Noncommutative Hodge-to-de Rham comparison conjecture

About 16 years old · traced to

Let kk be a ring containing Q{\mathbb Q}, and let A \circle*1.5A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}} be a smooth and proper DG algebra over kk. Write K \circle*1.5st(A \circle*1.5)K^{st}_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}}) for its semitopological KK-theory, with the Z[β]{\mathbb Z}[\beta]-module structure described in the paper, and write HP \circle*1.5(A \circle*1.5)HP_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}}) for its periodic cyclic homology. Let uu denote the periodicity map. Noncommutative Hodge-to-de Rham comparison conjecture. There exists a functorial map

c:K \circle*1.5st(A \circle*1.5)→HP \circle*1.5(A \circle*1.5)c:K^{st}_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}})\to HP_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}})

such that c(β(α))=u(c(α))c(\beta(\alpha))=u(c(\alpha)) for every α∈K \circle*1.5st(A \circle*1.5)\alpha\in K^{st}_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}}), and such that the induced map

K \circle*1.5st(A \circle*1.5)⊗Z[β]k[β,β−1]→HP \circle*1.5(A \circle*1.5)K^{st}_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}})\otimes_{{\mathbb Z}[\beta]}k[\beta,\beta^{-1}]\to HP_{\:\raisebox{1pt}{\text{\circle*{1.5}}}}(A^{\:\raisebox{3pt}{\text{\circle*{1.5}}}})

is an isomorphism. The conjecture supplies a comparison between semitopological KK-theory and periodic cyclic homology, and is relevant because tensoring semitopological KK-theory with kk gives structures analogous to those on the de Rham cohomology of an algebraic variety. Its resolution status is not specified in the supplied text.

References

Primary source

D. Kaledin, “Motivic structures in non-commutative geometry”, arXiv:1003.3210 (2010).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.