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Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Hodge theory \\[15mm] \small lecture 5: Fredholm operators} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, February 7, 2018 } \end{center} \newpage {\bf \blue Operators with closed image} \theorem {\bf \blue (``Banach inverse mapping theorem'')}\\ Let $A:\; H_1 \arrow H_2$ be an operator on Hilbert spaces with closed image. {\bf \red Then $A^{-1}:\; \im A \arrow H_1/\ker A$ is continuous.} \pstep Suppose that $A^{-1}$ is not continuous. This means that $\sup_{x\in \im A} \frac {|A^{-1}(x)|}{|x|}=\infty$, equivalently, that $\inf_{x\in (\ker A)^\bot} \frac {|A(x)|}{|x|}=0$. In other words, {\bf \purple there exists a sequence of vectors $v_1, ..., v_n, ...\in (\ker A)^\bot$ such that $|A(v_i)|=0$.} {\bf \green Step 2:} Choose a subsequence $w_1,..., w_n, ...$ in $\{v_i\}$ such that $|w_i|=1$ and $\sum |A(w_i)|^2 < \infty$. Restrict $A$ to the space $W$ generated by $w_i$. Clearly, the vector $\sum A(w_i)$ has no preimage, hence $A$ is never surjective. {\bf \green Step 3:} Replacing $H_1$ by $H_1/\ker A$ and $H_2$ by $\im A$, we may assume that $\ker A=0$, $\im A= H_2$. Clearly, $x\bot A(W) \Leftrightarrow A^*(x)\bot W \Leftrightarrow A^*(x)\in W^\bot$ and $x\bot A(W^\bot) \Leftrightarrow A^*(x)\in W$. Applying $(A^*)^{-1}$ to the direct sum decomposition $H_1 = W \oplus W^\bot$, {\bf \purple we obtain the direct sum decomposition $H_2=\overline {A(W)}\oplus \overline {A(W^\bot)}$.} Since $A$ cannot be surjective onto $\overline {A(W)}$ (Step 2), this leads to the contradiction. \endproof \newpage {\bf \blue Fredholm operators} \definition A continuous operator $F:\; H_1\arrow H_2$ of Hilbert spaces is called {\bf\blue Fredholm} if its image is closed and kernel and cokernel are finite-dimensional. \remark {\bf \red ``Cokernel'' of a morphism $F:\; H_1\arrow H_2$ of topological vector spaces is often defined as $\frac{H_2}{\overline {\im F}}$.} \definition An operator $F:\; H_1\arrow H_2$ {\bf \blue has finite rank} if its image has finite rank. \claim An operator $F:\; H_1\arrow H_2$ {\bf \red is Fredholm} if and only if there exists $F_1:\; H_2\arrow H_1$ such that {\bf \red the operators $\Id-FF_1$ and $\Id - F_1 F$ have finite rank.} \proof This is because $F$ defines an isomorphism $F:\; H_1/\ker F \arrow \im F$ as shown above. \endproof \newpage \begin{center} {\small \bf \green Erik Ivar Fredholm \\ Born: 7 April 1866 in Stockholm, Sweden\\ Died: 17 August 1927 in Danderyd, County of Stockholm, Sweden} \epsfig{file=Fredholm_2.jpeg,width=0.38\linewidth} {\em \small ...As a young boy he played the flute, but later took up playing the violin. He particularly loved to play Bach. Again he combined his talents, applying his mechanical skills to music as well as to mathematics. Unlikely as it sounds, he built his first violin from half a coconut, while he also used his talents at building machines to make one to solve differential equations...} \end{center} \newpage {\bf \blue Fredholm operators: invertible up to a finite-dimensional summand} \remark For any operator $F:\; H_1\arrow H_2$ on Hilbert spaces, one has $\ker F^* = H_2/\overline{\im F}$. Also, $F$ is Fredholm if and only if $F^*$ is Fredholm, because {\bf \red Fredholm is the same as ``invertible after adding a finite-dimensional subspace and extending the operator to this finite-dimensional subspace in appropriate way''.} {\bf \green Lemma 1:} Let $F, G$ be operators on a Hilbert space $H$, with $F\circ G$ Fredholm and $G \circ F$ Fredholm. {\bf \red Then $F$ and $G$ are Fredholm.