User marco radeschi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:29:08Z http://mathoverflow.net/feeds/user/3465 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102643/linear-parabolic-equations-smooth-dependence-from-initial-data LINEAR Parabolic equations. Smooth dependence from initial data Marco Radeschi 2012-07-19T11:16:16Z 2013-05-25T08:22:00Z <p>I am looking for results that show smooth dependence of a solution to a parabolic equation, from the initial data.</p> <p>More specifically I have the following problem:</p> <p>CONSIDER spaces $P:=\mathbb{R}^k$ ("parameter space"), $\Omega:=\mathbb{R}^n$ ("actual space") and $T=[0,t_0]$ ("time").</p> <p>Consider also $u_0,f_0\in C^{\infty}(P\times \Omega)$, and $L$ a linear elliptic operator on $\Omega$, whose components depend smoothly on $P$. For every $p\in P$, there is a solution $u_p\in C^{\infty}(\Omega\times T)$ of the PDE</p> <p>${\partial\over \partial t}u_p+L_p u_p=f_p \qquad on\;\{p\}\times\Omega\times T$</p> <p>$u_p=(u_0)_p\qquad on\; \{p\}\times\Omega\times\{0\}$</p> <p>QUESTION: is the function $\overline{u}(p,x,t):=u_p(x,t)$ a function in $C^{\infty}(P\times\Omega\times T)$?</p> <p>The problem comes from the following, geometric situation: I have a space $P\times \Omega$ with a metric on it, and $L$ is just the Laplacian $\Delta$ of $P\times \Omega$, restricted to the levels $\{p\}\times \Omega$.</p> <p>Any reference will be very welcome. Thanks in advance.</p> http://mathoverflow.net/questions/12900/surjectivity-of-the-normal-exponential-map Surjectivity of the normal exponential map Marco Radeschi 2010-01-25T04:11:31Z 2013-05-22T17:44:13Z <p>Given an isometric (in the Riemannian way) immersion $f:N\rightarrow M$ between complete, smooth riemannian manifolds, are there conditions on $M$, $N$, $f$, such that the normal exponential map $\mathrm{exp}^{\nu}:\nu(N)\rightarrow M$ is surjective?</p> <p>I'm interested in the case of $f$ being not closed. An example of non surjectivity is given by $f:\mathbb{R}\rightarrow\mathbb{R}^2$, where f is the logarithmic spiral. In this case, the normal exponential map misses the origin.</p> http://mathoverflow.net/questions/120419/submersions-from-compact-flat-manifold/120471#120471 Answer by Marco Radeschi for Submersions from compact flat manifold Marco Radeschi 2013-01-31T22:51:17Z 2013-01-31T22:51:17Z <p>So, the question was answered by Alexander Lytchak. I am writing down the answer for the sake of completeness.</p> <p>Consider the fibration between the universal covers $F'\to\tilde{M}\to \tilde{N}$. $\tilde{M}$ is contractible and $\tilde{N}$ is simply connected, thus we can apply the Serre spectral sequence with integral coefficients, and from it we obtain that $F'$ and $\tilde{N}$ are contractible. In fact, if <code>$H^*(F')$</code> has cohomological dimension $a$, and <code>$H^*(\tilde{N})$</code> has cohomological dimension $b$, then <code>$H^*(\tilde{M})$</code> would have cohomological dimension $a+b$ and this has to be $0$.</p> <p>Then $N$ is acyclic, and from the exact sequence in homotopy so is $F$. Moreover we have and exact sequence</p> <p>$$1\to\pi_1(F)\to \pi_1(M)\to \pi_1(N)\to 1$$</p> <p>where $\pi_1(M)$ is a Bieberbach group, i.e. a torsion free group with a finite index normal abelian subgroup. As for $\pi_1(F)$, a subgroup of a Bieberbach group is again a Bieberbach group and therefore $F$ is homeomorphic to a flat manifold.</p> <p>As for $N$, one can prove that there exists a normal abelian subgroup of $\pi_1(N)$ of finite index. Moreover, $\pi_1(N)$ is torsion free, since otherwise there would be a finite cyclic subgroup acting on the contractible manifold $\tilde{N}$ without fixed points. This is not possible, and therefore $\pi_1(N)$ is a Bieberbach group. Again, this implies that $N$ is a Bieberbach manifold.