User acky - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:35:38Z http://mathoverflow.net/feeds/user/13244 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110987/interesting-results-for-open-riemann-surfaces Interesting results for open Riemann surfaces Acky 2012-10-29T11:50:40Z 2012-10-30T00:32:30Z <p>As far as I know, interesting results for open Riemann surfaces are quite rare. One of them is the theorem of Gunning and Narasimhan, which asserts that every connected open Riemann surface admits a holomorphic immersion into the complex plane. Another example is given by the theorem of Behnke and Stein, which says that every connected open Riemann surface is a Stein manifold. Is there any more interesting results for open Riemann surfaces?</p> http://mathoverflow.net/questions/107597/collapsing-of-riemannian-manifolds-with-a-group-action Collapsing of Riemannian manifolds with a group action Acky 2012-09-19T17:42:21Z 2012-09-20T17:20:36Z <p>Let $M$ be a complete Riemannian manifold with bounded sectional curvature and $G$ a compact connected Lie group acts smoothly on $M$. Consider the fixed point set $F$, it is of course a submanifold of $M$ by the slice theorem. Let ${F_i}$ be the connected components of $F$. Then for each $i$, is there a sequence of Riemannian manifolds ${M_j},j\in\mathbb{N}$ with $M_0=M$ such that ${M_j}$ collapses to $F_i$ while keeping their sectional curvatures uniformly bounded?</p> <p>If in general such a sequence does not exist, how about the case $G=T$? Here $T$ is a finite-dimensional torus.</p> http://mathoverflow.net/questions/79728/can-euler-class-be-defined-by-the-splitting-principle-for-real-vector-bundles Can Euler Class be defined by the Splitting Principle for Real Vector Bundles? Acky 2011-11-01T17:08:01Z 2011-11-01T17:50:20Z <p>Let <code>$M$</code> be a manifold and <code>$S$</code> its sphere bundle with fiber <code>$\mathbb{S}^n$</code>. As we know, the notion of the Euler class is raised from the problem of finding a global form on <code>$S$</code> which restricts on each fiber to a generator of <code>$H_{\textrm{dR}}^n(\mathbb{S}^n)$</code>. One overcomes two obstructions, namely the orientability of the sphere bundle and the Euler class <code>$e(S)$</code> to find such a form. Now let <code>$N$</code> and <code>$M$</code> be two manifolds and <code>$f:N\rightarrow M$</code> be smooth, <code>$E$</code> is a vector bundle over <code>$M$</code>. The following functorial property of <code>$e(E)$</code> is well-known:</p> <p>$e(f^\ast E)=f^\ast e(E)$</p> <p>On the other hand, the splitting principle for cmplex vector bundles is wedely-known for its power on computing characteristic classes. In the real case, although the splitting principle does not hold in full generality, a weak version of this type of theorems can still be obtained: (cf. Lecture Notes in Mathematics, 638)</p> <p>In fact, let <code>$E$</code> be an even dimensional oriented real vector bundle over <code>$M$</code>. Then there exist a manifold <code>$N$</code> and a map <code>$g:N\rightarrow M$</code> satisfying the following two conditions:</p> <p>(1) the homomorphism <code>$g^\ast: H_{\textrm{dR}}^\ast(M)\rightarrow H_{\textrm{dR}}^\ast(N)$</code> is injective;</p> <p>(2) <code>$g^\ast(E)$</code> is a direct sum of plane bundles.