3
$\begingroup$

Maybe this question is not suitable for this platform, I already put that same question in math.stackexchange and I find only vague answers.

I'm studying the book of Rick Miranda; Algebraic Curves and Riemann Surfaces. I'm studying about degree of projective curves and I find a term used very often and that is very important by the amount of times it appears:

"general hyperplane $H$ in $\mathbb{P}^n$"

I need to understand what this means, but in the context of the subject. I searched all over Rick Miranda's book, but I could not find it.

I read Geometry of Algebraic Curves-Volume I, I often encounter the placement "Let $H$ be a general hyperplane in $\mathbb{P}^n$.

Important facts follow from this fact, as for example the calculus of the degree of the map of Gauss constructed from the divisor theta, for the demonstration of the Theorem of Torelli.

So it may seem trivial, but it's quite important the meaning of:

"general hyperplane $H$ in $\mathbb{P}^n$".

It turns out that I can not find it in any book. Everyone just uses it, but I need to know. So I am also grateful for references.

Thank you!

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Algebraic geometers use "general" the way other people use "random". You have to understand it in context. $\endgroup$
    – Donu Arapura
    Dec 29, 2017 at 21:30
  • 2
    $\begingroup$ For an introductory textbook reference, see Harris "Algebraic geometry: a first course", Lecture 5, "General Objects" starting on p.53. There isn't much said there that hasn't already been explained in the answers and comments though. $\endgroup$
    – j.c.
    Dec 30, 2017 at 16:50

3 Answers 3

Reset to default
8
$\begingroup$

The sentence:

Let $H$ be a general hyperplane. Then do so and so...

Means:

Pick a hyperplane $H$. Then do so and so, while keeping in mind that the "so and so" might sometimes not work out/be false/be impossible to do.

The use of the word general means that the set of hyperplanes $H$ for which "so and so" works/is true/is possible to do is Zariski-open [edit: and non-empty!] (or contains a non-empty Zariski-open subset) inside the set of all hyperplanes.

Improve this answer
$\endgroup$
4
  • $\begingroup$ So from what you just said, I could interpret the statement: "The general hyperplane $H$ in $\mathbb{P}^n$ is such that its divisor $div(H)$ consists of $d$ distinct points $\{pi\}$, each having $div(H)(pi) = 1$. in other words, " the set of hyperplanes $H$ for which its divisor $div(H)$ consists of $d$ distinct points $\{pi\}$, each having $div(H)(pi) = 1$ is Zariski-open (or contains a Zariski-open subset) inside the set of all hyperplanes." Is correct? Is that what you meant? $\endgroup$
    – Manoel
    Dec 29, 2017 at 21:55
  • 2
    $\begingroup$ Yes. You understand things correctly. If you pick your hyperplane poorly, then the $d$ points might fail to all be distinct. But the set of hyperplanes for which these points are indeed distinct is Zariski open. $\endgroup$
    – André Henriques
    Dec 29, 2017 at 22:06
  • $\begingroup$ So it seems to me that it is a strong hypothesis that this question of $H$ be a general hyperplane in $\mathbb{P}^n$. For example, "observe that for a general plane $H$, there will be $2g - 2$ points at which $H$ intersects $C$,and any subset of size $g - 1$ among these $2g - 2$ points will be independent." Thus, the subset of $(\mathbb{P}^{g-1})^*$ for which this occurs is dense in $(\mathbb{P}^{g-1})^*$. $\endgroup$
    – Manoel
    Dec 29, 2017 at 22:20
  • 2
    $\begingroup$ That Zariski-open locus in the projective space of hyperplanes is required to be non-empty! There are many results in algebraic geometry on the openness or constructibility of the loci where a condition holds (Chevalley's theorem on constructible images is an important input in many such arguments); that is what makes the condition often checkable (with non-emptiness typically requiring a special construction, or dimension reasoning). This notion of "general" is useful because non-empty open sets in the Zariski topology of an irreducible variety are so huuuge (any two overlap in another!). $\endgroup$
    – nfdc23
    Dec 30, 2017 at 0:18
6
$\begingroup$

I've encountered the "general hyperplane" terminology only in the context of assertions that some statement $\phi(H)$ is true for a general hyperplane. Recall that the hyperplanes in a projective space are parametrized by points in the dual projective space. From this point of view, "$\phi(H)$ holds for a general hyperplane" means that the set of hyperplanes for which $\phi(H)$ is not true is covered by a variety (possibly reducible) of lower dimension in the dual space.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi? Who is $\phi (H) $? $\endgroup$
    – Manoel
    Dec 29, 2017 at 20:56
  • 2
    $\begingroup$ As Andreas Blass says, $\phi(H)$ is a statement depending on the hyperplane $H$. $\endgroup$
    – Alex Suciu
    Dec 29, 2017 at 21:11
4
$\begingroup$

This is really a long remark.
The concept of a general hyperplane is used here to prove that the degree of a curve is well defined. I.e. that there exists a non empty Zariski open set of hyperplanes whose intersection with the given curve consists of d distinct points of multiplicity one. But if one wants to actually compute the degree, this definition is insufficient. I.e. one can calculate only with a specific hyperplane, but the answer is correct only if it is also “general”. The knowledge that most hyperplanes are general is of little help. This is why one usually proves also that a hyperplane all of whose intersections have multiplicity one, is indeed general. This gives a computable criterion for generality.

E.g. when Mumford computes the number of lines on a smooth cubic surface in his little yellow book, he first shows that most cubic surfaces have the same finite number of lines, and then proves that smoothness is a sufficient condition for a surface to have this number of lines. Then he checks that the Fermat surface is smooth and computes on that one.

Sometimes it is so difficult to find a specific example that is general, that one is forced to compute with a non general one. In his book on algebraic curves, Fulton gives a formula for the local multiplicity of the intersection of two curves at a point, and shows that, at least theoretically, one can compute the Bezout number even if the curves do not intersect everywhere with multiplicity one.

B. Segre gives in his book on cubic surfaces a lovely example where he computes the number of lines on a general cubic surface by using as example the highly non general case of the union of 3 general planes! Here a small deformation, to a general cubic, is defined by another cubic which meets each of the three lines of intersections of a pair of the original planes in 3 points. He then computes that a line lying in one of the 3 planes is the specialization of a line on a general deformation surface, if and only if it meets both of the other two planes at one of the distinguished points. Thus there are 9 such lines in each of the three planes, for a total of 27 lines on a general cubic surface.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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