User alberto abbondandolo - MathOverflow

When is a submersion locally volume-expanding?

2012-09-26T09:47:32Z

I would like to characterize the smooth maps $\varphi: \mathbb{R}^n \rightarrow \mathbb{R}^k$, $n\geq k$, with the following property:

For every $x\in \mathbb{R}^n$ there exists a positive number $r_0=r_0(x)$ such that for every positive number $r\leq r_0$ there holds $$ \mathrm{vol}_k \bigl( \varphi(B_r(x)) \bigr) \geq \omega_k r^k. $$

Here $B_r(x)\subset \mathbb{R}^n$ is the ball of radius $r$ around $x$ and $\omega_k$ denotes the $k$-volume of the unit ball in $\mathbb{R}^k$. These maps are in a certain sense ``submersions which locally expand volume''.

Some comments. An obvious necessary condition is that the $k$-Jacobian of $\varphi$ at every point is at least one: $$ J_k \varphi (x) := \max_{V\in \mathrm{Gr}_k(\mathbb{R}^n)} |\det D\varphi(x)|_V | = \sqrt{ \det \bigl( D\varphi(x) D\varphi(x)^* \bigr)} \geq 1, \quad \forall x\in \mathbb{R}^n. $$ Here $\mathrm{Gr}_k(\mathbb{R}^n)$ is the Grassmannian of $k$-planes in $\mathbb{R}^n$ and $A^*$ denotes the adjoint of the linear map $A:\mathbb{R}^n \rightarrow \mathbb{R}^k$ with respect to the Euclidean inner product.

This condition is clearly sufficient when $k=n$ (by the inverse mapping theorem and the area formula).It is also sufficient when $k=1$ (in which case $J_1 \varphi$ is just the norm of the gradient of the scalar function $\varphi:\mathbb{R}^n \rightarrow \mathbb{R}$): this is a nice exercise for students (hint: use the gradient flow of $\varphi$).

However, simple examples show that this condition is not sufficient when $n>k>1$.

The above condition becomes sufficient when we add the following requirement: let $\mathcal{V}(x)$ be the set of all $k$-planes $V\subset \mathbb{R}^n$ such that $|\det D\varphi(x)|_V |\geq 1$ (by the above condition the above set is not empty). Then we require the "multi-valued distribution" $\mathcal{V}$ to be ``integrable'', in the sense that $\mathbb{R}^n$ has a smooth $k$-dimensional foliation such that at every $x\in \mathbb{R}^n$ the leaf through $x$ is tangent to some $k$-plane in $\mathcal{V}(x)$.

The proof that this condition is sufficient uses the following fact, of which I do not know an elementary proof (see my MO question http://mathoverflow.net/questions/96776 and its answers): a smooth $k$-foliation of the unit ball in $\mathbb{R}^n$ which is close enough to the affine foliation has a leaf whose $k$-volume is at least $\omega_k$. Then the claim follows from the area formula.

However, this sufficient condition seems to be far from necessary.

Does anybody have an idea on how to narrow the gap between necessary and sufficient? Or some bibliographical suggestions?

Existence of a large leaf in a foliation of the ball

2012-05-12T14:56:21Z

Consider a smooth $k$-dimensional foliation of the unit ball $B$ of $\mathbb{R}^n$, all of whose leaves are diffeomorphic to $k$-disks.

Question: Is there a leaf whose $k$-volume is at least $\omega_k$?

Here $\omega_k$ denotes the volume of the unit ball of $\mathbb{R}^k$.

Particular cases. If $k=1$, the leaf through the origin (in this case a curve) has the desired property. If $k=n-1$, the leaf which divides the ball into two regions of equal volume has the desired property, by a relative isoperimetric inequality.

A possible argument. Replace each leaf $F$ by a $k$-submanifold which minimizes the $k$-volume among those with boundary $F\cap \partial B$. Among the minimal submanifolds obtained in this way, consider one which passes through the origin: its $k$-volume is at least $\omega_k$ by the monotonicity formula, and the $k$-volume of the leaf with the same boundary is even larger.