</p> <p>Therefore one may expect that the Euler class can be defined for an arbitrary vector bundle <code>$E$</code> by first introducing the Euler class for plane bundles, then extending it to <code>$E$</code> using the splitting principle stated above. In fact, suppose that <code>$e(P_i)$</code> has been introduced for every plane bundle <code>$P_i$</code>, and <code>$g^\ast(E)$</code> is a direct sum of <code>$P_i$</code>, then <code>$e\big(g^\ast(E)\big)$</code> is well defined, therefore <code>$e(E)$</code> may be defined by</p> <p>$e\big(g^\ast(E)\big)=g^\ast e(E)$</p> <p>However, one still needs to show the existence of <code>$e(E)$</code> and its independence of the choices of <code>$N$</code> and <code>$g$</code>. It doenâ€™t seem easy.</p> http://mathoverflow.net/questions/65029/exceptional-set-in-egoroffs-theorem Exceptional Set in Egoroff's Theorem Acky 2011-05-15T07:47:34Z 2011-05-17T15:09:08Z <p>I'll use the version of this question I posted on Stakexchange to replace the former version. To simplify the situaton, we assume the measure space we are dealing with right now is a closed interval $I$ of the real line.</p> <p>When we are using the Egoroff's theorem, little attention has been paid to the exceptional set, i.e. the set $E\subset I$ on which $(f_n)$ fails to converge uniformly. The only thing we know about $E$ is that $m(E)$ can be arbitrary small. Taking the topology on the real line into account, we can also assume that $E$ is closed.</p> <p>Recall that in the proof of this theorem, we constructed the exceptional set $E_{\varepsilon}$ with respect to a fixed $\varepsilon>0$ in the following way: $$E_{\varepsilon}=\bigcup_{k=1}^{\infty}\bigcup_{i=n(k,\varepsilon)}^{\infty}{x\in I:|f_{i}(x)-f(x)|\geq 1/k},$$ where $n(k,\varepsilon)$ is chosen so that $m(\bigcup_{i=n(k,\varepsilon)}^{\infty}{x\in I:|f_{i}(x)-f(x)|\geq 1/k})&lt;\varepsilon/2^k$.</p> <p>Now assume $(f_n)$ is a sequence of smooth functions, and $f\in L^1(I)$ is the limit function. Since $f$ is only determined up to a null set, the set $E_{\varepsilon}$ can only be determined up to a null set. Thus it is very natural to require $f$ to be the "best choice" to make $E_{\varepsilon}$ the as small as possible.</p> <p>A satisfatory and notable case is that the family ${E_{\varepsilon}}$ be a sequence of nested closed intervals, thus we have $\bigcap_{\varepsilon}E_{\varepsilon}={x_{0}}$, where $x_{0}\in I$. Or more generally, we can ask <strong>in which case the set $\bigcap_{\varepsilon}E_{\varepsilon}$ is a union of isolated points?</strong> </p> <p>However, this is not true in general case, GTM 2 (p.38) contains an example in which case the set $I-E$ is nowhere dense, of course this coincides with the well-known result that every subset of the line can be represented as a union of a null set and a set of first category.</p> http://mathoverflow.net/questions/64348/on-the-existence-of-certain-fourier-series On the Existence of Certain Fourier Series Acky 2011-05-09T02:36:05Z 2011-05-13T13:52:32Z <p>Is there an <code>$f\in L^{1}(T)$</code> whose partial sums of Fourier series <code>$S_{n}(f)$</code> satisfies <code>$\|S_{n}(f)\|_{L^{1}(T)} \rightarrow \|f\|_{L^{1}(T)}$</code> but <code>$S_{n}(f)$</code> fails to converge to <code>$f$</code> in <code>$L^1$</code>-norm ?</p> http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm Convergence of Fourier series in L^{\infty}-norm Acky 2011-04-03T11:53:47Z 2011-04-03T13:44:43Z <p>As we know, for <code>$1&lt;p&lt;\infty$</code>, the Fourier series of <code>$f\in L^{p}(T)$</code> converges to <code>$f$</code> in <code>$L^{p}$</code>-norm. But is there any results concerning the convergence of Fourier series in <code>$L^{\infty}$</code>-norm? Since <code>$L^{\infty}(T)$</code> is not separable, the trigonometric system fails to form a Schauder basis of <code>$L^{\infty}(T)$</code>, this implies that the Fourier series of <code>$L^{\infty}(T)$</code>-functions fails to converge in <code>$L^{\infty}$</code>-norm. But does the Fourier series of <code>$f$</code> converge in <code>$L^{\infty}$</code>-norm for every $f\in C(T)$?