} \proof Consider a category $\cac$ with $\Ob(\cac)$ Hilbert spaces and $\Mor(H_1, H_2)= \Hom(H_1, H_2)/{\cal F}$ where ${\cal F}$ is the space of finite rank maps. Clearly, a morphism is invertible in $\cac$ if and only if it is Fredholm. Then $FG$ and $GF$ are invertible in $\cac$, giving $FG P=\Id$ and $PGF =\Id$ for some morphism $P$. This gives $F \circ G P=\Id$ and $PG\circ F =\Id$, hence $F$ is invertible in $\cac$. \endproof \newpage {\bf \blue Operator norm } \definition Let $V_1$, $V_2$ be vector spaces equipped with a norm. {\bf\blue Norm} (``operator norm'') of a linear operator $E:\; V_1 \arrow V_2$ is the number $\|E\|:=\sup_{|v\neq 0} \frac {|E(v)|}{|v|}$. \remark {\bf \purple An operator is continuous if and only if its norm is finite.} \claim The space $\Hom(V, W)$ of operators on Banach spaces with the operator norm {\bf \purple is a Banach space.} {\bf \red It is not a Hilbert space} even when $V, W$ are Hilbert spaces. \newpage {\bf \blue Limits of operators of finite rank } \remark Hausdorff limit of compacts is again compact. Therefore, {\bf \red compact operators are closed in norm topology}. \theorem Let $U, W$ be Hilbert spaces, and $K \subset \Hom(U, W)$ be the closure of the space of operators of finite rank in operator norm topology. Then {\bf \red $K$ is the space of compact operators.} \pstep Since finite rank operators are compact, and compact operators are closed in operator norm topology, {\bf \purple $K$ is contained in the space of compact operators.} {\bf \green Step 2:} Fix a compact operator $A:\; U \arrow W$. Let $U_0\subset U_1\subset ...$ be a sequence of finite-dimensional spaces such that $U = \overline {\bigcup U_i}$, and $A_i\in \Hom(U, W)$ be operator equal to $A$ on $U_i$ and to 0 on $U_i ^\bot$. Then $\|A-A_i\|= \sup_{x\in U_i^\bot} \frac{|A_i(x)|}{|x|}$. {\bf \purple Unless $\lim_i \|A-A_i\|=0$, we have an infinite sequence $x_i$ of orthogonormal vectors in $W$ such that $|A(x_i)|\geq \epsilon >0$,} which is impossible because $A$ is compact. Then $A= \lim A_i$. \endproof \newpage {\bf \blue Fredholm operators and compact operators} \theorem {\bf \red The set of Fredholm operators is open in the operator norm topology.}\\ \pstep Let $F:\; U \arrow V$ be a Fredholm operator, and $U_1:= (\ker F)^\bot$. Since $F$ is invertible on $U_1$, it satisfies $\inf_{x\in U_1} \frac{|F(x)|}{|x|} > 2\epsilon$. Then, for any operator $A$ with $\|A\| < \epsilon$, one has $\inf_{x\in U_1} \frac{|F+A(x)|}{|x|} > \epsilon$. {\bf \purple This implies that $F\restrict{U_1}$ is an invertible map to its image, which is closed.} In particular, $\ker(F+A)$ is finite-dimensional. {\bf \green Step 2:} To obtain that $\coker (F+A)$ is finite-dimensional for $\|A\|$ sufficiently small, we observe that $\coker (F+A)= \ker (F^* +A^*)$, and $F^*$ is also Fredholm. Then Step 1 implies that {\bf \purple $\ker(F^*+A^*)$ is finite-dimensional for $\|A\|$ sufficiently small.} \endproof \corollary Let $A$ be compact and $F$ Fredholm. {\bf \red Then $A+F$ is Fredholm.} \proof Let $A_i$ be a sequence of operators with finite rank converging to $A$. Then $F+(A-A_i)$ is Fredholm for $i$ sufficiently big, because the set of Fredholm operators is open. However, a sum of Fredholm operator and operator of finite rank is Fredholm, hence $F+A= F+(A-A_i)+A_i$ is also Fredholm. \endproof \newpage {\bf \blue Calkin algebra} \claim {\bf \red Compact operators form a two-sided ideal in the algebra $\Hom(H,H)$.} \\ \proof Under a continuous homomorphism of normed vector spaces, {\bf \purple an image of a bounded set is bounded, and an image of a compact set is compact.} Therefore, for any compact $K\in \Hom(H,H)$ and any $A\in \Hom(H, H)$, the operators $KA$ and $AK$ are compact. \endproof \definition {\bf \blue Calkin algebra} is the quotient of $\Hom(H,H)$ by the ideal of compact operators. \theorem An operator $F:\; H_1\arrow H_2$ {\bf \red is Fredholm if and only if its image in the Calkin algebra is invertible.