</p> http://mathoverflow.net/questions/120419/submersions-from-compact-flat-manifold Submersions from compact flat manifold Marco Radeschi 2013-01-31T14:28:32Z 2013-01-31T22:51:17Z <p>Let $M=\mathbb{R}^n/G$ be a closed flat manifold, and let $F\to M \to N$ be a locally trivial submersion, where $F$ and $N$ are closed manifolds.</p> <p>My question is simple: are $F$ and $N$ homeomorphic to flat manifolds?</p> <p>This question seems quite natural to me, and I would expect the answer to this fact to be well known.</p> <p>Any reference will be welcome. Thank you in advance.</p> http://mathoverflow.net/questions/116636/dense-subgroups-of-lie-groups Dense subgroups of Lie Groups Marco Radeschi 2012-12-17T19:28:53Z 2012-12-17T22:34:30Z <p>SETUP: Let $G$ be a connected Lie group, and $H\subset G$ be a FINITELY GENERATED dense subgroup.</p> <p>I am interested in knowing what kind of information one can infer on the complexity of $H$.</p> <p>I am especially interested in the case in which $G$ is simply connected, non compact, and non diffeomorphic to $\mathbb{R}^n$. After some research online, the only result I found in this direction is in "On dense free subgroups of Lie groups", by Breuillard, E. and Gelander, T.. Here the authors prove that if $G$ is not solvable, and $H\subset G$ is finitely generated and dense, then it contains a free group of rank $r=2\dim G$.</p> <p>Does anyone have other references of result in this direction? I hope to find results of the type "such a group $H$ needs to be at least this complicated".</p> <p>In the case I am interested in, $H$ is the fundamental group of a compact manifold, so I have an "upper bound" on the complexity of "H". Now I want a "lower bound", if this makes any sense.</p> <p>Thank you in advance!</p> http://mathoverflow.net/questions/8846/proofs-without-words/16064#16064 Answer by Marco Radeschi for Proofs without words Marco Radeschi 2010-02-22T15:25:59Z 2012-11-13T01:22:31Z <p>The sequence of pictures</p> <p><img src="http://img17.imageshack.us/img17/9169/img2qg.png" alt="intersection of 3 diangles"> <img src="http://img839.imageshack.us/img839/8629/img4z.png" alt="intersection of 2 diangles"> <img src="http://img255.imageshack.us/img255/5643/img5bb.png" alt="intersection of 2 diangles"> <img src="http://img844.imageshack.us/img844/4253/img6i.png" alt="intersection of 2 diangles"></p> <p>proves the area formula for spherical triangles $A=\hat{ABC}+\hat{BCA}+\hat{CAB}-\pi$.</p> http://mathoverflow.net/questions/111000/non-exoticness-of-a-diffeomorphism-of-a-sphere (Non)-exoticness of a diffeomorphism of a sphere Marco Radeschi 2012-10-29T14:11:07Z 2012-10-29T22:57:07Z <p>Suppose you have a standard sphere $S^n$ and a "standard" $S^{n-2}\subset S^n$. I am really thinking about $S^{n}\subset \mathbb{R}^{n+1}$ the usual sphere, and $S^{n-2}=S^n\cap \{x_0=x_1=0\}$. Let $S^1$ be the circle "orthogonal" to $S^{n-2}$, i.e. $S^1=S^n\cap span\{x_0, x_1\}$. Then $S^n$ gets decomposed by the hypersurfaces $S_t:=S^{n-2}(\cos t)\times S^1(\sin t)$, i.e. the distance tubes around $S^{n-2}$ and $S^1$.</p> <p>Suppose now that $\phi:S^n\to S^n$ is a diffeomorphism such that:</p> <ul> <li>$\phi$ fixes $S^{n-2}$ pointwise: <code>$\phi\big|_{S^{n-2}}= id\big|_{S^{n-2}}$</code>.</li> <li>$\phi$ sends the hypersurfaces $S_t$ to themselves (it is not the identity though).</li> </ul> <p><strong>Question 1:</strong> is it true that $\phi$ is homotopic to an isometry of $S^n$ in $Diff(S^n)$?