0

I always have a problem with the word generic in the literature of mathematics. Let me ask you a specific question about "non-degenerate $\mathbb{Z}$-graded lie algebras''. The definition I'm working with says:

A $\mathbb{Z}$-graded lie algebra (over $\mathbb{C}$) $\mathfrak{g}$, i.e. $\mathfrak{g}=\bigoplus_{n\in \mathbb{Z}}\mathfrak{g}_n$, is non-degenerate if the following are satisfied:

  • $\mathfrak{g}_n$ are finite dimensional for all $n\in \mathbb{Z}$.
  • $\mathfrak{g}_0$ is abelian.
  • For any $n\in \mathbb{Z}_{>0}$, and generic $\lambda\in \mathfrak{g}_0^*$, the pairing $\mathfrak{g}_n\times \mathfrak{g}_{-n}\to \mathbb{C}$ given by $(x,y)\mapsto \lambda([xy])$ is non-degenerate.

Everything in this definition is perfectly clear to me except this word "generic". What is meant by a generic dual vector? I would ask what is meant by generic in general, but probably that's different according to the context.

Hamed
  • 6,793
  • Generic means "for essentially all". In you case, it probably means for all lambdas in an nonempty zariski open set. – Mariano Suárez-Álvarez Feb 07 '16 at 01:58
  • The set of $\lambda$ such that the pairing you mention is degenerate is a Zariski-closed subset of the complex vector space $\mathfrak{g}_0^$. So it is either equal to $\mathfrak{g}_0^$, or is small in many possible senses (closed subset of empty interior, subset with zero measure, etc.) So the author could have just said, instead "there exists $\lambda$ such that the form ... is non-degenerate". – YCor Feb 07 '16 at 22:31

2 Answers2

4

As you say, it depends on context. Vaguely, it means something like "every element except for the elements in a 'small' subset," where the meaning of 'small' depends on context. In this context it might mean either

  • every element except for the elements in a finite (or maybe countably infinite) union of affine subspaces, or
  • every element except for the elements in a finite (or maybe countably infinite) union of Zariski closed subvarieties.

I don't know which the author intends.

Qiaochu Yuan
  • 419,620
  • I think it is the the second one. But how do you define a Zariski topology on $\mathfrak{g}_0^$? Do you treat $\mathfrak{g}_0^$ as a commutative ring (with value-wise multiplication and addition) and then put the Zariski topology on the spectrum of that ring? – Hamed Feb 07 '16 at 04:12
  • Also you said every element except for elements in a countable union of Zariski closed varieties. But which ones? How do you determine that in general? – Hamed Feb 07 '16 at 04:13
  • @Hamed: any finite-dimensional vector space $V$ can be regarded as an affine space in the sense of algebraic geometry. The closed subsets in the topology are zero sets of polynomials, meaning elements of the symmetric algebra $S(V^{\ast})$. The condition is just that some such set exists; it doesn't specify what it is (but if you work through an example you should be able to write down what the set ends up being explicitly; probably it is the vanishing locus of a determinant or a set of minors or something like that). – Qiaochu Yuan Feb 07 '16 at 04:15
0

Here is the heuristic I use: roughly, there are two notions of "generic" in mathematics, each dual to the other. If you're working in some structure $M$ with a distinguished collection of substructures $N_i \hookrightarrow M$, one can either say that an $m \in M$ is generic if it lies outside of all the $N_i$, or one can say that an $m \in M$ is generic if it lies in all of the $N_i$.

For example, if you're working in a topological space, one notion of a "generic point" could be one that lies in every closed subset. This corresponds to the latter notion.

For another example, take a definable set of finite Morley rank $n$ in a first-order structure. A generic point of this set is one that lies in no definable subset of rank less than $n$. This corresponds to the former notion.

mad_algebraist
  • 860
  • A "generic point" is one that lies in every nonempty open subset, at least in algebraic geometry. (And on an irreducible scheme.) I think it is pretty rare for an interesting topological space to have the property that the intersection of all non-empty closed sets is nonempty... – Elle Najt Feb 07 '16 at 03:12