# Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal{D}_X$-module?

Let $R$ be a regular algebra over a field $k$ of char 0. Let $D$ be its corresponding algebra of differential operators.

As in the general setting of non-commutative algebra we can tensor right $D$-modules with left $D$-modules to get $R$-modules. However in this case we have more operations available to us.

Let $M$ and $N$ be left $D$-modules. One can define using the leibniz rule a structure of a $D$-module on the tensor product $M \otimes_R N$. The same can be done if we replace one of the factors with a right $D$-modules and flip some signs and similar statements exist for internal Homs over $R$ (i'm not so sure about right tensor right - although I assume that at least in the derived setting one can always use duality to define this structure).

Now, my question is rather vague. I'm trying to understand conceptually what properties of $D$ makes it possible to give a $D$-module structure on $M \otimes_R N$ and maybe understand in what way is this construction canonical (since so far all i've seen is a formula in this context). So to summarize:

Why do $M \otimes_R N$ and $\mathrm{Hom}_R(M,N)$ have a natural structure of $D$-modules?

Edit: After being confused by some conflicting answers I've posted a more detailed and exaustive question here: What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it?

EDIT: Some time has passed and i'm still not satisfied with the my current understanding of this. The original question still remains a mystery: What kind of algebraic object is $\mathcal{D}$? A suitable answer would give a definition of an algebraic object $D$ over a ring $R$ for which all the following holds

1. The opposite $D^{op}$ is canonically morita equivalent to $D$ (Canonically in the sense that the equivalence should be induced from the algebraic structure on $D$).

2. The ability to form tensor products and hom modules over $R$ between left and right $D$-modules except in two cases:

• Tensor product of a right $D$-module with a right $D$-module.
• Hom module (over $R$) from a right $D$-module to a left $D$-module

3. The forgetful from $D$-modules to $R$-modules is monoidal w.r.t. above tensor product.

• Your claim in the second paragraph that tensoring a right D-module with a left D-module gives an R-module is false (at least if the tensor product is over D). In general, this will just be a vector space over $k$. – Avi Steiner Dec 21 '16 at 17:44
• In case you are interested, there is a conceptual explanation that was not mentioned here so far. $D$-modules are the same as crystals of $O$-modules (or, if you prefer, quasi-coherent sheaves on the de Rham stack), which makes the tensor structure obvious. This is similar to the differential-geometric question/answer: Why does a tensor product of two bundles with a flat connection carry a flat connection? Answer: Because flat connection can be viewed as the data of parallel transport in fibers of a vector bundle (making the bundle into a local system), which makes tensor product clear. – t3suji Dec 23 '16 at 16:44
• @t3suji The case of flat vector bundles is clear. Anything which involves right D modules is a lot trickier. I realize a right D module should be something like a distribution but I don't have a precise statement. Do you know a natural category coming from differential geometry that is equivalent to the category of O-coherent D-modules (together with the forgetful functor obviously). – Saal Hardali Dec 23 '16 at 17:19
• @SaalHardali I am not sure what you are asking here. Are you worried about right vs left D-modules or about D-modules that are quasicoherent vs coherent as O-modules – t3suji Dec 23 '16 at 17:28
• I may just be repeating earlier comments, but the properties of $D$-mod that you are asking about are equivalent to saying that $D$ is a bi-algebroid which is Morita-Hopf. The bi-algebroid structure is precisely what it means for $D$-mod to have a monoidal structure such that the forgetful functor to $\mathcal O$-mod is monoidal. The term Morita-Hopf, I just made up, but it means that it looks like a Hopf algebra, except that that antipode is given by a bimodule rather than a map. – Sam Gunningham May 22 '17 at 22:51

OK, I'll give it a shot. The bi-algebra structure on $D$ is something that I found very confusing too, so I will try to spell it out as best I understand. These ideas were explained to me by Pavel Safronov, and I found these notes by Gabriella Bohm be helpful https://arxiv.org/abs/0805.3806 (though they deal with a more general case than we need here). See also the original papers by Sweedler and Takeuchi from the 70's.

## The $D$-module set-up

Suppose $X$ is a smooth algebraic variety, and $D=D_X$. The situation we have is the following: the category $D-mod$ and the forgetful functor to $\mathcal O-mod$, are equipped with (symmetric) monoidal structures (the duality of $D$-modules will be discussed later).

There are many ways to understand why this should be the case, as some of the other answers indicate. For example, the category $D-mod$ can be understood as quasi-coherent sheaves on the de Rham space $X_{dR}$ (and the ring $D$ expresses the descent data on the pullback to quasi-coherent sheaves on $X$). Alternatively, if one thinks of $D$ as a deformation quantization of $T^\ast X$, then the monoidal structure arises from the fact that the cotangent bundle is a symplectic group(oid) acting (trivially) on $X$. One can think of $T^\ast X$ as being a commutative group object in the category of symplectic varieties and lagrangian correspondences (I find this last persepective helpful in unpacking the notion of bialgebroid).

