User samuele - MathOverflow

Every continuous function is homotopic to a locally Lipschitz one

2013-04-05T17:57:27Z

I would like to know for which category/class/set of metric spaces the following holds: for any two metric spaces $X$, $Y$, for any continuous function $f:X\to Y$ there exists a locally Lipschitz continuous function $g:X\to Y$ which is homotopic to $f$.

EDIT: One could also ask a class of metrizable topological spaces such that each one of them can be given a metric so that the above property holds. Actually, I am more interested in the underlying topological space than in the actual metric space.

In general, the metric spaces I am considering are complete and weakly separable (there exists a sequence $(\phi_h)$ of $1$-Lipschitz functions such that for any two point $x,\ y$ $d(x,y)=\sup_h|\phi_h(x)-\phi_h(y)|$).

I don't know if this is a known fact among experts or not; in that case, I apologize for the standard question and would ask only for a reference.

ADDENDUM: Although I also have an interest for the general question as it is posed above, I could try to highlight some classes of metrizable spaces I have particular interest in knowing if they fulfill the request or not: manifolds, singular spaces (which singularities are allowed), spaces which are manifolds outside a "small" (in some sense) set, compact manifolds of infinite dimension or manifolds modeled on some "nice" linear space (Banach, Hilbert, Fréchet, ...).

Thanks.

Answer by Samuele for Linear (in)dependence of minors of a matrix

2012-12-16T15:05:41Z

Consider the map $\phi: V\to K^{M}$ where $M=\binom{n+1}{r+1}$, which sends a vector $y\in V$ to the $M-$tuple of minors (ordered as you wish) of the matrix $$A_y=(y\ \vert\ v_1\ \vert\;\cdots\;\vert\ v_{r+1})$$ then $y\in W$ if and only if $\phi(y)=0$, because $y\in W$ iff it is linearly dependent on $\{v_1,\ldots, v_{r+1}\}$, fact that happens iff $\mathrm{rk}A_y=r+1$ iff all the $(r+2)$-minors of $A_y$ vanish. Therefore, $W=\ker \phi$. Now, write $\phi=(\phi_1,\ldots, \phi_M)$, with $\phi_j\in V^*$. We have that $W=\{\phi_1=\ldots=\phi_M=0\}=\left(\mathrm{Span}\{\phi_1,\ldots,\phi_M\}\right)^0$, but $\dim W=r+1$, so $\dim\mathrm{Span}\{\phi_1,\ldots,\phi_M\}=(n+1)-(r+1)=n-r$.

Function of the incremental ratio tends weakly to a distribution

2012-12-16T14:55:13Z

I posted the following on StackExchange some time ago, but received no answer. I try to repost here.

Let $g:\mathbb{R}^3\to\mathbb{R^2}$ be a continuous function. Suppose that there exists $\Omega$ a neighborhood of $0$ where $Xg, Yg \in L^\infty(\Omega)$, with $$X=\partial_x-\frac{y}{2}\partial_z\qquad Y=\partial_y+\frac{x}{2}\partial_z$$ and $Z=[X,Y]=\partial_z$.

Now, let $J=\left(\begin{array}{cc}0&-1\\ 1&0\end{array}\right)$ and set $D_g=(Xg\ \vert Yg)$. The distribution $$[X((Yg)^tJg)-Y((Xg)^tJg)]-2\det D_g=T_g$$ is well-defined and one can check that, if $g\in\mathcal{C}^1$, then $T_g$ can be represented by integration against $(Zg)^tJg$.

My question is the following: is it true that the functions $$G_\lambda(x,y,z)=\frac{1}{\lambda}(g(x,y,z+\lambda)-g(x,y,z))^tJg(x,y,z)$$ converge (in the sense of distributions) to $T_g$ as $\lambda$ goes to $0$?

I tried to approximate $g$ by smooth functions $g_\epsilon$, but then I find myself dealing with a double limit: $G_{\epsilon,\lambda}\to T_g$ as $(\epsilon,\lambda)\to(0,0)$ and I don't know how to do it.

Obviously, the main problem is that the limit does not involve the incremental ratio alone, but rather a function of it, and there is no distributional meaning of $(Zg)^tJg$ for $g$ merely continuous (to my best knowledge, at least).

Answer by Samuele for Puiseux series expansion for space curves?