</p> http://mathoverflow.net/questions/59393/inversion-of-fourier-transformation Inversion of Fourier Transformation Acky 2011-03-24T06:40:11Z 2011-03-24T14:51:24Z <p>As we know, the inversion formula of Fourier transformation holds pointwise for Schwartz class. We also have a general result concerning the inversion of Fourier transformation on locally compact abelian groups, which says that if <code>$f$</code> belongs to the intersection of the <code>$L^1$</code>-algebra and the Fourier-Stieltjes algebra on a locally compact abelian group <code>$G$</code>, then the inversion formula holds a.e. for <code>$f$</code>. And the above result can be generalized in special cases. For example, If <code>$G$</code> is <code>$R$</code> or <code>$R/Z$</code>, the Carleson-Hunt theorem says the inversion formula holds a.e. for <code>$f$</code> in <code>$L^p$</code> with <code>$1&lt;p&lt;\infty$</code>.</p> <p>My question is, is there any other version of generalization of inversion of Fourier transformation concerning a given locally compact abelian group <code>$G$</code>? For example, <code>$G$</code> is an abelian Lie group, or <code>$G$</code> is a compact group?</p> http://mathoverflow.net/questions/56651/on-a-decomposition-of-l1g On a decomposition of L^1(G) Acky 2011-02-25T17:16:34Z 2011-02-28T13:06:41Z <p>[<strong>EDITED</strong> by Y. Choi - I have attempted to paraphrase the original question into something a bit terser and more precise; if this is not what the original poster intended, they should make corrections themselves.]</p> <p>Let $G$ be a locally compact abelian (LCA) group and let $f\in L^1(G)$. Can we always find $g\in L^2(G)$ such that $h=f-g$ lies in $L^1(G)\cap B(G)$, where $B(G)$ is the Fourier-Stieltjes algebra of $G$?</p> <p>($B(G)$ consists of all Fourier transforms of complex-valued regular Borel measures on $\Gamma$, the dual group of $G$.)</p> <p>If there are counterexamples, are there counterexamples with $G={\mathbb R}^n$?</p> <p>In the case $G={\mathbb R}^n$, as we know, the Calderon-Zygmund decomposition theorem asserts that every $f\in L^1({\mathbb R}^n)$ is the sum of its good part $g$ and bad part $b$. Since $g$ is bounded and belongs to $L^1({\mathbb R}^n)$, it is not hard to verify that $g$ belongs to $L^p({\mathbb R}^n)$ for every $p\ge 1$. But it is easy to see that there exists an $f$ such that the inversion formula of Fourier transform fails for $b$. That is to say, the Calderon-Zygmund decomposition is not the decomposition of $L^1({\mathbb R}^n)$ that I want.</p> http://mathoverflow.net/questions/77255/when-is-a-holomorphic-tangent-bundle-stable Comment by Acky Acky 2013-03-07T13:21:06Z 2013-03-07T13:21:06Z Maybe you can use the Yang-Mills interpretation of a Hermitian-Einstein connection. http://mathoverflow.net/questions/107597/collapsing-of-riemannian-manifolds-with-a-group-action Comment by Acky Acky 2012-09-19T18:17:59Z 2012-09-19T18:17:59Z Of course. The sequence $M_j$ is obtained by changing the metric on $M$. http://mathoverflow.net/questions/107597/collapsing-of-riemannian-manifolds-with-a-group-action Comment by Acky Acky 2012-09-19T18:05:38Z 2012-09-19T18:05:38Z Collapsing in the Gromov-Hausdorff sense. http://mathoverflow.net/questions/65029/exceptional-set-in-egoroffs-theorem/65055#65055 Comment by Acky Acky 2011-05-15T23:20:03Z 2011-05-15T23:20:03Z I think to solve the problem, a notable case is when the $F_\varepsilon$ form a family of nested closed intervals. The