The images of the singular sets for two dimensional quadratic maps.

Richard P. McGehee Chia-Hsing Nien University of Minnesota

Structure of the talk

1. Definitions

2. The phenomena we observed from pisces.

3. Basic facts

4. The proof ( Case 1 - 5 )

5. Some possible generalizations

6. An application in singularity theory

1. Definitions

Definition: Let F be a differentiable two dimensional map, then the singular set, J_{0}, of F is defined by the set

J_{0} = {x ∈ ℝ^{2}| det (DF (x)) = 0}. Definition: J_{1} = F (J_{0}). ———————————————————

\begin{matrix}ℝ^{2} & ⟶^{F} & ℝ^{2} \\ h↓ & & ↓k \\ ℝ^{2} & ⟶_{G} & ℝ^{2} \end{matrix}

2. The phenomena we observed from pisces (quadmapfig1.ps)

• If J_{0} is a pair of parallel lines then J_{1} is a point union a line.

• If J_{0} is a pair of cross lines then J_{1} is a ray union a parabola and touch at the starting point of the ray or two rays with the same starting point.

• If J_{0} is an ellipse then J_{1} is a close curve with three cusps.

• If J_{0} is a hyperbola then J_{1} has two branches with only one cusp in one branch.

• If J_{0} is a parabola then J_{1} has only one cusp.

3. Basic facts

• Lemma: Let h and k be two diffeomorphims from ℝ^{2} to ℝ^{2}, and F : ℝ^{2} → ℝ^{2} a smooth two dimensional map. Define the map G : ℝ^{2} → ℝ^{2} to be kF h^{−1}. Then J_{0}^{G} = h(J_{0}^{F}) and J_{1}^{G} = k(J_{1}^{F}) where J_{i}^{ρ} denotes the set J_{i} of the map ρ.

In particular, if A is a nonsingular affine map from ℝ^{2} to ℝ^{2} and h = A then J_{0}^{G} = A(J_{0}^{F}). We can then use an affine transformation to simplify the function f = det (DF).

• The mappings F and F + c share the same singular set J_{0}, and J_{1} of F is the same as that of F + c up to a translation.

J_{0}^{G} = {x| det (D(kF h^{−1})) = 0}
= {x| det (Dk) det (DF) det (Dh^{−1}) = 0}
= {x| det (DF) = 0}
= {x|h^{−1}(x) ∈ J_{0}^{F}}
= {x|x ∈ h(J_{0}^{F})}
= h(J_{0}^{F}).

J_{1}^{G} = G(h(J_{0}^{F}))
= kF h^{−1}h(J_{0}^{F})
= kF (J_{0}^{F})
= k(J_{1}^{F}).

• Hence we may assume that F is of the form

F = ( \begin{matrix}a_{0}x^{2} + a_{1}xy + a_{2}y^{2} + a_{3}x + a_{4}y \\ b_{0}x^{2} + b_{1}xy + b_{2}y^{2} + b_{3}x + b_{4}y \end{matrix} )

• we denote a_{i}b_{j} − a_{j}b_{i} by X_{ij} for i ≠ j ∈ {0, 1, 2, 3, 4}. it is easy to verify that for distinct i, j, k, l,

\begin{aligned}X_{ij} & = & −X_{ji} \\ a_{k}X_{ij} + a_{i}X_{jk} + a_{j}X_{ki} & = & 0 \\ b_{k}X_{ij} + b_{i}X_{jk} + b_{j}X_{ki} & = & 0 \\ X_{ij}X_{kl} − X_{ik}X_{jl} + X_{il}X_{jk} & = & 0. \end{aligned}

a_{2}X_{01} + a_{0}X_{12} + a_{1}X_{20} = 0
a_{3}X_{01} + a_{0}X_{13} + a_{1}X_{30} = 0
a_{4}X_{01} + a_{0}X_{14} + a_{1}X_{40} = 0
a_{3}X_{02} + a_{0}X_{23} + a_{2}X_{30} = 0
a_{4}X_{02} + a_{0}X_{24} + a_{2}X_{40} = 0
a_{4}X_{03} + a_{0}X_{34} + a_{3}X_{40} = 0
a_{3}X_{12} + a_{1}X_{23} + a_{2}X_{31} = 0
a_{4}X_{12} + a_{1}X_{24} + a_{2}X_{41} = 0
a_{4}X_{13} + a_{1}X_{34} + a_{3}X_{41} = 0
a_{4}X_{23} + a_{2}X_{34} + a_{3}X_{42} = 0
X_{01} X_{23} − X_{02} X_{13} + X_{03} X_{12} = 0
X_{01} X_{24} − X_{02} X_{14} + X_{04} X_{12} = 0
X_{01} X_{34} − X_{03} X_{14} + X_{04} X_{13} = 0
X_{02} X_{34} − X_{03} X_{24} + X_{04} X_{23} = 0
X_{12} X_{34} − X_{13} X_{24} + X_{14} X_{23} = 0

Replace a by b identities (1) ∼ (10) hold true.