</p> <p>Here is a (probably) much stronger assumption on $\phi$: fix a basis $x_0,\ldots x_n$ of $\mathbb{R}^{n+1}$, and suppose that $\phi:S^n\to S^n$ preserves any subsphere "main subsphere" $S^{n-k}=S^n\cap \{x_{i_1}=\ldots x_{i_k}=0\}$.</p> <p><strong>Question 2:</strong> is it true that $\phi$ is homotopic to an isometry of $S^n$ in $Diff(S^n)$?</p> <p>Regarding this second question, my approach was to start deforming $\phi$ to be an isometry on the smallest "main subspheres", and hopefully going up in dimension, but this requires me to know that $\pi_i(Diff(T^k))=0$, $i>0$, where $T^k$ is a $k$-dimensional torus. So here is a third, kind of related, question:</p> <p><strong>Question 3:</strong> is it true that $\pi_i(Diff(T^k))=0$ for every torus $T^k$ and every $i>0$?</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover Euler characteristic and universal cover Marco Radeschi 2012-10-09T12:03:18Z 2012-10-14T20:58:08Z <p>Let $M$ be a compact manifold, let $\tilde{M}$ be its universal cover, and suppose that the Euler characteristic $\chi(\tilde{M})=0$. My question is: does this imply that $\chi(M)=0$? This is clear if $\pi_1(M)$ is finite, but I am interested in the case $|\pi_1(M)|=\infty$.</p> <p>It might not feel right, but I can't think of any counterexample, either.</p> <p>Thank you very much in advance!</p> <hr> <p>EDIT: I was rightfully asked what I mean by Euler characteristic of the (non compact) manifold $\tilde{M}$. My answer right now is: the one you want!</p> <p>What I am thinking of, is $\chi(\tilde{M})=\sum_i (-1)^i\dim H_i(\tilde{M},k)$, with $k=\mathbb{Q}$ or $\mathbb{R}$, and $H_i$ are either the usual or the compactly supported cohomology groups.</p> <p>In my case, $\tilde{M}$ retracts to a compact Lie group.</p> http://mathoverflow.net/questions/108975/how-do-you-intepret-kill-a-cohomology-class-intuitively-for-attaching-an-n-cell/108989#108989 Answer by Marco Radeschi for How do you intepret "kill a cohomology class" intuitively for attaching an n-cell? Marco Radeschi 2012-10-06T10:56:56Z 2012-10-06T10:56:56Z <p>First of all, by Morse theory you know that $\mathcal{M}^+$ is obtained by attaching an $m$-cell to $\mathcal{M}^-$, i.e. $\mathcal{M}^+=\mathcal{M}^-\cup_{\phi}e$, where $e=D^m$ is an $m$-cell and $\phi:\partial(e)=S^{m-1}\to \mathcal{M}^-$ is some attaching map. Therefore you can compute $H^{*}(\mathcal{M}^+,\mathcal{M}^-)$ by excision (you excise $\mathcal{M}^-\setminus\phi(S^{m-1})$), and get that $H^{*}(\mathcal{M}^+,\mathcal{M}^-)\simeq H^{*}(D^m,S^{m-1})$ that has exactly one nonzero cohomology in degree $m$.</p> <p>As for the second question, I think it's more intuitive looking at homology (it's the same since we are looking at coefficients in a field). There you have the sequence</p> <p>$0\to H_m(\mathcal{M}^-)\to H_m(\mathcal{M}^+)\to H_m(\mathcal{M}^+,\mathcal{M}^-)\stackrel{\partial}{\to} H_{m-1}(\mathcal{M}^-)\to M_{m-1}(\mathcal{M}^+)\to 0$.</p> <p>Moreover, the generator of the unique nonzero class of $H_m(\mathcal{M}^+,\mathcal{M}^-)$ can be thought of the fundamental class $[e]$ of the cell you attached, and $\partial[e]=\phi_*[\partial e]$. In other words, the boundary operator sends $[e]$ essentially to the class of its topological boundary. If $\partial[e]=0$, it means that $\phi_{*}[\partial e]$ is a boundary in $\mathcal{M}^-$, i.e. there exists an $m$-cycle $e_-$ entirely contained in $\mathcal{M}^-$ such that $\partial[e]=\partial[e_-]$. In particular $\partial[e-e_-]=0$ and you can say that <em>$e$ can be completed to a cycle, contained in $\mathcal{M}^-$ away from the critical point}</em>.