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 $\ker T_x =\ker B_x= \eU_r(x)$, $\forall x$.

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 $[0,1]$.

An inequality involving moments

 $\newcommand{\bR}{\mathbb{R}}$  Suppose that $w:\bR\to \bR$ is a nonnegative, even smooth function  decaying fast at $\infty$, $w\in\mathscr{S}(\bR)$.

 Define

$$s_m(w)= \int_{\bR^m} w(|x|) dx,\;\;  d_m(w):=\int_{\bR^m} x_i^2 w(|x|) dx,\;\;\forall i $$

$$ h_m(w) = \int_{\bR^m} x_i^2x_j^2 w(|x|) dx,\;\;\forall i<j. $$

Is it true that


$$ d_m(w)^2\geq s_m(w) h_m(w)  \tag{A}$$


for any $m\geq 2$ and any $w$  satisfying the above restrictions?

Example 1.   Observe first that

$$ s_n(w)= \left(\int_{S^{m-1}} dA\right)\int_0^\infty r^{m-1} w(r) dr $$

$$ d_n(w)= \left( \int_{S^{m-1}}x_1^2 dA(x)\right)\int_0^\infty r^{m+1} w(r) dr, $$

$$ h_n(w)= \left( \int_{S^{m-1}}x_1^2x_2^2 dA(x) \right)\int_0^\infty r^{m+3} w(r) dr, $$

and


$$a_m:=\int_{S^{m-1}} dA = \frac{2\pi^{\frac{m}{2}}}{\Gamma(\frac{m}{2})},\;\;  b_m:=\int_{S^{m-1}}x_1^2 dA(x)=   \frac{\pi^{\frac{m}{2}}}{\Gamma(1+\frac{m}{2})}=\frac{a_m}{m}, $$

$$c_m:= \int_{S^{m-1}}x_1^2x_2^2 dA(x)=   \frac{\pi^{\frac{m}{2}}}{2\Gamma(2+\frac{m}{2})} = \frac{b_m}{m+2}. $$

Thus

$$ d_m^2= b_m^2 \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2=\frac{a_m^2}{m^2}   \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2, $$

and

$$s_m d_m= \frac{a_m^2}{m(m+2)} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right), $$

so that the inequality (A)  is equivalent to

$$ \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2\geq \frac{m}{m+2} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right). \tag{B} $$

A. Let us now choose $w(t)=t^{2k} e^{-t^2}$, $k$ nonnegative integer. Then for any $a>0$ we have

$$ \int_0^\infty t^a w(t) dt=\int_0^\infty t^{a+2k} e^{-t^2}  dt $$

($s=t^2$)

$$= \frac{1}{2}\int_0^\infty s^{\frac{a+2k-1}{2}} e^{-s} ds = \frac{1}{2}\Gamma\left(k+\frac{a+1}{2}\right). $$

For this choice of weight   the inequality (A) becomes

$$\Gamma(k +1+\frac{m}{2})^2\geq \frac{m}{m+2}\Gamma(k+\frac{m}{2})\Gamma(k+2+\frac{m}{2}). $$

This is equivalent to

$$ k+\frac{m}{2}=\frac{\Gamma(k +1+\frac{m}{2})}{\Gamma(k+\frac{m}{2})}\geq \frac{m}{m+2} \frac{\Gamma(k+2+\frac{m}{2})}{\Gamma(k+1+\frac{m}{2})}= \frac{m}{m+2}\left(k+1+\frac{m}{2}\right) . $$

One can easily verify that the last inequality  holds for any $m\geq 2$, $k\geq 0$. It   turns into an equality when $k=0$.

B.  Suppose that $w(t)=(1+t^{2k}) e^{-t^2}$. Then

$$ \int_0^\infty r^a w(t) dr=  \int_0^\infty r^a e^{-r^2} dr +  \int_0^\infty r^{a+2k} e^{-r^2} dr $$

$$= \frac{1}{2}\left( \Gamma\left(\frac{a+1}{2}\right)+\Gamma\left(k+\frac{a+1}{2}\right)\;\right).$$

In this case the inequality (A)  has the equivalent form

$$ \frac{\Gamma(1+\frac{m}{2})+\Gamma(k +1+\frac{m}{2})}{\Gamma(\frac{m}{2})+\Gamma(k+\frac{m}{2})}\geq \frac{m}{m+2}\times  \frac{\Gamma(2+\frac{m}{2})+\Gamma(k+2+\frac{m}{2})}{\Gamma(1+\frac{m}{2})+\Gamma(k+1+\frac{m}{2})},$$

or

$$ \frac{\Gamma(1+\frac{m}{2})}{ \Gamma(\frac{m}{2})}\times \frac{1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (j+\frac{m}{2})} \geq  \frac{m}{m+2}\times \frac{\Gamma(2+\frac{m}{2})}{\Gamma(1+\frac{m}{2})} \times \frac{1+ \prod_{j=0}^{k-1} (2+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})},$$

which further simplifies to

$$  \frac{1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (j+\frac{m}{2})}\geq  \frac{1+ \prod_{j=0}^{k-1} (2+j+\frac{m}{2})}{ 1+\prod_{j=0}^{k-1} (1+j+\frac{m}{2})}. $$


For $k=1$ this reduces to

$$ \frac{2+\frac{m}{2}}{1+\frac{m}{2}}\geq \frac{3+\frac{m}{2}}{2+\frac{m}{2}}. $$

This is obviously true since $(m+4)^2> (m+2)(m+6)$.  $\Box$


We  can reformulate  the inequality (A)  as a convexity inequality as follows. For $a>1$ define

$$T_a(w)=\frac{\int_0^\infty t^{a-1} w(t) dt}{ \int_0^\infty t^{a-1} e^{-t^2} dt}. $$

Then the inequality (A) can be rewritten as

$$\frac{T_{m+2}}{T_m}\geq \frac{T_{m+4}}{T_{m+2}} \Longleftrightarrow  T_{m+2}(w)^2\geq T_m(w) T_{m+4}(w). \tag{C}$$

Update 1. The inequality  (B) is false in  this generality. It suffices to choose $w(t)$ that approximates the  singular measure

$$\mu_n =\delta_1+\frac{1}{n^{m-2}}\delta_n, $$

where $\delta_c$ denotes the Dirac delta measure on $\bR$ concentrated at $c$. I  wonder if the inequality does hold if $(t)$ is log-concave in a rather special way

$$w(t)=e^{-U(t)}$$

where  $U(t)$ is smooth, even, convex and $\lim_{t\to\infty} U(t)=\infty$.

Also,  what are the asymptotics of

$$ R_m=\frac{\left(\int_0^\infty r^{m+1} w(r) dr\right)^2}{\left(\int_0^\infty r^{m} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right)} $$

as $m\to \infty$?

My talk at Banff (rudely interrupted)

Video from 12w5121: Geometric Structures on Manifolds | Banff International Research Statio

Live Blogging from Banff.

The conference is about to start.

This is only the first step

I'm starting this math blog and I hope it will work.      I hope you can see this beautiful equation

$$e^{\pi \boldsymbol{i}} +1=0,\;\;\boldsymbol{i}=\sqrt{-1}, $$

or this wonderful equality

$$\int_{\mathbb{R}} e^{-\frac{x^2}{2}} dx=\sqrt{2\pi}. $$