Tuesday, May 29, 2012

Vector bundles and Fredholm operators.

Problem $\newcommand{\bR}{\mathbb{R}}$  Suppose that $E\to X $ is a real vector bundle of rank $r$  over the compact CW-complex $X$. $\DeclareMathOperator{\Gr}{\boldsymbol{Gr}}$ and $H$ is a separable real Hilbert space. Construct a  continuous family of surjective Fredholm operators $T_x: H\to H$, $x\in X$,  such that the bundle $(\ker T_x)_{x\in X}$ is isomorphic to $E$.


Solution.  We may as well assume that   $X$ is the Grassmannian $\Gr_{r,N}$ of $r$-dimensional   subspaces of $\bR^N$, where $N$ is a large positive integer, and $E$ is the tautological bundle $\newcommand{\eU}{\mathscr{U}}$  $\eU_r\to\Gr_{r, N}$. Denote by $\newcommand{\eQ}{\mathscr{Q}}$ $\eU^\perp_r$ the orthogonal complement  of $\eU_r$ in the trivial bundle  $\underline{\bR}^N\to \Gr_{r, N}$ so that we have a short exact sequence of bundles

$$ 0\to\eU_r\stackrel{A}{\hookrightarrow} \underline{\bR}^N\stackrel{B}{\twoheadrightarrow} \eU^\perp_r\to 0. \tag{S} $$

First method. Consider a separable real Hilbert space $H$ and  form the Hilbert bundles  over

$$H_0: = H\oplus \underline{\bR^N}\to {\Gr}_{r, N},$$

$$ H_1: = H\oplus \eU_r^\perp \to  {\Gr}_{r, N}.$$

We can now define a surjective bundle  morphism $T: H_0\to H_1$,  $\newcommand{\one}{\boldsymbol{1}}$  $T=\one_H\oplus B$ whose kernel is $\eU_r$.  The group  of unitary transformations of  a Hilbert space  is contractible (Kuiper's theorem) so that the bundles $H_0, H_1$ are trivializable. By choosing such trivializations we can view $T$ as a  continuous family of  surjective Fredholm operators parametrized by $\Gr_{r,N}$ whose kernels from the tautological bundle $\eU_r$.

Second method.   Consider the trivial Hilbert bundle over $\Gr_{r, N}$ $\newcommand{\eH}{\mathscr{H}}$

$$ \eH=\bigoplus_{k=0}^\infty \underline{\bR}^N. $$

For $k\geq 0$  denote by $\underline{\bR}^N_k$ the $k$-th summand in the above direct sum of vector bundles. From the short exact sequence (S) we deduce

$$\eH = \bigoplus_{k=0}^\infty \eU_r\oplus \eU_r^\perp. $$

The fiber of $\eH$ over $x\in\Gr_{r, N}$ is

$$\eH_x=\bigoplus_{k=0}^\infty\Bigl( \eU_r(x)\oplus \eU_r(x)^\perp\Bigr)=\bigoplus_{k=0}^\infty\bR^N. $$

Thus an element in $h_x\in\eH_x$ is represented as a convergent series

$$h_x = u_0(x)+ v_0(x)+ u_1(x)+v_1(x)+\cdots=\sum_{k=0}^\infty\bigl(\; u_k(x)+v_k(x)\;\bigr), $$
where $u_k(x)\in \eU_r(x)$,  $v_k(x) \in \eU_r(x)^\perp$, and $\eU_r(x)\subset\bR^N$ denotes the fiber of $\eU_r$ over $x$. Define $T_x:\eH_x\to \eH_x$ by setting

$$ T_x \sum_{k=0}^\infty\bigl(\; u_k(x)+v_k(x)\;\bigr)=\underbrace{ B_x\bigl(\; u_0(x)+ v_0(x) \;\bigr) +u_1(x)}_{\in\bR^N_0} $$

$$  \oplus\; \underbrace{\bigl (\; v_1(x)+u_2(x) \;\bigr) }_{\in\bR^N_1}\oplus\; \underbrace{ \bigl (\; v_2(x)+u_3(x) \;\bigr)}_{\in\bR^N_2} \oplus\; \underbrace{\bigl (\; v_3(x)+ u_4(x) \;\bigr)}_{\in\bR^N_3}\oplus\cdots  . $$

Above  the bundle endomorphism $B$ is defined in (S). Note that $T_x$ is surjective and $\ker T_x =\ker B_x= \eU_r(x)$, $\forall x$.

