Fredholm operators in Hilbert spaces

Question

Suppose $T_r$ and $T_l$ are the left and the right translations in $l_2$. $T_l$ maps $(x_1,x_2,x_3,...)$ to $(x_2,x_3,x_4,...)$, $T_r$ maps $(x_1,x_2,x_3,...)$ to $(0,x_1,x_2,...)$. It can be easily seen that $T_r$ and $T_l$ are fredholm operators and $\mathrm{Ind}(T_l)=1$ and $\mathrm{Ind}(T_r)=-1$. Moreover it is easy to verify that $\mathrm{Ind}(T_l^k)=k$ and $\mathrm{Ind}(T_r^k)=-1$. So, for all $k\in \mathbb{Z}$ there is a bounded fredholm operator $l_2\longrightarrow l_2$ whose index is $k$. Could you tell me how to prove that statement with an arbitrary hilbert space?

Answer 1

Given an arbitrary (infinite-dimensional) Hilbert space $\mathcal H$, you can find an orthonormal sequence $x_n$ in it (inductively, by choosing a unit vector $x_n$ from $\{x_1,\dots,x_{n-1}\}^\perp$). Let $M$ be the closed linear span of $\{x_n:n=1,2,\dots\}^\perp$, and $N=M^\perp$. On the space $M$ you have the natural analogues of $T_l$ and $T_r$. They can also act on $\mathcal H$ by keeping the $N$-component of a vector unchanged. The Fredholm indices stay the same.