ be a compact manifold with many symmetries (i.e., a "big" Lie group acts isometrically), and consider embeddings of some manifold $N$ into $M$. Then if $N\subset M$ is CMC, it must have certain symmetries (and perhaps cannot have other symmetries)." In other words, are there any analogues of (or questions similar to) the conjecture of Pinkall and Sterling that CMC tori in $S^3$ are rotationally symmetric, but for other (perhaps higher dimensional) compact manifolds? Perhaps some of these questions are disguised as questions about Mean Curvature Flow?</p> <p>Notice also that if $N\subset M$ is an orbit of an isometric group action, then it is automatically CMC. However, there could be other CMC embeddings of $N$ into $M$, not congruent to any orbits (e.g., with less symmetries, as the tori above). Should one expect any general behavior for this other CMC embeddings? Are there any known results, questions or conjectures in this direction?</p> <hr> <p>I apologize for the somewhat vague question, but none of the above references seems to risk any further conjectures or indicates natural extensions of their results. Also, a quick google search and inverse citations on MathSciNet don't give much in terms of open questions. I was wondering if there is a reason behind it... Any references to recent survey-type papers related would also be very appreciated! </p> http://mathoverflow.net/questions/106902/a-result-of-soul-theorem-right/106939#106939 Answer by Renato G Bettiol for a result of soul theorem,right? Renato G Bettiol 2012-09-11T18:31:19Z 2012-09-11T23:26:22Z <p>If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:</p> <blockquote> <p>If $X$ is a closed manifold of positive sectional curvature and $Y\subset X$ is a codimension one totally geodesic submanifold that disconnects $X$, then $X$ is homeomorphic to a sphere.</p> </blockquote> <p>This follows, as the OP suggests, from the Soul Argument of Cheeger-Gromoll, extended to Alexandrov spaces by Perelman (see, e.g., <a href="http://www.math.psu.edu/petrunin/papers/alexandrov/perelmanASWCBFB2+.pdf" rel="nofollow">Section 6</a> of Perelman's notes). As mentioned in the comments, Cheeger-Gromoll's version of the argument actually suffices to get the conclusion.</p> <p>A few details: denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere.</p> <hr> <p><strong>edit</strong> (to answer GB's comment): As discussed above, if $C$ a compact Alexandrov space with curvatures $\geq k>0$, then the soul $S=\{p\}$ is a point. Moreover, according to Perelman, the pairs $(C,\partial C)$ and $(\overline K(\Sigma_S),\Sigma_S)$ are homeomorphic (see <a href="http://www.math.psu.edu/petrunin/papers/alexandrov/perelmanASWCBFB2+.pdf" rel="nofollow">6.2</a> for proof), where $\Sigma_S$ is the space of directions at the soul and $\overline K(\Sigma_S)$ is the closure of the topological cone over $\Sigma_S$, i.e., the join of $\Sigma_S$ and a point. If $C$ is a manifold, the space of directions are spheres, so $(\overline K(\Sigma_S),\Sigma_S)$ is simply a pair $(D,\partial D)$, where $D$ is a disk.</p> http://mathoverflow.net/questions/106360/examples-of-manifolds-with-effective-circle-actions/106376#106376 Answer by Renato G Bettiol for Examples of manifolds with effective circle actions? Renato G Bettiol 2012-09-04T19:13:02Z 2012-09-04T19:31:35Z <p>As already pointed out, whenever you have a compact group $G$ acting on a manifold $M$, this action can be made isometric by constructing a metric on $M$ via a standard averaging process. In particular, if you have a (effective) circle action on $M$, you can find a metric $g$ on $M$ such that this action is isometric. In other words, the isometry group of $(M,g)$ contains a circle. This should provide you with plenty of examples: intuitively, a manifold that carries a circle action is one that has (at least) a "circle" worth of symmetries.</p> <p>For example, take $M$ to be any rotationally symmetric surface in $\mathbb R^3$, e.g., a round sphere $S^2$ or a torus $S^1\times S^1$. Then $M$, by definition, has an isometric (effective) circle action. Moreover, these are the only (orientable) manifolds of dimension $2$ with an effective circle action. This follows from the fact that if we have a torus action on $M$, then the Euler characteristic of $M$ is $\chi(M)=\chi(M^T)$, where $M^T$ is the fixed point set of the torus. Since $M^T$ is a totally geodesic submanifold of $M$, it can be a circle or a finite collection of points. Thus, $\chi(M)\geq 0$, which means $M$ is homeomorphic to $S^2$, $\mathbb RP^2$, $S^1\times S^1$ or the Klein bottle $K^2$, according to orientability and $\chi(M)=2,1,0$, respectively. This is the classification you ask for in dimension equal to $2$.