Let $A$ be a smooth commutative $k$-algebra, for $k$ a commutative ring. By the Hochschild-Kostant-Rosenberg theorem, we have that $HH^*_k(A)\cong \Lambda^* \mathrm{Der}_k(A,A)$, where $\mathrm{Der}_k(A,A)$ is the $A$-module of $k$-linear $A$ derivations. Recall that $HH^*_k(A)$ is an graded-commutative algebra under cup product (it is actually a Gerstenhaber algebra, but we only need the cup product for now).

If $k$ is a field with characteristic not 2, it is easy to see that the relation $[f]\smallsmile[f]=0$ for $[f]\in HH^1_k(A)$ is satisfied, as $[f]\smallsmile[f]$ is 2-torsion. On the other hand, some more work seems to be required in the characteristic 2: even given $A=\mathbb{F}_2[x,y]$, it is not immediately obvious to me (and perhaps I just haven't played around with this enough) how to decompose $[\frac{\partial}{\partial x}]\smallsmile[\frac{\partial}{\partial x}]$ as a sum of Hochschild 2-boundaries, ie functions of the form $f(a,b)=a\cdot g(b)- g(ab)+g(a)\cdot b$. Does anyone know such a decomposition?


If you want an explicit coboundary, take the boundary of the function $g$ sending $[x^a y^b]$ to $x^{a-2} y^b$ times the image of $a(a-1)/2$ in $\mathbb{F}_2$. It works because the cup square of $\partial / \partial x$ sends $[x^i y^j | x^k y^l]$ to $ikx^{i+k-2}y^{j+l}$, and $dg ( [x^i y^j | x^k y^l]) = x^i y^j g(x^k y^l) - g(x^{i+k}y^{j+l}) + g(x^i y^j) x^k y^l$, and the coefficient of $x^{i+k-2} y^{j+l}$ is the image in $\mathbb{F}_2$ of $i(i-1)/2 + k(k-1)/2 + (i+k)(i+k-1)/2 = ik$.

The structure of this Hochschild cohomology ring can be seen in other ways. One is to write down the minimal bimodule resolution explicitly -- it has length two -- and compute products using this. It should turn out that $\operatorname{HH}^*(\mathbb{F}_2[x,y]) \cong \mathbb{F}_2[x,y] \otimes \mathbb{F}_2[X,Y]/(X^2,Y^2)$.

EDIT: There is a standard way to find such coboundaries. Suppose $f,g$ are cocycles and $\circ$ is the `composition' used to define the Gerstenhaber bracket -- see Gerstenhaber, "The cohomology structure of an associative ring" p.281, or any notes on deformation theory. Then $$ f \cup g - (-)^{|f||g|} g \cup f = d(f \circ g) $$ (sign conventions will differ...)

  • $\begingroup$ Excellent. Thank you for your answer. $\endgroup$ – KReiser Nov 2 '12 at 23:19

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