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$X$, $Y$ are Banach spaces. Let $S \in L(X, Y)$, $T \in L(Y, X)$, where $L(X, Y)$ denotes the Banach algebra of bounded linear operators from $X$ to $Y$. If we have that $Id_Y - ST \in \mathbb{K}(Y)$ and that $Id_X - TS \in \mathbb{K}(X)$. Is it then possible to define Index(T) to be $dim \ker(ST) - dim \ker(TS)$?

The intuition of this definition is that this is true for the case when $X=Y$ is some Hilbert space and $T=S^*$, the adjoint of $S$. I don't know how (if possible) to generalize this to the more general case.

If it is not true in the most general case, can you give a counterexample and a sufficient condition? Thank you!

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    $\begingroup$ Please don't cross post. Better, flag the moderators here, and ask then to migrate this version. Or just go ahead and delete it, and leave only the version on MSE. $\endgroup$
    – Andrés E. Caicedo
    Commented May 10, 2014 at 22:04
  • $\begingroup$ Andres, thanks for your comment. What would you suggest if I am trying to maximize the exposure this question get? I had posted my questions at MSE for 48 hours with no response before I posted it here. $\endgroup$
    – Clark Chong
    Commented May 11, 2014 at 0:49
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    $\begingroup$ It is true for some choices of $S$, for arbitrary Banach spaces $X$, $Y$. When $X=Y$ are Hilbert spaces it is false for many choices of $S$. $\endgroup$
    – Liviu Nicolaescu
    Commented May 11, 2014 at 8:18
  • $\begingroup$ Liviu: Thank you! Would you mind sharing with me one of the many examples where this fail in the Hilbert space? I am still trying to get some intuition about this. $\endgroup$
    – Clark Chong
    Commented May 11, 2014 at 13:02
  • $\begingroup$ Thanks all for your comments. I have deleted my question on MSE to conform to this requirement. $\endgroup$
    – Clark Chong
    Commented May 11, 2014 at 19:23

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