User kofi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:12:24Z http://mathoverflow.net/feeds/user/16702 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131420/dirac-measures-dense-in-space-of-measures Dirac measures dense in space of measures? Kofi 2013-05-22T08:42:45Z 2013-05-22T09:09:43Z <p>Let $I$ be a compact interval and $\mathcal{M}(I)$ the space of (signed) Borel measures. We equip it with the weak topology, i.e. a sequence $\mu_n$ converges to zero if and only if $$\left|\int_I f(x) \mathrm{d}\mu_n(x)\right| \longrightarrow 0$$ for all $f \in C(I)$.</p> <p><strong>Now the question is the following: Let $V \subset \mathcal{M}(I)$ be the vectorspace of all finite linear combinations of Dirac measures supported at different points in $I$. Is $V$ dense in $\mathcal{M}(I)$?</strong></p> <p>For example if $I = [0,1]$, the sequence $$\mu_n = \frac{1}{N}\sum_{j=1}^N \delta_{j/N},$$ $\delta_{j/N}$ being the Dirac measure supported at $j/N$, weak*-converges to the Lebesgue measure as $\mu_n$ is just the approximation by Riemann sums. Hence one can easily get all measures that are absolutely continuous w.r.t. the Lebesgue measure. </p> <p>However, there are more measures (singular measures) that are neither point measures nor Lebesgue measures and I don't have an idea how to reach those.</p> http://mathoverflow.net/questions/131211/the-first-eigenvalue-of-the-schrodinger-operator-is-simple/131216#131216 Answer by Kofi for The first eigenvalue of the Schrödinger operator is simple. Kofi 2013-05-20T12:26:19Z 2013-05-20T17:27:30Z <p>Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$</p> <p>I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods in Nonlinear Partial Differential Equations, Vol. 1: Maximum Principles and Applications.":</p> <p><em>Let $X$ be a Banach space, $C \subset X$ a solid cone (i.e. a cone with nonempty interior) and $T : X \longrightarrow X$ a compact linear operator which is strongly positive, i.e. $Tu \in C$ if $u \in C$. Then the spectral radius $r(T)$ fulfills $r(T) > 0$ and is a simple eigenvalue admitting an eigenvector $\psi \in \mathrm{int} C$ and there is no other eigenvalue that admits an eigenvector in $C$. Furthermore, all other eigenvalues $\lambda$ fulfill $|\lambda| &lt; r(T)$.</em></p> <p>By partial integration, you easily show that the operator is bounded from below, so $L + \alpha$ is strongly positive from $C^2$ to $C^0$ for some $\alpha$ big enough. It is also well-known that the inverse $T := (L + \alpha)^{-1}$ exists and is a compact operator on $C^0$. By the strong maximum principle, $Lu>0$ implies $u>0$, so $T$ is a strongly positive operator.</p> <p>For the strong maximum principle, you can consult Evans: Partial Differential Equations, for example. To get the statement on a manifold instead of an area in $\mathbb{R}^n$, use a partition of unity.</p> <p>Now it is easy to show that the set of positive functions is actually a solid cone in $C^0$ (even though it has empty interior in $L^2$), so we can apply the Krein-Rutmann theorem.</p> <p>By elliptic regularity, every $L^2$ eigenfunction is $C^\infty$ and because the manifold is compact, is bounded, hence in $C^0$. Conversely, every $C^0$ Eigenfunction is in $L^2$, again because the domain is bounded. Hence the eigenvectors and eigenfunctions of $L$ are the same, whether viewed as operator on $L^2$ or on $C^0$</p> http://mathoverflow.net/questions/130732/special-kind-of-operators Special kind of operators Kofi 2013-05-15T15:35:03Z 2013-05-15T19:17:39Z <p>Consider an operator $A: H \longrightarrow X$ ($H$ is a Hilbert space and $X$ is a Banach space) that has a representation $$A = \sum_{j=0}^\infty a_j \langle \cdot, e_j\rangle \cdot x_j,$$ where $(e_j)$ is an orthonormal basis of $H$ and $(x_j)$ is a family of vectors in $X$ with $\|x_j\| = 1$ and $(a_j) \in \ell_p(\mathbb{N})$.</p> <p>Is there a special name for such operators? For a while I thought that these were just the absolutely $2$-summing operators between $H$ and $X$, but this seems to be wrong.</p> <p>To give some background, if we have such an operator with $p=2$ and a bounded bilinear form $L$ on $X$, then the bilinear form $M$ on $H$ defined by $$M(v, w) = L(Av, Aw)$$ is trace-class, which I am interested in.</p> http://mathoverflow.net/questions/130400/triangle-area-on-surfaces-of-constant-curvature Triangle area on surfaces of constant curvature Kofi 2013-05-12T10:44:28Z 2013-05-14T21:00:03Z <p>I am looking for an elementary derivation of the formula for the area of a geodesic triangle lying in a surface of constant curvature $\kappa$, depending on the angles and side length.</p> <p>Of course, the formula can easy be derived from the Gauss-Bonnet formula to be $$A = \frac{1}{\kappa}(\alpha + \beta + \gamma - \pi)$$ for $\kappa \neq 0$. However, I would like to have an elementary geometric proof.</p> <p>Does anybody know a reference?