I know how to make this argument rigorous when the foliation is close enough to the foliation by parallel affine subspaces (but a reference or a more elementary proof also for this perturbative case would be very useful). In general, besides for possible singularities of the minimizers (which should not disturb), the problem I see is how to guarantee that at least one of them passes through the origin.

Answer by Alberto Abbondandolo for Space Derivatives of the Flow of a vector field

2012-05-24T20:16:38Z

This is certainly true if you choose $\lambda$ to be strictly smaller than the smaller eigenvalue of $DX(0)$. You may prove it inductively, by noticing that for a given $y$ the function $t\mapsto D^{\alpha}_y \Phi_t(y)$ solves a linear equation.

For instance, the first step goes as follows: the path of matrices $W(t):= D_y^{\alpha} \Phi_t(y)$ solves the ODE $$ W'(t) = DX(\Phi_t(y)) W(t), \quad W(0)=I, $$ where $\|DX(\Phi_t(y)) - DX(0)\| \leq C_0 e^{\lambda_0 t}$ for all $t\leq 0$. Then for every $\lambda_1<\lambda_0$ you can find $C_1$ such that $\|DX(\Phi_t(y))\| \leq C_1 e^{\lambda_1 t}$ for all $t\leq 0$.

A useful lemma for proving this and getting the uniformity you need is the following: given a continuous bounded path of matrices $t\mapsto A(t)$, $t\geq 0$, denote by $W_A(t)$ the solution of the linear Cauchy problem $$ W_A'(t) = A(t) W_A(t), \quad W_A(0) = I. $$ Assume that $\|W_A(t)W_A(s)^{-1}\|\leq c e^{\lambda (t-s)}$ for every $t\geq s\geq 0$. Then for every continuous bounded path of matrices $t\mapsto H(t)$, $t\geq 0$, there holds $$ \| W_{A+H}(t)W_{A+H}(s)^{-1}\|\leq c e^{\mu (t-s)}, \quad \forall t\geq s\geq 0, $$ with $\mu := \lambda + c \|H\|_{\infty}$.

(Sorry if here I switched to positive time, that's just because I am more used to work with stable manifolds).

Answer by Alberto Abbondandolo for What is a Lagrangian submanifold intuitively?

2012-05-02T22:12:40Z

More on the dynamical relevance of Lagrangian submanifolds for autonomous Hamiltonian systems:

Any Lagrangian submanifold which is contained in a regular energy level is automatically invariant for the dynamics.

When applied to Lagrangian graphs inside cotangent bundles, this observation leads to the stationary Hamilton-Jacobi equation.

Answer by Alberto Abbondandolo for Flow of a Hamiltonian vector field

2012-04-29T13:23:38Z

The diffeomorphisms which are generated by (time-dependent) Hamiltonian vector fields are said to be Hamiltonian diffeomorphisms. Hamiltonian diffeomorphisms form a subgroup of the group of symplectic diffeomorphisms (actually, they are a subgroup of the connected component of the identity).

As you observe, locally they cannot be distinguished from symplectic diffeomorphisms. But they are a much smaller class. For instance, the Hamiltonian diffeomorphisms of $\mathbb{T}^2$ are exactly those symplectic (i.e. area-preserving) diffeomorphisms which have a lift $\varphi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $\varphi(x) = x + \psi(x)$ where $\psi$ is $\mathbb{Z}^2$-periodic and satisfies $$ \int_{[0,1]^2} \psi(x) dx =0. $$ In particular, nontrivial translations on $\mathbb{T}^2$ are symplectic but not Hamiltonian (the latter fact is true also for $\mathbb{T}^{2n}$).

As for your doubts related to local existence: if $M$ is a compact manifold you of course have global existence, but these definitions make sense also on non-compact manifolds. Indeed, the non-exactness of $\imath_X \omega$ can be detected on a compact subset $K$ of $M$ (a circle is enough) and you can find $\tau>0$ such that the flow of a neighborhood of $K$ exists up to time $\tau$.

