User acky - MathOverflow

Interesting results for open Riemann surfaces

2012-10-29T11:50:40Z

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?

Collapsing of Riemannian manifolds with a group action

2012-09-19T17:42:21Z

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?

If in general such a sequence does not exist, how about the case $G=T$? Here $T$ is a finite-dimensional torus.

Can Euler Class be defined by the Splitting Principle for Real Vector Bundles?

2011-11-01T17:08:01Z

Let $M$ be a manifold and $S$ its sphere bundle with fiber $\mathbb{S}^n$ . As we know, the notion of the Euler class is raised from the problem of finding a global form on $S$ which restricts on each fiber to a generator of $H_{\textrm{dR}}^n(\mathbb{S}^n)$ . One overcomes two obstructions, namely the orientability of the sphere bundle and the Euler class $e(S)$ to find such a form. Now let $N$ and $M$ be two manifolds and $f:N\rightarrow M$ be smooth, $E$ is a vector bundle over $M$ . The following functorial property of $e(E)$ is well-known:

$e(f^\ast E)=f^\ast e(E)$

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)

In fact, let $E$ be an even dimensional oriented real vector bundle over $M$ . Then there exist a manifold $N$ and a map $g:N\rightarrow M$ satisfying the following two conditions:

(1) the homomorphism $g^\ast: H_{\textrm{dR}}^\ast(M)\rightarrow H_{\textrm{dR}}^\ast(N)$ is injective;

(2) $g^\ast(E)$ is a direct sum of plane bundles.

Therefore one may expect that the Euler class can be defined for an arbitrary vector bundle $E$ by first introducing the Euler class for plane bundles, then extending it to $E$ using the splitting principle stated above. In fact, suppose that $e(P_i)$ has been introduced for every plane bundle $P_i$ , and $g^\ast(E)$ is a direct sum of $P_i$ , then $e\big(g^\ast(E)\big)$ is well defined, therefore $e(E)$ may be defined by

$e\big(g^\ast(E)\big)=g^\ast e(E)$

However, one still needs to show the existence of $e(E)$ and its independence of the choices of $N$ and $g$ . It doen’t seem easy.

Exceptional Set in Egoroff's Theorem

2011-05-15T07:47:34Z

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.

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.

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: \begin{equation}E_{\varepsilon}=\bigcup_{k=1}^{\infty}\bigcup_{i=n(k,\varepsilon)}^{\infty}{x\in I:|f_{i}(x)-f(x)|\geq 1/k},\end{equation} where $n(k,\varepsilon)$ is chosen so that $m(\bigcup_{i=n(k,\varepsilon)}^{\infty}{x\in I:|f_{i}(x)-f(x)|\geq 1/k})<\varepsilon/2^k$.

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.

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 in which case the set $\bigcap_{\varepsilon}E_{\varepsilon}$ is a union of isolated points?

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.

On the Existence of Certain Fourier Series

2011-05-09T02:36:05Z

Is there an $f\in L^{1}(T)$ whose partial sums of Fourier series $S_{n}(f)$ satisfies $\|S_{n}(f)\|_{L^{1}(T)} \rightarrow \|f\|_{L^{1}(T)}$ but $S_{n}(f)$ fails to converge to $f$ in $L^1$ -norm ?

Convergence of Fourier series in L^{\infty}-norm

2011-04-03T11:53:47Z

As we know, for $1<p<\infty$ , the Fourier series of $f\in L^{p}(T)$ converges to $f$ in $L^{p}$ -norm. But is there any results concerning the convergence of Fourier series in $L^{\infty}$ -norm? Since $L^{\infty}(T)$ is not separable, the trigonometric system fails to form a Schauder basis of $L^{\infty}(T)$ , this implies that the Fourier series of $L^{\infty}(T)$ -functions fails to converge in $L^{\infty}$ -norm. But does the Fourier series of $f$ converge in $L^{\infty}$ -norm for every $f\in C(T)$?

Inversion of Fourier Transformation

2011-03-24T06:40:11Z

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 $f$ belongs to the intersection of the $L^1$ -algebra and the Fourier-Stieltjes algebra on a locally compact abelian group $G$ , then the inversion formula holds a.e. for $f$ . And the above result can be generalized in special cases. For example, If $G$ is $R$ or $R/Z$ , the Carleson-Hunt theorem says the inversion formula holds a.e. for $f$ in $L^p$ with $1<p<\infty$ .

My question is, is there any other version of generalization of inversion of Fourier transformation concerning a given locally compact abelian group $G$ ? For example, $G$ is an abelian Lie group, or $G$ is a compact group?

On a decomposition of L^1(G)

2011-02-25T17:16:34Z

[EDITED 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.]

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 $G$?

($B(G)$ consists of all Fourier transforms of complex-valued regular Borel measures on $\Gamma$, the dual group of $G$.)

If there are counterexamples, are there counterexamples with $G={\mathbb R}^n$?

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.

Comment by Acky

2013-03-07T13:21:06Z

Maybe you can use the Yang-Mills interpretation of a Hermitian-Einstein connection.

Comment by Acky

2012-09-19T18:17:59Z

Of course. The sequence $M_j$ is obtained by changing the metric on $M$.

Comment by Acky

2012-09-19T18:05:38Z

Collapsing in the Gromov-Hausdorff sense.

Comment by Acky

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.

Comment by Acky

2011-05-15T18:22:05Z

Thank you. It's a mistake, I'll correct it.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

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.

Comment by Acky

2011-04-03T12:14:03Z

It only implies almost everywhere pointwise convergence.

Comment by Acky

2011-04-03T12:10:49Z

You misunderstood me. I'm mot asking pointwise convergence but convergence in L^{\infty}-norm.

Comment by Acky

2011-03-24T08:17:07Z

Thanks, I'll sesrch for the relevant information.