Cotangent cohomology of Stanley-Reisner rings

Klaus Altmann  Jan Arthur Christophersen

Abstract

Simplicial complexes X provide commutative rings A(X) via the Stanley-Reisner construction. We calculated the cotangent cohomology, i.e., T^{1} and T^{2} of A(X) in terms of X. These modules provide information about the deformation theory of the algebro geometric objects assigned to X.

1 Introduction

Denote [n] := {0, …, n} and let Δ_{n} := 2^{[n]} be the full simplex. A simplicial complex X ⊆ Δ_{n} with vertex set [X] ⊆ [n] gives rise to an ideal

I_{X} := ⟨x^{p} | p ∈ Δ_{n} \ X⟩ ⊆ ℂ[x_{0}, …, x_{n}] =: P.

The Stanley-Reisner ring is then A_{X} = P/I_{X}. We can associate the schemes 𝔸(X) = Spec A_{X} and ℙ(X) = Proj A_{X} with these rings. The latter looks like X itself – its simplices have just been replaced by projective spaces.

For each ℂ-algebra A, there is a cohomology theory providing modules T_{A}^{i}, see e.g. [And74] or [Lau79]. However, only T_{A}^{1} and T_{A}^{2} are relevant for the deformation theory of Spec A. The module T_{A}^{1} collects the infinitesimal deformations of A and T_{A}^{2} contains the obstructions for lifting these deformations to decent parameter spaces. Eventually, Der_{ℂ}(A, A) will be called T_{A}^{0}.

The main result of the present paper is Theorem 9; it provides the modules T^{i} (i = 1, 2) for Stanley-Reisner rings A_{X} in terms of the geometry of the original simplicial complex X: The T^{i} are ℤ^{[n]}-graded, i.e., each degree c corresponds to a monomial of the quotient field of the ambient ring P. Splitting c = a − b in its positive and negative part, one obtains disjoint subsets a, b ⊆ [n] as their respective supports. They give rise to certain subsets N_{a−b}, Ñ_{a−b} ⊆ X, cf. Section 2 (right before Lemma 3), and it is the cohomology of their geometric realizations which provides the homogeneous part T_{c}^{i} of T^{i}.

One has to be a little careful with the geometric realization of a subset N ⊆ X which is not necessarily a subcomplex; in particular it depends on whether ∅ ∈ N or not – see Section 3 for the definition. In Theorem 13, we present a version of our T^{i}-formula that uses only open subsets of X or certain nice subcomplexes. We have chosen a non-trivial example (introduced as Example 1 in Section 2) to illustrate the theory. In particular, it is spread (in eight parts) throughout the text.

In the case that |X| is a homological sphere, Ishida and Oda have proven Theorem 9 in [IO81] using torus embeddings. Moreover, in [Sym97], Symms computes Hom(I_{X}, A_{X}) and T_{0}^{2} when |X| is a 2-dimensional manifold possibly with boundary. Our method is straightforward and allows us to get the T^{i} for all Stanley-Reisner algebras. We also compute the cup product T^{1} × T^{1} → T^{2} and the localization maps. Theorem 15 states that they are always injective.

Information about the T^{i}, the cup product, and their behavior under localization makes it possible to investigate the deformation theory of 𝔸(X) and ℙ(X). In fact, this paper was originally motivated by a question from Sorin Popescu about the smoothability of ℙ(X) when |X| ≈ S^{n} as this would have applications for degenerations of Calabi-Yau manifolds. In the forthcoming paper [AC02] we apply our results to the case when X is a combinatorial manifold, e.g., a sphere. Here we can give very explicit results and a good understanding of the deformations of 𝔸(X) and ℙ(X).

Acknowledgements. The first author is grateful for financial support from the University of Oslo during his visits. The second author is grateful for financial support from the universities FU and HU Berlin during his visits.

2 Cotangent cohomology in terms of the complex

Notation. We will often work in the polynomial ring ℂ[x] = ℂ[x_{0}, …, x_{n}]. Monomials are written as x^{p} ∈ ℂ[x] with exponents p ∈ ℕ^{n+1}. The support of p is defined as p := {i ∈ [n] | p_{i} ≠ 0}. On the other hand, subsets p ⊆ [n] will allways be identified with their characteristic vector p ∈ {0, 1}^{n+1}.

Let X ⊆ Δ_{n} be a simplicial complex and denote by {e_{p} | p ∈ Δ_{n} \ X} a basis for P^{|Δ_{n}\X|} parametrizing the generators x^{p} of I_{X}. The generating relations among them are

R_{p,q} := x^{q\p} e_{p} − x^{p\q} e_{q} .

The relations among these relations are

R_{p,q,r} : x^{r\(p∪q)} e_{p,q} − x^{q\(p∪r)} e_{p,r} + x^{p\(q∪r)} e_{q,r} .

Remark. What we have just described is a special case of the so called Taylor resolution – a construction of a free, but in general not minimal, resolution of any monomial ideal. For a description and proof of exactness see e.g. [BPS98].

Example 1. The following simplicial complex D will serve as a running example throughout the text: With vertex set {x, y, z, x′, y′, z′}, we define D ⊆ Δ_{5} to be the union of the octahedron with the 8 maximal faces (x^{(′)}y^{(′)}z^{(′)}) and the 3 diagonals (xx′), (yy′), and (zz′). Hence, the set Δ_{5} \ D providing the generators of I_{D} consists of all p ⊆ {x, y, z, x′, y′, z′} with #p ≥ 3 and containing at least one letter twice.

In general, for a finitely generated ℂ-algebra A, the modules T_{A}^{i} allow the following ad hoc definitions: Let P = ℂ[x] mapping onto A so that A ≃ P/I for an ideal I. Then T_{A}^{1} is the cokernel of the natural map Der_{ℂ}(P, P) → Hom_{P} (I, A). Moreover, if

0 → R → P^{m} →^{j} P → A → 0

is an exact sequence presenting A as a P module and R_{0} := ⟨j(f)e − j(e)f | e, f ∈ P^{m}⟩ ⊆ R denotes the so-called Koszul relations, then R/R_{0} is an A module and we obtain T_{A}^{2} as the cokernel of the induced map Hom_{P} (P^{m}, A) → Hom_{A}(R/R_{0}, A).

If A = A_{X} is a Stanley-Reisner ring, then A_{X} itself, its resolution, and all interesting A_{X}-modules such as the T_{A}^{i} are ℤ^{n+1}-graded; just set deg e_{p} = p, deg R_{p,q} = p ∪ q, and deg R_{p,q,r} = p ∪ q ∪ r. For an element c ∈ ℤ^{n+1}, we denote by

Hom(I_{X}, A_{X})_{c} and T_{c}^{i}(X) := T_{A_{X},c}^{i}

the homogeneous summands of the corresponding modules. Let c = a − b be the decomposition of c in its positive and negative part, i.e., a, b ∈ ℕ^{n+1} with both elements having disjoint supports a and b, respectively. This gives rise to the sets

M_{a−b} := {p ∈ (Δ_{n} \ X) | (p ∪ a) \ b ∈ X}

and

M_{a−b}^{(2)} := {(p, q) ∈ M_{a−b} × M_{a−b} | (p ∪ q ∪ a) \ b ∈ X} .