</p> http://mathoverflow.net/questions/130519/absolutely-2-summable-operator-on-a-hilbert-space Absolutely 2-summable operator on a Hilbert space Kofi 2013-05-13T21:01:02Z 2013-05-14T20:47:09Z <p>An bouneded linear operator $A \in L(X, Y)$ ($X$, $Y$ Banach spaces) is called absolutely $2$-summable if there exists a $C>0$ such that <code>$$\left( \sum_{j=1}^N \| A x_j\|_X^2 \right)^{1/2} \leq C \cdot \sup \left\{ \left(\sum | \langle x_j, \omega \rangle| \right)^{1/2} \mid \omega \in X^', \|\omega\|_{X^'} = 1\right\}$$</code> for any finite set of vectors $x_1, \dots x_N \in X$. The $2$-summable norm is the infimum of all such constants $C$.</p> <p>Those operators are the natural analog to Hilbert-Schmidt-Operators: If $X$ and $Y$ are Hilbert spaces, than $A$ is absolutely $2$-summable if and only if $A$ is Hilbert-Schmidt, and the norms coincide.</p> <p><strong>Now the question is:</strong> If $X$ is a Hilbert space (but $Y$ isn't), can we restrict to elements $x_j$ of some orthonormal basis? That is, if we allow only elements of a given ONB to take for $x_1, \dots, x_N$, can the resulting $C$ be strictly smaller than the $2$-summable norm?</p> http://mathoverflow.net/questions/130171/fourier-transform-of-a-bounded-function Fourier transform of a bounded function Kofi 2013-05-09T14:04:32Z 2013-05-09T14:36:44Z <p>This should really be well-known, but I was not able to find a definite answer to this question:</p> <p><em>Is the Fourier transform of a bounded function always a borel measure (i.e. an order 0 distribution)?</em></p> <p>In some sense, the distributional order corresponds to the order of a bounding polynomial and a bounded function can be bounded by an order zero polynomial. But I could not find any reference.</p> <p>If the above statement is false, what is a counterexample?</p> <p>If the statement is true: How does bounded variation correspond to all this? I.e. are there criterions for the Fourier transform to be of bounded variation?</p> http://mathoverflow.net/questions/128980/linearization-of-vector-fields/128989#128989 Answer by Kofi for Linearization of vector fields Kofi 2013-04-28T09:44:37Z 2013-04-28T09:44:37Z <p>The result of Poincaré is in fact not about the real case but about the complex case. Suppose a vector field of the form $$X = \sum_i\lambda_i x^i \partial_i + \text{higher order terms}$$ Then there is an analytic linearization diffeomorphism in a neighborhood of the origin, provided $$\text{all} ~\lambda_i~ \text{lie in the same half plane about the origin}$$ and $$\lambda_i \neq \sum_j m_i \lambda_j ~ \text{for any non-negativ integral} ~m_i~ \text{such that} \sum_i m_i> 1.$$</p> <p>This result is cited in "Local Contractions and a Theorem of Poincaré" by Shlomo Sternberg.</p> <p>Note that the semisimplicity of your matrix is not sufficient; I think you cannot get away without some diophantic conditions on the eigenvalues as stated above. This maybe seems somewhat strange at first, but it is not hard to construct counterexamples in the other case.</p> http://mathoverflow.net/questions/127863/heat-kernel-asymptotics-with-low-regularity Heat Kernel Asymptotics with low regularity Kofi 2013-04-17T15:36:33Z 2013-04-17T15:36:33Z <p>Let $M$ be a smooth manifold with Riemannian metric $g$, which is not smooth but only continuous.</p> <p><strong>Question:</strong> Is there still an asymptotic expansion of the heat kernel of the form $$p_t(x, y) \sim (4 \pi t)^{-n/2} e^{-\frac{1}{4t}d(x, y)^2}\sum_{j=0}^\infty t^j \Phi_j(x, y)$$ where the correction terms $\Phi_j$ are only continuous (or some other regularity that makes sense)?</p> <p><strong>Assume the following "mild form" of irregularity of $g$:</strong> Assume $g$ is smooth everywhere except at a submanifold $N$, where it is still smooth in direction of $N$ and only non-smooth in directions transversal to $N$, but maybe still Lipschitz-continuous.</p> http://mathoverflow.net/questions/126104/difference-between-parallel-transport-and-derivative-of-the-exponential-map Difference between parallel transport and derivative of the exponential map Kofi 2013-03-31T14:55:52Z 2013-04-02T02:04:48Z <p><em>This is a crosspost from <a href="http://math.stackexchange.com/questions/345623/difference-between-parallel-transport-and-derivative-of-the-exponential-map" rel="nofollow">math.stackexchange</a></em></p> <p>Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then we have to ways to map $T_pM$ to $T_{c(t_0)} M$ isomorphically. One is the parallel transport along $c$, let's call it $P_{c, 0, t_0}$ and the other is given by <code>$$d \exp_p|_{tX}: T_{tX}T_pM \cong T_p M \longrightarrow T_{\exp_p(tX)}M = T_{c(t_0)}.$$</code></p> <p><strong>My question is: What is the relation between those two? Are there formulas which relate the two concepts with curvature terms?