Let $H$ be an infinite-dimensional Hilbert space. The (Fredholm) index of a bounded operator $T : H \to H$ is defined as $$ \mathrm{index}(T) = \dim \ker(T) – \dim \mathrm{coker}(T), $$ whenever the kernel and cokernel of $T$ are finite-dimensional. Operators satisfying these conditions are called Fredholm operators. When $H$ is separable (equivalently, has a countably infinite orthonormal basis), $F_n(H) \neq \emptyset$ for all $n \in \mathbb Z$, which can be seen by picking a countable orthonormal basis and calculating the index of left and right shift operators with respect to this basis.
If $H$ is non-separable, is it possible that for some $n \in \mathbb Z$, the set $F_n(H)$ of Fredholm operators of index $n$ is empty?
Edit: I was able to simplify the problem with a little algebra. Any bounded invertible operator is a Fredholm operator, so $F_0(H) \neq \emptyset$. If there is a Fredholm operator $T$ of index $1$, then, given an integer $n \neq 0$, we can (much like the separable case with the shift) apply powers of $T$ and its adjoint to find an operator of index $n$. Specifically,
- If $n > 0$, then $$ \mathrm{index}(T^n) = \mathrm{index}(T \circ \cdots \circ T) = n \cdot \mathrm{index}(T) = n.$$
- If $n < 0$, then $$ \mathrm{index}(T^{*(-n)}) = -n \cdot \mathrm{index}(T^*) = -n \cdot (-1) = n. $$
So my question will be rephrased as follows: If $H$ is non-separable, is it possible that $F_1(H) \neq \emptyset$?
Best Answer
If $H$ is infinite dimensional then it may be written as $H_0\oplus H_1$, with $H_0$ separable infinite dimensional, and for any Fredholm operator $T$ on $H_0$ one has that $T\oplus I$ is Fredholm on $H$ and $\text{ind}(T\oplus I)= \text{ind}(T)$.
Picking a Fredholm operator $T$ on $H_0$ with a given index $n$, one then gets a Fredholm operator $T\oplus I$ on $H$ with the same index.