</p> <p>Think of the usual example of the torus, with the morse function being the heigth $z$. Let $p_1$ be the second critical point (the first critical point of index one) with critical value $z_1$, let $\mathcal{M}^-=z^{-1}([0,z_1-\epsilon])$ and $\mathcal{M}^+=z^{-1}([0,z_1+\epsilon])$. Then $\mathcal{M}^+$ is obtained by attaching a $1$-cell $e$, that is a little segment passing through $p_1$ and "going down". Its boundary $\partial[e]$ is just two points in $\mathcal{M}^-$. You can check that $\partial:H_{1}(\mathcal{M}^+,\mathcal{M}^-)\to H_0(\mathcal{M}^-)$ is zero, and in fact $\partial[e]$ is the boundary of an arc $e_-$ entirely contained in $\mathcal{M^-}$ (for example the arc through the minimum point). it's quite clear that we just "completed" $[e]$ to a cycle $[e]-[e_-]$ (that you can see as a closed nontrivial closed curve in the torus) that is "almost entirely contained in $\mathcal{M}^-$ away from the critical point.</p> <p>If $\partial\neq 0$, this means that we are killing some cohomology, that we created at some point before. Take for example the sphere $S^2$, embedded in $\mathbb{R}^3$ in the shape of a "$\bigcap$", and again with $z$ as a Morse function (just wiggle it a bit so that the two minima have different values). Then from bottom-up, you first see two index $0$ points (the two minima), that give you two 0-cells. Then, as you go to the third critical point (the first critical point of index 1), you notice that $\partial \neq 0$ and so the contribution of that critical point is to kill something we created before. In some sense then, this morse function is not "efficient", in the sense that creates too many critical points, that will be killed later anyways. An "efficient" Morse function only creates critical points when they really contribute to the homology of the space. "Efficient" Morse functions are called <em>perfect</em>.</p> <p>Here is the formal definition of perfect morse function: as you said, every critical point of index $m$, might give rise to at most one new generator of $H_{*}(\mathcal{M}_+)$. So if $c_m$ denotes the number of critical points of index $m$, we have that $c_m\geq \dim H_{_*}(M)$, where $M$ is our manifold. The Morse function is called <em>perfect</em> if the inequality above is an equality, for every $m$.</p> http://mathoverflow.net/questions/91249/spaces-of-matrices-with-same-eigenvalue-great-circles-in-on-orbits Spaces of matrices with same eigenvalue/Great circles in O(n)-orbits Marco Radeschi 2012-03-15T04:49:25Z 2012-03-16T15:52:50Z <p>Let $Sym^2(V)$ be the set of symmetric matrices of a real $n$-dimensional vector space $V$. Given an element $\underline{\lambda}=[\lambda_1,\ldots \lambda_n]\in \mathbb{RP}^n$, where $\lambda_1\leq\lambda_2\leq\ldots\lambda_n$, I say that a subspace $W\subset Sym^2(V)$ is a <em>$\underline{\lambda}$-space</em> if every matrix $A\in W$ has eigenvalues $\lambda_1(A)\leq\ldots \lambda_n(A)$ such that $[\lambda_1(A), \ldots,\lambda_n(A)]=\underline{\lambda}$. Let $R(\underline{\lambda})$ be the maximal dimension of a $\underline{\lambda}$-space. What I am interested about is the following question:</p> <p><strong>Question-version 1:</strong> Compute $R(\underline{\lambda})$. If not possible, find upper bounds on $R(\underline{\lambda})$.