2012-03-16T23:06:00Z

If $(C,0)$ is a germ of complex curve in $\mathbb{C}^n$, then you can find coordinates $(z_1, z')$ and a polydisc $V=V_1\times V'$ centered in $0$ such that the canonical projection $V\ni (z_1,z')\mapsto \pi(z_1,z')=z_1\in V_1$ is a ramified covering when restricted to $C$ with $p$ sheets. Let $S$ be the ramification locus, then $S$ is an analytic set in $V_1$, that is, a discrete set of points. Therefore, upon taking a smaller $V_1$, we can think that $S={0}$ or it is empty. Let us suppose $S$ is not empty, otherwise the projection is a local biholomorphism with the unit disc and this gives the thesis with $z_1=t$, $z_k=f_k(t)$, $f_k\in\mathcal{O}(D )$. Now, $C^*=C\cap \pi^{-1}(V_1\setminus S)$ is a $p-$sheeted covering of $V_1\setminus S=D\setminus{0}$, therefore it is isomorphic to the standard $p-$sheeted covering: $$D\setminus{0}\ni t\mapsto t^p\in D\setminus{0}.$$ This means that there exists a map $g:D\setminus{0}\to C^*$ such that $\pi\circ g (t)=t^p$; this map extends clearly to a holomorphic bijection by setting $g(0)=0$ (it is continuous in $0$ and holomorphic outside). Therefore, we have $$z_1=t^p$$ $$z_k=g_k(t)=\sum_{j=0}^\infty a_{kj}t^j.$$

The existence of such a system of coordinates is a standard result in local complex geometry, sometimes called local parametrization theorem.

Young inequality in weighted spaces

2012-03-16T11:46:50Z

Let $U$ be a bounded open set in $\mathbb{R}^2$, $g\in L^1_{\mathrm{loc}}(\mathbb{R}^2)$. Let moreover $w$ be a weight (i.e. a non vanishing locally integrable function) on $U$ and $p\geq2$. Does there exist a constant $C=C(p,U,g)$ such that $$\int_{U}|f\ast g|^pwd\lambda\leq C \int_{U}|f|^pwd\lambda$$ for every $f\in \mathcal{C}^\infty_c(U)$ (compactly supported smooth functions), with $d\lambda$ the Lebesgue measure on the plane?

I think that this cannot hold for a general weight $w$. What conditions on $w$ can we require in order to obtain such an estimate?

Edit: I'm not asking for this to hold for every $g$ locally integrable. I am interested in some particular functions. A meaningful example could be $g=1/z$ (identifying $\mathbb{R}^2$ with $\mathbb{C}$).

Lipschitz contractibility of complex spaces

2011-08-20T21:36:36Z

Well, the title says pretty much all :)

Given a complex space $X$ (possibly singular), is it true that for every point $x\in X$ there exist a neighborhood $U_x$ and a map $\phi:[0,1]\times U_x\to U_x$ which is Lipschitz in both variables?

We take the usual distance on $[0,1]$. On $(U_x)\cap X_\mathrm{reg}$, we consider the metric induced by an embedding into $\mathbb{C}^n$ and take the associated distance.

Reference: Right now I have no access to it, but I remember that the local contractibility of an analytic space is showed in: Dan Burghelea and Andrei Verona Local homological properties of analytic sets, Manuscripta Mathematica, vol 7, n.1, pp 55-66 (1972).

Holomorphic vector fields with growth conditions on $X_\mathrm{reg}$

2011-08-17T18:19:25Z

Let $M$ be a complex manifold with a hermitian metric (volumes and distances will be wrt this metric). Let $X\subset M$ be a complex analytic subspace of $M$ and $Y\subset X$ an analytic set containing $X_\mathrm{sing}$. Set $n=\dim X_\mathrm{reg}$.

Define $W_p$ as the space of holomorphic $p-$vectorfields $\xi\in\Omega_p(X\setminus Y)$ such that:

$$\int_{X\setminus Y}\langle \xi, \phi_\epsilon\rangle dV_{X\setminus Y}\xrightarrow[\epsilon\to0]{}0$$
with $\phi_\epsilon\in \mathcal{D}^{p,n}(M)$ supported in $Y_\epsilon=\{x\in X\ :\ d(x,Y)\leq \epsilon\}$;
$$\int_{X\setminus Y}\xi, \overline{\partial}\langle\psi_\epsilon\rangle dV_{X\setminus Y}\xrightarrow[\epsilon\to0]{}0$$ with $\psi_\epsilon\in\mathcal{D}^{p,n-1}(M)$ supported in $Y_\epsilon$.

What can we say about $W_p$?