Answer by Alberto Abbondandolo for When LCS is isomorphic to subspace of some function space?

2012-04-29T12:39:38Z

If I interpret the question correctly, Yaoliang would like to know which LCTVS are isomorphic to $\mathbb{C}^X$, where $X$ is a set (no topology), and $\mathbb{C}^X$ is given the product topology. If this is so, the answer is: very few. Actually, spaces of this form are fully determined by the cardinality of $X$.

Just to make an example: no infinite dimensional normed space can be isomorphic to a space of this form. Indeed, in this case the set $X$ would have to be infinite, but then every neighborhood of $0$ in $\mathbb{C}^X$ would contain a proper vector subspace, and this is never true for normed spaces.

What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces?

2012-04-28T22:49:14Z

One often reads (and writes) that an exact sequence of finite dimensional vector spaces $$ 0 \rightarrow X_1 \rightarrow X_2 \rightarrow \dots \rightarrow X_n \rightarrow 0 $$ induces a canonical isomorphism $$ \bigotimes_{i \; \mathrm{odd}} \Lambda^{\max} (X_ i) \cong \bigotimes_{i \; \mathrm{even}} \Lambda^{\max} (X_i), $$ where $\Lambda^{\max}(X)$ denotes the top exterior power of the vector field $X$. My problem is that there seem to be too many choices for the sign of this ``canonical'' isomorphism. For instance, to the exact sequence $$ 0 \rightarrow X \stackrel{A}{\rightarrow} Y \stackrel{B}{\rightarrow} Z \rightarrow 0 $$ it seems equally canonical to associate the isomorphism $$ \Lambda^{\max}(X) \otimes \Lambda^{\max}(Z) \cong \Lambda^{\max} (Y), \qquad x \otimes B_* (y) \rightarrow A_*(x) \wedge y $$

or the isomorphism $$ x \otimes B_* (y) \rightarrow y \wedge A_*(x). $$

Here $x$ is a generator of $\Lambda^{\max}(X)$ and $y\in \Lambda^{\dim Z}(Y)$ is such that $A_*(x) \wedge y$ generates $\Lambda^{\max}(Y)$.

Since this canonical isomorphism is often used in the theory of determinant bundles in order to define orientations for geometric objects, I find this uncertainty on a sign disturbing.

Reasonable requirements that one should ask to this canonical isomorphism are:

1) to the exact sequence $0\rightarrow X \stackrel{A}{\rightarrow} Y \rightarrow 0$ one associates the isomorphism $x \mapsto A_*(x)$;

2) naturality with respect to isomorphisms of exact sequences.

However, these requirements do not determine the isomorphism uniquely.

My question is: is there a standard convention regarding the definition of the canonical isomorphism which is associated to exact sequences of arbitrary length? And if not, what would be reasonable requirements to add to 1) and 2) in order to have a good definition?

Answer by Alberto Abbondandolo for A characterization of Lagrange multiplier. Where to find a proof?

2012-04-28T13:05:05Z

I am sure you noticed this, but here is a simple counterexample which shows that you need to assume the continuity of the minimizer $x(s)$ at $s_0$:

$$F(x_1,x_2) = x_2 \cos x_1, \qquad G(x_1,x_2) = x_2.$$

Then $E(s)=-|s|$ and, for $s\neq 0$, $\lambda(s)= - \mathrm{sgn} \; s$. For $s=0$, $\lambda(0)$ can be any number bewteen -1 and 1, depending on which minimizer you choose for the constant function $F(x_1,0) \equiv 0$.

Answer by Alberto Abbondandolo for The orientation-preserving diffeomorphism of $\mathbb R^n$

2012-04-28T10:16:02Z

Yes. You may use the fact that f is isotopic to the identity to see it as the time-1 flow of a time-dependent vector field. Then you just have to modify the vector field so that it vanishes outside from a large ball.

Comment by Alberto Abbondandolo

2012-05-26T07:18:15Z

@Vitali. Thanks! This is really promising.