</strong></p> <p>The parallel transport is a linear isometry, and the derivative of the exponential map is a radial isometry by the Gauss lemma, meaning $$\langle d \exp_p|_{tX} \cdot Y, \dot{c}(t) \rangle = \langle Y, X \rangle$$ for all $Y \in T_pM$. </p> <p>In two dimensions, this means that the two mappings coincide up to scaling, as there is only on orthogonal direction to the radial one. However, in higher dimensions, this is not true, i guess. However, on $S^3$, I computed that the derivative of the exponential map conincides with the parallel transport except that vectors orthogonal to the direction of parallel transport are multiplied by $\frac{\sin r}{r}$.</p> http://mathoverflow.net/questions/122965/heat-kernel-asymptotics-on-manifold-with-boundary Heat Kernel Asymptotics on Manifold with Boundary Kofi 2013-02-26T10:16:38Z 2013-02-26T10:26:50Z <p><em>This is crosspost from math.stackexchange <a href="http://math.stackexchange.com/questions/311213/heat-kernel-asymptotics-on-manifold-with-boundary" rel="nofollow">http://math.stackexchange.com/questions/311213/heat-kernel-asymptotics-on-manifold-with-boundary</a> where the question did not yield any answer</em></p> <p>On a closed Riemannian manifold $M$, the heat kernel $k_t(x, y)$ of the Laplace-Beltrami operator (or more general of any generalized symmetric Laplace-type operator acting on sections of a vector bundle) admits an asymptotic expansion of the form $$k_t(x, y) \sim \exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^\infty t^j \Phi_j(x, y)$$ where $d(x, y)$ denotes the Riemannian distance and $\Phi_j$ are appropriate smooth functions, not depending on $t$. This is meant in the sense that for each $N \in \mathbb{N}$, there exists a constant $C>0$ such that for all $x, y \in M$, we have $$\left| k_t(x, y) - \chi(x, y)\exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^Nt^j \Phi_j(x, y) \right| &lt; C t^{N+1}$$ where $\chi(x, y)$ is an appropriate cutoff function that is $\equiv 1$ near the diagonal.</p> <p>In the case that $M$ is still compact but has a boundary, in many books there can be found an asymptotic expansion of the trace, but I could not find an asymptotic expansion of the kernel itself, uniform on $M \times M$. Is there such an expansion?</p> <p><strong>Remark 1:</strong> When $M$ has a totally geodesic boundary, such an expansion is easy to get with help of the "Riemannian double". But of course, this case is quite unlikely (it is not even fulfilled for domains in Euclidean space).</p> <p><strong>Remark 2:</strong> I did not specify any boundary conditions, but one can assume that we are in the simplest case, i.e. the Laplace-Beltrami operator acting on functions with either Dirichlet or Neumann boundary conditions, whichever you like.</p> http://mathoverflow.net/questions/118314/trace-formula-for-psdos Trace formula for PSDOs Kofi 2013-01-07T22:12:36Z 2013-01-16T12:42:56Z <p>In Getzler's famous paper "Pseudodifferential Operators on Supermanifolds and the Atiyah-Singer Index Theorem", he states that for a (trace-class) pseudo-differential operator $P$ on a Riemannian manifold $M$, one has the formula $$\mathrm{Tr} P = \int_{T^*M} \sigma(P),$$ where $\sigma(P)$ is the full symbol of $P$, which is calculated using "the exponential map of $M$ to pull back the pseudodifferential operator near $x \in M$ to $T_xM$ and calculate its symbol on $T^*_xM$ using the Euclidean structure" (quote of p.164). </p> <p>However, as far as I know, the full symbol of a $\Psi$DO can be defined using the Levi-Civita connection, but only up to a smoothing symbol. And of course in applications, $P$ is usually a smoothing operator.Also, I could not get my hands on a full copy of the paper of Widom ("A complete symbolic calculus for pseudodifferential operators") cited there, and the preview of it on google books does not seem to help me.</p> <p>So, my question is: How does one define $\sigma(P)$ to make the above formula true?</p> http://mathoverflow.net/questions/118920/asymptotic-expansion-of-the-schrodinger-kernel Asymptotic Expansion of the Schrödinger kernel? Kofi 2013-01-14T20:24:56Z 2013-01-15T17:43:51Z <p>My stackexchange post [http://math.stackexchange.com/questions/275830/schrodinger-kernels-on-manifolds] was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was). So here it goes!</p> <p>Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and has a smooth integral kernel $k_t(x, y) \in C^\infty(M \times M)$. Furthermore, $k_t$ has an asymptotic expansion $$k_t(x, y) \sim \underbrace{(4 \pi t)^{-n/2} \exp \left( -\frac{1}{4t} \mathrm{dist}(x, y)^2 \right)}_{:= e_t(x, y)} \sum_{j=0}^\infty t^j \Phi_j(x, y)$$ meaning that $$\left| k_t(x, y) - e_t(x, y) \sum_{j=0}^N t^j \Phi_j(x, y) \right| \leq C t^{N+1}$$ uniformly in $x$ and $y$ in a neighborhood of the diagonal.</p> <p>Now by by formally substituting $t \rightarrow it$, one gets the formal asymptotic series $$e_{it}(x, y) \sum_{j=0}^\infty (it)^j \Phi_j(x, y),$$ which has the property that it formally (i.e. termwise, as asymptotic series in $t$) solves the Schrödinger equation $\left(i \frac{\partial}{\partial t} + \Delta\right)k_t = 0.