## 2.1 Basic properties of Fredholm operators

Let $H$ be a Hilbert space. A bounded linear operator $T: H \to H$ is *Fredholm* if $\ker T$ and $\operatorname{coker}T$ are both finite-dimensional.

$T$ induces an isomorphism between $\ker T^\perp$ and $\operatorname{im}T$. In particular, $\operatorname{im}T$ is closed and (if $\ker T \not= 0$) $0$ is an isolated point in the spectrum of $T$.

The *index* of a Fredholm operator $T: H \to H'$ is

If $H = H'$ is finite-dimensional, then all operators are Fredholm with index $0$, by the rank-nullity theorem.

Let $\mathrm{shift}: \ell ^2 \to \ell ^2$ be the map

$\mathrm{shift}(x_1, x_2, x_3, \dots ) = (x_2, x_3, x_4, \dots ).$Then this has index $1$. The adjoint of $\mathrm{shift}$, written $\mathrm{shift}^{-1}$, is given by

$\mathrm{shift}^{-1}(x_1, x_2, x_3, \dots ) = (0, x_1, x_2, \dots ).$which has index $-1$.

One can similar see that $\operatorname{idx}(\mathrm{shift}^k) = k$ for all $k \in \mathbb {Z}$. So in particular, every integer is the index of some operator.

The fact that the index of $\mathrm{shift}^k$ is the negative of the index of its adjoint is not a coincidence.

Let $T: H \to H$ be Fredholm. Then so is $T^*$, and $\operatorname{idx}T^* = - \operatorname{idx}T$.

We shall prove some basic properties of Fredholm operators. They all follow from the following result:

Consider the commutative diagram

If the rows are short exact, and any two of $T$, $S$ and $R$ are Fredholm, then so is the third, and

$\operatorname{idx}S = \operatorname{idx}T + \operatorname{idx}R.$

If $T: H \to H$ and $S: H' \to H'$ are Fredholm, then so is $T \oplus S: H \oplus H' \to H \oplus H'$, and we have

$\operatorname{idx}(T \oplus S) = \operatorname{idx}T + \operatorname{idx}S.$Let $T: H \to H'$ and $S : H' \to H”$ be Fredholm. Then $ST$ is Fredholm, and

$\operatorname{idx}(ST) = \operatorname{idx}T + \operatorname{idx}S.$

Let $I: H \to H$ be the identity and $K : H \to H$ be a compact operator. Then $I+ K$ is Fredholm and has index $0$.

where the vertical maps are all restrictions of $I+ K$. One sees that the maps are well-defined.

Now note that the left-hand map is one between finite-dimensional vector spaces, so is Fredholm with index $0$. On the other hand, the right-hand map is in fact the identity, since we quotiented out by the image of $K$, and is in particular Fredholm with index $0$. So we are done.

For general compact $K$, we can approximate $K$ arbitrarily well by finite-rank operators
^{1}
. So pick $K'$ such that $K'$ has finite rank and $\| K - K'\| < 1$. Thus $I+ (K - K')$ is invertible, and we can write

The first term is invertible, hence Fredholm with index $0$. The second is $I$ plus something of finite rank, so is Fredholm with index $0$. So the general result follows.

Let $T$ be a bounded linear operator. Then $T$ is Fredholm iff there exists $S: H \to H$ such that $TS - I$ and $ST - I$ are compact.

Note that $T$ gives an isomorphism between $(\ker T)^\perp$ and $\operatorname{im}T$. So we can make sense of $(T|_{\operatorname{im}T})^{-1}: \operatorname{im}T \to (\ker T)^{\perp }$. Then we can take $S$ to be

Then $ST - I$ and $TS - I$ have finite rank, as they vanish on $(\ker T)^\perp$ and $\operatorname{im}T$ respectively.

Using the previous corollary, we know $ST$ and $TS$ are Fredholm. Since $ST$ is Fredholm, $\ker T$ is finite-dimensional. Since $TS$ is Fredholm, $\operatorname{coker}T$ is finite-dimensional. So we are done.

If $T$ is Fredholm and $K$ is compact, then $T + K$ is also Fredholm.