When is an algebra defined by generators and relations finite-dimensional and satisfies Poincaré duality?

EDIT: My question is not very concrete. Rather I am wondering if there is anything known in the following direction.

Assume we are given a commutative algebra $A$ (say over complex numbers) which is graded and is given by its homogeneous generators and homogeneous relations: $$A=\mathbb{C}[x_1,\dots,x_k]/\langle f_1,\dots,f_l\rangle,$$ where $f_i$ are homogenous polynomials in $x_1,\dots,x_k$.

1) Are there sufficient conditions on $f_i$'s to guarantee $A$ to be finite-dimensional?

2) Are there sufficient conditions on $f_i$'s to guarantee $A$ satisfies Poincaré duality? (Poincaré duality means that $A_j=0$ for $j>N$, $A_0=\mathbb{C}, A_N\simeq \mathbb{C}$, and the product $A_j\times A_{N-j}\to A_N\simeq \mathbb{C}$ is a perfect pairing for any $j$.)

  • 1
    $\begingroup$ One way of making this question more concrete would be to describe more explicitly what you mean by "satisfies Poincaré duality" ? Do you just want any isomorphism $A_i \cong A_{n-i}$, i.e. do you only want equality of dimensions? Or do you want something more, like the isomorphism being a (member of a) specific (class of) map(s)? What about the top-dimension $n$ ? Do you want $n$ to be a pre-determined value (like the dimension of the manifold) or just whatever the last highest non-zero degree happens to be? $\endgroup$ Dec 2, 2019 at 19:31
  • $\begingroup$ @JohannesHahn: Thanks, added. $\endgroup$
    – asv
    Dec 2, 2019 at 19:39
  • 10
    $\begingroup$ $A$ is finite dimensional if and only if the radical of the ideal $I=(f_1,\dotsc,f_l)$ is the maximal graded ideal $(x_1,\dotsc,x_k)$. Assume $A$ is finite dimensional. Then $A$ has Poincaré duality if and only if $I$ is Gorenstein, equivalently there exists a homogeneous polynomial $F$ such that the ideal of $g=g(x_1,\dotsc,x_k)$ satisfying $g(\partial/\partial x_1,\dotsc,\partial/\partial x_l)(F) = 0$, is precisely $I$. There is a book by Meyer and Smith, Poincaré duality algebras, Macaulay's dual systems, and Steenrod operations. $\endgroup$ Dec 2, 2019 at 19:50

1 Answer 1


Say $I = (f_1,\dotsc,f_l)$ is the ideal generated by the $f_i$. The $f_i$ are homogeneous; let’s add an assumption that none of the $f_i$ are constant (degree zero). The following conditions are equivalent:

  1. $A$ is finite-dimensional (as a vector space over $\mathbb{C}$).
  2. The radical of $I$ is the maximal graded ideal $(x_1,\dotsc,x_k)$.
  3. For each $i$ there’s an $n_i$ so that $x_i^{n_i} \in I$, or there’s an $n$ so that for each $i$, $x_i^n \in I$.

This list can be extended. Perhaps some of the experts can suggest additional, simpler, conditions that are sufficient but not necessary.

Now suppose that $A$ is finite-dimensional. The following conditions are equivalent:

  1. $A$ has Poincaré duality as you described it ($A_N \cong \mathbb{C}$ and for every $n$, the multiplication map $A_n \times A_{N-n} \to A_N \cong \mathbb{C}$ is a perfect pairing).
  2. The socle of $A$ is one-dimensional (as a $\mathbb{C}$ vector space).
  3. $A$ is Gorenstein; $I$ is a Gorenstein ideal.
  4. There is a homogeneous polynomial $F$ so that $I$ is the ideal of polynomials $g$ such that $g(\partial/\partial x_1,\dotsc,\partial/\partial x_k)(F) = 0$ (the ideal of annihilators of $F$).

If $I$ is a complete intersection ($l=k$ and $A$ is finite-dimensional) then $A$ is Gorenstein, has Poincaré duality, etc. This is an example of a sufficient, but not necessary, condition.

There are various sources for this, including:

It's not sufficient to only have a symmetric Hilbert function ($\dim A_n = \dim A_{N-n}$). For example $A = \mathbb{C}[x,y]/(x^3,xy,y^4)$ has Hilbert function $1,2,2,1$. But the multiplication map $A_1 \times A_2 \to A_3$ is not a perfect pairing because multiplication by $x \in A_1$ is zero on $A_2$.


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