$</p> <p><strong>Now my question is the following: Does this asymptotic series have any relation to the solution operator $e^{it\Delta}$ of the Schrödinger equation, or to its distribution kernel?</strong></p> http://mathoverflow.net/questions/116250/heat-kernel-proof-of-poincare-hopf Heat kernel proof of Poincaré Hopf Kofi 2012-12-13T08:00:02Z 2012-12-13T08:00:02Z <p>I am familiar with Witten's proof of the Morse inequalities by semiclassical analysis. Hey uses semiclassical expansions of the first eigenfunctions to construct the Morse complex from it, which implies Poincaré-Hopf.</p> <p>Is there any proof of the theorem that uses a similar approach as in the atiyah-singer index theorem, by replacing the usual heat kernel by a parameter dependent one?</p> http://mathoverflow.net/questions/115810/combinatorics-product-rules Combinatorics: Product Rules. Kofi 2012-12-08T16:57:47Z 2012-12-09T21:17:50Z <p>I couldn't find a way to figure this out, though it is a somewhat basic question that came up when studying the stationary phase expansion of an integral. The abstract version is the following:</p> <p>I have the homogeneous polynomial function $$f(X) = \sum_{u_1, \dots u_n = 1}^n X_{u_1} \cdots X_{u_n}$$ where $n$ is even, and the differential operator $$L = \sum_{j=1}^n \lambda_j \frac{\partial^2}{\partial X_j^2},$$ where $\lambda_j$ are some nonzero numbers.</p> <p>Problem: Calculate $L^{n/2} f(X)$. Obviously, this is constant and of the form $$L^{n/2} f(X) = C(n) \sum_{u_1, \dots u_{n/2}=1}^{n/2} \lambda_{u_1} \cdots \lambda_{u_{n/2}},$$ for some number $C(n)$. But what is $C(n)$?</p> http://mathoverflow.net/questions/115553/index-theorems-and-orientability Index theorems and orientability Kofi 2012-12-05T22:19:53Z 2012-12-06T03:45:01Z <p>Given a Dirac operator $D$ acting on some Clifford bundle $\mathcal{E}$ over a compact, even-dimensional, oriented manifold $M$, the Atiyah-Singer index theorem states that its index is given by pairing some characteristic class (which are elements of $H^n(M, \mathbb{R})$) with the fundamental class of the manifold, i.e. $$\mathrm{ind}(D) = \langle \hat{A}(M) \wedge \mathrm{ch}_{\mathcal{E}/S}(\mathcal{E}), [M] \rangle$$ in the case of a Dirac operator. In this case, we could use Chern-Weil theory, such that the characteristic classes are (equivalence classes) of differential forms, which can be integrated over the manifold, as it is oriented.</p> <p>However, the theorem is true in the non-oriented case as well. In fact, the characteristic class above can be interpreted as a volume density, which can be integrated over $M$. </p> <p>Of course, I am aware that one can just go to the oriented double cover and use that fact that the index is multiplicative with respect to Riemannian coverings, so the "non-oriented Atiyah-Singer" follows easily from the regular one. However, I find this somewhat unsatisfactory.</p> <p>Isn't there some (co-)homology theory in which the terms on the right side of the equation above can be expressed? I know that one often uses $\mathbb{Z}_2$-valued (co-)homology to deal with non-oriented manifolds; however, the index can be any integer, not just zero or one, so that this does not seem useful here.</p> http://mathoverflow.net/questions/115160/fourier-transform-for-entire-function/115169#115169 Answer by Kofi for Fourier Transform, for entire function Kofi 2012-12-02T10:46:12Z 2012-12-02T10:46:12Z <p>The Fourier Transform $\mathcal{F}$ is at first defined on the Schwartz space $\mathcal{S}$ and is a linear isomorphism there. As always, there is a dual operator $\mathcal{F}^\prime$ that is an isomorphism on the dual space $\mathcal{S}^\prime$ of tempered distributions.</p> <p>This operator $\mathcal{F}^\prime$ is the extension you are looking for, as $\mathcal{S}$ and also more general functions (for example polynomials) can be regarded as subset of $\mathcal{S}^\prime$.</p> <p>However, as far as I know, not ANY entire function has a Fourier transform, only those that also lie in $\mathcal{S}^\prime$.</p> http://mathoverflow.net/questions/114697/an-invariant-method-of-stationary-phase An invariant method of stationary phase Kofi 2012-11-27T19:55:33Z 2012-11-28T08:41:05Z <p>The method of stationary phase is very well-known and employed in many areas of physics and mathematics, and, of course, included in various versions as theorem in textbooks, especially on pseudors and microlocal analysis.</p> <p>However, it always is somewhat dependent on local coordinates and the Fourier transform, despite being a quite invariant problem. To be precise, the question would be the following.