Example 1.2 (continued) Let p := (xx′yy′) and q := (xyy′z). Then p, q ∈ M_{∅−(yy′)}, but (p, q) ∉ M_{∅−(yy′)}^{(2)}. Moreover, (xx′yz), (xyz) ∉ M_{∅−(yy′)}, but for different reasons.

Lemma 2. Let c = a − b as before. The modules Hom(I_{X}, A_{X})_{c} and T_{c}^{2}(X) vanish unless b ∈ {0, 1}^{n+1}, i.e., b = b. Assuming b = b, these modules only depend on the supports a, b.

(i) Hom(I_{X}, A_{X})_{c} = \{μ : M_{a−b} → ℂ | \begin{matrix}μ(p) = 0 if b ⊈ p, \\ μ(p) = μ(q) if (p, q) ∈ M_{a−b}^{(2)}\} . \end{matrix}

(ii) Elements of Hom(I_{X}, A_{X})_{c} yield trivial deformations, i.e., belong to the image of Der_{ℂ}(P, P)_{c}, iff #(b) = 1 and μ(p) is a constant function. (iii) T_{c}^{2}(X) is the factor of

\{μ : M_{a−b}^{(2)} → ℂ | \begin{matrix}μ is antisymmetric, μ(p, q) = 0 if (p ∩ q) ∪ ((p ∪ q ∪ a) \ b) ∈ X \\ or if b ⊈ p ∪ q, μ(p, q) − μ(p, r) + μ(q, r) = 0 if (p ∪ q ∪ r ∪ a) \ b ∈ X\} \end{matrix}

by the subspace of functions μ(p, q) = μ′(p) − μ′(q) with μ′(m) = 0 if b ⊈ m.

Proof. An element φ ∈ Hom(I_{X}, A_{X})_{c} maps the generating monomials x^{p} to some μ(p) x^{p+a−b} with μ(p) ∈ ℂ. The condition that (p + a − b) ∈ ℕ^{n+1} yields that μ(p) = 0 unless b = b and b ⊆ p. On the other hand, if x^{p+a−b} ∈ I_{X}, then φ(x^{p}) = 0, and the value μ(p) does not matter at all. Hence, we may restrict the knowledge of μ to M_{a−b} ⊆ (Δ_{n} \ X). The linearity of φ translates into the last condition in (i). Eventually, the trivial deformations are spanned by φ = ∂/∂x_{i}. (iii) One obtains the description of Hom_{A}(R/R_{0}, A_{X})_{c} with the same arguments. We should only remark that it is the Koszul relations x^{q}e_{p} − x^{p}e_{q} ∈ R_{0} that are responsible for the vanishing of μ(p, q) in case of (p ∩ q) ∪ ((p ∪ q ∪ a) \ b) ∈ X. Afterwards, to get T^{2}, one needs to divide out the canonical generators D_{m} ∈ Hom_{ℂ[x]}(ℂ[x]^{Δ_{n}\X}, A_{X}) (m ∈ Δ_{n} \ X). They have degree −m, and applied to R_{p,q}, they yield non-trivial values x^{q\p} and −x^{p\q} only if p = m or q = m, respectively. Hence, in degree c, the map x^{m+a−b}D_{m} yields μ(p, q) = μ_{m}(p) − μ_{m}(q) with μ_{m} denoting the characteristic function of m. On the other hand, this contribution requires b ⊆ m. □

Of course, we are building some sort of cohomology to describe the graded pieces of T^{i}(X). However, in the previous lemma, the Koszul condition does not seem to fit. Surprisingly, this problem will be overcome by performing kind of a c-shift. Let

N_{a−b}(X) := {f ∈ X | a ⊆ f, f ∩ b = ∅, f ∪ b ∉ X}

and

Ñ_{a−b}(X) := {f ∈ N_{a−b} | ∃ b′ ⊂ b with f ∪ b′ ∉ X} .

Lemma 3. Let Φ : M_{a−b} → X be the application Φ(p) = (p ∪ a) \ b.

(i) Im Φ = N_{a−b}. Moreover, an element f ∈ N_{a−b} has a pre-image that does not contain b if and only if f ∈ Ñ_{a−b}. (ii) If f, g ∈ N_{a−b} and f ∪ g ∈ X, then f ∪ g ∈ N_{a−b}. If, moreover, g ∈ Ñ_{a−b}, then we even obtain f ∪ g ∈ Ñ_{a−b}.

Proof. (i) If p ∈ M_{a−b}, then Φ(p)∪b ⊇ p ∉ X, hence Φ(M_{a−b}) ⊆ N_{a−b}. On the other hand, let f ∈ N_{a−b} with f ∪b′ ∉ X for some b′ ⊆ b. It follows that f ∪b′ ∈ M_{a−b} and Φ(f ∪ b′) = f. With b′ := b, this implies N_{a−b} ⊆ Φ(M_{a−b}); with b′ being a proper subset of b, we obtain the Ñ_{a−b} statement. The claims in (ii) are obvious. □

Example 1.3 (continued) Considering the previous p = (xx′yy′) and q = (xyy′z) from M_{∅−(yy′)}, we obtain Φ(p) = (xx′) ∈ Ñ_{∅−(yy′)}(D) (note that Φ(p) = Φ(xx′y)), but Φ(q) = (xz) ∈ N_{∅−(yy′)}(D) \ Ñ_{∅−(yy′)}(D).

The map Φ : M_{a−b} → → N_{a−b} of the previous lemma can easily be extended to pairs. That is, with N_{a−b}^{(2)} := {(f, g) ∈ N_{a−b} × N_{a−b} | f ∪ g ∈ X} , we also have a surjective application Φ : M_{a−b}^{(2)} → → N_{a−b}^{(2)}.

Proposition 4. Let c = a − b as before. The modules Hom(I_{X}, A_{X})_{c} and T_{c}^{2}(X) vanish unless b ∈ {0, 1}^{n+1}, i.e., b = b. Assuming b = b, these modules only depend on the supports a, b.

(i) Hom(I_{X}, A_{X})_{c} = \{λ : N_{a−b} → ℂ | \begin{matrix}λ(f) = 0 if f ∈ Ñ_{a−b}, \\ λ(f) = λ(g) if f ∪ g ∈ X\} . \end{matrix}

(ii) Elements of Hom(I_{X}, A_{X})_{c} yield trivial deformations, i.e., belong to the image of Der_{ℂ}(P, P)_{c}, iff #(b) = 1 and λ(f) is a constant function. (iii) T_{c}^{2}(X) is the factor of the vector space of the antisymmetric maps λ : N_{a−b}^{(2)} → ℂ such that λ(f, g) = 0 if f, g ∈ Ñ_{a−b} and λ(f, g) − λ(f, h) + λ(g, h) = 0 if (f ∪ g ∪ h) ∈ X

by the subspace {λ(f) − λ(g)} with λ = 0 on Ñ_{a−b}.

