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removed unilluminating example.

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edited May 12, 2010 at 23:49

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Here's a concrete example of the flex/cusp correspondence, and I think that if you understand this then you also have intuition for how it works in general. I'm going to think in terms of the real picture throughout. Let C be the affine plane curve $y = x^3$. To choose local coordinates for C*, note that any tangent line to C can be written as $ax + y = b$, i.e. no tangent line is vertical. Now the geometric meaning of a and b is not so hard to figure out: for a point $p \in C$, the a-value is the (negative of the) slope of the tangent line $T_pC$, and the b-value is a measure of the (signed) distance between the tangent line and the origin. (In fact the distance between the line and the origin is given by $b/\sqrt{1+a^2}$, so the distance is not an algebraic function. But near the origin we can just as well approximately take the distance to be equal to b.)

Now think about what happens as we slide a point p along C past the origin. The slope is positive, shrinks to zero which is a stationary point, and then increases again. The distance is negative, shrinks to zero, and becomes positive. If you squint a bit, you can see that the dual curve really traces out the "standard cusp" at the origin.

A purely coordinate-free argument might be to think about what happens when you take a curve with a bitangent, and then deform the curve so that the two tangent points come together -- clearly this should produce a flex. We already understand how bitangents give rise to simple nodes on the dual curve. On the side of the dual curve, this deformation would correspond to taking a nodal curve and letting the "nodal part" shrink to a single point. Again, squintSquint and you get a cusp.

(There used to be also a big computation/handwave in this answer, but on a second reading it was really not illuminating at all.)

Here's a concrete example of the flex/cusp correspondence, and I think that if you understand this then you also have intuition for how it works in general. I'm going to think in terms of the real picture throughout. Let C be the affine plane curve $y = x^3$. To choose local coordinates for C*, note that any tangent line to C can be written as $ax + y = b$, i.e. no tangent line is vertical. Now the geometric meaning of a and b is not so hard to figure out: for a point $p \in C$, the a-value is the (negative of the) slope of the tangent line $T_pC$, and the b-value is a measure of the (signed) distance between the tangent line and the origin. (In fact the distance between the line and the origin is given by $b/\sqrt{1+a^2}$, so the distance is not an algebraic function. But near the origin we can just as well approximately take the distance to be equal to b.)

Now think about what happens as we slide a point p along C past the origin. The slope is positive, shrinks to zero which is a stationary point, and then increases again. The distance is negative, shrinks to zero, and becomes positive. If you squint a bit, you can see that the dual curve really traces out the "standard cusp" at the origin.

A purely coordinate-free argument might be to think about what happens when you take a curve with a bitangent, and then deform the curve so that the two tangent points come together -- clearly this should produce a flex. We already understand how bitangents give rise to simple nodes on the dual curve. On the side of the dual curve, this deformation would correspond to taking a nodal curve and letting the "nodal part" shrink to a single point. Again, squint and you get a cusp.

A purely coordinate-free argument might be to think about what happens when you take a curve with a bitangent, and then deform the curve so that the two tangent points come together -- clearly this should produce a flex. We already understand how bitangents give rise to simple nodes on the dual curve. On the side of the dual curve, this deformation would correspond to taking a nodal curve and letting the "nodal part" shrink to a single point. Squint and you get a cusp.

(There used to be also a big computation/handwave in this answer, but on a second reading it was really not illuminating at all.)

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created May 12, 2010 at 9:35

40.2k
2
114
201

Here's a concrete example of the flex/cusp correspondence, and I think that if you understand this then you also have intuition for how it works in general. I'm going to think in terms of the real picture throughout. Let C be the affine plane curve $y = x^3$. To choose local coordinates for C*, note that any tangent line to C can be written as $ax + y = b$, i.e. no tangent line is vertical. Now the geometric meaning of a and b is not so hard to figure out: for a point $p \in C$, the a-value is the (negative of the) slope of the tangent line $T_pC$, and the b-value is a measure of the (signed) distance between the tangent line and the origin. (In fact the distance between the line and the origin is given by $b/\sqrt{1+a^2}$, so the distance is not an algebraic function. But near the origin we can just as well approximately take the distance to be equal to b.)

Now think about what happens as we slide a point p along C past the origin. The slope is positive, shrinks to zero which is a stationary point, and then increases again. The distance is negative, shrinks to zero, and becomes positive. If you squint a bit, you can see that the dual curve really traces out the "standard cusp" at the origin.

A purely coordinate-free argument might be to think about what happens when you take a curve with a bitangent, and then deform the curve so that the two tangent points come together -- clearly this should produce a flex. We already understand how bitangents give rise to simple nodes on the dual curve. On the side of the dual curve, this deformation would correspond to taking a nodal curve and letting the "nodal part" shrink to a single point. Again, squint and you get a cusp.