$H$ 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)


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


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


  • $\begingroup$ you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem? $\endgroup$ – Picaud Vincent Nov 30 '18 at 23:16
  • $\begingroup$ @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 $\endgroup$ – user89940 Nov 30 '18 at 23:18
  • $\begingroup$ $coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ? $\endgroup$ – user89940 Dec 1 '18 at 0:00

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


$\newcommand{\co}{\operatorname{coker}}$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, $H/TH\cong (TH)^\perp$.

  • 1
    $\begingroup$ thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed $\endgroup$ – user89940 Dec 1 '18 at 3:10
  • $\begingroup$ For any subset $S$ of $H$, $H/\overline{span{S}}=S^\perp$. The very usage of $T(H)$ is closed is $T(H)=\overline{span{T(H)}}$ $\endgroup$ – C.Ding Dec 1 '18 at 4:13
  • $\begingroup$ A better answer has been updated. $\endgroup$ – C.Ding Dec 1 '18 at 4:23
  • $\begingroup$ 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 math.stackexchange.com/questions/1107449/… $\endgroup$ – user89940 Dec 1 '18 at 16:15
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    $\begingroup$ 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. $\endgroup$ – C.Ding Dec 5 '18 at 7:53

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