Clifford Algebras and Bott Periodicity: Basic 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

$\begin{useimager} \[ \begin{tikzcd} 0 \ar[r] & H \ar[r] \ar[d, "T"] & H' \ar[r] \ar[d, "S"]& H'' \ar[r] \ar[d, "R"] & 0\\ 0 \ar[r] & H \ar[r] & H' \ar[r] & H'' \ar[r] & 0 \end{tikzcd} \] \end{useimager}$

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

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Apply the snake lemma to obtain a long exact sequence

$\begin{useimager} \[ \begin{tikzcd}[column sep=small] 0 \ar[r] & \ker T \ar[r] & \ker S \ar[r] & \ker R \ar[r] & \coker T \ar[r] & \coker S \ar[r] & \coker R \ar[r] & 0 \end{tikzcd}.\qedhere \] \end{useimager}$

Proof

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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

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$\begin{useimager} \[ \begin{tikzcd} 0 \ar[r] & H \ar[r, "{(\id, T)}"] \ar[d, "T"] & H \oplus H' \ar[r, "T - \id"] \ar[d, "ST \oplus \id"]& H' \ar[r] \ar[d, "S"] & 0\\ 0 \ar[r] & H' \ar[r, "{(S, \id)}"] & H'' \oplus H' \ar[r, "\id - S"] & H'' \ar[r] & 0 \end{tikzcd}\qedshift \] \end{useimager}$

Proof

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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

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First consider the case where

K

has finite-dimensional image. Then consider the short exact sequences

$\begin{useimager} \[ \begin{tikzcd} 0 \ar[r] & \im K \ar[r] \ar[d] & H \ar[r] \ar[d]& \coker K \ar[r] \ar[d] & 0\\ 0 \ar[r] & \im K \ar[r] & H \ar[r] & \coker K \ar[r] & 0 \end{tikzcd} \] \end{useimager}$

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

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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

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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

$\begin{useimager} \[ \begin{tikzcd}[column sep=2cm] H \cong \im T \oplus (\im T)^\perp \ar[r, "(T|_{\im T})^{-1} \oplus 0"] & (\ker T)^\perp \oplus \ker T \cong H \end{tikzcd}. \] \end{useimager}$

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

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Corollary

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

Proof

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We can use the same

S

, since the algebra of compact operators is a two-sided ideal in

\operatorname{End}(H)

Proof

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