</p> <blockquote>Let $M$ be a manifold and $\phi: M \longrightarrow \mathbb{C}$ be a smooth function with values in the closed right half plane. Let $u$ be an volume density on $M$ with compact support in $M$. Determine an asymptotic expansion as $t \rightarrow \infty$ of the integral $$I(\phi, u, t) = \int_M e^{-\phi t} u$$ under some nondegeneracy conditions on $\phi$. </blockquote> <p>(For example, one could require $\phi$ to be Morse or, more general, require that the set where it vanishes is a submanifold $C$ of $M$ and that at a point $p \in C$, the Hessian of $\phi$ is non-degenerate on the space $T_pM/T_pC$.)</p> <p>It is well-known that in these cases $I(\phi, u, t)$ has an asymptotic expansion of the form $$I(\phi, u, t) = (t/\pi)^{-(n-k)/2}\sum_{j=0}^\infty t^{-j} \int_C s_j,$$ where $k$ is the dimension of $C$ and the $s_j$ are certain volume densities on $C$. <strong>In fact, they have to be certain universal terms, depending only on the $2j$-th jets of $\phi$ and $u$ at $C$.</strong> This is not stated in most textbooks. </p> <p>I wonder if it is possible to find these terms $s_j$ using Invariance theory alone. I would like if someone ever thought about this and knows a reference to this more invariant, geometric approach.</p> <hr> <p>/Edit: To clarify my question: I was wondering if it is possible to determine the constants by <strong>invariance theory</strong>, i.e. some argument like "there is only one polynomial on the $2j$-jets of $u$ and $\phi$ that is invariant under coordinate transformation" or so. For the first term, this goes like this, supposed that $\phi$ is purely real:</p> <p>Define the $n-k$-density $\mathrm{H}\phi$ on $C$ by setting $$\mathrm{H}\phi[X_1, \dots, X_{n-k}] := \sqrt{\left|\det \bigl( D^2\phi[X_i, X_j] \bigr)_{ij}\right|},$$ where $D^2\phi$ is the (on $C$ well-defined) Hessian of $\phi$. Now $u/\mathrm{H}\phi$ is a $k$-density on $C$ -- this is $s_0$.</p> <p>Now there should be similar characterizations of the higher $s_j$ (which obviously can get arbitrarily complicated).</p> http://mathoverflow.net/questions/109919/invariance-group-of-morse-charts Invariance group of Morse charts Kofi 2012-10-17T15:40:11Z 2012-10-18T14:13:54Z <p>Suppose I have a smooth function $\varphi$ that vanishes at $p$ and has a positive definite Hessian at that point (suppose that we are on a smooth manifold of dimension $M$). Then the Morse lemma tells us that we can find a chart $x$ (let us call it Morse chart) such that $$\varphi = (x^1)^2 + \dots + (x^n)^2 = \langle x, x \rangle.$$ What is the transformation group of Morse charts? </p> <p>To be more precise, I am looking for a group that acts freely and transitively on the set of Morse charts. </p> <p>Obviously, the group $O(n)$ acting on the set of Morse charts via $(Q, x) \mapsto Q\cdot x$ is a subgroup of this group. But are there more such transformations?</p> http://mathoverflow.net/questions/108923/euler-characteristics-and-the-difference-bundle-construction Euler characteristics and the difference bundle construction Kofi 2012-10-05T14:48:18Z 2012-10-08T22:10:06Z <p>I am reading on K theory in Lawson and Michelson (Spin Geometry). One has the "exact sequence spaces" $L(X,Y)$ and there is the theorem that there is a unique equivalence of functors $\chi$ between $L$ and $K$ (the Euler characteristic) such that in the case $Y = \emptyset$ $$\chi([V_0, \dots, V_n]) = \sum_{k=0}^n (-1)^k [V_k] \in K(X, \emptyset).$$ To give a more explicit characterization of the map in the case $n=1$, they make quite a complicated construction by defining a space $Z$ by gluying two copies of $X$ together along $Y$, defining bundles on $Z$ and using the isomorphism $K(Z, X_1) \cong K(X, Y)$. I compared to the original paper of Atiyah, Bott and Shapiro, and they make the same construction. However, I feel that one does not need all those surgical methods.</p> <p>Instead, to define $\chi([V_0, V_1])$, look at the exact sequence $$0 \rightarrow K(X, Y) \stackrel{i}{\rightarrow} K(X) \stackrel{j}{\rightarrow} K(Y) \rightarrow 0$$ and notice that the element $[V_0] - [V_1] \in K(X)$ is in the kernel of $j$ since $V_0$ and $V_1$ are isomorphic when restricted to $Y$. Now define the Euler characteristic as the preimage of $[V_0] - [V_1]$ under $j$, which is well-defined by exactness.</p> <p>Did I miss something or are the constructions that I referenced needlessly complicated? </p> http://mathoverflow.net/questions/109062/index-formula-for-pseudors Index formula for Pseudors Kofi 2012-10-07T14:00:25Z 2012-10-07T23:32:27Z <p>For elliptic differential operators $P$ on a compact manifold $M$, we have the formula</p> <p>$$\mathrm{ind}(D) = \mathrm{tr}(e^{-tP^*P}) - \mathrm{tr}(e^{-tPP^{\star}})$$</p> <p>I would think that this holds for Pseudo-Differential operators of positive order as well, but no text book states that. Is it not true? If not, what goes wrong?