5
$\begingroup$

Let $k$ be a commutative ring, $A$ a $k$-algebra and $HH^n(A, M)$ the $n$-th Hochschild cohomology of $A$ with coefficients in the $A$-bimodule $M$. In the book Cyclic Homology by Loday the following curious few lines appear on page 40 when talking about the form that the coefficients can take:

For $M=A$ the groups $HH^n(A, A)$ have been extensively studied in the literature because they are related to deformation theory. But one should note that they are not functors of $A$. However if $M = A^* = \mbox{Hom}_k(A, k)$, then the groups $HH^n(A, A^*)$ are indeed functors of $A$.

Now I'm quite confused by this assertion that $HH^n(A, A)$ isn't a functor of $A$, because I know that you can interpret the $n$-th cohomology as the derived functor

$$HH^n(A, M) = \mbox{Ext}_{A^e}^n(A, M)$$

where $A^e = A\otimes_k A^{op}$ is the enveloping algebra of $A$. In fact, it's mentioned on the very next page of the book. So clearly each $HH^n(A, M)$ is a functor if I've understood this correctly. So presumably the important part is that $HH^n(A, A)$ is not a functor "of $A$", but I'm not entirely sure what this means. I'd be grateful if someone could help me understand how it is that $HH^n(A, A)$ isn't a functor of $A$.

Edit:

Thanks to the comments beneath the question (thanks for your help guys, I appreciate it) I think I have a better idea of what's going on. So using the example of $\mbox{Hom}_k(A, A)$, the reason why this isn't a functor of $A$ is because given another $k$-algebra $B$, and a map $f:A\to B$, there is no real way of assigning $f$ to a map

$$\mbox{Hom}_k(A, A)\to\mbox{Hom}_k(B, B)$$

However, now that I understand this, it seems to me $\mbox{Hom}_k(A, A^*)$ suffers from the same issue. So how is it that (as mentioned in the paragraph I cite above) $HH^n(A, A^*)$ is a functor of $A$? Am I just missing some obvious way of assigning $f:A\to B$ to $\mbox{Hom}_k(A, A^*)\to\mbox{Hom}_k(B, B^*)$?

$\endgroup$
14
  • 4
    $\begingroup$ $A$ appears in the formula $\textrm{Ext}^n_{A^e} (A, A)$ several times, but not all with the same variance. So it isn't obviously a functor of $A$. (Recall that a functor acts on objects and morphisms. Can you define the action on morphisms here?) $\endgroup$
    – Zhen Lin
    Commented Aug 4, 2020 at 11:16
  • 4
    $\begingroup$ Well, for example, $\textrm{Hom} (X, Y)$ is contravariant in $X$ and covariant in $Y$. Thus $X$ appears in the expression $\textrm{Hom} (X, X)$ with mixed variance – and this is essentially the reason why $\textrm{End} (X)$ is not a functor of $X$. Hochschild cohomology is just a souped up version of this. $\endgroup$
    – Zhen Lin
    Commented Aug 4, 2020 at 12:05
  • 5
    $\begingroup$ @SeraPhim Maybe it's easier to see in the case of $\mathrm{Hom}(X,X)$ in some category, which is basically the same issue. If you want to make this a functor, then given $f:X\to Y$ you have to define a map in one direction or the other between $\mathrm{Hom}(X,X)$ and $\mathrm{Hom}(Y,Y)$, and in general there is no reasonable way to do this. $\endgroup$ Commented Aug 4, 2020 at 16:50
  • 4
    $\begingroup$ @SeraPhim: Given a map $f \colon X \rightarrow Y$, you have an induced pullback map $f^{*} \colon \operatorname{Hom}_k(Y,Y^{*}) \rightarrow \operatorname{Hom}_k(X,X^{*})$ given by $((f^{*}(\phi))(x))(x') = (\phi(f(x)))(f(x'))$. Similarly, you can define a map $f^{*} \colon C^n(B,B^{*}) \rightarrow C^n(A,A^{*})$ which commutes with the differential and induces a map on the Hochschild cohomology. $\endgroup$
    – levap
    Commented Aug 6, 2020 at 11:36
  • 2
    $\begingroup$ The point is that dualisation $(-)^*$ is a contravariant functor. Thus $\textrm{Hom} (X, X^*)$ is purely contravariant in $X$. More to the point, functors can be composed, and we have functors $X \mapsto (X, X^*)$ and $(X, Y) \mapsto \textrm{Hom} (X, Y)$, so we have the composite functor $X \mapsto \textrm{Hom} (X, X^*)$. By contrast $X \mapsto (X, X)$ and $(X, Y) \mapsto \textrm{Hom} (X, Y)$ cannot be composed because of incompatible variance. $\endgroup$
    – Zhen Lin
    Commented Aug 6, 2020 at 11:59

1 Answer 1

2
$\begingroup$

$A$ appears in the formula $\textrm{Ext}^n_{A^e} (A, A)$ several times, but not all with the same variance. So it isn't obviously a functor of $A$.

[...] $\textrm{Hom} (X, Y)$ is contravariant in $X$ and covariant in $Y$. Thus $X$ appears in the expression $\textrm{Hom} (X, X)$ with mixed variance – and this is essentially the reason why $\textrm{End} (X)$ is not a functor of $X$. Hochschild cohomology is just a souped up version of this.

Maybe it's easier to see in the case of $\mathrm{Hom}(X,X)$ in some category, which is basically the same issue. If you want to make this a functor, then given $f:X\to Y$ you have to define a map in one direction or the other between $\mathrm{Hom}(X,X)$ and $\mathrm{Hom}(Y,Y)$, and in general there is no reasonable way to do this.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .