Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 35 (1994), No. 1, 141-145.

Transnormal Graph Surfaces

A. M. d’Azevedo Breda and F. J. Craveiro de Carvalho

Departamento de Mathemática, Universidade de Aveiro P-3800 Aveiro, Portugal Departamento de Mathemática, Universidade de Coimbra P-3000 Coimbra, Portugal

Graphs of smooth maps very often are examples of transnormal submanifolds of euclidian space R^{n}. This fact gives us some motivation to look for conditions on the map which ensure us that the graph is transnormal or not.

In this note we study maps f: R^{2} → R. No doubt that our results (with the possible exception of Theorem 3.2 in 3.) still hold when we replace R^{2} by R^{n}. However what we gain from working in a more general context is very little. Our results depend heavily on results which have been established previously in [1] and [3].

This research was carried out under “Projecto de Investigação PBIC/C/CEN/1060/92”.

1.Introduction

Let f: R^{2} → R be a smooth (= C^{∞}) map. We consider F: R^{2} → R^{3} given by F (x, y) = (x, y, f(x, y)). Obviously F (R^{2}) is the graph G_{f} of f. The normal line to F at (x, y) will be denoted by N_{(x,y)} and we say that G_{f} is transnormal if whenever (x_{1}, y_{1}, f(x_{1}, y_{1})) ∈ N_{(x,y)} we have N_{(x,y)} = N_{(x_{1},y_{1})}. The extension of this notion to maps from Rto R^{2} is obvious. We remark that the property of being transnormal is invariant under affine isometries.

In what follows (.|.) and ∥.∥ will denote the standard inner product in R^{2} and the norm that it induces. Partial derivatives will be denoted by D_{i}f, i=1,2.

2. Transnormal graphs

Let f: R^{2} → R be smooth. Assume that (x_{0}, y_{0}, f(x_{0}, y_{0})) is such that D_{1}f(x_{0}, y_{0}) ≠ 0. Let π denote the affine plane which passes through (x_{0}, y_{0}, f(x_{0}, y_{0})) and whose direction is ⟨(0, 0, 1), (D_{1}f(x_{0}, y_{0}), D_{2}f(x_{0}, y_{0}), −1)⟩, where ⟨⋯⟩ means spanned by. Then π ∩ G is the graph of h: R → R^{2} given by h(t) = (μ(t), f(t, μ(t))), with μ(t) = y_{0} + (t − x_{0})\frac{D_{2}f(x_{0},y_{0})}{D_{1}f(x_{0},y_{0})}. Mention to h will be made several times below.

Proposition 2.1 Let (x_{0}, y_{0}) be such that D_{1}f(x_{0}, y_{0}) ≠ 0. If G_{f} is transnormal then G_{h} is also transnormal.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1994 Heldermann Verlag, Berlin

142 A.M. d’Azevedo Breda, F.J. Craveiro de Carvalho: Transnormal Graph Surfaces

Proof. Assume that we have t_{0}, t_{1} such that (t_{1}, h(t_{1})) ∈ N_{t_{0}}. Since N_{t_{0}} ∩ π = N_{(t_{0},h(t_{0}))} it follows from (t_{1}, h(t_{1})) ∈ π that (t_{1}, h(t_{1})) ∈ N_{(t_{0},h(t_{0}))} and consequently N_{(t_{1},h(t_{1}))} = N_{(t_{0},h(t_{0}))}. From this we conclude that N_{t_{1}} = N_{t_{0}}.

It is known that the graph of a smooth curve R → R^{n} is transnormal iff every normal n-plane meets the graph in just one point ([4] or [2] for an elementary proof). We then have

Theorem 2.1 G_{f} is transnormal iff every normal line meets G_{f} in just one point.

Proof. The if part is trivial.

Let (x_{0}, y_{0}) ∈ R^{2}. If D_{1}f, D_{2}f both vanish at (x_{0}, y_{0}) then G_{f} ∩ N_{(x_{0},y_{0})} = {(x_{0}, y_{0}, f(x_{0}, y_{0}))}. Suppose then that one of the partial derivatives does not vanish. Without loss of generality we may assume that D_{1}f(x_{0}, y_{0}) ≠ 0. Consider the curve h: R → R^{2} defined above. Then the assumption (x_{1}, y_{1}, f(x_{1}, y_{1})) ∈ N_{(x_{0},y_{0})} leads to (x_{1}, y_{1}) = (x_{0}, y_{0}) due to the transnormality of G_{h}.

Theorem 2.1 also follows from [5].

The next two results deal with sufficient conditions for the transnormality of G_{f}.

Theorem 2.2 Let f: R^{2} → R be such that ∥grad f(x, y)∥ ≤ 1, (x, y) ∈ R^{2}. Then G_{f} is transnormal.