</p> <p><strong>Example</strong>: if all the $\lambda_i$'s are nonzero, then a $\underline{\lambda}$-space $W$ has the property that any (nonzero) $A\in W$ is nonsingular. The problem of finding the maximal dimension $R_H(n)$ of subspaces of nonsingular $n\times n$ symmetric matrices was studied in <em>"On matrices whose real linear combinations are non-singular"</em>, by J. F. Adams, Peter D. Lax and Ralph S. Phillips. It turns out that $R_H(n)=\rho(n/2)+1$, where $\rho(n)$ is the Radon-Hurwitz function $\big($$\rho(n)-1 is the number of linearly independent vector fields on S^{n-1}$$\big)$. In particular this says that $$R(\underline{\lambda})\leq \rho(n/2)+1$$ if $\lambda_i\neq 0\quad \forall i$.</p> <hr> <p>Another version of the same question is maybe more geometric: Consider the space $Sym^2(V)$, with the (polar) action of the orthogonal group $O(V)$ given by conjugation: $$P\cdot A:=PAP^t,\quad P\in O(V),\, A\in Sym^2(V).$$</p> <p>The orbits of this action consist precisely of those matrices which share the same eigenvalues. If $W\subset Sym^2(V)$ is a $\underline{\lambda}$-space, then the unit sphere $W^1$ of $W$ is contained in a $O(V)$-orbit. So it is equivalent to ask:</p> <p><strong>Question-version 2</strong>: Given a $O(V)$-orbit $\mathcal{O}$, find the maximal dimension of a round sphere contained in $\mathcal{O}$</p> <p>On the one hand it feels that these spaces should have "good upper bounds", meaning it looks hard to come up with examples. Even finding 1-dimensional examples is a bit less trivial than (I) expected. In fact, a matrix $A$ spans a 1-dimensional $\underline{\lambda}$-space, for some $\underline{\lambda}$, if and only if the eigenvalues of $A$ are symmetric around $0$, i.e. if $\lambda$ is an eigenvalue then $-\lambda$ is an eigenvalue as well, with the same multiplicity. On the other hand there are geometric situation (where unfortunately it is hard to compute things directly) that prove the existence of high dimensional $\underline{\lambda}$-subspaces, for very specific $\underline{\lambda}$. Very likely the choice of $\underline{\lambda}$ must be special, and the elements of a $W$ should have special relationships among each other. This leads to my last (very vague) question:</p> <p><strong>Question 2</strong>: Given a n integer $k$, and supposing $R(\underline{\lambda})\geq k$, what can I say about $\underline{\lambda}$, and $W$ being a $\underline{\lambda}$-space?</p> <p>I am pretty much looking for anything that could help: references, observations for special cases, or why not? complete answers to my questions. :)</p> <p>Many thanks in advance!</p> <p><strong>EDIT</strong> Even though it doesn't feel too illuminating to me, here is a family of examples (all the ones I know, essentially).</p> <p><strong>Example 2</strong>: Let $0&lt; a_1\leq a_2\ldots \leq a_r$ be real numbers, and consider an embedding of $S:\mathbb{R}^n\to Sym^2(\mathbb{R}^{r(n+1)})$, $\underline{x}\mapsto S_{\underline{x}}$, as follows: if $\underline{w}=(\underline{w}_1,t_1,\underline{w}_2,t_2,\ldots \underline{w}_r,t_r)\in \mathbb{R}^{r(n+1)}$, where $\underline{w}_i\in \mathbb{R}^n$ and $t_i\in \mathbb{R}$, then $$S_{\underline{x}}=\big(a_1t_1\underline{x}, a_1 \langle \underline{x},\underline{w}_1\rangle, \ldots, a_rt_r\underline{x}, a_r\langle\underline{x},\underline{w}_r\rangle\big),$$ where $\langle,\rangle$ is the canonical scalar product in $\mathbb{R}^n$. One can check that the image of this space is a $\underline{\lambda}$-space, where $$\underline{\lambda}=[-a_r,-a_{r-1},\ldots, -a_1,0,0,\ldots,0,0,a_1,a_2,\ldots, a_r].$$ The number of zeroes in $\underline{\lambda}$, is $r(n-1)$.</p> <p>This example is a slight generalization of a family of examples arising from a geometric situation. Explicitly, I have a submanifold $M$, of a sphere, a point $p\in M$, and a linear subspace $W$ of the normal space $\nu_pM$ such that in all directions of $W$ the focal points of $M$ arise at the same distances. This means that all the shape operators $S_x$ (which are symmetric endomorphisms of $T_pM$) have the same eigenvalues, so they generate a $\underline{\lambda}$-space.