Timeline for Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?
Current License: CC BY-SA 2.5
6 events
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Feb 23, 2010 at 3:56 | comment | added | David Ben-Zvi | This is discussed in some detail in my new paper with David Nadler, which appeared as arxiv.org/abs/1002.3636 | |
Oct 24, 2009 at 2:39 | comment | added | Tyler Lawson | This is definitely char. 0 - and let me doubly emphasize Quillen's paper, it's a great place to learn "derived" thinking in general. | |
Oct 23, 2009 at 20:18 | comment | added | David Ben-Zvi | Ilya -- yes, there's a very cute (IMHO) very short proof of this there. Basically Hochschild homology is (by definition - once you understand the definition) functions on the derived loop space of Spec R. But maps into an affine variety always factor through the affinization of your space (Spec of global functions). The affinization of the circle (in char zero) is Spec k[e]/e^2=0, where degree(e)=1. So maps from the affinization are just the tangent bundle of Spec R, with a shift (tangent complex, if R is singular). Hence functions on it are the symmetric algebra of one-forms with a shift. | |
Oct 23, 2009 at 20:14 | comment | added | David Ben-Zvi | (let's assume char=0 before I get into trouble) | |
Oct 23, 2009 at 20:11 | comment | added | Ilya Nikokoshev | Could this be explained in detail in your new paper, please? | |
Oct 23, 2009 at 20:08 | history | answered | David Ben-Zvi | CC BY-SA 2.5 |