</p> <p>Classifying compact connected manifolds of any dimension with a circle action and nothing else seems rather too general. More interesting problems are, for example, classifying manifolds with an isometric circle action that also have, e.g., some curvature condition. For example, if $\dim M=4$ and $M$ has a metric of positive curvature that has an isometric circle action, then <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214443064" rel="nofollow">Hsiang-Kleiner</a> proved that $M$ must be homeomorphic to $S^4$ or $\mathbb C P^2$.</p> http://mathoverflow.net/questions/105292/minimal-distance-spheres-in-complex-projective-spaces Minimal distance spheres in complex projective spaces Renato G Bettiol 2012-08-23T02:41:30Z 2012-08-23T17:36:58Z <blockquote> <p>My question has to do with distance spheres in $\mathbb CP^{n+1}$. I am interested in knowing what is the radius $r$ of a distance sphere $S(r)$ around a point that makes it a <em>minimal submanifold</em> of $\mathbb CP^{n+1}$. It is known that, in this type of geometric situation, the distance spheres have constant mean curvature and some distance sphere will be minimal, and my problem is determining at what exact radius $r_0$ that happens. I thought of two possible approaches to this problem, and they seem to provide different answers; the first gives $r_0=\arctan\sqrt{2n+1}$ and the second $r_0=\pi/4$. Although I tend to believe the second is correct, I would like to understand what is going wrong here, or what I am missing...</p> </blockquote> <h2>Distance spheres in $\mathbb CP^{n+1}$:</h2> <p>Consider $\mathbb CP^{n+1}$ with the standard Fubini-Study metric $g_{FS}$ and distance spheres around a point $p\in\mathbb CP^{n+1}$. These are isometric embeddings $f_r\colon S^{2n+1}\to\mathbb CP^{n+1}$, parameterized by their radius $r\in ]0,\pi/2[$, that foliate $\mathbb CP^{n+1}$, and as $r\to 0$ collapse to the (round) point $p$ and as $t\to\pi/2$, these spheres collapse to $\mathbb CP^n\subset\mathbb CP^{n+1}$. They also happen to be (principal) orbits of a cohomogeneity one action of $\mathrm{SU}(n+1)$, whose singular orbits are the fixed point $p$ and $\mathbb CP^n$.</p> <p>It is well-known that the metric in such distance spheres is a <em>Berger metric</em>, i.e., obtained by shrinking the round metric in the direction of the Hopf fibers, see e.g. Petersen's book. In fact, one can compute the metric on the distance sphere of radius $r$ to be $$g_r:=f_r^*(g_{FS})=\sin^2 r (g+(\cos^2 r) h),$$ where we decompose the round metric as $g_{S^{2n+1}}=g+h$, such that $h$ is the component in the direction of the Hopf fiber and $g$ is the component in the directions orthogonal to the Hopf fiber.</p> <hr> <h2>First approach:</h2> <p>To find a minimal $(S^{2n+1},g_r)$ inside $\mathbb CP^{n+1}$, we compute its mean curvature, by computing its second fundamental form. According to multiple sources (e.g. <a href="http://www.math.shimane-u.ac.jp/memoir/33/maeda.pdf" rel="nofollow">Maeda's paper, p. 38</a> or <a href="http://people.math.uni-bonn.de/karcher/RiemCompConst.pdf" rel="nofollow">Karcher's survey p. 220</a>), the second fundamental form $A_r$ of $(S^{2n+1},g_r)$ is given by: $$A_r(\xi)=(2\cot 2r)\xi, \quad A_r(u)=(\cot r)u,$$ where $\xi$ is the vector tangent to the Hopf fiber and $u$ is any tangent vector orthogonal to $\xi$. Therefore, since we can form an orthonormal basis of eigenvectors of $A_r$ with $1$ vector in the direction of $\xi$ and $2n$ vectors orthogonal to $\xi$, we have that the mean curvature is the sum of the corresponding eigenvalues: $$H=2\cot2r+2n\cot r=0 \Leftrightarrow r=\arctan\sqrt{2n+1}.$$</p> <h2>Second approach:</h2> <p>According to a theorem of Hsiang (<a href="http://www.pnas.org/content/56/1/5.full.pdf" rel="nofollow">Hsiang, p. 6</a>), a $G$-orbit is minimal iff it has extremal volume among orbits of the same type. The volume of $(S^{2n+1},g_r)$ can be computed using Fubini's Theorem, as $$Vol(S^{2n+1},g_r)=Vol(\mathbb CP^n) \ Vol(\mbox{Hopf fiber}),$$ and the Hopf fiber is a circle of length $2\pi\sin r\cos r$. Differentiating the above with respect to $r$, since the only factor depending on $r$ is the volume of the Hopf fiber, we find that $(S^{2n+1},g_r)$ has extremal volume (and is hence minimal) iff $r=\pi/4$.