</p> http://mathoverflow.net/questions/80373/smoothness-of-vector-fields-with-smooth-flow Smoothness of vector fields with smooth flow Marco Radeschi 2011-11-08T10:40:09Z 2011-11-08T16:25:34Z <p>Suppose you have a Vector field $X$ on a smooth (complete) manifold $M$, whose flow $\phi_X^t$ is, for each time $t\in (-\varepsilon,\varepsilon)$, smooth (of class $C^k$).</p> <p>Questions: Is $X$ smooth (of class $C^{k-\textrm{something}}$)? Can I control that "something"? What about the case $M=\mathbb{R}^n$?</p> <p>Thank you in advance!</p> http://mathoverflow.net/questions/69187/multiplicity-of-eigenvalues-in-2-dim-families-of-symmetric-matrices Multiplicity of eigenvalues in 2-dim families of symmetric matrices Marco Radeschi 2011-06-30T14:22:33Z 2011-07-01T06:22:49Z <p>Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of course be obtained if $A$, $B$ have a common eigenspace of multiplicity at least $m$.</p> <p>My question is: is it the only possibility?</p> <p>A way to proceed is the following: the characteristic polynomia of the generic matrix is $\det(xA+yB-tI)$, and its discriminant $\Delta$ (with respect to $t$) is a homogeneous polynomial in $x,y$ of degree $n^2-n$, where $n$ is the number of rows and columns in $A$ and $B$. Since every matrix in the family has some eigenvalue of multiplicity $>1$, the polynomial $\Delta$ vanishes identically, hance all the $n^2-n+1$ coefficients in $\Delta$ vanish. This gives $n^2-n+1$ polynomial conditions on the $n^2+n$ coefficients of $A$ and $B$, and this might help somehow.</p> <p>Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/36565/smoothness-and-fullness-of-determinantal-varieties Smoothness and fullness of determinantal varieties Marco Radeschi 2010-08-24T16:11:26Z 2010-08-24T17:33:24Z <p>So, I'm in the following situation:</p> <p>I have vector spaces $H,V$, and a map $A:H\longrightarrow \hom(H,V)$, sending $x$ to $A_x$ (notation); i need to consider the variety</p> <p>$\Sigma_1(A)=\{[x]\in \mathbb{P}(H)|\; rank(A_x)\leq 1\}$.</p> <p>Mostly, I'm interested in wether this variety is smooth, reducible, and/or full (by full, i mean not contained in any proper projective subspace).</p> <p>Are there any conditions on $A$ that allow me to know any of these properties?</p> <p>I don't know what $A$ is, but here is a couple of things i know:</p> <ol> <li><p>A has no kernel;</p></li> <li><p>If I define $\hat{A}:V\longrightarrow \hom(H,H)$ by $\hat{A}_v(x):=A_x(v)$, then $\hat{A}_v$ is skew-symmetric for all $v\in V$.</p></li> </ol> <p>Many thanks in advance!</p> http://mathoverflow.net/questions/24411/sub-tori-of-a-torus-generated-by-1-dimensional-subgroup sub-tori of a torus, generated by 1-dimensional subgroup Marco Radeschi 2010-05-12T16:28:58Z 2010-05-12T18:11:28Z <p>Ok the question is pretty dumb: suppose you have a torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ and a vector $\bar{v}=(v_1,\ldots,v_n)\in\mathbb{R}^n$.</p> <p>Consider the torus $T_{\bar{v}}$ given by the closure of the one parameter group in $T^n$ generated by $\bar{v}$:</p> <p>$T_{\bar{v}}=\overline{ \{t\cdot\bar{v}\mod\mathbb{Z}^n|\phantom{a}t\in\mathbb{R}\}}$</p> <p>My questions are:</p> <ol> <li>what is the dimension of $T_{\bar{v}}$?</li> <li>How can i find a basis of vectors spanning the tangent space of $T_{\bar{v}}$ at the origin?</li> </ol> <p>My guess for question 1. is $\dim T_{\bar{v}}=\dim_{\mathbb{Q}}\langle v_1,\ldots, v_n\rangle$, but i don't know what the answer to question 2. can be.