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In their paper "Integral Transforms and Drinfeld Centers in Derived Algebraic Geometry" the authors show that for perfect stacks $X$ and $Y$ over $k$, and their $k$-linear $\infty$-categories of unbounded chain complexes of modules $QC(X)$ and $QC(Y)$ there are equivalences of categories $$ QC(X \times_k Y) \simeq QC(X)\otimes_k QC(Y)$$ and $$QC(X \times_k Y) \simeq Fun^L_k(QC(X),QC(Y)).$$

My question is the following: How can this be done in the absence of the underlying geometric spaces ($X$ and $Y$)? Given two perfect $k$-linear $\infty$-categories $\mathcal{A}$ and $\mathcal{B}$ is there a natural integral transform $$\mathcal{A}\otimes_k \mathcal{B} \to Fun_k^L(\mathcal{A},\mathcal{B})?$$ Under what conditions on $\mathcal{A}$ and $\mathcal{B}$ is this map an equivalence?

Any reference will be very helpful.

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    $\begingroup$ In the case where $B$ is a point, this requires a natural map from $A$ to functors from $A$ to a point. Basically a map $A\times A\to D^b(Vect)$. Given one of these, you should be able to produce the described map by tensoring this map with $B$. So you want a self-duality statement. $\endgroup$
    – Will Sawin
    Commented Apr 5, 2015 at 22:19
  • $\begingroup$ In the commutative case note that $\text{QC}(X)$ is in a suitable sense self-dual, and there's no reason for this to be true in the noncommutative case; more simply, already something must be wrong with your formula because it's covariant in $A$ on the LHS but contravariant in $A$ on the RHS. Even in the simplest setting where we have an equivalence like that, namely finite-dimensional vector spaces, we need to take $A^{op} \otimes B$ on the LHS. $\endgroup$ Commented Apr 6, 2015 at 18:04
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    $\begingroup$ Here's a less ambitious way to think about this. All these fancy QC(X) categories are secretly A-Mod for some dg-ring A (A = End(G), for a generator G. all separated schemes admit one by Bondal-vdB if I'm not mistaken). So, if you are willing to restrict to dg-categories equivalent to A-Mod for some dg-ring A, then Functors are given by bi-modules (and you need to stick an op superscript as QY points out). $\endgroup$ Commented Apr 6, 2015 at 18:39
  • $\begingroup$ @QiaochuYuan, ofcourse, thanks for pointing out. $\endgroup$ Commented Apr 6, 2015 at 18:39
  • $\begingroup$ @bananastack, saturated dg-categories also come from dg-algebras, by the way (from dg-algebras of finite type, even). $\endgroup$
    – AAK
    Commented Apr 6, 2015 at 20:33

3 Answers 3

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Let $DGCat_k$ denote the $\infty$-category obtained by localizing the category of dg-categories at Morita equivalences. This is presented by the Morita model structure on the category of dg-categories. Recall that the fibrant objects are the idempotent complete dg-categories, or equivalently dg-categories for which the canonical functor $T \hookrightarrow D^c(T)$, into the dg-category of compact dg-modules, is an equivalence. Its dualizable objects are precisely the saturated dg-categories (Cisinski-Tabuada, Theorem 5.8). These correspond to schemes over $k$ which are smooth and proper.

Let $A$ and $B$ be small dg-categories and consider To\"en's formula $$Fun_k(A,B) = D_{rqr}(A^{op} \otimes B)$$ for the internal hom in the $\infty$-category of dg-categories, where $D_{rqr}$ is the dg-category of right quasi-representable bimodules. If $A$ and $B$ are saturated, this reduces to an equivalence of dg-categories $$Fun_k(A,B) = D_c(A^{op} \otimes B)$$ where $D_c$ is the dg-category of compact dg-modules. If $A$ and $B$ are idempotent complete, and $A$ has the property of self-duality, then this gives an equivalence of dg-categories $$ Fun_k(A, B) = A \otimes B. $$ It is clear that this recovers the integral transform formula for perfect complexes, in the case of smooth proper $k$-schemes.

