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Derived intersection of Lagrangians is shifted symplectic

Statement

Let $k$ be a characteristic-zero field, assume AC for the cotangent constructions inherited from Derived schemes and the cotangent complex of a morphism, and let $F, X, Y$ be derived Artin stacks in a common finite geometricity range, each with perfect cotangent complex over $k$ . Their tangent complexes are the duals of these perfect cotangent complexes; the relative cotangent complexes below are perfect as cofibers of maps of perfect complexes. Suppose $F$ has a closed $n$ -shifted symplectic form $ω$ , and maps $f : X \to F$ and $g : Y \to F$ carry Lagrangian structures. Here a Lagrangian structure on $f$ is a path $h : 0 ⇝ f^{*} ω$ in $A^{2, c l} (X, n)$ whose induced map $T_{X} \to L_{X / F} [n - 1]$ is a quasi-isomorphism, where $L_{X / F} = cofib (f^{*} L_{F} \to L_{X})$ is the relative cotangent complex. Equivalently, the induced relative tangent-to-cotangent map $fib (T_{X} \to f^{*} T_{F}) \to L_{X} [n - 1]$ is a quasi-isomorphism; use the same convention for $g$ . Then the derived homotopy fiber product $Z = X \times_{F}^{h} Y$ carries a canonical closed $(n - 1)$ -shifted symplectic structure, the residue of $(ω, h, k)$ .

Statement references

Closed n-shifted symplectic form

Derived Artin stack and cotangent complex

Derived Artin stacks are closed under homotopy fiber products

Facts & Assumptions

Given: AC, the derived Artin stacks $F, X, Y$ , the $n$ -shifted symplectic form $ω$ , and Lagrangian paths and nondegeneracy maps for $f$ and $g$ .

[F1]

Closed n-shifted symplectic form: closed 2-forms of degree $m$ are points of the weighted negative-cyclic complex in weight two, with underlying 2-form map $T \to L [m]$ ; nondegeneracy means this map is a quasi-isomorphism.

[F2]

Derived Artin stack and cotangent complex: cotangent complexes corepresent derived derivations, and tangent complexes are their duals when perfect.

[F3]

Derived Artin stacks are closed under homotopy fiber products: the derived homotopy fiber product of the stated Artin stacks is derived Artin.

[A1]

The isotropic paths for $f$ and $g$ induce quasi-isomorphisms $T_{X} \to L_{X / F} [n - 1]$ and $T_{Y} \to L_{Y / F} [n - 1]$ under the relative cotangent convention in the Statement.

[A2]

AC is the foundational premise in the cotangent-comparison construction inherited from Derived schemes and the cotangent complex of a morphism.

Proof

1.1F2F3A2given

By [F3], $Z$ is derived Artin, so the closed-form spaces and cotangent complex in the claim are defined. Write $p : Z \to X$ , $q : Z \to Y$ , and $π = f \circ p ≃ g \circ q : Z \to F$ ; the homotopy in the derived fiber product is the displayed equivalence.

1.2F1A2givenconstruct

Pulling back the paths $h$ and $k$ gives paths from zero to $p^{*} f^{*} ω$ and $q^{*} g^{*} ω$ . The homotopy $f \circ p ≃ g \circ q$ induces a path between those two endpoints in $A^{2, c l} (Z, n)$ . Concatenate the first path, this endpoint homotopy, and the reverse of the second path. The result is a loop at zero. This space is a simplicial abelian group represented by a shifted complex, so $Ω A^{2, c l} (Z, n) ≃ A^{2, c l} (Z, n - 1)$ . The loop therefore defines a closed form $ω_{Z} \in A^{2, c l} (Z, n - 1)$ .

2.1F1F2A2step 1.2given

At every affine test point of $Z$ , derived derivations identify its tangent complex with the homotopy fiber in the exact triangle $T_{Z} ⟶ p^{*} T_{X} \oplus q^{*} T_{Y} ⟶ π^{*} T_{F} .$ For the cotangent triangle write $L_{X / F} = cofib (f^{*} L_{F} \to L_{X})$ and $L_{Y / F} = cofib (g^{*} L_{F} \to L_{Y})$ . The pullback cotangent calculation gives the exact triangle $L_{Z} ⟶ p^{*} L_{X / F} \oplus q^{*} L_{Y / F} ⟶ π^{*} L_{F} [1] .$ Shift it by $n - 1$ . The isotropic homotopies from step 1.2 induce a commutative map from the tangent triangle to this shifted cotangent triangle: the left map is the adjoint of $ω_{Z}$ , the middle maps are those induced by the two Lagrangian structures, and the right map is $Θ_{ω}$ . The endpoint homotopy makes the component maps compatible with the common map to $F$ .

3.1F1A1A2step 2.1

The middle vertical map in this comparison is the direct sum of the two quasi-isomorphisms in [A1]. The right vertical map is a quasi-isomorphism by nondegeneracy of $ω$ in [F1]. In a map of exact triangles, if the second and third vertical maps are quasi-isomorphisms, then the first is as well: the long exact cohomology sequences and the five lemma give an isomorphism in every degree. Thus the adjoint $T_{Z} \to L_{Z} [n - 1]$ induced by $ω_{Z}$ is a quasi-isomorphism.

4.1F1A2step 1.2step 3.1∎

Step 1.2 gives closedness and step 3.1 gives nondegeneracy. By the definition of shifted symplectic structure, $ω_{Z}$ is therefore an $(n - 1)$ -shifted symplectic structure on $Z$ . Its construction used only the given paths and the canonical fiber-product homotopy, so it is canonical relative to those data.

Remarks

The exact tangent and relative-cotangent triangles compared in steps 2.1 and 3.1 are summarized here. The proof uses their exactness and the five-lemma argument, not the diagram as evidence.

The statement includes derived intersections, including empty classical intersections whose derived fiber product is still the relevant moduli object. It makes no assertion that the ordinary underived intersection is smooth or nonempty.

Derived intersection of Lagrangians is shifted symplectic

Statement

Statement references

Facts & Assumptions

Proof

Remarks

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