</p> http://mathoverflow.net/questions/100804/deforming-isometric-embeddings-in-low-codimension Deforming isometric embeddings in low codimension Renato G Bettiol 2012-06-27T20:19:04Z 2012-06-27T20:19:04Z <p>Let $F:M\to \mathbb R^N$ be an embedding. This embedding induces a metric $g_F=dF\cdot dF$ on $M$, that turns $F$ into an <em>isometric</em> embedding. Probably the hardest part of the proof of the Nash Embedding Theorem is to prove a deformation result of the following type (see e.g. Thm 1.1 <a href="http://maths-old.anu.edu.au/research.publications/proceedings/040/CMAproc40-andrews.pdf" rel="nofollow">here</a>):</p> <blockquote> <p><strong>Thm:</strong> If $F:M\to\mathbb R^N$ is a <em>free</em> global (analytic) embedding and $h$ is a $C^k$ $(2,0)$-tensor in a neighborhood of zero, then there exists a $C^k$ map $V:M\to\mathbb R^N$ such that $$g_{F+V}=g_F+h, \quad (*)$$ i.e., $F+V$ is a <strong>global</strong> isometric embedding of $(M,g_F+h)$ into $\mathbb R^N$.</p> </blockquote> <p>The <em>free</em> condition means that the vectors $\partial F/\partial x_i$ and $\partial^2 F/\partial x_i\partial x_j$ are linearly independent at every point. The use of the above result to prove the Nash Embedding Theorem is when the ambient dimension $N$ is large, more precisely, $N\geq n(n+3)/2$. In this case, the <em>free</em> condition implies that the linearized version of $(*)$ is solvable, and then the problem follows by, e.g., the Nash-Moser method. My interest, however, is when $N&lt; n(n+3)/2$. In this case, by <em>free</em> we mean that the vectors $\partial F/\partial x_i$ and $\partial^2 F/\partial x_i\partial x_j$ span a subspace with largest possible dimension (although, in the particular case I'm interested, although they do span such a subspace, some of them are identically zero!). In this case, the linearized system is overdetermined, and may not admit solutions...</p> <blockquote> <p>My question is if (regardless of the linearized system being solvable or not) there are any known conditions under which the Thm above also holds for free embeddings with $N&lt; n(n+3)/2$. In other words, when can one deform (global) isometric embeddings of $(M,g_0)$ in low codimension to (global) isometric embeddings for all metrics near $g_0$?</p> </blockquote> <p>The analogous question for <em>local</em> isometric embeddings is dealt with in a paper by Bryant, Griffiths and Yang (Duke Math, 1983), but I do not know of any results (or counter-examples) regarding the <em>global</em> problem.</p> http://mathoverflow.net/questions/100072/constant-mean-curvature-hypersurfaces-condensing-onto-a-minimal-submanifold Constant Mean Curvature hypersurfaces "condensing" onto a minimal submanifold Renato G Bettiol 2012-06-20T04:29:25Z 2012-06-20T16:07:45Z <p>Let $M$ be Riemannian manifold and $S\subset M$ a minimal submanifold, with $\dim S&lt;\dim M-1$. According to a few references (e.g., <a href="http://math.stanford.edu/~mazzeo/Web/Papers/cmcf2.pdf" rel="nofollow">Mahmoudi, Mazzeo &amp; Pacard</a>), it should not be hard to see that:</p> <blockquote> <p><strong>''The closer a constant mean curvature (CMC) hypersurface of $M$ is to $S$ (in the Hausdorff metric), the larger its mean curvature must be.''</strong></p> </blockquote> <p>I was wondering if this claim is indeed <em>not hard to see</em>, given that I am still unable to find a simple/elementary proof. Any suggestions?</p> <p>Moreover, I was wondering if anything is known about the <em>speed</em> in which the value of the mean curvature $H(t)$ of a family $N_t$ of CMC hypersurfaces of $M$ "condensing" (i.e., collapsing) onto $S$ diverges to $+\infty$. For instance, <strong>is it always the case that <em>both</em> $H$ and $H'$ diverge to $+\infty$</strong> as those hypersurfaces collapse onto the minimal limit submanifold? Perhaps, in some special situation (e.g., if the sequence of CMC hypersurfaces collapsing is a solution to the Mean Curvature Flow (MCF)), this is implied by some property of the MCF?</p> http://mathoverflow.net/questions/96125/positively-curved-metrics-on-s2-times-s2/96145#96145 Answer by Renato G Bettiol for Positively curved metrics on $S^2\times S^2$ Renato G Bettiol 2012-05-06T17:04:37Z 2012-05-06T21:51:17Z <p>As already observed above, if you consider a warped product metric $g_1 + f \cdot g_2$ on $S^2\times S^2$ obtained from a positively curved metric on each factor, it <strong>will not have positive curvature</strong>. This can be seen in the following way. The formula for the sectional curvature of "mixed planes" (i.e., spanned by vectors $X$ and $Y$, one from each factor) with respect to the warped metric is essentially (up to renormalization) the hessian of $f$ in the direction of $Y$, $\mathrm{Hess} \; f(Y,Y)$. Since the domain of $f$ is $S^2$ (which is compact), $f$ will have a minimum and a maximum, so its <strong>hessian cannot be always positive (or negative) definite</strong>. Thus, the curvature of mixed planes with respect to warped product metrics cannot be always $>0$ (or $&lt;0$). You can find a slightly more precise description of this fact on a paper by <a href="http://163.14.246.20/mp/pdf/S13N26.pdf" rel="nofollow">Leysen and Verstraelen "On warped products and a conjecture of H. Hopf." Soochow J. Math. 13 (1987), no. 2, 175–178.