This is only the small version of your question. In the presentable case all I can say at the moment is the following. If $A$ and $B$ are locally presentable dg-categories, compactly generated with $A_c$ and $B_c$ the full subcategories of compact objects, then $$ Fun^L_k(A,B) = Fun^L_k(\hat{A_c}, \hat{B_c}) = (A_c^{op} \otimes B_c)^{\wedge} = (A_c \otimes B_c)^{\wedge} $$ where the second equivalences uses Corollary 7.6 in To\"en's "Derived Morita theory". Here $(-)^\wedge$ denotes the free cocompletion, and I am assuming self-duality for $A$ again. This seems to recover the integral transform formula in the case of schemes, using $Perf(X \times_k Y) = Perf(X) \otimes Perf(Y)$.

(It has been a while since I looked at this stuff, so let me know if I'm making a mistake somewhere.)

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I am not an expert of this area and I put it as an answer rahter than a comment just because it's too long.

I think an important work on this topic is "The homotopy theory of dg-categories and derived morita theory" by Bertrand Toen. He uses dg-categories instead of $\infty$-categories, so does the paper "Integral Transforms and Drinfeld Centers in Derived Algebraic Geometry".

One of the main result (Corollary 4.8) in this paper is: If $\mathcal{C}$ and $\mathcal{D}$ are two small dg-categories, then there is a functorial bijection between the set of maps $[\mathcal{C},\mathcal{D}]$ in $Ho(dg-Cat)$, the homotopy category of small dg-categories, and the set of isormorphism classes of right quasi-representable objects in $Ho((\mathcal{C}\otimes \mathcal{D}^{op})-Mod)$, the homotopy category of $\mathcal{C}-\mathcal{D}$ bi-modules.

The (right quasi-representable) $\mathcal{C}-\mathcal{D}$ bi-modules, rather than $\mathcal{C}\otimes \mathcal{D}$, is natural to give functors from $\mathcal{C}$ to $\mathcal{D}$. Hence as Will Sawin comments above, you may need a self-dual statement to get the isomorphism you want.

By the way, the isomorphism $$ QC(X\times Y)\cong Fun^L(QC(X),QC(Y)) $$ is a (not very straightforward) consequence of the above result, which is Theorem 8.9 in Toen's paper, where he checks the self-duality of $QC(X)$.

Maybe you could investigate your categories $\mathcal{A}$ and $\mathcal{B}$ to see whether they fit into Toen's framework.

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    $\begingroup$ To avoid possible competing definitions/confusion, let's assume $X$ is a sufficiently nice scheme over a field $k$. The self-duality you need here is not a very complicated thing---- you take a perfect complex $P$ to it's linear dual $RHom(P,O_X)$ to obtain a functor $Perf(X) \to Perf(X)^{op}$. Passing to co-limit completions gives the required self-duality. Being able to perform this last step is one of the key points of the perfect definition. $\endgroup$ Commented Apr 6, 2015 at 14:17
  • $\begingroup$ Yes I agree. I think that it is a little bit more complicated to express the quasi-representability under the self-duality but I'm not sure about the details. $\endgroup$ Commented Apr 6, 2015 at 14:50
  • $\begingroup$ @ZhaotingWei, Thanks for the Toen reference! $\endgroup$ Commented Apr 6, 2015 at 16:38
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Here is a simple underived but noncommutative statement along these lines. Let $R$ and $S$ be two $k$-algebras, $k$ a commutative ring. The Eilenberg-Watts theorem asserts that there is an equivalence (of $k$-linear categories) between

  • $k$-linear cocontinuous functors $\text{Mod}(R) \to \text{Mod}(S)$, and
  • $k$-bimodules over the pair $(R, S)$.

The equivalence in one direction sends a bimodule $M$ to the functor $(-) \otimes_R M$, and in the other direction sends a functor $F$ to the bimodule $F(R)$. If you like, you can think of $k$-bimodules over $(R, S)$ as right $R^{op} \otimes_k S$-modules, and that gets you a statement which looks at least formally like what you want (with an appropriately inserted $^{op}$), provided that you believe there is a meaningful sense in which $\text{Mod}(R^{op} \otimes_k S)$ is $\text{Mod}(R^{op}) \otimes_k \text{Mod}(S)$.

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