Thursday, May 3, 2012

Moments again

I am now interested in a weaker  question than  in the previous post. Suppose $w:[0,\infty)\to[0,\infty)$  is smooth and fast decaying at $\infty$. Assume that $w$ is not identically zero. Consider the moments


$$ I_k(w)= \int_0^\infty t^k w(t) dt, \;\; k\in\mathbb{Z}_{\geq 0},$$

and set

$$ R_k(w):=\frac{I_k(w)^2}{I_{k-2}(w) I_{k+2}(w)}. $$

The Cauchy inequality implies that $R_k\leq 1$.

Question 1.   Is it true that  that $R_k$ has a limit as $k\to \infty$?     Are there any simple conditions on $w$ guaranteeing the existence of such limits? I will denote by $R_\infty(w)$ this limit, whenever it exists.

Question 2.    We know that $R_\infty(w)\in [0,1]$.  Are there additional constraints on $R_\infty$?

May 3, 2012    I have found out the following simple examples  from Mikael de  la Salle,  suggesting a negative answer to  Question 2.


Example 1. Suppose that

$$w(r)= e^{-(\log r)\log(\log r)},\;\;r\geq 1. $$

Then

$$I_k(w)\sim J_k:= \int_1^\infty r^k  e^{-(\log r)\log (\log r)} dr\;\; \mbox{as $k\to\infty$}. $$

Using the substitution  $r=e^t$ we deduce

$$J_k =\int_0^\infty e^{(k+1)t-t\log t} dt. $$

We  will investigate the large $\lambda$ asymptotics of the   integral

$$ T_\lambda=\int_0^\infty e^{-\phi_\lambda(t)} dt,\;\;\phi_\lambda(t)=\lambda t- t\log t. \tag{T} $$


Note that

$$\phi_\lambda'(t)=\lambda -\log t -1,\;\; \phi_\lambda''(t)=-\frac{1}{t}. $$

Thus $\phi_\lambda(t)$ has a unique critical point

$$ \tau=\tau(\lambda):= e^{\lambda-1}. $$

We make the change in variables $t=\tau s$ in (T). Observe that

$$ \lambda e^{\lambda-1}s-e^{\lambda-1}s \log(e^{\lambda-1} s)= e^{\lambda-1}s-(\lambda-1)e^{\lambda-1}s -e^{\lambda-1} \log s =e^{\lambda-1}s(1-\log s). $$

  and we deduce

$$ T_\lambda= \tau \int_0^\infty e^{-\tau h(s)} ds,\;\;h(s) = s(\log s-1) $$


The  asymptotics of the last integral can be determined   using  the Laplace method and we have

$$ T_\lambda \sim \tau e^{-\tau h(1)} \sqrt{\frac{2\pi}{\tau h''(1)}}=\sqrt{2\pi\tau} e^\tau. $$

This shows that  for this weight  we have

$$\lim_{k\to\infty} R_k(w) =0. $$

Example 2. Suppose that

$$w(r)= \exp\bigl(-C(\log r)^2\;\bigr), \;\; C>0, \;\;r>1 $$

Then  as $k\to \infty$

$$ I_k(w)\sim\int_0^\infty r^k  \exp\bigl(\;-C(\log r)^2\;\bigr)dr=\int_0^\infty e^{(k+1) t-C t^2} dt. $$

Again, set

$$ T_\lambda= \int_0^\infty e^{-\phi_\lambda(t)} dt,\;\;\phi_\lambda(t)=Ct^2-\lambda t. $$

Note that

$$ \phi_\lambda'(t)= 2Ct -\lambda. $$

The function $\phi_\lambda$ has a unique critical point

$$\tau(\lambda)=\frac{\lambda}{2C}.$$

Observe that

$$\phi_\lambda(\tau s)=\frac{\lambda^2}{4C}\left( s^2- 2 s\right). $$

$$ T_\lambda= \tau(\lambda)\int_0^\infty  e^{-\frac{\lambda^2}{4C}(s^2-2s)} ds,  $$

We set $g(s)=s^2-2s.$ Using Laplace method again we deduce

$$  T_\lambda\sim \tau(\lambda)e^{-g(1)\frac{\lambda^2}{4C}}\sqrt{ \frac{8C\pi}{\lambda^2 g_a''(1)}} = \sqrt{\frac{\pi}{C}} \times \exp\left(\; \frac{\lambda^2}{2C}\;\right).$$

We can now show that

$$\log R_k(w)\sim -\frac{4}{C}. $$ Thus $R_\infty(w)$ can have any value  in $[0,1]$.