</p> http://mathoverflow.net/questions/108523/closed-formula-for-heat-kernel Closed formula for heat kernel Kofi 2012-10-01T07:15:39Z 2012-10-01T19:09:35Z <p>Is there, similar to the Mehler kernel, a closed formula for the heat kernel of the heat equation associated to the Laplacian $$-\sum_j \frac{d^2}{dx_j^2} + 2\sqrt{-1} \sum_j \lambda_j \frac{d}{dx_j} + \sum_{ij} a_{ij}x_ix_j$$ on $\mathbb{R}^n$? Here, the matrix $(a_{ij})$ is supposed to be symmetric and positive definite, while the $\lambda_j$ can be arbitrary.</p> http://mathoverflow.net/questions/100513/maximal-solution-to-eiconal-equation Maximal solution to eiconal equation Kofi 2012-06-24T08:53:03Z 2012-06-24T08:53:03Z <p>Let the function $v$ have a nondegenerate minimum at 0. Then there exists a neighborhood $U$ of 0 such that there is a unique function $\varphi$ on $U$ with $\varphi>0$ and $\varphi(0)=0$ that solves the eiconal equation $$v = |d\varphi|^2$$ This can be obtained by taking the Hamiltonian vector field associated to this equation and applying the stable manifold theorem. See for example here (http://mathoverflow.net/questions/82227/solutions-to-the-eikonal-equation).</p> <p>My question is if this solution exists on the whole unstable manifold of the gradient field of $v$, or if there is a counterexample. This is not totally obvious to me from the proof.</p> http://mathoverflow.net/questions/100440/bounded-operators-and-axiom-of-choice Bounded operators and axiom of choice Kofi 2012-06-23T10:43:47Z 2012-06-23T16:28:07Z <p>In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of choice (called DC). </p> <p>So, if I want to prove that an operator A defined on a Hilbert space H is bounded, is it enough to just check that the axiom of choice was not used to define it?</p> <p>And a related question: To show that a set is measurable, is it enough to check that the definition of this set didn't use the axiom of choice? (As similarly, the statement "all subsets of R" are Lebesgue measurable is consitent with ZF without the axiom of choice.)</p> <p>~ Link: <a href="http://www.ams.org/journals/bull/1973-79-06/S0002-9904-1973-13399-3/S0002-9904-1973-13399-3.pdf" rel="nofollow">http://www.ams.org/journals/bull/1973-79-06/S0002-9904-1973-13399-3/S0002-9904-1973-13399-3.pdf</a></p> http://mathoverflow.net/questions/99929/asymptotic-number-of-invertible-matrices-with-integer-entries Asymptotic number of invertible matrices with integer entries Kofi 2012-06-18T18:21:21Z 2012-06-19T06:36:18Z <p>Let $\|\cdot \|$ be some matrix norm on the space of $n \times n$ matrices. Denote $$M(r) := { A \in \mathrm{Mat}_{n \times n}(\mathbb{Z}) \mid \| M \| \leq r }.$$ Denote by $p(r)$ the fraction of invertible matrices in $M_r$. </p> <p><strong>Question</strong>: Does $p(r)$ possess an asymptotic expansion in $r$ as $r \rightarrow \infty$, and if yes, what is it?</p> <p>Of course, this does depend on the norm used and the dimension. Taking in the simplest case $n=1$ (thus eliminating the question about which norm to take), one gets $p(r) = 2/(2r + 1)$. Of course, in general, $p(r) \longrightarrow 0$ as $r \rightarrow \infty$ as the invertible matrices are dense in the set of all matrices.</p> <p>Of course, it should be easy to check for example, how many of the matrices that only have the numbers $-10, \dots, 10$ as entries are invertible. But what about an asymptotic series? Did someone think about this?</p> http://mathoverflow.net/questions/97856/space-derivatives-of-the-flow-of-a-vector-field Space Derivatives of the Flow of a vector field Kofi 2012-05-24T18:08:01Z 2012-05-25T16:56:51Z <p>Suppose I have a smooth vector field that has the form $$X(y) = \sum_j \lambda_j y^j \partial_j + \text{higher order terms}$$ for $\lambda_j>0$. Let $\Phi_t$ be the flow of $X$. Then it follows that $\Phi_t(y) \longrightarrow 0$ for $y$ near $0$ as $t \longrightarrow - \infty$.</p> <p>I am now looking estimates on the $y$-derivatives. Precisely, suppose that $K$ is a compact neighborhood of $0$ that lies in the unstable manifold near the point $0$. I would like to have a statement like "For every multiindex $\alpha$, there exists a constant $C>0$ such that $$\sup_{y \in K} |D^\alpha_y \Phi_t(y)| \leq C e^{t\lambda}$$ for all $t&lt;0$ and $y \in K$, where $\lambda$ is the smallest eigenvalue of the linearization of $X$ at $0$"</p> <p>Is some statement like this true? Where to find it or how do I prove it?</p> http://mathoverflow.net/questions/97490/first-eigenvalue-of-schrodinger-operator-is-simple First eigenvalue of Schrödinger operator is simple Kofi 2012-05-20T16:57:24Z 2012-05-20T17:04:36Z <p>I once read that the first eigenvalue of a Schrödinger operator always is simple, together with an easy proof of it. But I cannot remember where. Does anybody know a reference?</p> http://mathoverflow.net/questions/94506/boundary-of-unstable-manifold Boundary of unstable manifold Kofi 2012-04-19T08:30:01Z 2012-04-19T11:57:39Z <p>Let $X$ be a vector field on a compact manifold $M$ that has the form $$X = \lambda_1 x^1 \partial_1 + \dots + \lambda_n x^n \partial_n + \dots$$ with respect to some chart $x$ around a point $p$. Also, let $\lambda_1, \dots, \lambda_n > 0$.