</a></p> <p>For a similar reason, a double-warped metric will also not have positive curvature, since the hessians of the warping functions are not definite. The Hopf conjecture is a nasty (but terribly intriguing) problem...</p> <hr> <p>EDIT (due to Anton's comment): One comment above mentioned using Hsiang-Kleiner's result (a positively curved metric on $S^2\times S^2$ cannot have an isometric circle action) to answer the question. This indeed works if we are warping a positively curved metric on one $S^2$ that <strong>has a circle in its isometry group</strong> (e.g. the round metric) with another positively curved metric on the other $S^2$ (and the warping function is defined on this second factor). In this way, there is a circle acting isometrically in the warped metric and Hsiang-Kleiner's result applies. Nevertheless, in general, a positively curved metric on $S^2$ does not have an isometric circle action, so this reasoning cannot be used.</p> http://mathoverflow.net/questions/96062/difference-between-s2-bundles-over-s2-and-cp2-sharp-cp2/96073#96073 Answer by Renato G Bettiol for Difference between $S^2$-bundles over $S^2$ and $CP^2\sharp CP^2$ Renato G Bettiol 2012-05-05T16:08:13Z 2012-05-05T21:07:18Z <p>Maybe a good geometric way to visualize $\mathbb C P^2\# \mathbb C P^2$ that can help you is the following. The $2$-torus $T^2$ acts on $\mathbb C P^2$ with cohomogeneity two. Recall the construction of this action can be seen as having $T^{n+1}$ act on $\mathbb C^{n+1}$ by multiplication coordinate-wise, then restrict the action to the unit sphere $S^{2n+1}\subset \mathbb C^{n+1}$ and take a quotient of $T^{n+1}$ acting on $S^{2n+1}$ by the circle in $T^{n+1}$ whose action on $S^{2n+1}$ is the Hopf action that gives $\mathbb C P^n=S^{2n+1}/S^1$. In this way we get $T^n$ acting on $\mathbb C P^n$ (and the example above is the case $n=2$). The orbit space of this $T^2$ action on $\mathbb CP^2$ is a spherical triangle with the three angles equal to $\pi/2$. Each of the vertices corresponds to a torus fixed point in the manifold (and the sides have singular isotropy a circle).</p> <p>Now, take two copies of this spherical triangle (corresponding to two copies of $\mathbb CP^2$ - and if we want to be careful I think they have to have opposite orientations), remove a neighborhood of one vertex in each triangle (corresponding to removing a ball in each copy of $\mathbb CP^2$, since the orbits of these vertices are points) and glue them together along this deleted neighborhoods. The resulting Alexandrov space is a rectangle (since the two remaining vertices on each of the triangles glued together had angle $\pi/2$); and this is exactly the orbit space of the cohomogeneity $2$ torus action on $\mathbb C P^2\#\mathbb CP^2$. [Since you were asking, in particular, you get lots of circle sub actions, but with larger orbit spaces.]</p> <p>This picture is similar to the one you had of the cohomogeneity one manifold $S^3\times_{S^1} S^2$ in the following way. Imagine we are foliating the rectangle (orbit space of $T^2$-action on $\mathbb C P^2\#\mathbb CP^2$) by vertical segments, from side to side. Each of the vertical (actually, any) sides of the rectangle is a cohomogeneity one manifold with singular orbits equal to points and principal isotropy group a circle. Indeed, they are equivariantly diffeomorphic to $S^2$ with the $S^1$ rotation action. In this way, you get two $S^2$'s on opposite sides, and in the "middle", we're left with vertical segments that correspond to cohomogeneity one manifolds with singular orbits equal to circles and trivial principal isotropy groups. They are in fact spheres $S^3$ with the standard $T^2$-action. So we have a similar picture: two $S^2$'s on opposite sides and $S^3$'s in the middle. But this is different from the cohomogeneity one manifold $S^3\times_{S^1} S^2$, where you have an $S^3$ action, even though it has a similar topological structure of two $S^2$'s on opposite sides and $S^3$'s filling the "middle". Sorry for the rather informal description, I hope it helps...