Proof. Denote, just for this proof, the spaces given by (i) and (iii) of the previous proposition by Hom(N) and T^{2}(N), respectively. Then we have to ascertain that pulling back via Φ induces isomorphisms Φ^{∗} : Hom(N) →^{∼} Hom(M) and Φ^{∗} : T^{2}(N) →^{∼} T^{2}(M) with Hom(M) and T^{2}(M) being the corresponding spaces from Lemma 2. Step 1. The maps Φ^{∗} are correctly defined: This is clear for the Hom case. For T^{2}, we set μ(p, q) := λ(Φp, Φq), and the only non-trivial task is to check the two conditions that should lead to the vanishing of μ(p, q). If b ⊈ p∪q, then both b ⊈ p and b ⊈ q, hence Φ(p), Φ(q) ∈ Ñ by Lemma 3(i), hence λ(Φp, Φq) = 0. On the other hand, if (p ∩ q) ∪ ((p ∪ q ∪ a) \ b) ∈ X, then we also obtain b ⊈ p and b ⊈ q: Otherwise, if b ⊆ p, it would follow that q ∩ b ⊆ p ∩ q and q \ b ⊆ (p ∪ q ∪ a) \ b, thus, q = (q ∩ b) ∪ (q \ b) ⊆ (p ∩ q) ∪ ((p ∪ q ∪ a) \ b) ∈ X, but this contradicts q ∈ M. Step 2. The maps Φ^{∗} are injective: The Hom case follows from the surjectivity of Φ. For T^{2}, assume that λ(Φp, Φq) = μ(p, q) = μ(p) − μ(q). In particular, if p, q belong to a common fiber Φ^{−1}(f), then μ(p) − μ(q) = 0. Hence, μ(p), μ(q) only depend on Φ(p) and Φ(q).

Step 3. The maps Φ^{∗} are surjective: Let {μ(p)} represent an element of Hom(M). Then, the property that μ(p) = μ(q) for (p, q) ∈ M^{(2)} implies that μ(p) only depends on Φ(p). In particular, {μ(p)} ∈ Φ^{∗}(Hom(N)). To check the T^{2} case, we would like to proceed similarily with elements {μ(p, q)} ∈ T^{2}(M). However, this requires a correction by coboundaries: By the cocycle property of the μ(p, q)’s, we have to find {μ(p)} such that μ̃(p, q) := μ(p, q)+(μ(p)−μ(q)) vanishes if p, q belong to a common fiber Φ^{−1}(f). Using the cocycle property again, we see that μ(p) := μ(m_{f}, p) will almost do the job for any fixed m_{f} ∈ Φ^{−1}(f); but we also have to ensure that μ(p) = 0 whenever b ⊈ p. This is done by proving the following Claim: Let m, p ∈ M with b ⊈ m, b ⊈ p, and Φ(m) ⊇ Φ(p). Then μ(m, p) = 0. With f := Φ(m), we have that (m ∩ p) ∪ f = (m ∩ p) ∪ ((m ∪ p ∪ a) \ b). Thus, (m ∩ p) ∪ f ∉ X (or the Koszul condition in Lemma 2(iii) immediatly implies μ(m, p) = 0). In particular, (m ∩ p) ∪ f ∈ Φ^{−1}(f), and since b ⊈ m ∪ [(m ∩ p) ∪ f] and b ⊈ [(m ∩ p) ∪ f] ∪ p, we obtain that

μ(m, p) = μ(m, (m ∩ p) ∪ f) + μ((m ∩ p) ∪ f, p) = 0 + 0 = 0 .

Eventually, it is possible to define λ(Φp, Φq) := μ̃(p, q) = μ(m_{Φ(p)}, m_{Φ(q)}), and it remains to show its vanishing for Φ(p), Φ(q) ∈ Ñ. Since, by Lemma 3(ii), Φ(p) ∪ Φ(q) ∈ Ñ, we may assume that Φ(p) ⊆ Φ(q). Now everything follows from applying the previous claim again. □

Definition 5. A subset Y ⊆ X of a simplicial complex X has property U, or is a

U subset, if f, g ∈ Y and f ∪ g ∈ X ⇒ f ∪ g ∈ Y .

If Y has this property, then we define the sets

Y^{(k)} := {(f_{0}, …, f_{k}) ∈ Y^{k+1} | f_{0} ∪ ⋯ ∪ f_{k} ∈ Y }

and the complex of ℂ-vector spaces

K^{k}(Y) := {λ : Y^{(k)} → ℂ | λ is alternating} ⊆ Λ^{k+1}(ℂ^{Y} )

with the usual differential d : K^{k−1}(Y) → K^{k}(Y) defined by

d(λ)(f_{0}, …, f_{k}) := ∑_{v=0}^{k}(−1)^{v}λ(f_{0}, …, f̂_{v}, …, f_{k}) .

By Lemma 3(ii), both N_{a−b} and Ñ_{a−b} are U subsets. Moreover, there is a canonical surjection of complexes K^{•}(N_{a−b}) ↠ K^{•}(Ñ_{a−b}) leading to Corollary 6. Assume c = a − b with disjoint a, b ∈ X. Then

Hom(I_{X}, A_{X})_{c} ≃ H^{0}(ker(K^{•}(N_{a−b}) ↠ K^{•}(Ñ_{a−b})))
T_{c}^{2}(X) ≃ H^{1}(ker(K^{•}(N_{a−b}) ↠ K^{•}(Ñ_{a−b}))) ,

and the trivial deformations inside Hom(I_{X}, A_{X})_{c}, i.e., those yielding 0 in T_{c}^{1}(X), form a one-dimensional subspace whenever #(b) = 1 (and are absent otherwise).

Example 1.4 (continued) We still consider the degree a = ∅, b = {y, y′} for D being the octahedron with diagonals. The set N_{∅−(yy′)}(D) consists of the 4 vertices x^{(′)}, z^{(′)} and all the 6 edges connecting them. However, only the interior of the edges xx′ and zz′ survive in Ñ_{∅−(yy′)}(D).

N_{∅−(yy^{|})}  Ñ_{∅−(yy^{|})}

To obtain elements of Hom(I_{D}, A_{D})_{c}, we have to consider maps λ : N → ℂ, i.e., each of the 4 vertices and 6 edges will be assigned a value. The two conditions encoded by “H^{0}” and “ker” in the previous corollary mean that λ has to be both constant along the graph and zero on int xx′ and int zz′. Hence, Hom(I_{D}, A_{D})_{c} = 0.

Remark. As we mentioned in the beginning, there is a general cohomological definition of the T^{i}. Hence, it is no surprise that we ended up with a cohomological description of these spaces in terms of X, too. Moreover, it should even be a challange to find a direct way to obtain the previous result (without touching elements). If this involved a description of the so-called cotangent complex, one would obtain important information about the deformation theory of both 𝔸(X) and ℙ(X).

3 Cotangent cohomology and the geometry of X

In the following, we will relate the previous description of T^{i}(X) with the geometry of the complex. Let us start with some notation. For g ⊆ [n], denote by ḡ := 2^{g} and ∂g := ḡ \ {g} the full simplex and its boundary, respectively. The join X ∗ Y of two complexes X and Y is the complex defined by

X ∗ Y := {f ∨ g : f ∈ X, g ∈ Y }

where ∨ means the disjoint union. If f ∈ X is a face, we may define • the link of f in X; lk(f, X) := {g ∈ X : g ∩ f = ∅ and g ∪ f ∈ X},

• the open star of f in X; st(f, X) := {g ∈ X : f ⊆ g}, and

• the closed star of f in X; \overline{st}(f, X) := \{g ∈ X : g ∪ f ∈ X\}.