Remarks

The two conditions can be reformulated like this:
$$\int_{Y_\epsilon}\langle \xi, dg_1\wedge\ldots\wedge dg_{n+p}\rangle dV\to0$$ as $\epsilon\to0$, for every $g_1,\ldots, g_{n+p}\in\mathcal{C}^\infty(M)$;
$$\int_{bY_\epsilon}\langle \xi\llcorner\nu^{0,1}, dg_1\wedge\ldots\wedge dg_{n+p-1}\rangle dV'\to0$$
as $\epsilon\to0$, for every $g_1,\ldots, g_{n+p}\in\mathcal{C}^\infty(M)$, with $\nu^{0,1}$ the $(0,1)$ component of the normal covector of $bY_\epsilon$ and $dV'$ the volume of $bY_\epsilon$.
On complex curves, the problem is purely local (as $Y$ has to be discrete); we can do all the computations and notice that $W_0$ consists of the $(0,1)-$vector fields which are holomorphic on the normalization (or better on a resolution of singularities) and $W_1$ consists of the $(1,1)-$vector fields which are, on the resolution of singularities, sections of $\mathcal{O}(|E|-E)$, with $E$ the exceptional divisor.
The main case I'm interested in is $M=\mathbb{CP}^m$, $Y=X_\mathrm{sing}$.
Morally, the two conditions relate the growth of $\xi$ near the singular set with the vanishing of differential forms due to the singularity, so it seemed to me a good idea to try and pull back the problem on the desingularization, but then I need to control, in some way, the vanishing of the maps induced by the desingularization morphism on the spaces of differential forms...

Thanks in advance.

Answer by Samuele for Existence of a symmetric matrix.

2011-08-14T23:46:51Z

Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then
$$f(\alpha H)=\alpha^2f(H)$$
for every real $\alpha$. Obviously $f(0)=0$. So the problem boils down to find an $H$ such that $f(H)>0$ and another one such that $f(H)<0$.

Now, denoting by $h_k$ the $k-$th column of $H$,
$$(HAH)_{ij}=(h_i)^tAh_j$$
Therefore $f(H)=\sum d_i (h_i)^tAh_i$, if $D=\mathrm{diag}(d_1,\ldots, d_n)$.

If, wlog, $d_1>0$ and $d_2<0$, then we are obviously done.
If every $d_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done.

If every $d_i$ is wlog positive (or null) and $A$ is semi-definite, then $f(H)\geq0$ for every $H$ symmetric and $f(S)\leq 0$ for every $S$ skew-symmetric, so, given $H$ such that $f(H)\neq0$, we can always find $\alpha$ such that $f(\alpha H)=-f(S)$.

Answer by Samuele for Every real function has a dense set on which its restriction is continuous

2011-08-13T00:13:02Z

It is a theorem due to Blumberg (New Properties of All Real Functions - Trans. AMS (1922)) and a topological space $X$ such that every real valued function admits a dense set on which it is continuous is sometimes called a Blumberg space.

Moreover, in Bredford & Goffman, Metric Spaces in which Blumberg's Theorem Holds, Proc. AMS (1960) you can find the proof that a metric space is Blumberg iff it's a Baire space.

Class of flat currents stable under $\overline{\partial}$ operator

2011-08-12T10:17:18Z

Given $U\subset\mathbb{C}^n$, open domain, a locally flat current on $U$ is a $k-$current $T$ such that for every $f\in\mathcal{D}(U)$ (smooth functions with compact support in $U$) there exist a compactly supported $k-$vectorfield $\xi$ of class $L^1(U, dV)$ and a compactly supported $(k+1)-$vectorfield $\eta$ also of class $L^1(U, dV)$ such that
$$T(f\omega)=\int_{U}\langle \xi,f\omega\rangle dV+(-1)^{k+1}\int_{U}\langle\eta,d(f\omega)\rangle dV$$
for every $\omega\in\mathcal{D}^k(U)$. In other words, $T=\xi\llcorner dV + b(\eta\llcorner dV)$, where $b$ is the boundary operator between currents.

The flat currents are obviously stable under $b$: if $T$ is locally flat, then $bT$ is too.

I would like to know if there is a subclass of flat currents which is stable under the $\overline{\partial}$ operator or, which is the same, stable under $b$ and under taking $(p,q)-$components.

The class of currents associated with $\mathcal{C}^\infty$ vectorfields obviously works, but I would be interested in larger classes.

The "obvious" idea $\xi+\overline{\partial}\eta$ does not work as, for example, $$\overline{\partial}\left(z^{-1}\frac{\partial}{\partial z}\wedge\frac{\partial}{\partial \bar{z}}\right)=\delta_0\frac{\partial}{\partial z}$$
and this isn't flat.