</p> http://mathoverflow.net/questions/95637/connected-compact-semisimple-lie-group-finite-fundamental-group/95683#95683 Answer by Renato G Bettiol for connected compact semisimple lie group finite fundamental group Renato G Bettiol 2012-05-01T18:04:50Z 2012-05-02T16:39:40Z <p>I think you can get a much faster (and maybe easier...) proof using Riemannian geometry, as follows: </p> <hr> <p>First, recall that a semi-simple connected Lie group $G$ is compact if and only if its Killing form $B$ is negative-definite (the proof is easy, see, e.g., Thm 2.28 in <a href="http://arxiv.org/pdf/0901.2374.pdf" rel="nofollow">these notes</a>). The side we will use ($G$ compact semi-simple $\Rightarrow$ $B$ neg.-def.) actually follows directly from $B(X,X)=tr(ad(X)\cdot ad(X))$ using an orthonormal basis with respect to an auxiliary bi-invariant metric to compute this trace.</p> <p>Now, the Ricci curvature of any bi-invariant metric on $G$ (that exists because $G$ is compact) can be computed as: $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same <a href="http://arxiv.org/pdf/0901.2374.pdf" rel="nofollow">notes</a>. By the observation above, since $G$ is compact and semi-simple, its Killing form $B$ is negative-definite. Hence the above formula gives $Ric>0$. So, by the Bonnet-Myers Theorem, $G$ must have finite fundamental group. Q.E.D.</p> http://mathoverflow.net/questions/94479/does-every-vector-bundle-allow-a-finite-trivialization-cover/94488#94488 Answer by Renato G Bettiol for Does every vector bundle allow a finite trivialization cover? Renato G Bettiol 2012-04-19T03:37:21Z 2012-04-19T14:11:15Z <p>The answer (to both questions (a) and (b)) is <strong>YES</strong> (assuming $B$ is a <strong>smooth manifold</strong>). A proof can be found on Walschap's book "<a href="http://www.amazon.com/Metric-Structures-Differential-Geometry-Walschap/dp/038720430X" rel="nofollow">Metric Structures in Differential geometry</a>", p. 77, Lemma 7.1.</p> <p>For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in topology, this (and in fact any) cover of an $n$-dim manifold $B$ admits a refinement $\{ V_\alpha\}_{\alpha\in A}$ such that any point in $B$ belong to at most $n+1$ $V_\alpha$'s. Let $\{\phi_\alpha\}$ be a partition of unity subordinate to this cover and denote by $A_i$ the collection of subsets of $A$ with $i+1$ elements. Given $a=\{\alpha_0,\dots,\alpha_i\}\in A_i$, denote by $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b)\lt\phi_{\alpha_0}(b),\dots,\phi_{\alpha_i}(b)$ for all $\alpha\neq\alpha_0,\dots,\alpha_i$. Then the collection of $n+1$ open subsets $U_i:=\cup_{a\in A_i} W_a$ covers $B$ and is such that your bundle restricted to each $U_i$ is trivial.</p> http://mathoverflow.net/questions/75864/invariant-metrics-on-the-sphere/79678#79678 Answer by Renato G Bettiol for Invariant Metrics on the Sphere Renato G Bettiol 2011-11-01T04:21:17Z 2012-03-28T18:53:10Z <h2>Classification of $SU(n+1)$-homogeneous metrics on $S^{2n+1}$:</h2> <p>First, note that for any $n\geq1$, the round metric on the sphere $S^{2n+1}$ can be scaled by $t^2$ in the direction of the Hopf fibers $S^1\to S^{2n+1}\to \mathbb C P^n$, giving rise to a one-parameter family $g_t$ of $SU(n+1)$-homogeneous metrics (so that $g_1$ is the original round metric).</p> <blockquote> <p><strong>Prop.</strong> Any $SU(n+1)$-homogeneous metric on $S^{2n+1}$ is isometric to some $g_t$, $t>0$.</p> </blockquote> <p>In other words, they are always parameterized by (positive) real numbers. Alternatively, these $SU(n+1)$ homogeneous metrics on $S^{2n+1}$ are isometric to distance spheres in the complex projective space $\mathbb C P^{n+1}$.</p> <p>Both of the above statements follow from a more general result, namely W. Ziller's classification of all homogeneous metrics on spheres, see [<em>Homogeneous Einstein metrics on spheres and projective spaces.</em> Math. Ann. 259 (1982), no. 3, 351–358].</p> <hr> <h2>A few more details:</h2> <p>An explicit formula for any $G$-homogeneous metric on the homogeneous space $G/H$, in terms of an $Ad(H)$-invariant decomposition $g= h\oplus p$ of the Lie algebra of $G$ is: $$\langle ,\rangle=h|_{p_0} + \sum_{i=1}^r\alpha_i B|_{p_i},$$ where $p= p_0\oplus\dots\oplus p_r$ is a decomposition so that the $H$ representations on $p_i$ are not equivalent, $H$ acts trivially on $p_0$ and irreducibly on $p_i$, $i=1,\dots r$; $h|_{p_0}$ is any inner product on $p_0$; $B$ is a bi-invariant metric on $G$ and $\alpha_i>0$ are real parameters. Any $G$-homogeneous metric is defined by choosing these parameters. In the case of $SU(n+1)$, this decomposition is $p=p_0\oplus p_1$ and $\dim p_0=1$, so (up to renormalization) there is only one parameter, $\alpha_1$ to be chosen (that I called $t$ above).