Proof. It is well-known that for such an f we have

|f(x, y) − f(x′, y′)| ≤ ∥(x, y) − (x′, y′)∥, (x, y), (x′, y′) ∈ R^{2}.

Assume that, for (x_{1}, y_{1}) ≠ (x_{0}, y_{0}), (x_{1}, y_{1}, f(x_{1}, y_{1})) belongs to N_{(x_{0},y_{0})}. We then have

\begin{aligned}x_{1} − x_{0} & = (f(x_{0}, y_{0}) − f(x_{1}, y_{1}))D_{1}f(x_{0}, y_{0}), \\ y_{1} − y_{0} & = (f(x_{0}, y_{0}) − f(x_{1}, y_{1}))D_{2}f(x_{0}, y_{0}) \end{aligned}

and consequently

\frac{(x_{1} − x_{0})^{2} + (y_{1} − y_{0})^{2}}{|f(x_{0}, y_{0}) − f(x_{1}, y_{1})|^{2}} = D_{1}f(x_{0}, y_{0})^{2} + D_{2}f(x_{0}, y_{0})^{2}.

The left-hand side is greater than or equal to 1 while the right hand side is less or equal than 1. It the follows that ∥(x_{0}, y_{0})−(x_{1}, y_{1})∥^{2} = |f(x_{0}, y_{0})−f(x_{1}, y_{1})|^{2}. Therefore (x_{0}, y_{0}) is a maximum for g with g(x, y) = |f(x, y) − f(x_{1}, y_{1})|^{2} − ∥(x, y) − (x_{1}, y_{1})∥^{2}. Accordingly we must have D_{1}g(x_{0}, y_{0}) = D_{2}g(x_{0}, y_{0}) = 0. On the other hand it is easy to obtain

D_{1}g(x_{0}, y_{0}) = 4(x_{1} − x_{0}), D_{2}g(x_{0}, y_{0}) = 4(y_{1} − y_{0}),

and at least one of them is nonzero. The contradiction shows that G_{f} is transnormal.

Theorem 2.3 Let f: R^{2} → R be such that, for i=1,2, D_{i}f ≥ 0 or D_{i}f ≤ 0. Then G_{f} is transnormal.

Proof. Without loss of generality we may assume that D_{i}f ≥ 0, i = 1, 2. Assume that G_{f} is not transnormal. Then there exist distinct (x_{1}, y_{1}), (x_{2}, y_{2}) such that

x_{2} − x_{1} = (f(x_{1}, y_{1}) − f(x_{2}, y_{2}))D_{1}f(x_{1}, y_{1})

and

y_{2} − y_{1} = (f(x_{1}, y_{1}) − f(x_{2}, y_{2}))D_{2}f(x_{1}, y_{1}).

Suppose, for instance, that x_{2} − x_{1} ≠ 0. Then \frac{y_{2}−y_{1}}{x_{2}−x_{1}} ≥ 0. Using the mean value theorem we have

x_{2} − x_{1} = (D_{1}f(z_{1}, z_{2})(x_{1} − x_{2}) + D_{2}f(z_{1}, z_{2})(y_{1} − y_{2}))D_{1}f(x_{1}, y_{1}),

that is, 1 = (−D_{1}f(z_{1}, z_{2}) − D_{2}f(z_{1}, z_{2})\frac{y_{2}−y_{1}}{x_{2}−x_{1}} ) D_{1}f(x_{1}, y_{1}). The right-hand side is ≤ 0 and we obtain a contradiction.

3. Nontransnormal graphs

Again we take a smooth map f: R^{2} → R and assume that D_{1}f(x_{0}, y_{0}) ≠ 0. Take the map h: R → R^{2}, with h(t) = (μ(t), f(t, μ(t))), already considered at the beginning of 2. We will write h_{2}(t) = f(t, μ(t)).

Theorem 3.1 If D_{1}f(x_{0}, y_{0}) < 0 (resp. > 0) and lim_{t→∞} h′_{2}(t) = ∞ (resp. − ∞) then G_{f} is not transnormal.

Proof. Assume that G_{f} is transnormal and define g : (x_{0}, ∞) → R by g(t) = (t − x_{0}) + (h(t) − h(x_{0})|h′(x_{0})). The transnormality of G_{f} implies that g never vanishes and since g < 0 leads to a contradiction we must have g > 0. Therefore

\frac{h_{2}(t) − h_{2}(x_{0})}{t − x_{0}} < \frac{1}{h′_{2}(x_{0})} (−1 − (\frac{D_{2}f(x_{0}, y_{0})}{D_{1}f(x_{0}, y_{0})})^{2}) .

On the other hand lim_{t→∞} h′_{2}(t) = ∞ implies lim_{t→∞} h_{2}(t) = ∞. By L’Hôpital’s rule lim_{t→∞} \frac{h_{2}(t)−h_{2}(x_{0})}{t−x_{0}} = lim_{t→∞} h′_{2}(t) = ∞. This contradicts the inequality above.