</p> <p>By the stable manifold theorem, there is an $n$-dimensional unstable submanifold $N$ of $M$ around $p$, i.e. a manifold with $\lim_{t \rightarrow -\infty} \Phi_t(p^\prime) = p$ for each $p^\prime \in N$, where $\Phi_t$ is the flow of $X$.</p> <p>Thus, $N$ is just an open set in $M$. My question is: Is the boundary of $N$ smooth (I suspect yes) and if so, how to prove it or where to look it up? </p> http://mathoverflow.net/questions/94038/is-f-a-polynomial-provided-that-it-is-partially-smooth/94051#94051 Answer by Kofi for is f a polynomial provided that it is "partially" smooth? Kofi 2012-04-14T18:36:02Z 2012-04-15T21:17:23Z <p>The following is wrong, but the comments are really nice :)</p> <hr> <p>Suppose That on $(a_1, b_1)$, $f$ is a polynomial of degree $N$. This means that in its Taylor series at any point $x \in (a_1, b_1)$, every coefficiant past the $(N+1)$st vanishes. </p> <p>This must also be true for the border points, $x = a_1$ and $x=b_1$. However, because the intervals are dense in $O$, $a_1$ or $b_1$ lies in the closure of some other interval. The rest is something like an induction: It follows that on every set $(a_n, b_n)$, the coefficients past the $(N+1)$st vanish.</p> <p>This shows that $f$ is a polynomial, by the fundamental theorem of calculous.</p> http://mathoverflow.net/questions/93543/the-spectrum-of-schrodinger-equation/93580#93580 Answer by Kofi for The spectrum of Schrodinger Equation Kofi 2012-04-09T16:01:30Z 2012-04-09T16:01:30Z <p>Since $u$ tends to $0$ as $x$ goes to infinity and it is apparently supposed to be $C^1$, it is bounded. Since the Laplace operator is a negative operator, the operator $L = \Delta - u$ is bounded from above, meaning $(Lu, u) \leq C$ for some constant constant $C$. Here, one can choose $C:= \min u$. It is a classical theorem that such operators, when densely defined, have a self-adjoint extension (this can be found in many books, if necessary I can give a citation). In this case, a dense domain would be the set of Schwartz functions, for example, and the theorem states that $L$ is essentially self-adjoint here.</p> <p>Now, the spectrum of a self-adjoint operator is real, and clearly also bounded from above by $C$. </p> <p>Regarding eigenvalues, however, I am afraid that your operator is not very well conditioned in general. If $u$ is a positive function, then there will be no eigenvalues at all, except possibly zero. However, if $C = \min u &lt; 0$, then there can be eigenvalues in the interval [0, C], but this is not necessary.</p> <p>To give some explanation, there is a theorem that states in your case, if $u$ tends to $0$ when $x \rightarrow \infty$, then the essential spectrum is bounded from above by $0$.</p> <p>So, there is in general no reason, why this solution should have any solution in $L^2$, but if it does, it automatically fulfills your condition ii, as does every function in $L^2$. Also, as far as I know, there is little to no hope to write down any solution analytically.</p> http://mathoverflow.net/questions/90509/linearization-of-a-gradient-field Linearization of a gradient field Kofi 2012-03-07T21:51:58Z 2012-03-08T14:08:11Z <p><strong>Setup:</strong> Suppose we are given a smooth function $\phi$ that has a nondegenerate minimum at $x=0$. Then we can choose a coordinate system $x$ such that the gradient is given by $$X = \mathrm{grad} \phi = \sum_i a_i x^i \frac{\partial}{\partial x^i} + O(|x|^2)$$ where the $a_i>0$ are the eigenvalues of the Hessian of $\phi$ at $x=0$.</p> <p>Now we look for smooth functions $a$ and numbers $\lambda$ such that $$X^i \frac{\partial a}{\partial x^i}= \lambda a.$$</p> <p>If we can somehow (smoothly) linearize $X$, then it is straight forward that the only solutions are $\lambda = k^1 a_1 + \dots + k^n a_n$, together with $a = x^k$, and we also have the expected multiplicities of eigenvalues, meaning, to each multiindex $k$ corresponds exactly one eigenvalue.</p> <p><strong>Question:</strong> The question is now if a linearization is possible. There is a theorem by Sternberg and Poincaré that gives a positive answer under the additional assumption that the condition $$a_i \neq m^1 a_1 + \dots + m^{i-1} a_{i-1} + m^{i+1} a_{i+1} +\dots + m^n a_n$$ with all $m^i\geq 0$ is satisfied, but does not use that $X$ is a gradient field. So I am hoping that one can maybe drop the condition in this case?</p> <p><strong>Motivation:</strong> The question arose when following the WKB method outlined in the literature. Helffer (Semi-Classical Analysis for the Schrödinger Operator and Applications) does cite the Sternberg theorem but doesn't even mention the condition above. Dimassi and Sjöstrand (Spectral Asymptotics in the Semiclassical Limit) use other methods than linearization but require instead that all numbers of the form $$m^1 a_i + \dots + m^n a_n$$ are different, which is even a stronger condition. The whole situation seems a bit curious here.