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

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 $$j$$.)

• 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? – Johannes Hahn Dec 2 '19 at 19:31
• @JohannesHahn: Thanks, added. – MKO Dec 2 '19 at 19:39
• $A$ 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. – Zach Teitler Dec 2 '19 at 19:50

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 $$A_2$$.