4
$\begingroup$

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 $j$.)

$\endgroup$
  • 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? $\endgroup$ – Johannes Hahn Dec 2 '19 at 19:31
  • $\begingroup$ @JohannesHahn: Thanks, added. $\endgroup$ – MKO Dec 2 '19 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. $\endgroup$ – Zach Teitler Dec 2 '19 at 19:50
8
$\begingroup$

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 $A_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.