</p> <p>EDIT. Since I claim the metrics are parameterized by one positive real number (and another answer above claims there must be two real parameters), a clarification is in order here. The point is that, indeed, there are two parameters ($h|_{p_0}$ and $\alpha_1$ in the notation above), nevertheless it is always possible to divide the entire metric by the first one (the number that determines the metric $h$ on the 1-dim space $p_0$), which leaves us with just one parameter. Treating the family as having the $2$ parameters is slightly ambiguous because lots of metrics will be simple rescaling of other ones. In my description above, they are pairwise non-conformal.</p> http://mathoverflow.net/questions/91766/where-can-i-find-a-full-proof-of-the-chern-gauss-bonnet-theorem/91776#91776 Answer by Renato G Bettiol for Where can I find a full proof of the Chern-Gauss-Bonnet theorem ? Renato G Bettiol 2012-03-21T00:57:15Z 2012-03-21T00:57:15Z <p>One reference that seems fairly good and that I just found by googling those key words is <a href="http://www.math.upenn.edu/~alina/GaussBonnetFormula.pdf" rel="nofollow">http://www.math.upenn.edu/~alina/GaussBonnetFormula.pdf</a></p> <p>The first time I learnt this, however, was with these lecture notes: F. Mercuri, P. Piccione, D. V. Tausk, <em>Notes on Morse theory</em>, Publicações Matemáticas do IMPA, Rio de Janeiro, 2001, ISBN 85-244-0178-8; which maybe a little hard to find, but are very nicely written and I like them very much. Though, the proof you are looking for should be widely available elsewhere (google gives thousands of results, and I only looked at the first ones)...</p> http://mathoverflow.net/questions/91397/exotic-spectrum-of-laplace-operator/91416#91416 Answer by Renato G Bettiol for Exotic spectrum of Laplace operator Renato G Bettiol 2012-03-16T21:43:43Z 2012-03-18T15:37:40Z <p>Regarding the asymptotic behavior of the spectrum of the Laplacian (or, as the OP puts it, the behavior at infinity), the most basic result is <strong>Weyl's asymptotic formula</strong> (see <a href="http://www.amazon.com/Eigenvalues-Riemannian-Geometry-Applied-Mathematics/dp/0121706400" rel="nofollow">Chavel's book</a>, p.172): let $(M,g)$ be a compact manifold with $\dim M=n$ and $0=\lambda_0&lt;\lambda_1\leq \lambda_2\leq\dots$ be the eigenvalues of the Laplacian, each distinct eigenvalue repeated according to its multiplicity. Denote by $N(\lambda)=\sum_{\lambda_j\leq\lambda} 1$ the number of eigenvalues (counted with multiplicity) that are $\leq\lambda$. Then</p> <p>$$N(\lambda)\sim vol(M,g)\frac{vol(B^n)}{(2\pi)^n}\lambda^{n/2}, \quad \mbox{as} \quad\lambda\to+\infty,$$</p> <p>where $vol(B^n)=\frac{\pi^{n/2}}{\Gamma(n/2+1)}$ is the volume of the unit ball of $\mathbb R^n$. In particular,</p> <p>$$(\lambda_k)^{n/2}\sim\frac{(2\pi)^n}{vol(B^n)}\frac{k}{vol(M,g)}, \quad \mbox{as}\quad k\to+\infty.$$</p> <blockquote> <p>Thus, the asymptotic behavior of the eigenvalues <strong>cannot be prescribed</strong> - it has to satisfy the above.</p> </blockquote> <hr> <p>Also, as far as I understand, Colin de Verdière's result is stronger than stated. Namely, given <strong>any</strong> compact connected manifold M, with $\dim M\geq 3$, and any finite sequence $0\leq a_1\leq\dots\leq a_k$, there exists a Riemannian metric on $M$ such that the first eigenvalues in the spectrum of its Laplacian are $0\leq a_1\leq\dots\leq a_k$.</p> http://mathoverflow.net/questions/85481/the-first-eigenvalue-of-the-laplacian-for-complex-projective-space/85604#85604 Answer by Renato G Bettiol for The first eigenvalue of the laplacian for complex projective space Renato G Bettiol 2012-01-13T18:05:57Z 2012-03-16T15:33:16Z <p>The spectrum of the Laplacian of $\mathbb C P^n$ with the Fubini-Study metric is</p> <blockquote> <p>$$Spec(\Delta_{\mathbb C P^n})=\{4k(n+k):k\in\mathbb N\} \quad\quad(*)$$</p> </blockquote> <p><strong>So, the first non-zero eigenvalue of $\mathbb C P^n$ is $\lambda_1=4n+4$.</strong></p> <p>Note this matches with the fact that $\mathbb C P^1$, with the FS metric, is isometric to the $2$-sphere of radius $1/2$, whose first non-zero eigenvalue is $\lambda_1=8$.