Clifford Algebras and Bott PeriodicityBasic properties of Fredholm operators

## 2.1 Basic properties of Fredholm operators

Definition

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.

Exercise

$T$ 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 $T$.

Definition

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

$\operatorname{idx}T = \dim \ker T - \dim \operatorname{coker}T.$

Example

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

Example

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.

Exercise

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:

Lemma

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

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

Proof
Apply the snake lemma to obtain a long exact sequence

Proof

Corollary

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

$\operatorname{idx}(T \oplus S) = \operatorname{idx}T + \operatorname{idx}S.$
Of course, we could have proved this directly.

Corollary

Let $T: H \to H'$ and $S : H' \to H”$ be Fredholm. Then $ST$ is Fredholm, and

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

Proof

Proof

Corollary

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 $0$.

Proof
First consider the case where $K$ has finite-dimensional image. Then consider the short exact sequences

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

$I+ K = (I+ (K - K')) (I+ (I+ K - K')^{-1} K').$

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.

Proof

Theorem

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.

Proof

• 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.

Proof

Corollary

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

Proof
We can use the same $S$, since the algebra of compact operators is a two-sided ideal in $\operatorname{End}(H)$.
Proof