Today we think about plane curves. But first a bit on projective space. Recall that the Grassmannian is the parameter space of k-planes in n-dimensional space, and so \mathbb{P}^n=Gr(1,n+1). We noted at the time that Gr(k,n)\cong Gr(n-k,n). So that means that we also get a projective space Gr(n,n+1). So for \mathbb{P}^n, we have a dual projective space, denoted (\mathbb{P}^n)^* consisting of all the hyperplanes in \mathbb{P}^n.

In coordinates for n=2, take a point (y_0,y_1,y_2)\in (\mathbb{P}^2)^*. This gives the line y_0x_0+y_1x_1+y_2x_2=0 in \mathbb{P}^2, whose coordinates are given by the x_i. Now this lets us define the dual of a curve by saying that C^*=\{L\in (\mathbb{P}^2)^*\mbox{ with }L\mbox{ tangent to }C\mbox{ at some point}\}. Now, we’re going to show that this is in fact an algebraic curve, that if C was irreducible, so will C^* be. It is also true that C^{**}=C, but that we leave as an exercise. In fact, it’s true that \deg C^*\geq 2. But we will make the assumption that C has no components which are lines, so if we assume irreducibility, that just means that \deg C\geq 2 in the first place.

Let F be the equation of C and let L=V(y_0X_0+y_1X_1+y_2X_2) be an arbitrary line. On the open set in (\mathbb{P}^2)^* where y_2\neq 0, we can use this to eliminate the variable X_2 from F, and get G(X_0,X_1)=y_2^n F(X_0,X_1,-\frac{1}{y_2}(y_0X_0+y_1X_1))= b_0X_1^n+b_1X_1^{n-1}X_0+\ldots+b_nX_0^n with the b_i homogeneous polynomials of degree n in the y_i. Zeros of this polynomial are the intersections of C with the line L. Let D be the discriminant of g(X_1)=G(1,X_1). It will be homogeneous of degree 2n^2-n in the variables (y_0,y_1,y_2). Now, the discriminant is nonzero, and so we have C'=V(D)\subset (\mathbb{P}^2)^*. This isn’t quite the dual curve yet, but it’s getting close.

So now we’re going to kill linear factors of D. The only reason we might not have C^* turns out to be that there are extra lines. So let’s look at what sorts of lines we might get before we actually check this. Look at point in C\cap V(X_0). We can assume, for simplicity, that (0:0:1)\notin C and so X_1 is nonzero at these points, which are of the form x=(0:x_1:x_2). Let y be a line through x, it will have the form y=(y_0:-x_2:x_1). Now b_0(y)x_1^n=G(0,x_1)=0, so we have b_0(y)=0. Thus, x_1Y_1+x_2Y_2 divides b_0, and so does all of D.

The other type of linear factor that can arise is from singularities. Assume x=(x_0:x_1:x_2) is a singular point, and we can assume x_0\neq 0 and so is 1, because we already handled all points with x_0=0. Now we let y be a line through x and see that G(1,X_1) has a multiple zero at X_1=x_1, so the discriminant must vanish. So the line V(x_0Y_0+x_1Y_1+x_2Y_2)\subset C' is another irreducible component. Now there’s only finitely many lines of either type.

So now look at C' minus all of these lines. We claim that C^* is the closure of this set. Now, this set is the same as the set of all tangent lines to smooth points of C. That its closure is C^* follows from the fact that a line through a point is a tangent line if and only if there is a sequence of points converging to the point whose tangent lines converge to the line. So, \overline{C'\setminus(L_1\cup\ldots\cup L_k)}=C^*, which is thus algebraic.

To check irreducibility, we notice that for the set of smooth points of C, we have an injective map. In fact, this map has image C'\setminus (L_1\cup\ldots\cup L_k), and has closure C^*. Thus, as the original curve is irreducible, the image of an open subset is, and so is its closure, so C^* is irreducible.

Now, the problem with examples is that you get high degree polynomials most of the time. However, we can work out x_2^2y_x-x_1^3, the cuspidal cubic, explicitly. We jump right to the polynomial g(1,x) which we take the discriminant of, and it gives us y_0^2y_2+2y_0y_1y_2 x+y_1^2y_2 x^2-y_2^3x^3. It has discriminant -y_0^3 y_2^9(27y_0y_2^2+4y_1^3). The part we want, 27y_0y_2^2-3y_1^3 just happens to be another cuspidal cubic. Now, if you try a general elliptic curve, you’ll get a higher degree dual curve, so you don’t normally get a curve like what you started with back.

Now, the dual curve does strongly reflect the geometry of the original curve. For instance, we define a multiple tangent to be a line which is tangent to C at more than one point. These will correspond to singularities of the dual curve, which are ordinary r-fold points if r is the number of points of tangency. By this we mean that the point on the dual curve will have r distinct tangent directions and around that point the equation of the dual curve can be written to have smallest term of degree r. Now, as the dual is algebraic, it has finitely many singularities, so only finitely many lines can be multiple tangents to C.

So the point is, that the dual curve is useful for working out the geometry of your curve, and that you can explicitly compute what it is, so long as you have powerful enough computational tools (generally you’ll get bored or make a mistake trying to do these computations by hand…or at least I do).

An aside: I believe that this procedure can be generalized as follows: let V\subset\mathbb{P}^n be a variety. Then there is a dual variety V^* sitting in Gr(n+1-k,n+1) which consists of the k-planes tangent to V, though I’ve never seen mention of this anywhere, and it’s not obvious how to get the equations for it on the Grassmannian. Has anyone encountered this?

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