</p> <hr> <p>Let me quote a brief justification of (*) that I had written <a href="http://mathoverflow.net/questions/70900/multiplicity-of-eigenvalues-of-the-laplacian-on-quaternionic-projective-space" rel="nofollow">here</a>:</p> <p><em>Using the classic spherical harmonics theory, one obtains the $k$-th eigenvalue of the $n$-dimensional round sphere $S^n$ to be $k(k+n-1)$, and its multiplicity is $\binom{n+k}{k}-\binom{n+k-1}{k-1}$, see e.g. [Berger, Gauduchon,Mazet, "Le spectre d'une variété riemannienne", Lecture Notes in Mathematics, Vol. 194 Springer-Verlag]. By looking at eigenfunctions of the Laplacian on $S^n$,$S^{2n+1}$ and $S^{4n+3}$ (note they are the unit spheres of $\mathbb R^{n+1}$, $\mathbb C^{n+1}$ and $\mathbb H^{n+1}$) that are respectively invariant under the natural actions of</em> $\mathbb Z_2$, $S^1$ <em>and</em> $S^3$, <em>one can obtain the eigenfunctions hence the $k$-th eigenvalue of the projective spaces $\mathbb R P^n$, $\mathbb C P^n$ and $\mathbb H P^n$, respectively. These are, respectively, $2k(n+2k-1)$, $4k(n+k)$ and</em> $4k(k+2n+1)$.</p> <p>If you can understand some French, you will find a thorough explanation of the above in the book by Berger, Gauduchon, Mazet, "Le spectre d'une variete Riemannienne", Lecture Notes in Math, Springer, vol 194.</p> http://mathoverflow.net/questions/58936/torus-minimizer-of-willmore-energy/89902#89902 Answer by Renato G Bettiol for Torus minimizer of Willmore energy Renato G Bettiol 2012-02-29T20:35:07Z 2012-02-29T20:35:07Z <p>Just a few days ago, Fernando C Marques and Andre Neves posted a <a href="http://arxiv.org/abs/1202.6036" rel="nofollow">preprint</a> on the arxiv in which they (claim to) provide a complete proof of Willmore's Conjecture. I have no idea how much of it has been verified, especially given it is so recent, but the geometric ideas used (min-max theory for minimal submanifolds) seem very elegant.</p> http://mathoverflow.net/questions/128850/positively-curved-manifold-with-almost-extreme-diameter Comment by Renato G Bettiol Renato G Bettiol 2013-05-02T02:14:55Z 2013-05-02T02:14:55Z @J Ge: This is an open problem, if memory doesn't fail me. I actually remember talking about it with Grove in one of our group meetings. Also, I believe that the recent work of Curtis Pro might shed some light on that, although, if I remember well, he assumes almost maximal volume. Perhaps you should contact those two people to figure out the state of the art regarding this question. http://mathoverflow.net/questions/90933/possible-isometries-of-a-positively-curved-s2-times-s2/129325#129325 Comment by Renato G Bettiol Renato G Bettiol 2013-05-02T02:00:48Z 2013-05-02T02:00:48Z @Luis: Thanks for the reference. I'm not sure I have read it before, I'll definitely check it out. By the way, welcome to MathOverflow -- it's great to see more differential geometers around here! http://mathoverflow.net/questions/122987/proper-subgroups-of-rmsud-that-act-transitively-on-rmcpd-1/123147#123147 Comment by Renato G Bettiol Renato G Bettiol 2013-02-28T15:05:23Z 2013-02-28T15:05:23Z @Aakumadula: Sorry, there were a couple typos... I've just fixed them and double checked the list -- it should be fine now. Thanks! http://mathoverflow.net/questions/96062/difference-between-s2-bundles-over-s2-and-cp2-sharp-cp2/96073#96073 Comment by Renato G Bettiol Renato G Bettiol 2013-02-23T03:32:04Z 2013-02-23T03:32:04Z For more, see: <a href="http://www.math.binghamton.edu/somnath/Notes/S2S2.pdf" rel="nofollow">math.binghamton.edu/somnath/Notes/S2S2.pdf</a> http://mathoverflow.net/questions/96062/difference-between-s2-bundles-over-s2-and-cp2-sharp-cp2/96073#96073 Comment by Renato G Bettiol Renato G Bettiol 2013-02-23T03:29:30Z 2013-02-23T03:29:30Z Perhaps another quick comment that may be useful for future reference is that $\mathbb C P^2\#\mathbb C P^2$ does not admit a cohom 1 action, only $\mathbb C P^2\#-\mathbb C P^2$ does (the difference is that the latter has the orientation opposite from the one given by the complex structure in one of the summands). The manifold $\mathbb C P^2\#\mathbb C P^2$ can be written as a biquotient $S^3\times S^3//T^2$, and is <i>not</i> diffeomorphic to the nontrivial $S^2$-bundle over $S^2$, which coincides with $\mathbb C P^2\#-\mathbb C P^2$. http://mathoverflow.net/questions/121107/banach-manifold/121115#121115 Comment by Renato G Bettiol Renato G Bettiol 2013-02-07T22:32:06Z 2013-02-07T22:32:06Z @Pietro: Tracing back, this approach with exponential maps is what Ebin &amp; Marsden use in their Ann. of Math. paper of 1970. That's probably where the whole thing came from. http://mathoverflow.net/questions/121107/banach-manifold/121115#121115 Comment by Renato G Bettiol Renato G Bettiol 2013-02-07T22:22:12Z 2013-02-07T22:22:12Z @Pietro: Although most classic constructions do require an auxiliary