Geodesic
A note on the index of $B$-Fredholm operators
Mathematica Bohemica, Tome 129 (2004) no. 2, pp. 177-180

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From Corollary 3.5 in [Berkani, M; Sarih, M.; Studia Math. 148 (2001), 251–257] we know that if $S$, $ T$ are commuting $B$-Fredholm operators acting on a Banach space $X$, then $ST$ is a $B$-Fredholm operator. In this note we show that in general we do not have $\operatorname{\text{ind}}(ST)= \operatorname{\text{ind}}(S) +\operatorname{\text{ind}}(T)$, contrarily to what has been announced in Theorem 3.2 in [Berkani, M; Proc. Amer. Math. Soc. 130 (2002), 1717–1723]. However, if there exist $ U, V \in L(X) $ such that $S$, $T$, $U$, $V$ are commuting and $ US+ VT= I$, then $\operatorname{\text{ind}}(ST)= \operatorname{\text{ind}}(S)+\operatorname{\text{ind}}(T)$, where $\operatorname{\text{ind}}$ stands for the index of a $B$-Fredholm operator.
From Corollary 3.5 in [Berkani, M; Sarih, M.; Studia Math. 148 (2001), 251–257] we know that if $S$, $ T$ are commuting $B$-Fredholm operators acting on a Banach space $X$, then $ST$ is a $B$-Fredholm operator. In this note we show that in general we do not have $\operatorname{\text{ind}}(ST)= \operatorname{\text{ind}}(S) +\operatorname{\text{ind}}(T)$, contrarily to what has been announced in Theorem 3.2 in [Berkani, M; Proc. Amer. Math. Soc. 130 (2002), 1717–1723]. However, if there exist $ U, V \in L(X) $ such that $S$, $T$, $U$, $V$ are commuting and $ US+ VT= I$, then $\operatorname{\text{ind}}(ST)= \operatorname{\text{ind}}(S)+\operatorname{\text{ind}}(T)$, where $\operatorname{\text{ind}}$ stands for the index of a $B$-Fredholm operator.
DOI : 10.21136/MB.2004.133905
Classification : 47A53, 47A55
Keywords: $B$-Fredholm operators; index of the product of Fredholm operators 
Berkani, M.; Medková, D. A note on the index of $B$-Fredholm operators. Mathematica Bohemica, Tome 129 (2004) no. 2, pp. 177-180. doi: 10.21136/MB.2004.133905
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