For the next theorem we need the following lemma

Lemma 3.1 Let f : R^{2} → R be polynomial of positive degree m, depending on both variables, and given by f(x, y) = ∑_{i+j≤m} a_{ij}x^{i}y^{j}. Then there exists (x_{0}, y_{0}) such that \frac{D_{2}f(x_{0},y_{0})}{D_{1}f(x_{0},y_{0})} is nonzero and it is not a zero of p(y) = ∑_{i+j=m} a_{ij}y^{j}.

Proof. Start by choosing a point (x_{1}, y_{1}) such that \frac{D_{2}f(x_{1},y_{1})}{D_{1}f(x_{1},y_{1})} is nonzero. Take then the connected component A of R^{2} − {(x, y) ∈ R^{2} : D_{1}f(x, y) = 0} which contains (x_{1}, y_{1}).

If \frac{D_{2}f}{D_{1}f} is not constant in A then it is clear that we can find (x_{0}, y_{0}) as required.

Assume then that \frac{D_{2}f}{D_{1}f} is constant with value c and that p(c) = 0. Since in A D_{2}f = cD_{1}f it follows that f(x, y) = g(cy + x). A simple argument using Taylor’s formula then shows that, in an open neighbourhood of (x_{1}, y_{1}), f(x, y) = ∑_{i=0}^{m} λ_{i}(cy+x)^{i}, with λ_{m} ≠ 0. Keeping in mind that p(c) = 0 and equating coefficients we arrive at c^{2m} + \binom{m}{1}c^{2(m−1)} + ⋯ + \binom{m}{m−1}c^{2} + 1 = 0 which is obviously impossible.

Theorem 3.2 Let f: R^{2} → R be polynomial of positive even degree. Then G_{f} is not transnormal.

Proof. If f depends only on one variable, say f(x, y) = ∑_{i=0}^{m} a_{i}x^{i}, choose (x_{0}, y_{0}) such that D_{1}f(x_{0}, y_{0}) ≠ 0 . Then if G_{f} were transnormal the corresponding curve h: R → R^{2} would have positive even degree and transnormal graph. This cannot happen [1].

If f depends on both variables choose (x_{0}, y_{0}) as in the lemma. This ensures again that h is a polynomial curve of positive even degree and consequently G_{h} cannot be transnormal [1]. Consequently G_{f} is not transnormal.

4. Transnormal isotopy

Next we show how to deform a map with transnormal graph to a constant map through maps with transnormal graph.

Theorem 4.1 Let f: R^{2} → R have transnormal graph. Then, for 0 ≤ t ≤ 1, f_{t}: R^{2} → R given by f_{t}(x, y) = tf(x, y) has also transnormal graph.

Proof. We assume that there exists 0 < t < 1 such that the graph of f_{t} is not transnormal. Then there are (x_{0}, y_{0}), (x_{1}, y_{1}), (x_{1}, y_{1}) ≠ (x_{0}, y_{0}), such that (x_{1}, y_{1}, f_{t}(x_{1}, y_{1})) belongs to the normal line to F_{t} at (x_{0}, y_{0}), with F_{t}(x, y) = (x, y, f_{t}(x, y)). Without loss of generality we may assume that f(x_{0}, y_{0}) ≠ 0. Then the normal lines to F and F_{t} at (x_{0}, y_{0}) determine a plane parallel to the z-axis. Using a convenient affine isometry of R^{3} if necessary we may suppose that such a plane is determined by y = 0.

Let now h and h_{t} denote the maps relative to f and f_{t} and to the point (x_{0}, 0) mentioned at the beginning of 2. It follows that h_{t}(x) = th(x). Since G_{f} is transnormal G_{h} is transnormal and G_{h_{t}} is also transnormal [3]. We have now a contradiction because obviously our initial assumption implies that G_{h_{t}} is not transnormal.

References

[1] A.M. d’Azevedo Breda and F.J. Craveiro de Carvalho: Transnormal graphs of polynomial curves, Journal of Geometry 49 (1994), 36 - 41.

[2] F.J. Craveiro de Carvalho: Gráficos transnormais de funções polinomiais, Boletim Soc. Portuguesa Mathemática 22 (1992), 33 - 36.

[3] F.J. Craveiro de Carvalho: Transnormal graph deformations, Contributions to Algebra and Geometry 34 (1993), 287 - 290.

A.M. d’Azevedo Breda, F.J. Craveiro de Carvalho: Transnormal Graph Surfaces 145 [4] B. Wegner: Transnormale Isotopien und transnormale Kurven, Manuscripta Math. 4

(1971), 361 - 372.

[5] B. Wegner: Einige Bemerkungen zur Geometrie transnormaler Mannigfaltigkeiten, J. Diff. Geom. 16 (1981), 93 - 100.

Received 22.4.94