Notice that the closed star is the subcomplex \overline{st}(f, X) = f̄ ∗ lk(f, X). Recall that the geometric realization of X, denoted |X|, may be described by

|X| = {α : [n] → [0, 1] | {i | α(i) ≠ 0} ∈ X and ∑_{i} α(i) = 1} .

To every non-empty f ∈ X, one assigns the relatively open simplex ⟨f⟩ ⊆ |X|;

⟨f⟩ = {α ∈ |X| | α(i) ≠ 0 if and only if i ∈ f} .

On the other hand, each subset Y ⊆ X determines a topological space

⟨Y ⟩ := \begin{cases}⋃_{f∈Y} ⟨f⟩ & if ∅ ∉ Y , \\ cone (⋃_{f∈Y} ⟨f⟩) & if ∅ ∈ Y . \end{cases}

In particular, ⟨X \ {∅}⟩ = |X| and ⟨X⟩ = | cone(X)| where cone(X) is the simplicial complex Δ_{0} ∗ X.

Any subset Y of X is a poset with respect to inclusion and we may construct the associated (normalized) order complex Y′: The vertices of Y′ are the elements of Y and the k-faces of Y′ are flags f_{0} ⊂ f_{1} ⊂ ⋯ ⊂ f_{k} of Y -elements. If Y is a complex, Y′ is the barycentric subdivision of cone(Y).

A complex and its barycentric subdivision have the same geometric realization, so if Y is a subcomplex of X, we have |Y′| = ⟨Y ⟩. This identity is obtained by sending a vertex f of Y′ to the barycenter of f in ⟨f⟩ if f ≠ ∅ and the vertex corresponding to ∅ ∈ Y′ to the vertex of the cone. For a general subset Y ⊆ X, we only know that |Y′| ⊆ ⟨Y ⟩ inside |X′| = ⟨X⟩.

Lemma 7. If Y ⊆ X, then |Y′| is a deformation retract of ⟨Y ⟩. In particular, both sets have the same cohomology.

Proof. If f ∈ cone(X), then we may identify ⟨f⟩ ⊂ ⟨X⟩ with the union of all ⟨F ⟩ in |X′| where F = (f_{0} ⊂ f_{1} ⊂ ⋯ ⊂ f_{k}) and f_{k} = f. Thus, ⟨Y ⟩ as subset of |X′| is the union of such ⟨F ⟩ with f_{k} ∈ Y. For such an F let F_{Y} ≤ F be the maximal subflag consisting only of faces in Y. Now we can continuously retract ⟨F ⟩∪⟨F_{Y} ⟩ onto ⟨F_{Y} ⟩ and this can be done simultaneously for all F belonging to the above union. □

For a subset Y ⊆ X, let C^{•}(Y) be the cochain complex of Y′. We have an obvious inclusion of the k-flags in Y into Y^{(k)} given by f_{0} ⊂ f_{1} ⊂ ⋯ ⊂ f_{k} ↦ (f_{0}, f_{1}, …, f_{k}). This induces a surjection of complexes K^{•}(Y) → C^{•}(Y) when Y has property U.

Lemma 8. The surjection K^{•}(Y) → C^{•}(Y) is a quasi-isomorphism, i.e., it induces an isomorphism in cohomology.

Proof. We will prove the dual statement in homology using the method of acyclic models (see e.g. [Spa66, 4.2]). If Y has property U, consider the simplicial complex Y^{∗} where the vertices are the elements of Y and a set of vertices {f_{0}, …, f_{k}} is a face if f_{0} ∪ ⋯ ∪ f_{k} ∈ Y. If C_{•}(Y^{∗}) is the chain complex of Y^{∗}, then our K^{•}(Y) is the dual of C_{•}(Y^{∗}) ⊗ ℂ, and we are finished if we can prove that C_{•}(Y^{∗}) is chain equivalent to C_{•}(Y′). To this end, consider the set Y of U subsets of X as a category with inclusions as morphisms. Let the models in Y be M = {f̄∩Y | f ∈ Y, Y ∈ Y}. Finally, define the two functors from Y to chain complexes by F′(Y) = C_{•}(Y′) and F^{∗}(Y) = C_{•}(Y^{∗}). We must show that both F′ and F^{∗} are free and acyclic with respect to these models. Now a basis element of F′(Y), (f_{0} ⊂ f_{1} ⊂ ⋯ ⊂ f_{k}), comes from F′(\overline{f_{k}} ∩ Y) and a basis element \{f_{0}, …, f_{k}\} of F^{∗}(Y) comes from F^{∗}(\overline{f_{0} ∪ ⋯ ∪ f_{k}} ∩ Y), so both functors are free. If f ∈ Y, one may check that (f̄∩ Y)^{∗} is a simplex and (f̄∩ Y)′ is a cone over the vertex {f}. Thus, in both cases the chain complexes are acyclic. □

We may apply Lemma 7 and 8 to N_{a−b} and Ñ_{a−b}. Via the 5-Lemma, the result of Corollary 6 translates into

Theorem 9. Assume c = a−b with a, b ∈ ℕ^{n+1} having disjoint supports a, b ⊆ [n]. The homogeneous pieces in degree c of the cotangent cohomology of the Stanley-Reisner ring A_{X} vanish unless b ∈ {0, 1}^{n+1}, i.e., b = b. Assuming b = b, these modules only depend on the supports a, b, and we have isomorphisms

T_{c}^{i}(X) ≃ H^{i−1}(⟨N_{a−b}⟩, ⟨Ñ_{a−b}⟩, ℂ) for i = 1, 2

unless b consists of a single vertex. If b consists of only one vertex then the above formulae become true if we use the reduced cohomology instead. Moreover, replacing T_{c}^{1}(X) by Hom(I_{X}, A_{X})_{c} creates true formulae without any need to distinguish between several cases.

Example 1.5 (continued) In the previous session of Example 1, we have seen that the pair (N_{∅−(yy′)}, Ñ_{∅−(yy′)}) equals (complete graph on 4 vertices, 2 opposite edges). Now, we may use that this is homotopy equivalent to ( [figure], [figure]), i.e., T_{c}^{1}(D) = 0, but dim_{ℂ} T_{c}^{2}(D) = 4.

The previous theorem also provides information about T^{0}(X) := Der_{ℂ}(A_{X}, A_{X}). While this module has already been described in [BS95], we would like to demonstrate its relation to our techniques.

Corollary 10. T^{0}(X) = ⨁_{v=0}^{n} 𝔞_{v} ∂/∂x_{v} where 𝔞_{v} is the ideal of A_{X} generated by the monomials x^{a} with \overline{st}(a, X) ⊆ \overline{st}(v, X). In particular, T^{0}(X) is generated, as a module, by δ_{v} := x_{v} ∂/∂x_{v} if and only if every non-maximal a ∈ X is properly contained in at least two different faces.

