If $ST$ is a Fredholm operator, then show that $T$ is Fredholm if and only if $S$ is Fredholm.

  • $\begingroup$ @Aaron I tried to use the fact that $T$ is Fredholm iff $T=I+F$, where $F$ is a finite dimensional operator, it worked in one direction but not the other. $\endgroup$ – i.a.m Jan 30 '13 at 7:00

$\DeclareMathOperator{\coker}{coker}\DeclareMathOperator{\index}{ind}$Using the definition that a Fredholm operator is an operator $F \colon X \to Y$ for which both $\ker{F}$ and $\coker{F} = Y/F(X)$ are finite-dimensional this really is an exercise in linear (or homological) algebra. Notice that the requirement that that $F(X)$ be closed in $Y$ is automatic: Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?

For every composition $ST$ of linear maps $S$ and $T$ there is an exact sequence of vector spaces $$ 0 \to \ker{T} \to \ker{ST} \to \ker{S} \to \coker{T} \to \coker{ST} \to \coker{S} \to 0. $$ Since $ST$ is Fredholm, $\ker{ST}$ and $\coker{ST}$ are finite-dimensional. Therefore $\ker{T}$ and $\coker{S}$ are finite-dimensional, being a subspace and a quotient of a finite-dimensional space.

We are given the additional information that either $\ker{S}$ or $\coker{T}$ is finite-dimensional, depending on whether $S$ or $T$ is Fredholm. Thus, we know that five out of six of the vector spaces in the exact sequence are finite-dimensional, hence so is the sixth.

In other words, we can read off this sequence:

If two out of the three operators $S$, $T$ and $ST$ are Fredholm then so is the third.

Incidentally, this also gives a quick proof of the additivity of the Fredholm index: recall that the index of a Fredholm operator is $\index{F} = \dim{\ker{F}} - \dim{\coker{F}}$.

From the above sequence and the fact that an exact complex has zero Euler characteristic we get $$ 0 = \dim{\ker{T}} - \dim\ker{ST} + \dim{\ker{S}} - \dim{\coker{T}}+\dim{\coker{ST}} -\dim{\coker{S}} $$ which is immediately rearranged to $$ \index ST = \index S + \index T. $$

  • $\begingroup$ Thank you for solving the question $\endgroup$ – i.a.m Jan 31 '13 at 10:12
  • $\begingroup$ "Thus, we know that five out of six of the vector spaces in the exact sequence are finite-dimensional, hence so is the sixth." Could you please explain why that happens? Also, if we only knew that $S$ and $T$ were Fredholm, could we conclude that $ST$ is? $\endgroup$ – Guillermo Mosse Jun 30 '17 at 23:36
  • $\begingroup$ @SeñorBilly: The key fact is that if $0 \to X \to Y \to Z \to 0$ is exact and if $X$ and $Y$ are finite dimensional (f.d.), then so is $X$. (Indeed, $Y \cong X \oplus Z$.) One can apply this, for example, to see that $\ker ST$ is finite if both $\ker S$ and $\ker T$ are finite. Indeed, since $\ker S$ is f.d., the image of $\ker ST$ is f.d. and since the sequence is exact, the sequence $0 \to \ker T \to \ker ST \to {\rm image}(\ker ST) \to 0$ is exact. Other dimensions can seen to be finite using similar tricks. $\endgroup$ – Chris Judge Jul 11 '17 at 20:30

What you want to use, if possible, is the definition that $A \in B(H)$ is Fredholm if and only if there exists some $B \in B(H)$ such that $AB = I \bmod K(H)$ and $BA = I \bmod K(H)$. If you don't mind a little bit more machinery, recalling that $K(H)$ is a closed, two-sided $\ast$-ideal in the $C^\ast$-algebra $B(H)$, we can equivalently say that $A$ is Fredholm if and only if the image $[A]$ of $A$ in the Calkin algebra $Q(H) := B(H)/K(H)$ is invertible (i.e., with inverse $[A]^{-1} = [B]$).

With all this in mind, then, consider why the following observation is true:

Let $R$ be a unital ring, let $a$, $b \in R$, and suppose that $ab$ is invertible. Then $a$ is invertible if and only if $b$ is invertible.

EDIT: In the more general case where $A \in B(H_1,H_2)$, we have that $A$ is Fredholm if and only if there exists some $B \in B(H_2,H_1)$ such that $AB = I_{H_2} \bmod K(H_2)$ and $BA = I_{H_1} \bmod K(H_1)$. Given that $K(H_i)$ is a two-sided ideal of $B(H_i)$ for $i=1,2$, the essential elementary observation is the following:

Let $t : E \to F$ and $s : F \to G$, and suppose that $st : E \to G$ is bijective. Then $s$ is bijective if and only if $t$ is bijective.

Of course, you can't apply this literally, but rather have to think about invertibility modulo compacts on $H_1$ and on $H_2$.

  • $\begingroup$ cacic What if $T\in B(X,Y)$ and $S\in B(Y,Z)$ can we should be able to do it. $\endgroup$ – i.a.m Jan 30 '13 at 22:55
  • $\begingroup$ @i.a.m Indeed you can---see my expanded answer. $\endgroup$ – Branimir Ćaćić Jan 30 '13 at 23:09
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    $\begingroup$ The same proof works for Fredholm operators on Banach spaces (Riesz-Schauder theorem). Both the $\ast$-structure and the Hilbert space structure are irrelevant. $\endgroup$ – Martin Jan 30 '13 at 23:51

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