Remark An idea only a little less obvious works, but it is still not satisfactory for me:
denote by $W_{p,q}(U)$ the set of $(p,q)-$vectorfields $\xi$ which are in $L^1(U, dV)$ and such that $\overline{\partial}\mathrm{iv}\xi$ belongs to $L^1(U, dV)$ as well (here $\overline{\partial}\mathrm{iv}$ is the $(p,q-1)-$ component of the divergence);
given $\xi\in W_{p,q}(U)$, the current
$T_\xi(\omega)=\int_{U}\langle\xi,\omega\rangle dV$
is such that $\overline{\partial} T$ is associated to $\overline{\partial}\mathrm{iv}\xi$, which is obviously in $L^1(U, dV)$, but then for every current we know that $\overline{\partial}\overline{\partial}T=0$;
this means that the current $S=\overline{\partial}T$ is in $L^1(U, dV)$ and also $\overline{\partial}S$ is (because the latter is $0$).

Why is this not satisfactory? Well, this works for the $\overline{\partial}$, but not for a decomposition in $(p,q)-$components (if we define $W_k(U)$ in the same way, it does not hold that $W_k=\bigoplus W_{p,q}$); moreover, I would like to see if there exists a larger class, inside flat currents, stable under $\overline{\partial}$.

Answer by Samuele for Behavior of essential singularities in an 'open cone'

2011-08-12T14:48:50Z

Maybe it's not exactly what you are asking for (and maybe you know it already), but a related concept to what you are asking is that of Julia line.

For sake of simplicity, consider an entire function $f$ with an essential singularity at $\infty$; let
$S(\phi,\epsilon)=\{z\ :\ |\mathrm{arg}(z)-\phi|<\epsilon\}$
be a sector around the line $R(\phi)=\{re^{i\phi}\ :\ r\geq0\}$. We call $R(\phi)$ a Julia line if, for every $\epsilon>0$, $f$ takes on every complex value in $S(\phi,\epsilon)$ with possibly one exception infinitely many times.

$R(\phi)$ is a weak Julia line if for every $\epsilon>0$, every $r>0$, the image of $S(\phi,\epsilon)\cap\{|z|>r\}$ is dense in the complex plane.

Both notions are stronger than what you are asking for and both deal with entire functions rather than local behaviour around isolated essential singularities, but it could be a starting point.

Results

Every trascendental function has a weak Julia line (I don't have any reference for this, but it is more an exercise in one complex variables)
Every trascendental function has a Julia line (you can look up in Cartwright, Integral functions)
If $f(z)=\sum a_k z^{n_k}$ and $n_k/k\to \infty$, then $f$ takes on every complex value infinitely many times in every $S(\phi,\epsilon)$ (Hayman, Angular value distributions of power series with gaps)

Another reference I know about this stuff is Anderson, Clunie, Entire functions of finite order and lines of Julia.

Warning -- I don't know of any example of a weak Julia line which isn't a Julia line, so the two concepts could very well coincide. But I think it is an open problem.

Answer by Samuele for A dual theory to the theory of currents?

2011-08-12T10:36:16Z

There exists a notion of generalized form: cochains. These are linear functionals on a certain class of currents; the problem is that you would like to put on your class of currents a meaningfull topology (maybe induced by a norm) and you would also like this class to be stable under boundary.

Smooth $k-$surfaces with boundary have the second property, but not the first one. In order to put some topological structure on them, you have to enlarge the space and consider integer rectifiable currents, where you have a norm-induced topology, compactness theorems and stability under the boundary operator.

The cochains have been defined on flat currents (which contain the previous ones); you could look up for flat cochains into Federer's Geometric Measure Theory (4.1.19). They ultimately are functionals $\ell$ associated to a couple of measurable forms $(\alpha,\beta)$ of degree $k$ and $k-1$ so that for every flat current $T$ you have
$$\ell(T)=T(\alpha)+ \partial T(\beta)=T(\alpha)+T(d\beta)$$
They were also studied by Whitney, I believe you can find something in his Geometric Integration Theory (Chapter 9 and following).

I don't know if this was what you were looking for, but that's what I know about dualizing currents.

Comment by Samuele

2013-04-07T15:18:06Z

Yes, I wrote in a hurry: 1. the sequence should be the same for any couple of points 2. compact infinite dimensional spaces or manifolds modeled on nice linear spaces (the or was missing). Anyway, could you give a reference for the statement in your first comment?

Comment by Samuele

2013-04-07T06:45:46Z

Well, I assume my metric space to be complete and with a weak property of separability (for any two point $x,\ y$ there exists a sequence of $1$-Lip maps $(\phi_h)$ such that $d(x,y)=\sup_h |\phi_h(x)-\phi_h(y)|$). But obviously I don't expect that every of each spaces has a metric such that my request is satisfied. I think I could ask the following: is it true for manifolds? does it remain true if we allow singularities? which ones? is it true for infinite dimensional manifolds (maybe compact)? for Banachian or Hilbertian compact manifolds, at least?