Proof. T^{0}(X) is the kernel of Der_{ℂ}(ℂ[x], A_{X}) → Hom(I_{X}, A_{X}). Hence, since Ñ_{a−{v}} = ∅, the previous theorem implies that an element x^{a} ∂/∂x_{v} belongs to T^{0}(X) if and only if H^{0}(⟨N_{a−{v}}⟩, ℂ) = 0, i.e., iff N_{a−{v}} = ∅. On the other hand, this means that, for every f ∈ X, the conditions a ⊆ f and v ∉ f imply f ∪ v ∈ X. Since the assumption v ∉ f can be omitted, this translates into st(a, X) ⊆ \overline{st}(v, X). Finally, T^{0}(X) is generated, as a module, by δ_{v} = x_{v} ∂/∂x_{v} if and only if \overline{st}(a, X) ⊆ \overline{st}(v, X) cannot happen for faces a with v ∉ a. But this is equivalent to the condition formulated in the corollary. □

4 Reduction to the a = ∅ case and localization

The set N_{a−b} is empty unless b ≠ ∅ and a ∈ X is a face. Moreover, by the next proposition, we may reduce all the calculations of N_{a−b} and Ñ_{a−b}, and therefore the T^{i}, to the case of a = ∅ on a smaller complex. See Example 17 for a demonstration of a consequent usage of this method.

Proposition 11. T_{c}^{i}(X) = 0 for i = 1, 2 unless a ∈ X and b ⊆ [lk(a)]. If b ⊆ [lk(a)], then the map f ↦ f \ a is a bijection N_{a−b}(X) →^{∼} N_{∅−b}(lk(a)) inducing isomorphisms T_{∅−b}^{i}(lk(a)) ≃ T_{a−b}^{i}(X) for i = 1, 2.

Proof. First assume that there is a vertex v ∈ b\[lk(a)]. If f ∈ N_{a−b}, then f ∪v ∉ X (otherwise a ∪ v ∈ X and v ∈ lk(a)). Thus, N_{a−b} = Ñ_{a−b} unless b = {v}. If b = {v}, then Ñ_{a−b} = ∅ and N_{a−b} = st(a). Thus, ⟨N_{a−b}⟩ may be contracted to ⟨a⟩ and its reduced cohomology is trivial, so T_{a−b}^{i}(X) = 0 by Theorem 9. It is a simple matter to check that the map between the N sets is a bijection (with inverse g ↦ g ∪ a) and that it restricts to a bijection of the Ñ subsets. Since it clearly preserves inclusions, it induces a simplicial isomorphism on the complexes N_{a−b}(X)′ ≃ N_{∅−b}(lk(a))′. From Lemma 7, it follows that we get an isomorphism in the relative cohomology. □

Example 1.6 (continued) Assume that a = {x} for D from Example 1. The link lk(x) equals the boundary of the rectangle (yzy′z′) plus the isolated point x′. If #b ≥ 3, then there is always a proper subset b′ ⊂ b with b′ ∉ X. In particular, N_{∅−b}(lk(x)) = Ñ_{∅−b}(lk(x)), hence T_{(x)−b}^{1}(D) = T_{(x)−b}^{2}(D) = 0. The case #b = 2 does not provide any T^{i} either, but if b = {x′}, then N_{∅−(x′)}(lk(x)) is the boundary of the rectangle and Ñ_{∅−(x′)}(lk(x)) = ∅. Hence, T_{(x)−(x′)}^{1}(D) = 0 and dim_{ℂ} T_{(x)−(x′)}^{2}(D) = 1. Similarily, we obtain dim_{ℂ} T_{(x)−∗}^{1}(D) = 1 and T_{(x)−∗}^{2}(D) = 0 where ∗ stands for any of the vertices y^{(′)}, z^{(′)}.

The subsets ⟨N_{a−b}⟩ and ⟨Ñ_{a−b}⟩ are in general neither open nor closed in |X|. In the case a = ∅ though, we may find open sets retracting onto them. These sets are easier to define, but they are not always easier to handle. However, their openess often allows use of standard tools for calculating cohomology. Let

U_{b} = U_{b}(X) := {f ∈ X | f ∪ b ∉ X}

and

Ũ_{b} = Ũ_{b}(X) := {f ∈ X | (f ∪ b) \ {v} ∉ X for some v ∈ b} .

Notice that U_{b} = Ũ_{b} = X unless ∂b is a subcomplex of X. Moreover, if ∂b ⊆ X, then with L_{b} := ⋂_{b′⊂b} lk(b′, X) we have

X \ U_{b} = \begin{cases}∅ \\ \overline{st}(b) \end{cases} and X \ Ũ_{b} = \begin{cases}∂b ∗ L_{b} & if b is a non-face, \\ (∂b ∗ L_{b}) ∪ \overline{st}(b) & if b is a face. \end{cases}

Example 1.7 (continued) Going back to the degree ∅ − (yy′) from Example 1.5, we see that D \ U_{b} is the (closed) subcomplex induced from the edge (yy′). On the other hand, L_{b} is the boundary of the rectangle (xzx′z′), hence ∂b ∗ L_{b} is the octahedron (without the diagonals). Thus, while U_{b} is D with the closed diagonal yy′ being removed, Ũ_{b} consists of the (disjoint) union of the open diagonals (xx′) and (zz′). Comparing this with the easy (N, Ñ) in Example 1.5, we feel that one has to pay for the advantage of getting open sets.

U_{(yy^{|})}

Lemma 12. H^{•}(⟨N_{∅−b}⟩, ⟨Ñ_{∅−b}⟩, ℂ) ≃ H^{•}(⟨U_{b}⟩, ⟨Ũ_{b}⟩, ℂ).

Proof. If f is in U (respectively Ũ), then f \ b is in N (respectively Ñ). Now, if f \ b ≠ ∅ for all f ∈ U, then there is a standard retraction taking α ∈ ⟨f⟩ to an α′ ∈ ⟨f \ b⟩ which fits together to make (⟨N⟩, ⟨Ñ⟩) a strong deformation retract of (⟨U⟩, ⟨Ũ⟩). (See e.g. [Spa66, Proof of 3.3.11].) If b is a face, then f \ b = ∅ is impossible, since then f ∪ b = b ∈ X. If b is a non-face and ∅ ∈ Ñ, i.e., ∂b ⊈ X, then all four spaces are cones and there is nothing to prove. If b is a non-face and ∅ ∉ Ñ, then both ⟨N⟩ and ⟨U⟩ are cones, so H^{i}(⟨N⟩, ⟨Ñ⟩) ≃ H̃^{i−1}(⟨Ñ⟩) and H^{i}(⟨U⟩, ⟨Ũ⟩) ≃ H̃^{i−1}(⟨Ũ⟩) and the above retraction works again. □

When we plug this into Theorem 9 and use Proposition 11 we get as a corollary the following description of the graded pieces of T_{A_{X}}^{i}.

Theorem 13. The homogeneous pieces in degree c = a − b (with disjoint supports a and b) of the cotangent cohomology of the Stanley-Reisner ring A_{X} vanish unless a ∈ X and b = b ≠ ∅. If these conditions are satisfied, we have isomorphisms

T_{c}^{i}(X) ≃ H^{i−1}(⟨U_{b}(lk(a, X))⟩, ⟨Ũ_{b}(lk(a, X))⟩, ℂ) for i = 1, 2

unless b consists of a single vertex. If b consists of only one vertex, then the above formulae become true if we use the reduced cohomology instead.