} \pstep Invertibility in Calkin algebra means that there exists $G:\; H_2\arrow H_1$ such that {\bf \purple the operators $\Id-FG$ and $\Id - G F$ are compact.} {\bf \green Step 2:} By definition, $F$ is Fredholm if and only if there exists $G:\; H_2\arrow H_1$ such that $\Id-FG$ and $\Id - G F$ have finite rank. {\bf \purple Therefore, $F$ is invertible in the Calkin algebra.} \\ {\bf \green Step 3:} Assume that $\Id-FG$ and $\Id - G F$ are compact. A sum of Fredholm and compact is compact, hence $FG$ and $GF$ are Fredholm. Then $F$ and $G$ are Fredholm by Lemma 1. \endproof \newpage {\bf \blue Index of Fredholm operators} \definition Let $F:\; H \arrow H_1$ be a Fredholm operator. Define its {\bf \blue index} as $\ind(F):= \dim \ker F - \dim \coker F$. \claim Let $H_1 \stackrel F\arrow H_2\stackrel G\arrow H_3$ be Fredholm operators. {\bf \red Then $\ind (FG)=\ind(F)+ \ind(G)$.} \proof For $F$, $G$ finite-dimensional this is clear, because then $\ind(F) = \dim H_1- \dim H_2$, $\ind(G)= \dim H_2-\dim H_3$ and $\ind(FG) = \dim H_1- \dim H_3$. To reduce everything to the finite-dimensional case, chose $W_1\subset H_1, W_2 \subset H_2, W_3 \subset H_3$ of finite codimension. Then $\ind (FG)$ is index of the map $H_1/W_1 \stackrel {FG}\arrow H_3/W_3$, and $\ind F$, $\ind G$ are indices of the maps $H_1/W_1 \stackrel F\arrow H_2/W_2\stackrel G\arrow H_3/W_3$. \endproof \claim Let $F:\; H \arrow H_1$ be a Fredholm operator, and $A:\; H \arrow H_1$ a finite rank operator. {\bf \red Then $\ind(F+A)= \ind(F)$.} \proof Find a closed subspace $W\subset \ker A$ of finite codimension such that $F\restrict W\arrow F(W)$ is an isomorphism. Then $\ind(F+A)$ is index of the map $F+A:\; H/W\arrow H_1/F(W)$, and $\ind(F)$ is index of $F:\; H/W\arrow H_1/F(W)$. However, for finite-dimensional spaces $P, Q$ and a map $G:\; P \arrow Q$, $\ind(G) = \dim P - \dim Q$, hence $\ind(F+A)= \ind(F)$. \endproof \newpage {\bf \blue Index of a sum of Fredholm operator and compact} \corollary {\bf \red For any compact operator $K:\; H \arrow H$, one has $\ind(K+\Id)=0$.} \proof Decompose $K$ onto a sum $K= K_1 + K_0$, with $K_1$ of finite rank, and $\|K_0\| <1$. Then $(\Id+ K_0)^{-1} = \Id- K_0 + K_0^2 - ...$, hence $\ind(\Id+ K_0)=0$, and $\ind(\Id+K)=0$ because $\Id +K$ is a sum of $\id+K_0$ and an operator of finite rank. \endproof \theorem Let $F$ be a Fredholm operator, and $K$ compact. {\bf \red Then $\ind (F+K) = \ind(F)$.} \proof There exists a Fredholm operator $G$ such that $FG= \Id+ A$, $GF= \Id+ B$, and $A, B$ are compact operators. Then $(F+K)G = \Id + A - KG$, which is a sum of identity and a compact operator, giving $\ind((F+K)G)=0$. Using $\ind(FG) = \ind(F)+\ind(G)$, we obtain $\ind(F) = - \ind(G)$ and \[ 0 = \ind((F+K)G)= \ind(G) + \ind(F+K) \] giving $\ind(F+K)=-\ind(G)= \ind(F)$. \endproof \newpage {\bf \blue Index of Fredholm operator is locally constant} \remark Recall that the set of Fredholm operators {\bf \purple is open in the operator norm topology.} \corollary {\bf \red Index is a locally constant function on the space of Fredholm operators} taken with the operator norm topology. \proof Let $F, G$ be Fredholm, $FG= \Id+ A$, $GF= \Id+ B$, where $A, B$ are compact, and $\|H\|< \epsilon$. Then $(F+H)G= \Id+ A + HG$. Choosing $\epsilon$ in such a way that $\|HG\| <1$, we obtain that $\Id+ HG$ is invertible, hence the index of $\Id+ A + HG$ is zero (it is a sum of invertible and a compact operator). This gives \[ 0 = \ind((F+H)G) = \ind(F+H) + \ind(G) = \ind(F+H)-\ind(F) \] hence $\ind(F+H)=\ind(F)$. \endproof \end{document}