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$H$ is a Hilbert space, and $T\in B(H)$ continuous and Fredholm operator. Same books in definition of Fredholm use (when work in Banach space)

$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\operatorname{coker}(T))$

and another's (when work with Hilbert space)

$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\ker(T^{*}))$ , $T^{*}$ is the adjoint operator

so how can i show that

$\dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$

$\operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ?

thanks

user89940
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  • you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem? – Picaud Vincent Nov 30 '18 at 23:16
  • @PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal – user89940 Nov 30 '18 at 23:18
  • $coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ? – user89940 Dec 01 '18 at 00:00

1 Answers1

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For any closed subspace $K$ of a Hilbert space $H$, we have $$H/K\cong K^\perp.$$

Proof.

$\newcommand{\co}{\operatorname{coker}}$Let \begin{align*}\Phi: & K^\perp\to H/K\\ & h\mapsto [h]\end{align*}

(1) Of course, $\Phi$ is linear bounded.

(2) $\Phi$ is injective. Suppose $h\in K^\perp$ such that $\Phi(h)=0$, then $h\in K^\perp\cap K$, so $h=0$.

(3) $\Phi$ is surjective. For any $h\in H$, there is some $h'\in K^\perp$ such that $h-h'\in K$, thus $\Phi(h')=[h']=[h]$.

Noting that $TH$ is closed subspace in the question, $H/TH\cong (TH)^\perp$.

C. Ding
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    thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed – user89940 Dec 01 '18 at 03:10
  • For any subset $S$ of $H$, $H/\overline{span{S}}=S^\perp$. The very usage of $T(H)$ is closed is $T(H)=\overline{span{T(H)}}$ – C. Ding Dec 01 '18 at 04:13
  • A better answer has been updated. – C. Ding Dec 01 '18 at 04:23
  • the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer https://math.stackexchange.com/questions/1107449/on-fredholm-operator-on-hilbert-spaces?rq=1 – user89940 Dec 01 '18 at 16:15
  • @user89940 An operator with finite coker has image closed in Hilbert version too. – C. Ding Dec 03 '18 at 14:47
  • but say $dimKer(T^{})<\infty$ and $dimKer(T)<\infty$ not imply $T(H)$ closed, so request $dimKer(T^{})<\infty$ is more weak than $dimH/(T(H))<\infty$ – user89940 Dec 04 '18 at 17:17
  • @user89940 Suppose $T^#$ is the ajoint operator of $T$ in the sense of Banach space version, then there is a conjugate linear bijection between $\ker T^#$ and $\ker T^$. So the dimension of $\ker T^#$ is equal to that of $\ker T^$. Hence if $dim \ker T^#<\infty$ implies $TH$ is closed, then $dim \ker T^*<\infty$ implies $TH$ is closed. – C. Ding Dec 05 '18 at 03:17
  • but here is an example an operator in hilbert with $dimKer(u)<\infty$ and $dimKer(u^{*})<\infty$ and with $u(H)$ not closed. https://math.stackexchange.com/questions/1107449/on-fredholm-operator-on-hilbert-spaces?rq=1 – user89940 Dec 05 '18 at 04:59
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    Yes. If you define coker of an operator $u:X\to Y$ between Banach spaces as $\ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version. – C. Ding Dec 05 '18 at 07:53