case of uniform convergence on every point of $E$ seems too trivial. http://mathoverflow.net/questions/65029/exceptional-set-in-egoroffs-theorem Comment by Acky Acky 2011-05-15T18:22:05Z 2011-05-15T18:22:05Z Thank you. It's a mistake, I'll correct it. http://mathoverflow.net/questions/64348/on-the-existence-of-certain-fourier-series/64369#64369 Comment by Acky Acky 2011-05-09T10:51:05Z 2011-05-09T10:51:05Z Yes, your information is correct. I haven't considered such a concrete example, but I knew that if $(a_{n})$ is an even, convex and positive sequence, then we are able to find an $f\in L^{1}(T)$ such that $\sum_{-\infty}^{infty}a_{n}e^{inx}$ is a Fourier series of $f$. And $\sum_{-\infty}^{infty}a_{n}e^{inx}$ is bounded in $L^1$-norm if and only if $\lim_{n\rightarrow\infty}a_{n}\log n=O(1)$. However, even if we know not only $\|S_{n}(f)\|$ is bounded but $\|S_{n}(f)\|$ converges, it is still a very tough work to compute the limit that $\|S_{n}(f)\|$ converges to. Anyway thank you very much. http://mathoverflow.net/questions/64348/on-the-existence-of-certain-fourier-series Comment by Acky Acky 2011-05-09T08:20:36Z 2011-05-09T08:20:36Z @Harper: Thank you for your suggestion. Unfortunately, I'm a junior and I don't have access to much literature either. I believe such examples may exist since it is very easy to find an $f\in L^1$ whose $S_{n}(f)$ is bounded in $L^1$ but $L^1$-convergence fails. And I also believe the example I want has already been appeared in some old papers or treatises, an expert in trigonometric series may know one. http://mathoverflow.net/questions/64348/on-the-existence-of-certain-fourier-series Comment by Acky Acky 2011-05-09T05:51:03Z 2011-05-09T05:51:03Z I tried hard to construct such an example, but I still can't find one. Howerver, it seems harder to give an existential proof. http://mathoverflow.net/questions/64348/on-the-existence-of-certain-fourier-series Comment by Acky Acky 2011-05-09T03:20:32Z 2011-05-09T03:20:32Z $\|S_{n}(f)\|\rightarrow \|f\|$ will not imply norm convergence and I ask an example for the failiure of norm convergence in this condition. http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm Comment by Acky Acky 2011-04-03T13:18:12Z 2011-04-03T13:18:12Z The Fourier series can even diverge on a null set, so the convergence in $L^\infty$ is different from the convergence in $sup$-norm. http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm/60432#60432 Comment by Acky Acky 2011-04-03T13:04:11Z 2011-04-03T13:04:11Z NO, it's not the answer to the question of mine. The result of duBois Reymond is well known and is a easy corollary of the Banach-Steinhaus theorem. But it is about pointwise convergence, not convergence in $L^\infty$-norm. http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm Comment by Acky Acky 2011-04-03T12:50:36Z 2011-04-03T12:50:36Z Since there is no need for the convergence to be uniform, the limit need not to be continuous. It is still possible for the Fourier series to diverge on a null set. http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm Comment by Acky Acky 2011-04-03T12:14:03Z 2011-04-03T12:14:03Z It only implies almost everywhere pointwise convergence. http://mathoverflow.net/questions/60427/convergence-of-fourier-series-in-l-infty-norm Comment by Acky Acky 2011-04-03T12:10:49Z 2011-04-03T12:10:49Z You misunderstood me. I'm mot asking pointwise convergence but convergence in L^{\infty}-norm. http://mathoverflow.net/questions/59393/inversion-of-fourier-transformation Comment by Acky Acky 2011-03-24T08:17:07Z 2011-03-24T08:17:07Z Thanks, I'll sesrch for the relevant information.