</p> <p>Thanks!</p> http://mathoverflow.net/questions/15106/can-projective-hypersurfaces-contain-linear-spaces-how-big Can projective hypersurfaces contain linear spaces? How big? Marco Radeschi 2010-02-12T14:20:35Z 2010-02-12T14:53:38Z <p>I am in this, rather friendly, situation:</p> <p>I have a complex projective space $\mathbb{P}^n$, and there i have a (possibly non-smooth) hypersurface $S$ defined by one irreducible polynomial $P$ of degree $d$.</p> <p>What i want is to get information about the existence or not of linear subvarieties of $S$, and their maximal dimension $m$. I seem to remember the existence of some ways to get bounds on $m$, given $n$ and $d$, but i don't remember anymore and i don't know where to look..</p> http://mathoverflow.net/questions/120419/submersions-from-compact-flat-manifold/120471#120471 Comment by Marco Radeschi Marco Radeschi 2013-02-01T13:49:06Z 2013-02-01T13:49:06Z The point is: if $H^a(\tilde{N})$ is the highest nonzero cohomology group of $\tilde{N}$, and $H^b(F')$ the highest cohomology group of $F'$, then in the second page of the spectral sequence the term $E_2^{a,b}$ is nonzero, and no nonzero differentials land on it, or leave it. For this fact to be true it is important that both cohomologies of $F'$ and $\tilde{N}$ have only a finite number of nonzero cohomology group, so this passage does not apply in the situations you described. It follows that $E^{a,b}_2$ survives to the infinity page, and in particular $H^{a+b}(\tilde{M})\neq 0$. http://mathoverflow.net/questions/120419/submersions-from-compact-flat-manifold/120471#120471 Comment by Marco Radeschi Marco Radeschi 2013-02-01T08:58:51Z 2013-02-01T08:58:51Z Igor: As you say, you precompose the pullback with the universal cover. By the exact sequence in homotopy the fiber F′ is connected. What is not working in the second paragraph? http://mathoverflow.net/questions/116636/dense-subgroups-of-lie-groups Comment by Marco Radeschi Marco Radeschi 2012-12-17T21:05:10Z 2012-12-17T21:05:10Z @Misha: thanks for the explanation. One possible concept of complexity is the growth of H. I am trying to understand what kind of compact manifolds can admit H as fundamental group (or a quotient of it). http://mathoverflow.net/questions/116636/dense-subgroups-of-lie-groups Comment by Marco Radeschi Marco Radeschi 2012-12-17T19:57:52Z 2012-12-17T19:57:52Z @Yves: you are right, I meant finitely generated dense subgroup, without &quot;discrete&quot;. http://mathoverflow.net/questions/116636/dense-subgroups-of-lie-groups Comment by Marco Radeschi Marco Radeschi 2012-12-17T19:54:24Z 2012-12-17T19:54:24Z @Ryan, Yves: I edited my question, I hope it makes more sense now... http://mathoverflow.net/questions/111000/non-exoticness-of-a-diffeomorphism-of-a-sphere/111023#111023 Comment by Marco Radeschi Marco Radeschi 2012-10-29T21:55:27Z 2012-10-29T21:55:27Z Hi Ryan, thanks for the answer first of all. By torus i really meant $(S^1)^n$, even though I didn't say how that came up. I thought it was going to make the question heavier, and all I really wanted was to get to question 3. http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover Comment by Marco Radeschi Marco Radeschi 2012-10-15T13:48:23Z 2012-10-15T13:48:23Z Hi Johannes! actually, my problem is that $\tilde{M}$ retracts to a lie group, but is not one. in particular, i don't see how i can make $\pi(M)$ act on the retraction, in a free fashion http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover/109607#109607 Comment by Marco Radeschi Marco Radeschi 2012-10-15T13:45:59Z 2012-10-15T13:45:59Z Mark, thanks a lot for this! Two questions though: 1) For which groups does this work? from the other comments, i seem to understand that it holds for $\pi$ finitely presented, is this right? 