Riemannian metric at some point, as far as I remember they don't use the exponential map. For example, the way Hirsch does it in his book works like that. Another way of looking at spaces of functions $f\colon M\to N$ is as sections of the trivial bundle $M\times N$ over $M$, using Palais' VBNs (vector bundle neighborhoods), also without exponential maps. I learnt the faster construction above from Gerard Misiolek, during some seminars he gave on diffeomorphism groups (so there maybe some limitations to this method). http://mathoverflow.net/questions/120977/topological-structure-of-lie-algebra Comment by Renato G Bettiol Renato G Bettiol 2013-02-06T15:02:53Z 2013-02-06T15:02:53Z I presume you are talking about a finite-dimensional Lie group $G$, in which case its Lie algebra $\mathfrak g$ is a finite-dimensional vector space, hence has a standard topology coming from (any) norm. Note that $\mathfrak g$ is canonically isomorphic to the tangent space to $G$ at the identity $T_1 G$, which also gives you a way to see its topology. http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Comment by Renato G Bettiol Renato G Bettiol 2013-01-29T21:25:55Z 2013-01-29T21:25:55Z @Otis: PS: You may want to replace the link to that problem set from last summer at the MSRI... It looks like even their own link from the MSRI summer school webpage is broken! I might have an offline copy of that worksheet if you want it. http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Comment by Renato G Bettiol Renato G Bettiol 2013-01-29T21:21:21Z 2013-01-29T21:21:21Z @Otis: Thanks for elaborating to such extent with many links and references! Hope to see you in April here at ND! http://mathoverflow.net/questions/115973/is-there-any-generalization-of-warp-product Comment by Renato G Bettiol Renato G Bettiol 2012-12-10T23:14:14Z 2012-12-10T23:14:14Z @GB: Conversely, these (the CROSS) are the only compact cohomogeneity one manifolds with a fixed point. However, it seems like your condition is more general, i.e., probably may hold in the case where the geodesic spheres are not homogeneous. I couldn't come up with a different example though... http://mathoverflow.net/questions/115973/is-there-any-generalization-of-warp-product Comment by Renato G Bettiol Renato G Bettiol 2012-12-10T23:10:52Z 2012-12-10T23:10:52Z @GB: Thanks for the explanations. I don't quite see a characterization yet, but for sure all the CROSS (compact rank one symmetric spaces) with their standard Fubini metric satisfy your condition. Namely, $S^n$, $\mathbb CP^n$ and $\mathbb HP^n$, $\mathbb{Ca}P^2$ have a (linear) cohomogeneity one action with a fixed point (the point you call $p$), and the geodesic spheres around $p$ are principal orbits of the action, which eventually collapse to a lower dimensional projective space via the Hopf map. Consequently, these spheres are homogeneous, hence the principal curvatures are constant... http://mathoverflow.net/questions/115973/is-there-any-generalization-of-warp-product Comment by Renato G Bettiol Renato G Bettiol 2012-12-10T17:03:39Z 2012-12-10T17:03:39Z @GB: Regarding a metric characterization, an equivalent (but silly) way of encoding the above property the doesn't (seem to) use shape operators is to say that &quot;the eigenvalues of the hessian of the distance function $d(x)=dist(p,x)$ are constant on the levelsets of $d$&quot;. This is silly because the Hessian of the distance function is precisely the second fundamental form of the geodesic sphere, but might help?... http://mathoverflow.net/questions/115973/is-there-any-generalization-of-warp-product Comment by Renato G Bettiol Renato G Bettiol 2012-12-10T17:00:00Z 2012-12-10T17:00:00Z @GB: Just to make sure I understand your question correctly, what you want is a characterization of Riemannian manifolds $M$ such that all the geodesic spheres $S_r=\partial B(p,r)$ around one point $p\in M$ (every point?) have the following property: at every point $x\in S_r$, all the eigenvalues of the second fundamental form of $S_r$ are equal and depend only on $r$. Is that what you are asking for? This would be much much stronger than any warped product -- actually, I suspect it implies $M$ is a space form. http://mathoverflow.net/questions/114825/homogeneity-of-the-hopf-fibration-s7-to-s15-to-s8 Comment by Renato G Bettiol Renato G Bettiol 2012-11-29T02:20:27Z 2012-11-29T02:20:27Z @Ryan: Thank you! I will vote to close the question then, and, if possible, leave it here for future reference. This issue confused me for a couple days, and I suppose it might be helpful for someone else...