</p> <p><strong>P.S.:</strong> Even if a Linearization is not possible, I would like to know if it is true that the eigenvalues are exactly $k^1 a_1 + \dots k^n a_n$ <em>counted with multiplicity</em> in the general setup or if one actually needs some of these diophantine-type conditions above. Easy linear examples show that the statement can still be true if Sternberg's condition is violated.</p> http://mathoverflow.net/questions/12900/surjectivity-of-the-normal-exponential-map/131489#131489 Comment by Kofi Kofi 2013-05-22T20:34:35Z 2013-05-22T20:34:35Z Do you mean that $x_n$ is in the inverse image under $f$ of the closed geodesic ball about $y$ of radius $\mathrm{dist}(y, f(N)) + 2$? Because otherwise, one can construct a counterexample with a circle of radius $&gt;1$ embedded in the plane. http://mathoverflow.net/questions/131420/dirac-measures-dense-in-space-of-measures/131427#131427 Comment by Kofi Kofi 2013-05-22T10:41:16Z 2013-05-22T10:41:16Z What an easy argument! Thanks! http://mathoverflow.net/questions/131211/the-first-eigenvalue-of-the-schrodinger-operator-is-simple/131216#131216 Comment by Kofi Kofi 2013-05-20T14:27:44Z 2013-05-20T14:27:44Z The argument in the last paragraph also works with $C^2$ instead of $C^\infty$, and the solutions are always $C^2$. You should check out some book about PDE (e.g. the mentioned one by Evans or Gilbarg-Trudinger) to get the exact statements about regularity (especially at the border), but in principle, the same argument should work in most related cases. http://mathoverflow.net/questions/130732/special-kind-of-operators/130753#130753 Comment by Kofi Kofi 2013-05-16T08:24:28Z 2013-05-16T08:24:28Z Thank You for the reference. It turned out, such an operator is (2,1,2) nuclear. http://mathoverflow.net/questions/130519/absolutely-2-summable-operator-on-a-hilbert-space/130620#130620 Comment by Kofi Kofi 2013-05-14T21:06:29Z 2013-05-14T21:06:29Z Wow, that is really interesting! Thank you for your answer! http://mathoverflow.net/questions/130400/triangle-area-on-surfaces-of-constant-curvature/130428#130428 Comment by Kofi Kofi 2013-05-13T07:12:05Z 2013-05-13T07:12:05Z I don't understand what you mean. I have a Riemannian metric of constant curvature on a Surface. This gives me curvature and a volume density, so I know both what curvature and area is and I can in theory calculate what the area of a geodesic triangle is. Where do I need an axiom? http://mathoverflow.net/questions/130171/fourier-transform-of-a-bounded-function/130174#130174 Comment by Kofi Kofi 2013-05-09T16:04:42Z 2013-05-09T16:04:42Z I suppose, continuity of the function does not help either? http://mathoverflow.net/questions/129706/zeta3-in-terms-of-derivatives-of-zeta-at-1-2-and-pi Comment by Kofi Kofi 2013-05-05T10:39:04Z 2013-05-05T10:39:04Z What do you mean with &quot;the last equality holds to precision 10^-4&quot;? If the result was true, shouldn't the equality hold up to any precision? http://mathoverflow.net/questions/129376/gap-between-first-two-nonzero-laplacian-eigenvalues-on-closed-compact-surface Comment by Kofi Kofi 2013-05-02T09:15:06Z 2013-05-02T09:15:06Z Probably one should skip the &quot;nonzero&quot; requirement as in this case, the gap is always positive. http://mathoverflow.net/questions/125248/sections-of-tensor-product-bundle-tensor-product-of-two-vector-bundles Comment by Kofi Kofi 2013-03-22T08:42:50Z 2013-03-22T08:42:50Z First, how is your actual question related to connections? Secondly, on any bundle, you can construct a section by patching it together locally with partitions of unity! http://mathoverflow.net/questions/125099/very-nice-polynomial-question Comment by Kofi Kofi 2013-03-20T22:26:09Z 2013-03-20T22:26:09Z In what way is this connected to mathematics-education, what is the motivation of this question and where does it come from? http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles Comment by Kofi Kofi 2013-02-07T11:53:56Z 2013-02-07T11:53:56Z I think what you want to hear in relation to (C) is the word &quot;Thom isomorphism&quot; http://mathoverflow.net/questions/119176/about-writing-of-mathematical-papers Comment by Kofi Kofi 2013-01-17T15:38:19Z 2013-01-17T15:38:19Z I almost never use $\Longrightarrow$, $\exists$ or $\forall$. http://mathoverflow.net/questions/118898/stochastic-neumann-problem Comment by Kofi Kofi 2013-01-14T19:26:43Z 2013-01-14T19:26:43Z Ok, I edited the question, maybe it should be closed. I will see if I find something now. http://mathoverflow.net/questions/118359/perturbed-morse-bott-function/118420#118420 Comment by Kofi Kofi 2013-01-09T08:00:15Z 2013-01-09T08:00:15Z This multiplication by a bump function does not make sense.What you want to do is to EXTEND the definition of the Morse function on the submanifold to a neighborhood of the manifold