Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 1.
FREDHOLM THEORY
Proof. Let ˜T be a bijection associated with T , X 0 , and Y 0 : i.e., X = T −1 (0) ⊕ X 0
and Y = T (X) ⊕ Y 0 . Suppose T 0 := T | X0 . Since ˜T is invertible, S ˜T is invertible and
index (S ˜T ) = index S. By identifying X 0 and X 0 × {0}, we see that ST 0 is a common
restriction of S ˜T and ST to X 0 . By Lemma 1.5.1, ST is Fredholm and
index (ST ) = index (ST 0 ) + dim X/X 0
(
)
= index(S ˜T ) − dim X 0 × Y 0 /X 0 × {0} + α(T )
= indexS − dim Y 0 + α(T )
= index S − β(T ) + α(T )
= index S + index T.
The converse of Theorem 1.5.2 is not true in general. To see this, consider the
following operators on l 2 :
T (x 1 , x 2 , x 3 , . . .) = (0, x 1 , 0, x 2 , 0, x 3 , . . .)
S(x 1 , x 2 , x 3 , . . .) = (x 2 , x 4 , x 6 , . . .).
Then T ad S are not Fredholm, but ST = I. However, if ST = T S then we have
ST is Fredholm =⇒ S and T are both Fredholm
because T −1 (0) ⊂ (ST ) −1 (0) and (ST )(X) = T S(X) ⊂ T (X).
Remark 1.5.3. For a time being, a Fredholm operator of index 0 will be called a
Weyl operator. Then we have the following question: Is there implication that if
ST = T S then
S, T are Weyl ⇐⇒ ST is Weyl
Here is the answer. The forward implication comes from the “Index Product Theorem”
without commutativity condition. However the backward implication may fail
even with commutativity condition. To see this, let
T =
[ U 0
0 I
]
and S =
[ ] I 0
0 U ∗ ,
where U is the unilateral shift on l 2 . Evidently,
[ ] U 0
index (ST ) = index
0 U ∗
but S and T are not Weyl.
= index U + index U ∗
= 0,
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