Stanley-Reisner ring of a simplicial complex is a functor?

Question

Let $K$ bea field and $[n]=\{1,\ldots,n\}$ and $K[x]=K[x_1,\ldots,x_n]$. For $\sigma=\{i_1,\ldots,i_k\}\subseteq [n]$, denote $x_\sigma=x_{i_1}\cdots x_{i_k}=\prod_{i\in\sigma}x_i\in K[x]$. Let $\Delta$ and $\Delta'$ be (abstract finite) simplicial complexes on $[n]$ and $[n']$ respectively. Let $f: \Delta\rightarrow \Delta'$ be a simplicial map, i.e. a map $f:[n]\rightarrow[n']$ for which $\forall \sigma\in\Delta: f(\sigma)\in\Delta'$ holds.

The Stanley-Reisner ideal of $\Delta$ is $I_\Delta:=\langle\langle x_\sigma;\; [n]\supseteq\sigma\notin\Delta\rangle\rangle$, and Stanley-Reisner ring of $\Delta$ is $$K[\Delta]:=K[x]/I_\Delta= K[x_1,\ldots,x_n \:|\: x_\sigma;\; [n]\supseteq\sigma\notin\Delta].$$

The coordinate ring of an affine variety $A\subseteq\mathbb{A}^n_K$ is $K[A]=K[x]/I_A$, where $I_A$ is the vanishing ideal $\{f\in K[x]; f|_A=0\}$. Thus the notation is the same as that of the Stanley-Reisner ring. Since the coordinate ring $K[-]$ is a functor from the category of affine varieties over an algebraically closed field to the category of finitely generated commutative unital reduced $K$-algebras, defined for $f: A\rightarrow A'$ by $K[f]: K[A']\rightarrow K[A]$ that sends $\alpha:A'\rightarrow K$ to $\alpha\circ f: A\rightarrow K$. This is in fact an antiequivalence of categories.

The immediate impulse is to try to make the Stanley-Reisner ring an antiequivalence of categories. But I'm having trouble making it even a functor. I didn't find anything in the literature (Bruns & Herzog, Stanley, Herzog & Hibi, Miller & Sturmfels) regarding functors.

1st try: We let $K[f]: K[\Delta]\rightarrow K[\Delta']$ send $x_i\mapsto x_{f(i)}$. But then for $\sigma\notin\Delta$, this map sends $x_\sigma\mapsto x_{f(\sigma)}$ (here $f(\sigma)$ is considered as a multiset, i.e. if $f(i)=f(j)$, then $x_{f(\sigma)}$ contains both $x_{f(i)}$ and $x_{f(j)}$), and we do not necessarily have $f(\sigma)\notin\Delta'$, meaning that $K[f]$ is not well defined on the quotient of $K[x_1,\ldots,x_n]$.

2nd try: We let $K[f]: K[\Delta']\rightarrow K[\Delta]$ send $x_{i'}\mapsto x_{f^{-1}(i')}$, where $f^{-1}$ denotes the preimage. Then for $\sigma'\notin\Delta'$, this map sends $x_{\sigma'}\mapsto x_{f^{-1}(\sigma')}$. We wish to have $f^{-1}(\sigma')\notin\Delta$. If this does not hold, $f^{-1}(\sigma')\in\Delta$, then $\sigma'\supseteq f(f^{-1}(\sigma'))\in\Delta'$, so I don't get any contradiction, and we don't have well-definedness.

Question: How can the Stanle-Reisner ring be made into a functor (preferrably in a way that it becomes an antiequivalence between the category of simplicial complexes and category of finitely generated commutative unital monomially related reduced $K$-algebras)?

Question: Can the Stanley-Reisner ideal be seen as an ideal of polynomial functions $\alpha: \Delta\rightarrow K$?

Answer 1

I think I'd want to deal with partially defined functions $f: [n] \to [n']$, with the property that if $F$ is a face of $\Delta$, then $f(F)$ is a face of $\Delta'$. The linear extension of such an $f$ is a linear map from $K^n \to K^{n'}$, taking the $i$th basis vector to the $f(i)$th, or to zero if $f(i)$ is not defined.

The Stanley-Reisner ideal is the functions vanishing on the union $X_\Delta \subseteq K^n$ of coordinate spaces $\bigcup_{F\in\Delta} K^F$, where $K^F$ denotes the linear span of {the $i$th basis vector : $i\in F$}. (Does that answer your second question?) Then the linear extension of $f$ takes $X_\Delta$ into $X_{\Delta'}$.

Answer 2

The answer to the first question is Proposition 3.1.5 on page 95 of the upcoming Toric Topology book by Buchstaber and Panov, which I recommend.

The functor $K[-]$ from the category of simplicial complexes to the category of finitely generated commutative unital monomially related reduced $K$-algebras is not an equivalence of categories: if $\Delta=\{0\}$ is the one-point simplicial complex, then there is only one simplicial map $\Delta\rightarrow\Delta$, but there are many homomorphisms $K[\Delta]=K[x]\rightarrow K[\Delta]=K[x]$, namely for any polynomial $f(x)$ there is a homomorphism of K-algebras $x\mapsto f(x)$. I suspect this functor is an anti-equivalence from the category of simplicial complexes to ... graded algebras.

Answer 3

The abstract simplicial complex $\Delta$ has a realization as a subspace $\lvert \Delta\rvert \subset \mathbb R^{n+1}$, where the vertex $j$ corresponds to the point $e_j=(0,…,0,1,0,…,0)$ with the 1 in j-th position, and the realization is the union (over all simplices) of the convex hull of the vertices of the simplex.

Now consider the projective closure of this subspace, that is, we have the standard inclusion $\mathbb R^{n+1}\to \mathbb RP^{n+1}$, $(x_0,…x_n)\mapsto [x_0:\dots :x_n:t=1]$, and we consider the closure of $\lvert \Delta\rvert$ in $\mathbb RP^{n+1}$ in the Zariski topology.

The homogeneous coordinate ring of this projective closure is exactly the Stanley-Reisner ring of $\Delta$.

Note that in Allen Knutson's space $X_\Delta$ one has an affine $k+1$-space for each $k$-simplex, whereas in the projective case the dimensions are right.

Note that there is one equation $x_0+…+x_n=1$ which holds in all of $\lvert \Delta \rvert$. Upon projectivizing, this becomes $x_0+…+x_n=t$, and $$\mathbb R[x_0,…,x_n,t]/(t-x_0-…-x_n)\cong R[x_0,…,x_n],$$ which explains why the additional variable $t$ is not present in the description of $\mathbb R[\Delta]$.

(Since the variety is already defined over $\mathbb Z$, it now makes sense to generalize to $K[\Delta]$ for any $K$.)

So I think this is the correct algebro-geometric description, and this gives a sort of answer to both questions.