The reduction to the a = ∅ case also appears in a completely different context. One of the main issues of the paper [AC02] is the deformation theory of ℙ(X) ⊆ ℙ^{n}. The relation to the deformation theory of its affine charts D_{+}(x_{v}) is governed by the localization maps T_{A_{X},0}^{i} → T_{D_{+}(x_{v})}^{i}; here “0” is meant with respect to the usual ℤ-grading of A_{X}, and the localization maps are obtained by dehomogenizing. Now, the point is that these affine charts also come from simplicial complexes. If v ∈ [n], then D_{+}(x_{v}) = 𝔸(lk(v, X)), and we can use the techniques developed so far to describe the localization maps.

Remark. Although, strictly speaking, we should consider lk(v, X) as a subcomplex of Δ_{n}, the T^{i} depend only on the complex itself. In particular, when we look at graded parts, Proposition 11 shows that T_{a−b}^{i}(lk(v)) = 0 if a or b contain non-vertices of lk(v, X).

Lemma 14. Let c = a − b belong to degree 0, i.e., deg a = #b with deg denoting the sum of entries. Fix a vertex v ∈ [n].

(i) Localization with respect to v maps T_{a−b}^{i} ⊆ T_{A_{X},0}^{i} into the graded summand T_{(a\v)−(b\v)}^{i}(lk(v, X)) ⊆ T_{D_{+}(x_{v})}^{i} where (a \ v) means cancellation of the v entry. (ii) The map of (i) is induced from ψ_{a−b}(v) : N_{(a\v)−(b\v)}(lk(v, X)) → N_{a−b}(X) defined by ψ_{a−b}(v)(g) = g ∪ (v \ b). It is compatible with the Ñ level. (iii) If v ∈ a, then the localization map is an isomorphism in degree a − b.

Proof. It is straightforward to check that ψ_{a−b}(v) is well defined and has the necessary properties to induce ψ_{a−b}^{∗}(v) : T_{a−b}^{i}(X) → T_{(a\v)−(b\v)}^{i}(lk(v, X)) by Theorem 9. Moreover, it is clear that this means exactly dehomogenization with respect to v, hence ψ_{a−b}^{∗}(v) coincides with the localization map. Finally, to prove that ψ_{a−b}(v) : g ↦ g ∪ v is an isomorphism in case of v ∈ a, we use that lk ((a \ v), lk(v, X)) = lk(a, X) and apply Proposition 11 to both N_{(a\v)−b}(lk(v, X)) and N_{a−b}(X). □

In fact, localizing with respect to all variables x_{v} with v running through the vertices of a given face a ∈ X is induced by the map ψ_{a−b}(a) : N_{∅−b}(lk(a, X)) → N_{a−b}(X) sending g to g ∪ a. This is the inverse of the a killing map of Proposition 11.

Theorem 15. The maps T_{A_{X},0}^{i} → ⨁_{v∈[n]} T_{D_{+}(x_{v})}^{i} are injective for i = 1, 2.

Proof. If a graded piece T_{a′−b′}^{i}(lk(v)) meets the image of T_{0}^{i}(X) → T^{i}(lk(v)), then its pre-image is a unique summand T_{a−b}^{i}(X) ⊆ T_{0}^{i}(X). Indeed, by Proposition 14, the only possibility to get degree 0 is a = a′ +(#b′ −deg a′) v, b = b′ if deg a′ ≤ #b′ and a = a′, b = b′ ∪ v if deg a′ > #b′. This means that images of elements with different multidegree c cannot cancel each other, and it remains to consider the multigraded pieces

⨁v∈[n]ψ_{a−b}^{∗}(v) : T_{a−b}^{i}(X) ⟶ _{⨁v∈[n]}T_{(a\v)−(b\v)}^{i}(lk(v))

with deg a = #b. Since, by Proposition 14(iii), every summand ψ_{a−b}^{∗}(v) with v ∈ a is an isomorphism, we obtain the injectivity of the above map whenever a has vertices at all. On the other hand, since deg a = #b, the face a cannot be empty. □

5 Examples

First, in Examples 16 and 17, we present the complete treatment of the easiest X of all, the triangulations of 0- and 1-dimensional manifolds. While the dimensions of T^{i} for, say, the cone over the n-gon are already well known, we can demonstrate how the multigrading comes in. Moreover, for higher-dimensional examples like the surfaces in Example 18, the smaller ones are needed because they occur as links.

Example 16. Let S^{0} = {∅, 0, 1} be the 0-dimensional complex consisting of two points only. It may be considered a triangulation of the 0-dimensional sphere. If b = {0}, then Ñ_{a−0} = ∅ and N_{a−0} = {f ∈ S^{0} | a ⊆ f, f ∪ 0 ∉ S^{0}} = {1} for both possibilities a = ∅ or a = {1}. In particular, T_{a−0}^{1}(S^{0}) = T_{a−0}^{2}(S^{0}) = 0. If b = {0, 1}, then a = ∅, hence N_{∅−{0,1}} = {∅} and Ñ_{∅−{0,1}} = ∅. This yields T_{∅−{0,1}}^{2}(S^{0}) = 0, but T_{∅−{0,1}}^{1}(S^{0}) is one-dimensional .

How does this infinitesimal deformation perturb the S^{0}-equation x_{0}x_{1} = 0? If ε denotes the infinitesimal parameter from ℂ[ε]/ε^{2}, then one obtains x_{0}x_{1} − ε = 0.

Example 17. Denote by E_{n} the simplicial complex representing an n-gon with n ≥ 3. Index the vertices cyclically with 0, …, n − 1; all addition is done modulo n. First, we will show how to use the a = ∅ reduction from Proposition 11. If a is an edge, then lk(a, E_{n}) = ∅, hence T_{a−b}^{i} = 0. If a = {1} is a vertex, then lk(a, E_{n}) = {∅, 0, 2} ≅ S^{0}, hence from Example 16 we obtain dim T_{{1}−{0,2}}^{1}(E_{n}) = 1

as the only non-trivial contribution; it translates into x_{0}x_{2} − εx_{1}^{≥1}. Let us now assume that a = ∅. If #b = 1, i.e., b is a vertex, then Ñ_{∅−b} = ∅, and ⟨N_{∅−b}⟩ equals |E_{n}| after removing the edges containing b. In particular, ⟨N_{∅−b}⟩ is contractible, and the corresponding T_{∅−b}^{i}(E_{n}) are trivial. If b is an edge, then ⟨N_{∅−b}⟩ looks similar, and ⟨Ñ_{∅−b}⟩ equals ⟨N_{∅−b}⟩ without the endpoints. We obtain

T_{∅−b}^{i}(E_{n}) = 0 for n ≥ 4, but dim T_{∅−b}^{1}(E_{3}) = 1 yielding x_{0}x_{1}x_{2} − εx_{i}.

The final case is that a = ∅ and b ∉ E_{n}. If #b ≥ 3, then, except for n = 3, the set b always contains proper subsets which are non-faces. In particular, Ñ_{∅−b} = N_{∅−b}, leading to T_{∅−b}^{i}(E_{n}) = 0. The exception is dim T_{∅−{0,1,2}}^{1}(E_{3}) = 1 yielding x_{0}x_{1}x_{2}−ε.

