How exactly is Hochschild homology a monad homology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:19:26Z http://mathoverflow.net/feeds/question/3078 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology How exactly is Hochschild homology a monad homology? Peter Arndt 2009-10-28T16:07:51Z 2010-02-03T22:54:37Z <p>Many texts which praise the generality of the bar construction associated to a monad, say that Hochschild homology is an example of this. </p> <p>What exactly is in this case the underlying endofunctor of the monad, on which category is it an endofunctor, what are the monad structure maps and, most important (since I think my confusion lies here), why do then the face maps look as on the <a href="http://en.wikipedia.org/wiki/Hochschild_homology" rel="nofollow">wikipedia page</a>?</p> http://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology/3094#3094 Answer by Yemon Choi for How exactly is Hochschild homology a monad homology? Yemon Choi 2009-10-28T17:14:36Z 2009-12-16T18:41:05Z <p><a href="http://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology/3085#3085" rel="nofollow">Tyler, are you sure about this?</a> I thought the bar construction comes from the adjunction between R-modules and k-modules for R a given k-algebra (i.e. relative Tor). Besides, what you say only makes sense if we're taking coefficients in R itself and not a general bimodule M.</p> <p>If memory serves correctly, starting with a k-algebra R and looking at a simplicial resolution for it via the adjunction k-modules -- k-algebras leads to <em>cyclic</em> homology as in the paper of Feigin-Tsygan.</p> <p>That wiki page also looks off to me: the Loday construction is for the Hochschild homology and decomposition for <em>commutative</em> algebras, and this isn't made very clear in the wiki.</p> <p><b>quick edit:</b> There's a good if terse discussion of the bar construction via monads in Weibel Chapter 8 (p.283). I suspect you could also extract the desired information out of the much more general machinery in <a href="http://www.tac.mta.ca/tac/reprints/articles/2/tr2abs.html" rel="nofollow">Jon Beck's thesis</a>, modulo some possible struggle with notation.</p> <p><b>encore une fois:</b> Consider the adjunction between k-mod and R-mod. If M is an object of R-mod then the simplicial construction provided by the adjunction looks like this</p> <p>M &lt;--- R\otimes M &lt;--- R \otimes R \otimes M &lt;----- etc</p> <p>where I've not been able to draw in all the face maps, but hopefully you get what I mean. Now by taking the alternating sum of face maps in each degree, we get a split exact sequence of R-module maps</p> <p>M &lt;--- R\otimes M &lt;--- R \otimes R \otimes M &lt;----- etc</p> <p>which is a resolution in the classical sense of M in R-mod-R by R-mod projectives - I'm assuming k is a field for sake of convenience. (So you can use it to calculate Tor^R if you wish.)</p> <p>Now take M=R and note that we have a resolution of R by R^e-projectives. Apply Hom<sub>{R^e}</sub>(__, X) where X is your coefficient module, and you get <em>precisely</em> the Hochschild chain complex as in the original papers.</p> <p>Of course, we didn't have to take sums of face maps before applying the Hom functor. So, if we start with R regarded as an object of R-mod, the canonical simplicial construction (for M=R) would give us a contractible simplicial object in R-mod with M=R at the bottom, this object would in fact live in R^e-mod, and so is eligible to be hit with Hom<sub>R^e</sub>(__,X). If we do this, we get a simplicial object in k-mod, and said object <em>should</em> be the one described in the wiki article, corresponding to the Hochschild chain complex.</p> http://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology/3166#3166 Answer by Yemon Choi for How exactly is Hochschild homology a monad homology? Yemon Choi 2009-10-28T23:12:26Z 2010-02-03T22:54:37Z <p>This is partly in response to Reid, but also intended as general clarification.</p> <p>As I understand it, Peter's original question was:</p> <p>-- here is the Hochschild chain complex for an algebra $A$ and bimodule $M$, as defined in Hochschild's original papers; -- it is the chain complex associated to a certain simplicial object as defined on the Wikipedia page; -- one is told that this object comes from the bar construction (or standard resolution) associated to some monad; -- where/what is the monad?