However, I think what you are after is not why $D$-modules have this structure, but what structure on the ring $D$ endows $D-mod$ with these structures. The answer (as has already been mentioned) is that $D$ is a bialgebroid over $\mathcal O$. Let me try my best to unpack what that means below.

## The categorical structure

(You can ignore this bit if you don't like it).

Consider the following situation: we have monoidal categories $\mathcal C$ and $\mathcal D$ and a monoidal functor $$F:\mathcal D \to \mathcal C$$ Suppose also that the functor $F$ is monadic, so that $\mathcal D$ can be expressed as modules for a monad $T$ acting on $\mathcal C$. The monoidal structures on $\mathcal D$, $\mathcal C$ and $F$ must then be reflected in the monad $T$. Such a structure on a monad (acting on a monoidal category $\mathcal C$) is called a bimonad. Rather than saying what this all this means in general, let's consider a special case.

## Bialgebroids (over a commutative base)

Suppose $R$ is a commutative ring, and let $\mathcal C = R-mod$. Then a (colimit preserving) monad acting on $R-mod$ is nothing more than a $R$-ring, i.e. a ring $B$ with a ring homomorphism $R\to B$ (note that $R$ need not be central in $B$). In the case we are interested in $B=D$ and $R=\mathcal O$.

Before giving an algebraic definition of a bialgebroid, we note that the point of all this is that a (left) bialgebroid structure on $B$ is exactly equivalent to data of a monoidal structure on $B-mod$ and on the forgetful functor to $R-mod$. Note that if $R$ is central in $B$, this is the usual Tannakian theory, and an $R$-bialgebroid is just an $R$-bialgebra.

So what is an $R$-bialgebroid? Well, we already know that $B$ is an $R$-ring, so there is a product: $$B_{\bullet} \otimes_{R} {}_\bullet B \to B$$ where the dots indicate on which side $R$ is acting on $B$. As one might expect, there is also a coproduct, which tells you how $B$ should act on the tensor product $M \otimes_R N$ of two left $B$-modules, but one has to be careful about which monoidal category the coalgebra structure on $B$ lives in. If you unwind the definitions, you see that the coproduct is given by a map $$B \to {}_\bullet B \otimes_R {}_\bullet B$$ Note that, unlike in the product map, $R$ is acting on the left on both factors. This is a little confusing at first, but perhaps not so surprising if you consider that in the category $B-mod$ we want to understand how to tensor two left $B$-modules.

Of course, there are some axioms. The one that I found hardest to digest involves something called the Takeuchi product. Let me try to motivate that a bit.

## Takeuchi Product

In the usual theory of bialgebras, there is an axiom which says that the coproduct is an algebra map. This doesn't make sense for bialgebroids as ${}_\bullet B\otimes_R {}_\bullet B$ is not an algebra under componentwise multiplication. The Takeuchi product is a certain subspace of this object, defined by: $$B {}_R \times B := \left\{ \sum b_i \otimes b_i' \in {}_\bullet B\otimes_R {}_\bullet B \mid \sum b_i r \otimes b_i' = b_i \otimes b_i'r \right\}$$ Note that the the $r$'s in the condition are acting on the right, whereas the relative tensor product is using multiplication on the left. Note also that if $R$ is central in $B$, then the condition is vacuous. One can check that $B {}_R \times B$ is ring under compoentwise multiplication. One of the axioms of a bialgebroid is that the coproduct map factors through the Takeuchi product and is a ring homomorphism. (There is another interesting bialgebroid axiom, which is about the counit map, but for brevity, I won't discuss that).

The Takeuchi product (which in the $R$ commutative case appears to be due to Sweedler?) seemed somewhat mysterious to me until I saw that there is a ring isomorphism: $$B {}_R \times B \simeq End_{B^{op}\otimes B^{op}} (_\bullet B \otimes_R {}_\bullet B )$$ Thus, the comultiplication map is nothing more than the structure of a left $B$, right $B\otimes B$ bimodule on ${}_\bullet B\otimes_R {}_\bullet B$. This fits well with the $D$-modules story: the coproduct on $\mathcal D$ is precisely the transfer bimpdule structure on $$\mathcal D_{X\to X\times X} = {}_\bullet \mathcal D \otimes_{\mathcal O} {}_{\bullet} \mathcal D$$ (as it should be, as the transfer bimodule represents the tensor product functor).

## The D-module structure on Hom

Let me come back to this another time...