With these notation,

f = 2X_{01} x^{2}

+4X_{02} xy +2X_{12} y^{2} +(2X_{04} − X_{13} )x +(X_{14} − 2X_{23} )y +X_{34}

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4. The proof ( Case 1 - 5 )

The proof is essentially pretty elementary, we’ll prove case 1. If we have time left, we can outline the proof for the other cases later.

Case 1

Assume f = x^{2} − 1 then

2X_{01} = 1  (1)
4X_{02} = 0  (2)
2X_{12} = 0  (3)
2X_{04} + X_{31} = 0  (4)
2X_{32} + X_{14} = 0  (5)
X_{34} = −1.  (6)

Note that a_{2} = b_{2} = 0 by (2) and (3). Hence X_{23} = 0. And then J_{1} = C_{1} ∪ C_{2} where C_{1} = {((a_{1} +a_{4})t+a_{0} +a_{3}, (b_{1} +b_{4})t+b_{0} +b_{3})} and C_{2} = {((a_{4} − a_{1})t + a_{0} − a_{3}, (b_{4} − b_{1})t + b_{0} − b_{3})}

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By (5),

X_{23} = 0 ⇒ X_{14} = 0.

Also Note that

\{ \begin{matrix}X_{01} X_{34} − X_{03} X_{14} + X_{04} X_{13} = 0 \\ 2X_{04} + X_{31} = 0 \end{matrix}  ⇒ X_{13}^{2} = 1

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Now we see that if

X_{13} = 1 ⇒ a_{1} − a_{4} = b_{1} − b_{4} = 0

and if

X_{13} = −1 ⇒ a_{4} + a_{1} = b_{4} + b_{1} = 0

simply by the fact

a_{4}X_{13} + a_{1}X_{34} + a_{3}X_{41} = 0

Finally, note that only one of then may occur. Since

\{ \begin{matrix}a_{1} − a_{4} = a_{1} + a_{4} = 0 \\ b_{1} − b_{4} = b_{1} + b_{4} = 0 \end{matrix}  ⇒ a_{1} = b_{1} = 0.

Hence X_{12} = 0 which is impossible.

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Case 5

Assume f = x^{2} − y then

2X_{01} = 1
4X_{02} = 0
2X_{12} = 0
2X_{04} + X_{31} = 0
2X_{32} + X_{14} = −1
X_{34} = 0.

J_{0} = {(t, t^{2})}

Then J′_{1} = {(a_{3} + 2a_{0}t + 2a_{4}t + 3a_{1}t^{2} + 4a_{2}t^{3}, b_{3} + 2b_{0}t + 2b_{4}t + 3b_{1}t^{2} + 4b_{2}t^{3})}

In this case we have a_{2} = b_{2} = 0, X_{23} = X_{24} = 0, X_{14} = −1, and X_{04} = \frac{1}{2}X_{13} .

So t = \frac{X_{13}}{3}

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Finally, check that it’s really a solution

\frac{1}{3}(3a_{3} + (2a_{0} + 2a_{4})X_{13} + a_{1}X_{13}^{2} )
= \frac{1}{3}((a_{3} + 2a_{0}X_{13} + a_{1}X_{13}^{2} ) + (2a_{3} + 2a_{4}X_{13} ))
= 0

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5. Some possible generalizations (quadmapfig2.ps)

• Similar situations hold for map of 3-1 type, one component of the map has degree 3 and the other degree 1, but more possibilities. ( Fig 2 ) And the proof is even simpler.

• As the degree of the map increases the more possible cases may occure. For example, the map

F = ( \begin{matrix}x \\ (y − p(x))^{2} + q(x) \end{matrix} )

has (x, p(x)), the graph of p, as its singular set which is mapped to the graph of q, (x, q(x)) under F. However from topological view point they are equivalent.

This might be the right place to say something about the geometric relation between the kernel field and precusps.

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6. An application in singularity theory Corollary The two germs

f(x, y) = (x^{2} + y^{2}, xy)

and

g(x, y) = (x^{2} + y^{2} + x^{3}, xy)

is not A-equivalent in any neighborhood of the origin. Proof Figure 3.(hotmfig1.ps, hotmfig2.ps) The point we want to make is that the truncation of higher order term needs more thought.

Acknowledgment

We like to thank Pisces team members. We also like to thank Rick Wicklin for many helpful conversations.

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