In the following, all varieties will be algebraic over $\mathbb{C}$.

I have some general problems with concepts like the "space of lines in $\mathbb{P}^5$", "space of lines on a surface in some projective space" and dual projective spaces. Can anyone provide a good reference on this? (All i know about it is what i found out myself by assuming the definitions to be what seems logical to me....)

In particular, i wonder about the following:

Take a cubic hypersurface $V \subset \mathbb{P}^5$, containing a line $l$. The set of lines in $V$ that intersect with $l$ is supposed to be a smooth algebraic surface. Can anyone tell me why?

For a surface $X \subset \mathbb{P}^3$, what is the definition of the "dual surface"?

Thanks a lot!

  • $\begingroup$ More generally, how to compute dimension of a space of $k$-planes in some quadric hypersurface (or any hypersurface) $Q \subset \mathbb{P}^n$ and whether it is irreducible or not? $\endgroup$
    – Joachim
    Oct 30, 2012 at 11:45

2 Answers 2


Take your cubic hypersurface $f(x,y,z) = 0$ and look for solutions linear in a parameter $$f(x(t),y(t),z(t))=0 \quad\text{ where }\quad x(t)=a t + b,\;y(t)=ct + d,\;z(t) = et+f$$ Then it becomes a matter of solving for $a,b,c,d,e,f$ giving you a "space of lines" in $V$.

If in addition you need them to intersect another line $l = \{ m x + ny + p z = c \}$, this is just another equation in your variety.

The result is a system of equations: \begin{eqnarray} f( a t + b,\;ct + d,\; et+f) &=& 0 \\ m( a t + b)+n(ct + d)+p( et+f )&=& c\\ \end{eqnarray} where we are solving for $a,b,c,d,e,f$.

The term "dual surface" could mean a number of things. Wikipedia has it as the variety of lines in space tangent to the surface. It really depends on where you look.

Have you looked at Joe Harris' algebraic geometry text? Or even Shafarevich? Or Miles Reid's Chapters of Algebraic Surfaces.

  • 1
    $\begingroup$ Hey @john, thanks for your answer. Sorry it took a while for me to reply. Normally i only work projectively but i guess it just carries over. I'd like to ask you some further questions. Consider $G$, the grassmanian of lines in $\mathbb{P}^3$, alternatively the planes in $\mathbb{A}^4$. It is exactly the decomposable elements $v_1 \wedge v_2$ in $\mathbb{P}\bigg(\bigwedge \mathbb{A}^4\bigg) \cong \mathbb{P}^5$. As a variety it is $X_0X_1 - X_2X_3 + X_4X_5 = 0$. $\endgroup$
    – Joachim
    Nov 6, 2012 at 12:07
  • $\begingroup$ Now my question, which is actually my main question, how do i carry over these equations you get in $a,b, \ldots, f$ to equations in this Grassmannian? It might be really straightforward but i do not see it yet. $\endgroup$
    – Joachim
    Nov 6, 2012 at 12:10
  • $\begingroup$ Sorry for not accepting yet but i'm not quite there yet. Hope you can help me out. I will check out the references later, hopefully tonight, i have an assigmnent to do now.. =D $\endgroup$
    – Joachim
    Nov 6, 2012 at 12:16

More generally, how to compute dimension of a space of k-planes in some quadric hypersurface (or any hypersurface) Q⊂ℙn and whether it is irreducible or not?

Let $X \subset \mathbb{P}^{n}_{k}$ be a (non-singular) projective variety over a closed field $k$.

Then $X = V(J)$ is the zero locus of some subset of polynomials $ J \subset k[X_{1}, \dots, X_{n}]$.

We want to study the relationship between the geometric properties of the surface $ X = V(S)$ and the algebraic properties of the commutative polynomial ring $R = k[X_{1}, \dots, X_{n}]$.

So $$V(J) = \{(a_{1}, \dots, a_{n}) \, \colon \,\forall f_{i} \in J \, , \, f_{i}(a_{1}, \dots, a_{n}) = 0\}.$$

This is a set-theoretic way of describing the thing, but with a bit of commutative algebra -- life becomes much easier.

The set of polynomials $J \subset R$ giving structure to our variety is, of course, an ideal in the of polynomials over $k$, which is (by definition) the subring containing all finite sums of the form $$\sum \, f_{i} \, g_{i} \, \, \textrm{for } \,f_{i} \in J \, , \, g_{i} \in R.$$

If the ideal can be written as the union of two non-trivial subsets $J = J_{1} \cup J_{2}$, then it is reducible -- in which case, the variety is said to be reducible. Otherwise, they are both irreducible.

This vocabulary corresponds to the topological notion of reducibility, which we make sense of by constructing the Zariski Topology.

Proposition: An algebraic variety $X$ is irreducible if and only if $I(X)$ is a prime ideal in the polynomial ring $R$.

Proving this will make it clear how to start answering your question about how to determine if a given variety is irreducible or not.

Corollary: Let $X = V(J)$, where $J = (f)$ for some irreducible polynomial $f$. Then $J = I(X)$ is a prime ideal in the polynomial ring $R$, and hence $X$ is an irreducible algebraic variety.

I recommend the following books:

(1) Undergraduate Algebraic Geometry [Miles Reid]

(2) Ideals, Varieties and Algorithms [Cox, Little, O'Shea]

(3) Computations in Algebraic Geometry with Macualay2 [Eisenbud, Grayson, Stillman]



It may also be instructive to use Macaulay2 for some actual computations of dimension, codimension and prime decomposition of ideals/varieties. This is a good way to "see for yourself" the contravariance of the functor between $V(J)$ (algebraic subsets) and $I(X)$ (ideals). You can use the decompose() function to check the decomposition of an ideal, and dim()/codim() to check dimension/codimension of some given space.

  • 3
    $\begingroup$ Dear Charles, How does this answer the OP's actual question, which is about determining the dimension/irreducibility of the space of $k$-planes in a given quadric hypersurface? Regards, $\endgroup$
    – Matt E
    Nov 4, 2012 at 2:31
  • $\begingroup$ He asked in a comment in his original question about reducibility and dimension of hypersurfaces, my response is intended to help the OP work towards an answer. $\endgroup$ Nov 4, 2012 at 3:11
  • 3
    $\begingroup$ Dear Charles, He asked about reducibility and dimension of spaces of $k$-planes contained in a hypersurface, which is quite a bit harder to work out then dimension and irreducibility of the hypersurface itself. Regards, $\endgroup$
    – Matt E
    Nov 4, 2012 at 3:15
  • $\begingroup$ Dear Charles, thanks for your response. However, it answers quite a different question and is indeed not the answer that i was looking for. $\endgroup$
    – Joachim
    Nov 4, 2012 at 20:49
  • 1
    $\begingroup$ Dear Charles, i just checked out your references and i'm actually quite happy to learn about $(2)$ and $(3)$, especially the Macaulay2.. Thanks for that!! $\endgroup$
    – Joachim
    Nov 6, 2012 at 12:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .