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