2) I seem to understand that for the definition of $\chi(\tilde{M})$ you used the &quot;usual&quot; cohomology groups (as opposed to the compactly supported ones), is this correct? again, thank so much! http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover/109223#109223 Comment by Marco Radeschi Marco Radeschi 2012-10-09T13:40:42Z 2012-10-09T13:40:42Z Hi Liviu, thanks for the answer. The thing is, my question is the other way round. I am assuming that $\tilde{M}$ has zero Euler characteristic, and I ask wether the same is true for $M$. http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover Comment by Marco Radeschi Marco Radeschi 2012-10-09T13:11:54Z 2012-10-09T13:11:54Z In the case I am interested in, $\tilde{M}$ turns out to be a vector bundle over a Lie group, so it has compact topology. To me the Euler characteristic is $\chi(M)=\sum_i (-1)^{i}\dim H_{i}(M,k)$, where $k$ is some field, say $\mathbb{R}$ or $\mathbb{Q}$. I don't know how much the question might change, but one could use compactly supported homology groups. I am very flexible at the moment. http://mathoverflow.net/questions/109217/euler-characteristic-and-universal-cover Comment by Marco Radeschi Marco Radeschi 2012-10-09T12:38:26Z 2012-10-09T12:38:26Z HW, I am not sure what you are asking. In your example $\chi(\tilde{M})$ is not 0... if $M_1\to M_2$ is a $n$-sheeted covering then $\chi(M_1)=n\cdot\chi(M_2)$, so $\chi(M_1)=0$ iff $\chi(M_2)=0$. Probably I didn't understand your comment though. http://mathoverflow.net/questions/102643/linear-parabolic-equations-smooth-dependence-from-initial-data Comment by Marco Radeschi Marco Radeschi 2012-07-19T16:04:05Z 2012-07-19T16:04:05Z Can you expand a little bit on that? In particular: when you differentiate wrt $p$, how does the term $(L_pu_p)'$ behave? It feels like it becomes $(L_p)'u_p+ L_p(u_p')$ (here by &quot;u'&quot; i mean differentiation wrt $P$ of course). But then i don't see how the resulting equation is still parabolic... http://mathoverflow.net/questions/91249/spaces-of-matrices-with-same-eigenvalue-great-circles-in-on-orbits Comment by Marco Radeschi Marco Radeschi 2012-03-16T20:06:41Z 2012-03-16T20:06:41Z Claudio: I totally see what you are saying, and you're right! Let me see if I got this straight: if I prescribe eigenvalues (in my case some coincide) i take the correponding orbit (singular, in my case). Now, this is isoparametric, and I know that some great circles in this orbit correspond to the integral manifolds of eigendistributions of some shape operator. Also, I guess that all great circles arise in this way? That would help a lot... Thanks! If it were an answer I would have probably accepted that already... :) http://mathoverflow.net/questions/80373/smoothness-of-vector-fields-with-smooth-flow/80395#80395 Comment by Marco Radeschi Marco Radeschi 2011-11-08T17:49:33Z 2011-11-08T17:49:33Z that is exact. I do not assume any regularity in the $t$ direction. http://mathoverflow.net/questions/80373/smoothness-of-vector-fields-with-smooth-flow/80395#80395 Comment by Marco Radeschi Marco Radeschi 2011-11-08T16:25:43Z 2011-11-08T16:25:43Z thank you Pietro. What you say suggests that &quot;the flow has regularity one more than the vector field&quot;. I am actually interested in understanding the converse: if i know regularity of the flow, what can i say about the regularity of X? I re-edited the question so that it might be clearer.