</p> <p>The last one seems to be Reid's underlying point/question. Tyler says you can get it, up to a dimension-shift, from the adjunction between k-modules and k-algebras (at least when $A=M$). My earlier recollection was that this more naturally leads to cyclic homology a.k.a. additive K-theory as defined by Feigin and Tsygan, but I have yet to check this against a copy of their paper. (The point is that in characteristic zero, the cyclic homology of a free tensor algebra on a given k-module, coincides with the cyclic homology of the ground field, so one can take free resolutions of a given $k$-algebra and then use spectral sequence arguments.) On reflecting a bit more, because the Hochschild homology of a free (=tensor) algebra is confined to degrees 0 and 1, perhaps one can also obtain $H_n(A,M)$ as Tyler suggests, by taking the free algebra resolution of A (in the category of k-algebras) and then hitting the resulting simplicial object with a suitable functor - but this seems trickier than in the commutative case (Andre-Quillen) and I can't get hold of a copy of Quillen's paper at the moment.</p> <p><em>Alors.</em> As I understand it, following Weibel's book (and the papers of Barr &amp; Beck et al), the simplicial object (in the category of $k$-modules) that yields the Hochschild chain complex, arises by applying a certain Hom-functor (namely ${}_A{\rm Hom}_A(\ \cdot\ ,X)$ ) to another simplicial object, say $\beta(A)$, in the category of $A$-bimodules.</p> <p>Now $\beta(A)$ is not contractible in the category of $A$-bimodules, in general, and doesn't come from a (co)monad on that category. However, $\beta(A)$ <em>can be identified with</em> another simplicial object $F(A)$, which lives in the category of $A$-modules.</p> <p>What is $F(A)$?</p> <p>Well, take a step back and consider the adjunction between $k$-modules and $A$-modules (maybe you need $k$ to be a field at this point, maybe not). That gives rise to a bar construction in $A$-mod, namely for any given $M$ in $A$-mod one obtains a simplicial object $F(M)$ which is given in each degree by</p> <p>$$F_{-1}(M)=M\quad,\quad F_n(M) = M \otimes A^{\otimes n+1} \ {\rm for }\ n \geq 0.$$</p> <p>Note that this is contractible in $A$-mod by the general machinery of the bar resolution associated to a monad. There was nothing to stop us taking $M=A$, that's a perfectly good $A$-module; and on doing so, lo and behold, we get the same simplicial object $F(A)$.</p> <p>Thus, Hochschild homology, regardless of the choice of coefficients, can be thought of as "coming from" a comonad - namely, that induced on $A$-mod by the forgetful functor from $A$-mod to $k$-mod. In my opinion, <em>that</em> is probably the (co)monad they are talking about.</p> <p>It so happens that, since $F(A)$ is contractible in $A$-mod and hence <em>a fotiori</em> in $k$-mod, the "chain-complex-ification" of $\beta(A)$ is, as a chain complex in $R$-bimod, a resolution of $R$ by $k$-relatively projective $R$-bimodules -- and hence applying ${}_R{\rm Hom}_R(\ \cdot \ ,X)$ to it and taking homology coincides with taking $k$-relative Tor of $R$ and $X$ as R-bimodules. Hence the point of view that Hochschild homology is a special case of relative Tor.</p> <p>Finally, I actually agree with Reid that this is not the best example to motivate (co)monad (co)homology. Group cohomology with coefficients in the ground field; or indeed Andr&eacute;-Quillen cohomology, which <em>is</em> given by a "free algebra" adjunction but only for <em>commutative</em> algebras, or sheaf cohomology, would be better. (No originality in my choices; I've cribbed them out of Weibel Section 8.6).</p> <p>(Apologies for the length and the tediousness, by the way.)</p> http://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology/9079#9079 Answer by Mariano Suárez-Alvarez for How exactly is Hochschild homology a monad homology? Mariano Suárez-Alvarez 2009-12-16T05:44:20Z 2009-12-16T05:44:20Z <p>Assuming that the base ring $k$ is a field (or that the algebra is projective over $k$), a functor whose monad computes Hochschild homology is the one which maps $k$-modules $M$ to the $A$-bimodule $A\otimes M\otimes A$, which is adjoint to the forgetful functor in the other direction.</p> <p>The bar complex corresponding to this adjunction is not exactly the one used by Hochschild to define his cohomology, but it is easily seen to give naturally isomorphic results, since it more or less evidently constructs gives projective resolutions of bimodules (not only of $A$, as is the case with the monad coming from one-sided extension of scalars)</p>