Comment by Samuele

2013-04-07T02:01:36Z

First of all, thank you. So, finite (geometric) simmplicial complexes work. Why do you need them to be finite? About the counterexample, that's interesting! But I was kind of expecting something like that: that's why I asked for a class of metric spaces, which come together with their distances. Another way to put the question could be to ask for a class of topological spaces which can be endowed with a distance (inducing their topology, hence metrizable spaces) so that the property holds. Could it be the case that CW-complexes do work?

Comment by Samuele

2012-12-17T11:44:16Z

You are welcome! Prego.

Comment by Samuele

2012-12-16T22:47:27Z

Hmm maybe the fact is (I didn't read the articles, so I'm just guessing) that for a non compact base manifold, the Fréchet structures you obtain are not tamely equivalent...

Comment by Samuele

2012-03-16T14:31:19Z

Whops, I should have said it more clearly... no, for me, $g$ is of a particular form. A usefull (for me) example could be $g=1/z$ (after identifying $\mathbb{R}^2$ with $\mathbb{C}$).

Comment by Samuele

2011-08-25T15:13:41Z

As for question 1, there is an article by J. Bourgain, Real isomorphic complex Banach spaces need not be complex isomorphic, in Proc. AMS, vol 96, n.2 (1986) where an example is given of two complex Banach spaces which are isometric but not linearly isomorphic (over complex numbers). I don't remember the details, so I'm not sure it has really something to do with your question, but it could be a start.

Comment by Samuele

2011-08-20T23:41:37Z

Just to check if I have understood the question correctly... let us say that the constraints are of the form $f(g_i(x))=f(g_j(x))$ with $g_i(0)=0$ for every $i$. (The only hypothesis is the last one: without any structure of $M$, we can only impose things like $f(\textrm{sth})=f(\textrm{sth})$). Let us suppose that $M$ has a transitive group of diffeomorphism; then we can check the condition of subbundle in some point $p$ and then carry the subspace from fiber to fiber. The constraints, in terms of Jacobian matrices in $0$, are $$XG_i=XG_j$$ with $X=\mathrm{Jac}(f)_0$. And this is a subspace.

Comment by Samuele

2011-08-16T20:33:48Z

@Hugo: map a fiber of $L$ into $C\times\mathbb{C}$ through the alg.var.isomorphism, then project down to $C$; it must be constant (genus is high), so any isomorphism between the two bundles sends fiber to fiber (possibly on a different point). Now you can produce a bundle isomorphism by composing with the appropriate automorphism of the basis and adding on each fiber the right vector to fix the origin.

Comment by Samuele

2011-08-16T16:18:39Z

On the example of a complex manifold carrying two different algebraic structures, take an affine curve $C$ of high genus and an algebraically non trivial line bundle $L$ on it. As $C$ is a Stein space, every line bundle on it is analytically trivial. But if $L$ and $C\times\mathbb{C}$ are algebraically isomorphic as varieties, then they are isomorphic as algebraic line bundles and that's impossible.

Comment by Samuele

2011-08-14T20:53:18Z

Just to make an example, take $M=\mathbb{CP}^1=\mathbb{C}\cup\\{\infty\\}$. Take $U_1=\mathbb{C}$ and $U_2=\mathbb{C}^*\cup\\{\infty\\}$, $\alpha_1=e^{z}dz$, $\alpha_2=e^{1/z}dz$. How could you glue them? What conditions of "compatibility" would you like to ask for? I mean, what should the link between the data and the result be, apart from the latter being a holomorphic (or meromorphic) form?

Comment by Samuele

2011-08-13T22:36:35Z

I don't get that $+\infty$ ... what is the meaning of a sequence of complex numbers having $+\infty$ as a limit?

Comment by Samuele

2011-08-13T17:21:19Z

@fedja: that is a well know theorem ... i wasn't even trying to exhibit a counterexample for that :P Back to the problem, the matrices $$\left(\begin{array}{rr}-1&-a&0\\1/a&0&1/a\\0&-a&1\end{array}\right)$$ with $a$ small enough are again counterexamples (they have eigenvalues $0$, $\pm i$).

Comment by Samuele

2011-08-13T14:32:10Z

A minimal counterexample to your stronger statement could be $$\left(\begin{array}{rr}1&-1\\2&-1\end{array}\right)$$ the eigenvalues are $i$ and $-i$, but they are both ouside the disk with center $1$ and radius $1$ ( or center $-1$ and radius $1$ if you want to use the columns).