It remains to take two non-adjacent vertices for b, say b = {u, v}. Since ∅ ∈ N_{∅−b}, the set ⟨N_{∅−b}⟩ is always a cone, hence contractible. In particular, T_{∅−b}^{1}(E_{n}) = 0 if Ñ_{∅−b} ≠ ∅, and this is always the case except for dim T_{∅−{v,v+2}}^{1}(E_{4}) = 1 ; in terms of equations: x_{v}x_{v+2} − ε. On the other hand, the long exact cohomology sequence for the pair (N_{∅−b}, Ñ_{∅−b}) yields T_{∅−b}^{2}(E_{n}) = H̃^{0}(Ñ_{∅−b}). Since the set ⟨Ñ_{∅−b}⟩ equals |E_{n}| with u and v and the adjacent edges being removed, we eventually obtain dim T_{∅−{u,v}}^{2}(E_{n}) = 1 whenever |u − v| ≥ 3 .

Adding up, we find that T^{2}(E_{n}) = 0 if n ≤ 5, and that dim T^{2}(E_{n}) = n(n − 5)/2 if n ≥ 6. In the latter case, we can even locate where the cup product takes place. Considering the coarse ℤ-grading, we see that T^{1}(E_{n}) spreads in degre ≥ −1, and T^{2}(E_{n}) sits in degree −2. Hence, the cup product lives in the pieces T_{−1}^{1}×T_{−1}^{1} → T_{−2}^{2} only. Using the ℤ^{n} multigrading, one obtains a finer result. The cup product splits into products

T_{{v}−{v−1,v+1}}^{1}(E_{n}) × T_{{v+1}−{v,v+2}}^{1}(E_{n}) ⟶ T_{∅−{v−1,v+2}}^{2}(E_{n})

with all the three vector spaces being one-dimensional. See the continuation on p. 17 for more information.

Example 18. Let us look at the degree 0 deformations when X is the triangulation of a two-dimensional manifold. The non-zero multigraded pieces T_{a−b}^{1}(X) of T_{0}^{1}(X) require a ≠ ∅, hence they are induced from lower-dimensional links. They are all 1-dimensional and are gathered in the following list.

(i) a is an edge; |a| = |b| = 2. a = a.

[figure: b; b; a; a]

(ii) a is a vertex with valency 4; |a| = 1, |b| = 2. a = 2 · a.

[figure: a; b; b]

(iii) a is a vertex with valency 3; |a| = 1, |b| = 3; a = 3 · a.

[figure: b; a; b; b]

(iv) a is a vertex with valency 3; |a| = 1, |b| = 2. a = 2 · a.

[figure: a; b; b]

The perturbation of the equations also comes from the corresponding one of the lower-dimensional link. E.g., denote the vertices in (i) such that a = {y, x_{1}} and b = {x_{0}, x_{2}}. Then we obtain x_{0}x_{2} − εx_{1}y.

Moreover, T_{0}^{2}(X) is only present for a being a vertex of valency at least 6, a = 2 · a, and b consisting of two vertices having exactly a as a common neighbor.

Example 1.8 (finished) Finally, we conclude our running Example 1; the simplicial complex D is the octahedron plus the three diagonals (xx′), (yy′), and (zz′). First, if a is a non-empty face, then one has the following types of links: lk(xyz) = ∅, lk(xx′) = ∅ (both yielding T^{1} = T^{2} = 0), lk(xy) = {z, z′, ∅} ≅ S^{0} yielding dim T_{(xy)−(zz′)}^{1}(D) = 1 . Moreover, we studied the case a = {x} in Exam-

ple 1.6; the non-zero results were dim T_{(x)−(x′)}^{2}(D) = 1 and, for any ∗ ∈ {y^{(′)}, z^{(′)}},

dim T_{(x)−(∗)}^{1}(D) = 1 . In terms of equations, the T^{1}-contributions look like yzz′ −

εxy^{2} in the first case, and like xx′y − εx^{2}x′, xyy′ − εx^{2}y′, yy′z − εxy′z in degree (x) − (y). Second, if a = ∅, then there are two major cases to distinguish. If b is not a face, then one knows in general that ∅ ∈ N_{∅−b}(X), i.e., ⟨N_{∅−b}(X)⟩ is contractible. Hence, by the long exact cohomology sequence for the pair (N, Ñ), dim T_{c}^{1} = 1 and T_{c}^{2} = 0 if Ñ_{∅−b}(X) = ∅, and, otherwise, T_{c}^{1} = 0, T_{c}^{2} = H̃^{0}(Ñ_{∅−b}(X)). In the case of X = D, this always yields T_{c}^{i} = 0. It remains to consider faces for b. While b = {x′} or {x′, y′} do not yield any T^{i}, we obtained in Example 1.5 that dim T_{∅−(yy′)}^{2}(D) = 4 .

6 Appendix: The cup product

The cup product T^{1} × T^{1} → T^{2} is an important tool to obtain more information about deformation theory than just the knowledge of the tangent or obstruction spaces T^{i} themselves. The associated quadratic form T^{1} → T^{2} describes the equations of the versal base space up to second order.

In the case of Stanley-Reisner rings A_{X}, we only managed to get a nasty description of this product using the language of Proposition 4. We have not yet found a relation to the geometry of the complex. However, since the cup product provides important information needed in some applications in [AC02], we have decided to present it in an appendix. It is suggested that overly-sensitive readers quit reading at this point.

The cup product can be defined in the following way (see [Lau79, 5.1.5]): Let A = P/I with I generated by equations f^{p}. If φ ∈ Hom(I, A), lift the images of the f^{p} obtaining elements φ̃(f^{p}) ∈ P. Given a relation r ∈ R, the linear combination ⟨r, φ̃⟩ := ∑_{p} r_{p} φ̃(f^{p}) vanishes in A, i.e. it is contained in I. If φ, ψ ∈ Hom(I, A) represent two elements of T^{1}, then we define for each relation r ∈ R

(φ ∪ ψ)(r) := ψ(⟨r, φ̃⟩) + φ(⟨r, ψ̃⟩) .  (6.1)

This determines a well defined element of T^{2}. If A = A_{X} is a Stanley-Reisner ring, then the cup product respects the multigrading, and, using Proposition 4, we can give a formula for ∪ : T_{a^{1}−b^{1}}^{1} × T_{a^{2}−b^{2}}^{1} → T_{a−b}^{2} with a − b = a^{1} − b^{1} + a^{2} − b^{2}.

Proposition 19. ∪ : T_{a^{1}−b^{1}}^{1} × T_{a^{2}−b^{2}}^{1} → T_{a−b}^{2} is determined by the following:

(i) If b^{1} ∩ b^{2} ≠ ∅, then ∪ = 0. (ii) If b^{1} ∩b^{2} = ∅, then b = (b^{1} \a^{2})∪(b^{2} \a^{1}) and a = (a^{1} \b^{2})∪(a^{2} \b^{1})∪(a_{≥2}^{1} ∪a_{≥2}^{2}) with a≥2^{•} := {v ∈ [n] | av^{•} ≥ 2} denoting the locus of higher multiplicities.

(iii) If (f, g) ∈ N_{a−b}^{(2)}, choose maximal subsets d, e ⊆ b such that f ∪ (b \ d) and g ∪ (b \ e) ∉ X. If φ ∈ T_{a^{1}−b^{1}}^{1} and ψ ∈ T_{a^{2}−b^{2}}^{1}, then the value of (φ ∪ ψ)(f, g) is

[φ([f \ b^{1}] ∪ [b^{2} \ d]) − φ([g \ b^{1}] ∪ [b^{2} \ e])] · ψ(f ∪ g ∪ a^{2} \ b^{2})

+ [ψ([f \ b^{2}] ∪ [b^{1} \ d]) − ψ([g \ b^{2}] ∪ [b^{1} \ e])] · φ(f ∪ g ∪ a^{1} \ b^{1})

with φ, ψ defined to be zero on non-elements of X.

