1
$\begingroup$

I was wondering about the signs of the $SBI$-sequence ("Connes' periodicity exact sequence") in equations $(5.2.7.2)$ and $(5.2.7.3)$ of 'Cyclic Homology' by Jean-Louis Loday. Why is the sequence given as $$ \cdots \to HH_{n}^- \to HD_n \to HD'_{n-2}\to HH^+_n \to HD'_{n-1}\to HD_{n-3}\to HH^-_{n-2}\to\cdots \tag{1} $$ instead of $$ \cdots \to HH_{n}^- \to HD_n \to HD_{n-2}\to HH^-_n \to HD_{n-1}\to HD_{n-3}\to HH^-_{n-2}\to\cdots\tag{2} $$ and an analoguous one with opposite sign?


In cyclic homology the $SBI$-sequence exists because of the inclusion of the two first columns (the Hochschild complex $C_*$ plus a contractible column, assuming our algebra is unital) into the cyclic double complex $CC_{**}$. The quotient complex is then the cyclic chain complex with the two first columns missing. This gives the short exact sequence $$ 0\to C \to \operatorname{Tot}CC \to (\operatorname{Tot} CC)[2]\to 0 $$ where $[2]$ indicates a shift in degree by $2$.

In the case of dihedral homology, with the sign conventions used in Loday's book, the negative Hochschild complex $C^-_*$is included in the positive cyclic complex $CC_{**}^+$ (the second column is still contractable in the split case, but the contracting homotopy $s$ given in the book needs a slight augmentation to work).

I would think that the short exact sequence in the positive case should be $$ 0 \to C^-\to \operatorname{Tot}CC^+\to (\operatorname{Tot}CC^+)[2]\to 0 $$ but according to the book, the correct exact sequence seems to be $$ 0 \to C^-\to \operatorname{Tot}CC^+\to (\operatorname{Tot}CC^-)[2]\to 0 $$ I can't figure out why that is.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.