Is there a "categorical" description of Grothendieck's algebra of differential operators? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:22:20Z http://mathoverflow.net/feeds/question/18203 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18203/is-there-a-categorical-description-of-grothendiecks-algebra-of-differential-op Is there a "categorical" description of Grothendieck's algebra of differential operators? Theo Johnson-Freyd 2010-03-14T19:53:12Z 2010-04-01T11:33:30Z <p>First, pick a commutative ring $k$ as the "ground field". Everything I say will be $k$-linear, e.g. "algebra" means "unital associative algebra over $k$". Then recall the following construction due to Grothendieck:</p> <p>Let $A$ be a commutative algebra, and $X$ an $A$-module. Then the <strong>differential operators on $X$</strong> is the filtered algebra $D = D(A,X)$ given inductively by: $$D_{\leq 0} = \text{image of A in }\hom_k(X,X)$$ $$D_{\leq n} = \{ \phi \in \hom_k(X,X) \text{ s.t. } [\phi,a] \in D_{\leq n-1}\, \forall a\in A\}$$ $$D = \bigcup_{n=0}^\infty D_{\leq n}$$ (<strong>Edit:</strong> In the comments, Michael suggests that $D_{\leq 0} = \hom_A(X,X)$ is the more standard definition, and the rest is the same.)</p> <p>Then the following facts are more or less standard:</p> <ul> <li>If $A$ acts freely on $X$, then $D_{\leq 1}$ acts on $A$ by derivations. (It's always true that $D_{\leq 1}$ acts on $D_{\leq 0}$ by derivations; the question is whether $D_{\leq 0} = A$ or a quotient.) If $X = A$ by multiplication, then $D_{\leq 1}$ splits as a direct sum $D_{\leq 1} = \text{Der}(A) \oplus A$.</li> <li>If $k=\mathbb R$, $M$ is a finite-dimensional smooth manifold, and $A = C^\infty(M) = X$, then $D$ is the usual algebra of differential operators generated by $A$ and $\text{Vect}(M) = \Gamma(TM \to M)$.</li> <li>If $A,X$ are actually sheaves, so is $D$.</li> </ul> <p>Thus, at least in the situation where $A = X = C^{\infty}(M)$, the algebra $D$ is acting very much like the universal enveloping algebra of $U (\text{Vect}(M))$; in particular, the map $U(\text{Vect}(M)) \to D$ is filtered and is (almost) a surjection: it misses only the non-constant elements of $A$. So when $A = X = C^\infty(-)$ are sheaves on $M$, it's very tempting to think of $D$ as a sheafy version of $U(\text{Vect}(-))$. Note that $U(\text{Vect}(-))$ is not a sheaf: its degree $\leq 0$ part consists of constant functions, not locally constant, for example, and there are non-zero elements in $U_{\leq 2}$ that restrict to $0$ on an open cover. I think that it cannot be true that the sheafification of $U(\text{Vect}(-))$ is $D$, as the sheafification of $U_{\leq 0}$ is the locally-constant sheaf, not $C^\infty$.</p> <p>So: is there a description of $D$ that makes it more obviously like a universal enveloping algebra? E.g. is there some adjunction or other categorical description? Is it really true that $D$ is a "sheafy" version of $U$ in a precise sense, or is this just a chimera?</p> http://mathoverflow.net/questions/18203/is-there-a-categorical-description-of-grothendiecks-algebra-of-differential-op/18204#18204 Answer by Ben Webster for Is there a "categorical" description of Grothendieck's algebra of differential operators? Ben Webster 2010-03-14T19:59:26Z 2010-03-15T00:12:51Z <p>The construction you're looking for is the universal enveloping algebra of a <a href="http://en.wikipedia.org/wiki/Lie_algebroid" rel="nofollow">Lie</a> <a href="http://ncatlab.org/nlab/show/Lie+algebroid" rel="nofollow">algebroid</a>.</p> <p>A Lie algebra is a Lie algebroid on a point. It's universal enveloping algebra as an algebroid is the same as the usual.</p> <p>Every smooth manifold has a Lie algebroid structure on its tangent bundle, which differential operators are the universal enveloping algebra of.</p> <p>Another cool examples is that TDOs (sheaves of twisted differential operators) can be constructed from central extensions of the tangent Lie algebroid by the structure sheaf (you take UEA, and then identify the two copies of the structure sheaf).</p> <p>(I'll just note: Mariano's answer and mine are essentially equivalent. See the nLab page linked above for brief explanation).</p> http://mathoverflow.net/questions/18203/is-there-a-categorical-description-of-grothendiecks-algebra-of-differential-op/18205#18205 Answer by Mariano Suárez-Alvarez for Is there a "categorical" description of Grothendieck's algebra of differential operators? Mariano Suárez-Alvarez 2010-03-14T20:00:07Z 2010-03-15T03:01:12Z <p>In the affine, characteristic zero, smooth case, you can obtain $\mathcal D$ as the envelopping algebra of the Lie-Rinehart pair $(\mathcal O_X, \mathrm{Der}(X))$ (I think this is precise in for the global sections only, but I do not recall having seen the notion of a sheaffy Lie-Rinehart pair...). </p> <p>In the general case, things are surely harder, since $\mathcal D$ is not generated in general from derivations and functions only: you need higher order generators. (In fact, a very famous conjecture of Nakai claims that an affine variety is smooth iff $\mathcal D$ is generated by functions and derivations.) In positive characteristic, things are worse, as you already need more generators on $\mathbb A^1$.</p> <p>PS: I cannot find Nakai's <em>Higher order derivations, I</em> in MathSciNet nor in Zentralblatt... I don't think lots of progress has happened with respect to that conjecture, sadly (Villamayor-Mount, Becker, &amp;c have special cases)</p> http://mathoverflow.net/questions/18203/is-there-a-categorical-description-of-grothendiecks-algebra-of-differential-op/20061#20061 Answer by Bugs Bunny for Is there a "categorical" description of Grothendieck's algebra of differential operators? Bugs Bunny 2010-04-01T11:33:30Z 2010-04-01T11:33:30Z <p>Linear maps from X to X form A-A-bimodule or A\otimes A-modules. Let I be the multiplication kernel. A linear map is a differential operator if it is annihilated by a power of I. It is essentially Grothendieck's definition, retold to sound functorially.</p>