• This is splendid so far (though i might need to come back to this several times to absorve everything) do you think that in this framework you could explain why there is no tensor product for right D modules? Or $Hom$ from right to left $D$-modules? – Saal Hardali May 23 '17 at 18:08
• It's not really in the spirit of my answer, but you can see why tensor product over $\mathcal O$ doesn't work for right $D$-modules as a special case of the fact that the pullback functor $\Delta^\dagger$ is `naively'' defined for left $D$-modules, but not for right (it involves the side-changing operations). – Sam Gunningham May 23 '17 at 18:23
• Another way of expressing this is that the forgetful functor from right $D$-modules naturally lands in the category of Ind-coherent sheaves equipped with the shriek tensor product (rather than quasi-coherent sheaves with the usual tensor product). In the case of a smooth variety, this is just quasi-coherent sheaves, but where the tensor product is twisted by $\omega$. Take a look at the papers on crystals by Gaitsgory and Rozenblyum for more on this. – Sam Gunningham May 23 '17 at 18:25
• This doesn't seem to rule out immediatly the possiblity of side changing both sides then side changing back... – Saal Hardali May 23 '17 at 18:25
• I'm not sure I understand. The category of right $D$-modules and left $D$-modules are equivalent (by tensoring with $\omega$), so you can transport the tensor structure from left modules to right modules. When you do that, it becomes the shriek tensor product on the underlying $O$-module of a right $D$-module. What are you trying to rule out? – Sam Gunningham May 23 '17 at 18:43

This is a replacement for an old confused answer. There is a related context in which I know a good answer. Suppose $A$ is a ring and $S$ is a central subring. If $M$ and $N$ are $A$ modules, then $M \otimes_S N$ is an $A \otimes_S A$-module and, if we have a map of $S$-algebras $\Delta: A \to A \otimes_S A$, then this makes $M \otimes_S N$ into an $A$-module again. The data of such a $\Delta$ makes $A$ into a bi-algebra.

Similarly, $\mathrm{Hom}(M,S)$ is an $A^{op}$-module and, if we have a map of $S$-algebras $r: A \to A^{op}$, then $\mathrm{Hom}(M,S)$ becomes an $A$-module again. Such an $r$ and $\Delta$, if they obey the correect compatabilities, make $A$ into a Hopf algebra.

But this is not the right description for $D$-modules. $R$ is not central in $D$. And I am confused about how to fix this. Indeed, $D \otimes_R D$ is not a ring at all! (If you think it is, do $(d/dx) \otimes 1$ and $1 \otimes x$ commute? What about $(d/dx) \otimes 1$ and $x \otimes 1$? But $x \otimes 1 = 1 \otimes x$!)

The best description I can give of the action of $D$ on $M \otimes_R N$ is to take the unique map of rings $D \to D \otimes_k D$ sending $X \mapsto X \otimes 1 + 1 \otimes X$ for $X$ a vector field. $D \otimes_k D$ acts on $M \otimes_k N$ and, for some unclear reason, the image of $D$ under this map passes to the quotient $M \otimes_R N$. Similarly, the action on $\mathrm{Hom}(M,N)$ uses the map $X \mapsto -X \otimes 1 + 1 \otimes X$ to $D \otimes_k D^{op}$.

I don't understand why this works. Other people have suggested that the term "bi-algebroid" is the right context to understand this, but I have to admit I don't understand the sources on bi-algebroids.

Finally, on a smooth projective variety, there need not be any map $D \to D^{op}$. For example, consider $\mathbb{P}^1$ with open chart $\mathrm{Spec}\ k[x]$ and suppose $\phi: D \to D^{op}$ is a map of rings and of $\mathcal{O}_{\mathbb{P}^1}$-modules. Then $\phi(d/dx) x - x \phi(d/dx) = -1$, so $\phi(d/dx)$ is of the form $-d/dx + h(x)$ for some $h(x) \in k[x]$. But $x^2 (d/dx)$ is defined on all of $\mathbb{P}^1$, so $\phi(x^2 d/dx) = - x^2 d/dx + 2x + x^2 h(x)$ must extend to all of $\mathbb{P}^1$, and $2x+x^2 h$ can not be a global regular function on $\mathbb{P}^1$.

