Invariance of the Fredholm index under finite-dimensional perturbations

Question

Let's call a linear map $f : V \to W$ between vector spaces over some field Fredholm if $\ker(f)$ and $\mathrm{coker}(f)$ are finite-dimensional. (Equivalently, it represents an isomorphism in the abelian quotient category (vector spaces)/(finite-dim. vector spaces).) We can then define the Fredholm index by $$\mathrm{ind}(f) := \dim(\ker(f))-\dim(\mathrm{coker}(f)).$$ Is it true that $\mathrm{ind}(f+g)=\mathrm{ind}(f)$ if $g : V \to W$ has finite-dimensional range?

I have tried to write down exact sequences and use that the Euler characteristic vanishes on such sequences, but this didn't work out. Another idea would be the following: We may assume wlog that $\mathrm{im}(g)$ is $1$-dimensional. Then write $g = w \otimes \alpha$ for some $\alpha \in V^*$ and $w \in W \setminus \{0\}$. Now two cases appear: 1) $w \in \mathrm{im}(f)$. 2) $w \notin \mathrm{im}(f)$. I have tried to compute $\dim(\ker(f+g))$ and $\dim(\mathrm{coker}(f+g))$ in each case, but didn't succeed.

PS: I have borrowed the terminology for linear operators between Hilbert spaces (Fredholm operator). I don't know if this is standard. Notice that the proof in the Wikipedia article that the index is invariant under compact perturbations is analytic and therefore probably cannot be used here.

Answer 1

The result perhaps can be proved inductively showing that $\text{index}\,( T- \delta) = \text{index}\,T$ for $\delta$ of rank $1$. A useful particular case is showing by hand that $\text{index}( I - \delta) = 0$, starting with rank of $\delta=1$. The different cases presented suggests that there could be a uniform proof. We sketch one below. Break it into several steps.

1. Show the additivity of the index

Given $$U\xrightarrow[]{S}V \xrightarrow[]{T} W$$

with $S$, $T$ Fredholm, then $T S$ is Fredholm and we have

$$\text{index}{\,TS} = \text{index}{\,S}+ \text{index}{\,T}$$

This follows right away from the long exact sequence:

$$0 \to \ker \,S\to \ker \,TS\to \ker\, T \to \text{coker}\,S \to \text{coker}\,T S \to \text{coker}\,T \to 0 $$

by using the fact that the Euler characteristic of the sequence is $0$.

2. Show that for every $\pi \colon V\to V$ of finite rank, $I- \pi$ is Fredholm of index $0$

$$\text{index} \, (I-\pi)=0$$

This follows from the existence of a natural isomorphism:

$$\frac{\text{im} (I - \pi)}{\ker \pi} \simeq \frac{\text{im} ( \pi)}{\ker(I- \pi)}$$

(proof is left for the reader)

3. If $\pi\colon V \to V$ is of finite rank, and $T\colon U \to V$, then $T$ is Fredholm if and only if $(I-\pi) T$ is, and if so,

$$\text{index}\,T = \text{index}(I -\pi)\, T$$

This follows easily from $1.$ and $2.$

4. Let $\delta \colon U \to V$ of finite rank and $T \colon U \to V $ Fredholm. Then $T - \delta$ is Fredholm and

$$\text{index}\,( T -\delta) = \text{index}\, T$$

Indeed, there exists $\pi \colon V \to V$ of finite rank so that $\pi \delta = \delta$, for instance the projection of $V$ onto the image of $\delta$, considered as a map from $V$ to $V$. Therefore $(I-\pi) \delta = 0$. We conclude $(I -\pi) T = (I-\pi) (T-\delta)$ and from $1.$ and $2.$ we get:

$$\text{index}\,(T-\delta) = \text{index}\, (I-\pi) (T-\delta) = \text{index}\, (I-\pi) T= \text{index}\, T$$