Comment by Alberto Abbondandolo

2012-05-25T17:15:38Z

@Kofi. $X_A$ was the same thing as $W_A$, I just edited my answer fixing this (sorry for the confusion). Unfortunately I do not know a reference where your statement is explicitly proved. What I wrote should be enough for the case of first order derivatives; for higher derivatives you need also the formula of variation of arbitrary constants (higher order derivatives solve a inhomogeneous linear equation). If you find difficulties in proving it I can try to write more details.

Comment by Alberto Abbondandolo

2012-05-16T18:10:00Z

@Ulrich: thanks!

Comment by Alberto Abbondandolo

2012-05-16T16:27:06Z

And do you have a simple proof of the fact that $U^{st*}(H)$ is contractible? Here I mean simpler than Kuiper's one for $U(H)$.

Comment by Alberto Abbondandolo

2012-05-13T07:40:34Z

Do you know how to find Almgren's manuscript? I saw it cited in a paper by Guth, where it is referred as "manuscript available in the Princeton math library", but it does not appear in the library's catalogue. Is it available somewhere on the web? Thanks!

Comment by Alberto Abbondandolo

2012-05-12T16:03:14Z

$B$ is closed. The foliation is a restriction of a smooth foliation of a neighborhood of $B$, and the intersection of each leaf with the interior of $B$ is either empty or it is an embedded $k$-disk with boundary on $\partial B$.

Comment by Alberto Abbondandolo

2012-05-03T20:26:40Z

You are welcome. Ciao!

Comment by Alberto Abbondandolo

2012-05-02T13:08:05Z

What do you mean by "truly" infinite dimensional Hilbert manifold? You can easily build a fiber bundle over $S^1$ with fiber $S^1 \times H$, $H$ a Hilbert space, which is non-trivial just by homotopy reasons.

Comment by Alberto Abbondandolo

2012-04-29T17:54:02Z

A bijection between $X$ and $Y$ induces an isomorphism (of TVS) between $\mathbb{C}^X$ and $\mathbb{C}^Y$, so only the cardinality of $X$ matters. Notice also that in your class of spaces $\mathbb{C}^X$ there is only one infinite dimensional separable space, the one which you obtain for $X=\mathbb{N}$. The problem is that you are considering ALL functions from $X$ to $\mathbb{C}$ (and if you replace $\mathbb{C}$ by a larger vector spaces things get even worse). If you want to represent interesting TVS you should put more structure on $X$ and restrict the class of functions.

Comment by Alberto Abbondandolo

2012-04-29T08:21:45Z

And no, I don't know a precise reference. But the statement of Palais which Igor cites in his answer is strongly related.

Comment by Alberto Abbondandolo

2012-04-29T08:15:41Z

To Anton. You are right, this is my first answer on MO and I still have to get familiar with its style. I should have added at least 2 things: 1) The fact that every orientation preserving diffeomorphism of $\mathbb{R}^n$ is isotopic to the identity is proved in Milnor, "Topology from the differentiable viewpoint", Chapter 6, Lemma 2. 2) The desired diffeomorphism is obtained by integrating up to time 1 the modified vector field.

Comment by Alberto Abbondandolo

2012-04-29T07:47:07Z

Thanks for the link to your book. If I understand well, the notion of "weighted line" allows $\Lambda^{\max}(X)$ to "remember" the dimension of the space $X$ and this helps when dealing with sign issues.

Comment by Alberto Abbondandolo

2012-04-28T23:30:05Z

To René: you are right, the first definition looks more "canonical". And this definition extends to sequences of arbitrary length. But is it the standard one? And how is it characterized? My reason to doubt that this is the correct definition comes from the construction of the determinant bundle on the space of Fredholm operators: in this construction one uses an exact sequence with 4 terms and the isomorphism which one associates to it is not this one (see D. QUILLEN, Determinants of Cauchy-Riemann operators over a Riemann surface, Functional Anal. Appl. 19 (1985), 31–34).