# Can I define Fredholm Index using $\dim \ker ST - \dim \ker TS$?

$X$, $Y$ 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!

• 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. May 10, 2014 at 22:04
• 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. May 11, 2014 at 0:49
• 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 $S$. May 11, 2014 at 8:18
• 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. May 11, 2014 at 13:02
• Thanks all for your comments. I have deleted my question on MSE to conform to this requirement. May 11, 2014 at 19:23