• Small correction: $D$ is not an $R$-algebra (an algebra object in $R$-modules), but an $R$-algebroid (an algebra object in $R$-bimodules). Correspondingly, $D$ is a bialgebroid. However, it is not a Hopf algebroid (manifested e.g. in the fact that right modules over $D$ are left modules over $D$ twisted by the canonical bundle). See e.g. Example 3.1 in arxiv.org/abs/math/9905192. – Pavel Safronov Dec 21 '16 at 13:07
• @PavelSafronov Thanks! There isn't an Example 3.1 in that paper, only a Definition 3.1. Is that what you mean? – David E Speyer Dec 21 '16 at 13:51
• @PavelSafronov I'm sorry if this is too elementary (I know almost nothing about Hopf algebras). I'm trying understand duality for $D$-modules and where it comes from. The formulas relating $D$ with its twisted opposite look like magic. Another confusing thing is that It's unclear where the "derived" line passes. For example if I want to take an $\mathcal{O}_X$ dual of a $\mathcal{D}_X$ module should i resolve it first? How then can I hope to construct the category of $D$ modules in canonical way? This is shouting $\infty$-cats at me which i'd really like to avoid. – Saal Hardali Dec 21 '16 at 14:31
• There are lots of questions here. What exactly do you mean by the $\mathcal{O}_X$-dual of a left $\mathcal{D}_X$-module $\mathcal{M}$? Do you mean the left $\mathcal{D}_X$-module $\mathrm{Hom}_{\mathcal{O}_X}(\mathcal{M},\mathcal{O}_X)$? What do you mean by 'construct the category of $D$-modules in a canonical way? What is non-canonical about the usual construction? I think you must mean more by 'construct the category' than I understand. – Simon Wadsley Dec 21 '16 at 14:51
• Go to under what operation? You now seem to be asking for much more than your original question. – Simon Wadsley Dec 21 '16 at 15:56

One way to think about this is that $D$ is the universal enveloping algebra $U(R,L)$ of the $(k,R)$ Lie-Rinehart algebra $\mathrm{Der}_k(R,R)$. Whenever one has such an enveloping algebra one may perform these kind of constructions. More details can be found in https://arxiv.org/abs/dg-ga/9702008. See section 2 in particular.

The well-known fact for $D$-modules that tensoring with the 'canonical sheaf', that is the top exterior power of $\Omega^1_{R/k}$, defines an equivalence between left $D$-modules and right $D$-modules also holds in this more general setting provided that $L$ is a projective $R$-module of finite rank (then the $R$-linear dual of the top exterior power of $L$ plays the role of the canonical sheaf).

• This is a little progress but it is still desirable to have an independent characterization of what kind of algebras (perhaps with extra structure) over a ring come from lie-reinhart pairs. Just as we know that a hopf algebra in char 0 is the enveloping algebra of some finite dimensional lie algebra iff it is a finitely primitively generated. – Saal Hardali May 22 '17 at 17:23
• @SaalHardali Your question is answered (with an unsurprisingly analogous answer) in Moerdijk, I.; Mrčun, J. On the universal enveloping algebra of a Lie algebroid. Proc. Amer. Math. Soc. 138 (2010), no. 9, 3135–3145. – David Ben-Zvi Jan 29 '18 at 16:03

There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x +1$ on $f$ by your formula and see that it does not act by zero. For this fellow to act by zero, one of $M,N$ needs to be a left $D$-module and the other one is a right $D$-module.

The stuff above is plain wrong. There is an action. See below.

To answer your question (with my left-right correction for the second part) you need to follow David Speyer's answer, forgetting all about the evil antipode (it just does not exist in any useful for you form). $D$ is an $R$-$R$-bimodule. Now $D\otimes_k D$ has 4 stuctures of $R$-module. I call $R$ and $R^\prime$ to distinguish them. The comultiplication is now a map $$\Delta : D \rightarrow \,_RD_{R^\prime} \otimes_R \,_RD_{R^\prime}$$ taking image in the equalizer of the two $R^\prime$-module structures. This is all you need this Christmas to define your actions.

• Thx! This at least shows that there is a some confusing issue here. Why is there an action for left and right modules? – Saal Hardali Dec 21 '16 at 15:30
• Did I answer in the edit? If not, just write the obvious Leibnitz Rule on $f$ using $\Delta$ but no antipodes... – Bugs Bunny Dec 21 '16 at 15:38
• Doesn't the claim in the first paragraph depend on the formula. The reference arxiv.org/abs/dg-ga/9702008 claims that $(\frac{d}{dx}(f))(m)=\frac{d}{dx}(f(m)) - f(\frac{d}{dx}m)$ works. Proposition 1.2.9 of this book math.columbia.edu/~scautis/dmodules/hottaetal.pdf agrees – Simon Wadsley Dec 21 '16 at 15:49
• I apologise for being scary; it is not intentional. Terms 3 and 4 should be $-((dx)(f))(m)=(-d(xf))(m)=-d((xf)(m))+(xf)(dm)$ and this is $-f(m)-xd(f(m))+(xf)(dm)$. The second of these cancels your Term 1 the third your Term 2 and the first would cancel your Term 1 if you had the correct relation in the Weyl algebra namely $[d,x]=1$ rather than $[x,d]=1$. – Simon Wadsley Dec 21 '16 at 16:28
• Sorry. Final sentence should read: The second of these cancels your Term 1 the third your Term 2 and the first would cancel your Term 5 if you had the correct relation in the Weyl algebra namely [d,x]=1[d,x]=1 rather than [x,d]=1[x,d]=1. – Simon Wadsley Dec 21 '16 at 17:18