In fact, the maximality of d and e is not quite necessary. The point is to choose them non-empty whenever possible.

Proof. (i) We know that b ∈ {0, 1}^{n} if T_{a−b}^{2} ≠ 0. This would not be the case if b^{1} ∩ b^{2} ≠ ∅. Statement (ii) is a straightforward calculation. For (iii), we must first recall what we did in Step 3 of the proof of Proposition 4. Antisymmetric functions μ(•, •) on the M level had been turned into antisymmetric functions λ(•, •) on the N level via λ(f, g) := μ(m_{f}, m_{g}) with certain elements m_{•} ∈ Φ^{−1}(•). Hence, setting p := m_{f} = f ∪ (b \ d) and q := m_{g} = g ∪ (b \ e), we may compute the value of (φ ∪ ψ)(f, g) by applying the expression 6.1 on the relation R_{p,q} described in the beginning of Section 2. Using φ̃(x^{p}) = φ(Φ_{a^{1}−b^{1}}(p)) · x^{p+a^{1}−b^{1}}, we obtain

⟨R_{p,q}, φ̃⟩ = [φ(Φ_{a^{1}−b^{1}}(p)) − φ(Φ_{a^{1}−b^{1}}(q))] · x^{(p∪q)+a^{1}−b^{1}}.

Plugging this into 6.1, we get

(φ ∪ ψ)(f, g) = [φ(Φ_{a^{1}−b^{1}}(p)) − φ(Φ_{a^{1}−b^{1}}(q))] · ψ(Φ_{a^{2}−b^{2}}[Φ_{a^{1}−b^{1}}(p ∪ q)])
+ [ψ(Φ_{a^{2}−b^{2}}(p)) − ψ(Φ_{a^{2}−b^{2}}(q))] · φ(Φ_{a^{1}−b^{1}}[Φ_{a^{2}−b^{2}}(p ∪ q)]) .

To finish the proof, it is still necessary to take a closer look at the occurring arguments, i.e., to calculate

Φ_{a^{1}−b^{1}}(p) = Φ_{a^{1}−b^{1}}(f ∪ (b \ d))
= [f ∪ (b^{1} \ a^{2} \ d) ∪ (b^{2} \ a^{1} \ d) ∪ a^{1}] \ b^{1}
= [f ∪ (b^{2} \ d)] \ b^{1} (since (a^{1} \ b^{2}) ⊆ f and a^{1} ∩ d = ∅)
= (f \ b^{1}) ∪ (b^{2} \ d)

and

Φ_{a^{2}−b^{2}}[Φ_{a^{1}−b^{1}}(p ∪ q)] = Φ_{a^{2}−b^{2}}[Φ_{a^{1}−b^{1}}((f ∪ g) ∪ (b \ (d ∩ e))]
= Φ_{a^{2}−b^{2}}[((f ∪ g) \ b^{1}) ∪ (b^{2} \ (d ∩ e))]
= [((f ∪ g ∪ a^{2}) \ (b^{1} \ a^{2})] \ b^{2}
= [f ∪ g ∪ a^{2}] \ b^{2} (since (b^{1} \ a^{2}) ∩ (f ∪ g) = ∅) .

One can check that all the arguments of φ and ψ are in N_{a^{i}−b^{i}} if they are in X. □

Example 17 (continued). We are going to calculate the cup product mentioned at the end of Example 17. With f_{1} := {v − 2}, f_{2} := {v} we choose representatives from both connected components of Ñ_{∅−{v−1,v+2}}. Since N_{∅−{v−1,v+2}} = Ñ_{∅−{v−1,v+2}} ∪ {∅}, we obtain that λ(f_{1}, ∅) and λ(f_{2}, ∅) suffice to know about a function λ : N_{∅−{v−1,v+2}} → ℂ from Proposition 4. Since T^{2} results from dividing out a subspace, we obtain [λ(f_{1}, ∅) − λ(f_{2}, ∅)] as the ultimate coordinate of T_{∅−{v−1,v+2}}^{2}(E_{n}). As auxillary elements d, e ⊆ b (cf. Proposition 19), we may choose d := {v − 1} for both f_{1} and f_{2} and e := ∅ for the second argument g := ∅. On the other hand, we know that N_{{v}−{v−1,v+1}} = {v} and N_{{v+1}−{v,v+2}} = {v + 1} with Ñ = ∅ in both cases. Hence, the T^{1} spaces are represented by maps φ and ψ yielding 1 on the faces {v} and {v + 1}, respectively. Applying Proposition 19, we find

\begin{aligned}(φ ∪ ψ)(f_{1}, ∅) & = [φ(\{v − 2, v, v + 2\}) − φ(\{v, v + 2\})] · ψ(\{v − 2, v + 2\}) \\ & + [ψ(\{v − 2, v + 2\}) − ψ(\{v − 1, v + 1\})] · φ(\{v − 2, v\}) = 0 \end{aligned}

and

\begin{aligned}(φ ∪ ψ)(f_{2}, ∅) & = [φ(\{v, v + 2\}) − φ(\{v, v + 2\})] · ψ(\{v + 1\}) \\ & + [ψ(\{v + 1\}) − ψ(\{v − 1, v + 1\})] · φ(\{v\}) = 1 . \end{aligned}

Thus, the cup product mentioned at the end of Example 17 yields φ ∪ ψ = 1.

Corollary 20. Let n ≥ 7. If t_{1}, …, t_{n} denote the coordinates of T_{−1}^{1}(E_{n}) = ⊕_{v∈ℤ/nℤ} T_{{v}−{v−1,v+1}}^{1}(E_{n}), then the equations of the negative part of the base space S of the versal deformation of E_{n} are t_{v}t_{v+1} = 0 for v ∈ ℤ/nℤ. In particular, E_{n} is not smoothable over S.

Proof. Via the cup product, we see that each part T_{∅−{v−1,v+2}}^{2}(E_{n}) is responsible for t_{v}t_{v+1} in the quadratic part of the obstruction equations. Moreover, since T^{2} is concentrated in degree −2, no higher order obstructions involving only degree −1 deformations can appear, i.e., S is described by the desired equations. Thus, in any flat deformation of degree −1, every other parameter must vanish. One directly checks that this means that any fiber is singular, in fact reducible. □

In contrast, if n = 6, then each of the three T^{2}-pieces is the common target of two different cup products. In particular, the negative part of the base space S is given by the equations t_{0}t_{1} − t_{3}t_{4} = t_{1}t_{2} − t_{4}t_{5} = t_{2}t_{3} − t_{5}t_{0} = 0. This yields the cone over the three-dimensional, smooth, projective toric variety induced by the prism over the standard triangle.

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Klaus Altmann FB Mathematik und Informatik, WE2 Freie Universität Berlin Arnimallee 3 D-14195 Berlin, Germany email: altmann@math.fu-berlin.de Jan Arthur Christophersen Department of Mathematics University of Oslo at Blindern Oslo, Norway email: christop@math.uio.no