This chapter collect some materials that didn't quite fit in the main development of the theory.
We recall that the closed unit ball of a Banach space is compact if and only if the space is finite-dimensional. This is a special case of the next lemma:
7 Lemma Let
be a closed densely defined operator. Then the following are equivalent.
- (i)
and the range of T is closed.
- (ii) Every bounded sequence
has a convergent subsequence when
is convergent.
Proof: We may assume T has dense range. (i)
(ii): Suppose
is a bounded sequence such that
is convergent. In view of the Hahn-Banach theorem, X is a direct sum of the kernel of
and some other subspace, say,
. Thus, we can write:
![{\displaystyle f_{j}=g_{j}+h_{j},(g_{j}\in \operatorname {ker} (T),h_{j}\in {\mathcal {W}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f50b97dbb5b6825d809f0f031e5ed0d12767302)
By the closed graph theorem, the inverse of
is continuous. Since
, the continuity implies that
is convergent. Since
contains a convergent subsequence by the paragraph preceding the theorem,
has a convergent subsequence then. (ii)
(i): (ii) implies the first condition of (i), again by the preceding paragraph. For the second, suppose
is convergent. Then by (ii)
has a subsequence
converging to, say,
. Since the graph of T is closed,
converges to
.
A bounded linear operator
between Hilbert spaces is said to be Fredholm if T and
both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of T and the quotient
are finite-dimensional. In fact, if
is finite-dimensional, then
is a complemented subspace; thus, closed. That
has closed range implies that
has closed range. For a Fredholm operator at least, it thus makes sense to define:
.
Because of the first isomorphism theorem, the index is actually independent of any operator T when T is a map between finite-dimensional spaces. This is no longer the case for operators acting on infinite-dimensional spaces.
7 Lemma Let
and
. If
and
are Fredholm operators, then
is a Fredholm operator with
.
Conversely, if
, and both
and
are Fredholm operators, then
is a Fredholm operator.
Proof:
Since
, and
,
we see that
is Fredholm. Next, using the identity
![{\displaystyle \dim X+\dim X^{\bot }\cap Y=\dim Y+\dim Y^{\bot }\cap X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb1c065c5fbb88454b4f3b132949dc52c02f7a11)
we compute:
![{\displaystyle \dim \operatorname {ker} (ST)=\dim \operatorname {ker} (S)\cap \operatorname {ran} (T)+\dim \operatorname {ker} (T)=\dim \operatorname {ker} (S)\cap \operatorname {ker} (T^{*})^{\bot }+\dim \operatorname {ker} (T)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d462b7d3275ba4d1f4fc4b64d46e245eec5fc24)
![{\displaystyle =-\dim \operatorname {ker} (T^{*})+\dim \operatorname {ker} (S)+\dim \operatorname {ker} (S)^{\bot }\cap \operatorname {ker} (T^{*})+\dim \operatorname {ker} (T)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5280af722f82943cea3da44c74137422c12dd372)
![{\displaystyle =\operatorname {ind} (T)+\operatorname {ind} (S)+\dim \operatorname {ker} (T^{*}S^{*}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98703c6b225cee12354e8d17ec98ff2c88d739cd)
For the conversely, let
be a bounded sequence such that
is convergent. Then
is convergent and so
has a convergent sequence when
is Fredholm. Thus,
and
has closed range. That
is a Fredholm operator shows that this is also true for
and we conclude that
is Fredholm.
7 Theorem The mapping
![{\displaystyle T\mapsto \operatorname {ind} (T)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ad977e7ad143f42adeaf18bac428efede3c5e8)
is a locally constant function on the set of Fredhold operators
.
Proof: By the Hahn-Banach theorem, we have decompositions:
.
With respect to these, we represent T by a block matrix:
![{\displaystyle T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/803423b77f734143ebf6c16c8189595c1a3b2594)
where
. By the above lemma,
is invariant under row and column operations. Thus, for any
, we have:
,
since
is invertible when
is small. A depends on S but the point is that
is a linear operator between finite-dimensional spaces. Hence, the index of A is independent of A; thus, of S.
7 Corollary If
is a Fredholm operator and K is a compact operator, then
is a Fredholm operator with
![{\displaystyle \operatorname {ind} (T+K)=\operatorname {ind} (T)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb77b9e03bcc0ce6f0a6135060dabde4e966f852)
Proof:
Let
be a bounded sequence such that
is convergent. By compactness,
has a convergent subsequence
such that
is convergent.
is then convergent and so
contains a convergent subsequence. Since
is compact, the same argument applies to
. The invariance of the index follows from the preceding theorem since
is Fredholm for any complex number
, and the index of
is constant.
The next result, known as Fredholm alternative, is now easy but is very important in application.
7 Corollary If
is a compact, then
and ![{\displaystyle {\mathfrak {H}}_{2}/\operatorname {ran} (K-\lambda I)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0793a6a4323640ae7896fd39d761a628673edb3f)
have the same (finite) dimension for any nonzero complex number
, and
consists of eigenvalues of K.
Proof: The first assertion follows from:
,
and the second is the immediate consequence.
7 Theorem Let
. Then
is a Fredholm operator if and only if
and
are finite-rank operators for some
. Moreover, when
and
are of trace class (e.g., of finite-rank),
![{\displaystyle \operatorname {ind} (T)=\operatorname {Tr} (I-ST)-\operatorname {Tr} (I-TS)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe61948cb761aa01fabc20a2a470b6f483296d1d)
Proof: Since the identities are Fredholm operators (in fact, any invertible operator) and since
![{\displaystyle ST=I_{1}+(I_{1}-ST)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3273003f11ac3a9ac4bd234ddccef73dae4a7bb1)
![{\displaystyle TS=I_{2}+(I_{2}-TS)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af37284c1cf2005dde68d45019fad3dc28848f3f)
and
are Fredholm operators, which implies T is a Fredholm operator. Conversely, suppose T is a Fredholm operator. Then, as before, we can write:
![{\displaystyle T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/803423b77f734143ebf6c16c8189595c1a3b2594)
where
is invertible. If we set, for example,
, then
has required properties. Next, suppose S is given arbitrary:
. Then
.
Similarly, we compute:
.
Now, since
, and
is invertible, we have:
. ![{\displaystyle \square }](https://wikimedia.org/api/rest_v1/media/math/render/svg/455831d58fa08f311b934d324adcff89a868b4e4)
Theorem Every irreducible unitary representation of a compact group is finite-dimensional.