Evolute Surfaces of a Two-Dimensional Dually Normalized $D_2$ in $E_4$
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Submitted 1962-01-01 | SovietRxiv: ru-196201.59016 | Translated from Russian

Abstract Generated abstract

This note studies two-dimensional dually normalized surfaces of class D2 immersed in four-dimensional Euclidean space. Using the differential equations of such surfaces, it analyzes congruences of normals, determines their focal points, and introduces axes of curvature in the normal plane, with associated invariants interpreted as reciprocal distances to these axes. The paper proves that the normal planes of a D2 surface possess an enveloping evolute surface of zero Gaussian curvature, and conversely that a surface whose normal planes admit such an envelope is dually normalizable. It concludes that finding D2 surfaces in E4 is equivalent to finding zero-curvature surfaces, with every D2 surface realized as an evolvent surface of such a surface.

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A. V. CHAKMAZYAN

EVOLUTE SURFACES OF A TWO-DIMENSIONAL DUALLY NORMALIZED \(D_2\) IN \(E_4\)

(Presented by Academician P. S. Aleksandrov on 8 II 1962)

  1. We shall say that a surface \(X_m\) in the projective space \(P_n\) is dually normalized if it is normalized in the sense of A. P. Norden \(\left({}^{1}\right)\) and its normal of the first kind contains the characteristic of the family of hyperplanes tangent to \(X_m\). A number of properties of dually normalized surfaces were obtained in the author’s papers \(\left({}^{2}\right)\). In this note we shall restrict ourselves to the case of a surface of two dimensions \(X_2\), immersed in the Euclidean \(E_4\).

Suppose that \(X_2\) can be completed to a hypersurface in such a way that the characteristics of the family of tangent hyperplanes are perpendicular to the tangent plane of the surface \(X_2\). It is obvious that then the natural normalization of \(X_2\) will at the same time also be dual. The surfaces \(X_2\) admitting a dual normalization form a certain class, which in what follows we shall denote by \(D_2\).

The basic differential equations of surfaces of the class \(D_2\) have the form \(\left({}^{2б}\right)\)

\[ \nabla_j r_i = h_{ij}X + k_{ij}Y,\qquad \mathbf{X}_j=-h_j^{\,l}r_l,\qquad \mathbf{Y}_j=-k_j^{\,l}r_l, \tag{A} \]

where \(h_{ij}=-\partial_j r\,\partial_i X=\mathbf{X}\nabla_j r_i\), \(k_{ij}=-\partial_j r\,\partial_i Y=\mathbf{Y}\nabla_j r_i\), and \(\mathbf{X}\) and \(\mathbf{Y}\) denote, respectively, the normal vector of the tangent hyperplane and the vector of the characteristic line.

  1. Let us construct the congruence of normals of a dually normalized surface, going in the direction of the unit vectors

\[ X^* = X\cos\alpha + Y\sin\alpha,\qquad \alpha=\mathrm{const}. \]

\[ \mathbf{R}=\mathbf{r}(u^1,u^2)+\rho X^*(u^1,u^2). \tag{1} \]

Let us pose the question: do focal surfaces of this congruence exist, i.e., such surfaces which are touched by all its rays? In this case there must exist a function \(\rho(u^1,u^2)\) such that the vectors \(\mathbf{R}_1,\mathbf{R}_2\) and \(X^*\) are coplanar. Taking (1) and (A) into account, we obtain

\[ \mathbf{R}_i=\left[\delta_i^s-\rho\left(h_i^s\cos\alpha+k_i^s\sin\alpha\right)\right]\mathbf{r}_s+\rho_iX^*. \]

Taking the latter into account, we must require the existence of numbers \(\mu^1,\mu^2,\gamma\), not all simultaneously equal to zero, such that

\[ \mu^i\mathbf{R}_i+\gamma X^*=0, \]

i.e.

\[ \mu^i\left[\delta_i^s-\rho\left(h_i^s\cos\alpha+k_i^s\sin\alpha\right)\right]\mathbf{r}_s+(\mu^i\rho_i+\gamma)X^*=0. \]

Hence it follows that

\[ \mu^i\left[\delta_i^s-\rho\left(h_i^s\cos\alpha+k_i^s\sin\alpha\right)\right]=0. \]

Thus, the desired function \(\rho(u^1,u^2)\) must be a quantity reciprocal to a root of the equation

\[ \operatorname{Det}\left\|h_i^{\,s}\cos\alpha+k_i^{\,s}\sin\alpha-\omega\delta_i^s\right\|=0. \tag{2} \]

Since the tensors \(h_{ij}, k_{ij}\) have common principal directions \((^{26})\), the roots of equation (2) are equal to

\[ \omega_i=\sigma_i\cos\alpha+\tau_i\sin\alpha, \]

where \(\sigma_i,\tau_i\) are the principal values of the tensors \(h_{ij}, k_{ij}\), respectively.
Thus, on the ray (1) there exist two desired focal points

\[ \underset{1}{\mathbf R} =\mathbf r+ \frac{1}{\sigma_1\cos\alpha+\tau_1\sin\alpha} (\mathbf X\cos\alpha+\mathbf Y\sin\alpha), \]

\[ \underset{2}{\mathbf R} =\mathbf r+ \frac{1}{\sigma_2\cos\alpha+\tau_2\sin\alpha} (\mathbf X\cos\alpha+\mathbf Y\sin\alpha). \tag{3} \]

If in (1) we put \(\alpha=0\) and \(\alpha=\pi/2\), then we obtain the congruences of normals \(D_2\) going in the directions of the unit vectors \(\mathbf X\) and \(\mathbf Y\), respectively.
Thus, on each normal to \(D_2\) there exist two focal points. Let us determine how the focal points are situated in the normal plane to \(D_2\). Introduce in this plane a rectangular coordinate system with origin at the point of the surface \(D_2\) and with axes going in the directions of the unit vectors \(\mathbf X,\mathbf Y\). Then from (3) it follows that the coordinates of the focal points \(F_1(\underset{1}{\mathbf R})\) are equal to

\[ x=\frac{\cos\alpha}{\sigma_1\cos\alpha+\tau_1\sin\alpha}, \qquad y=\frac{\sin\alpha}{\sigma_1\cos\alpha+\tau_1\sin\alpha}. \]

Eliminating the parameter \(\alpha\), we obtain the equation

\[ \sigma_1x+\tau_1y=1. \tag{4} \]

Thus, the focal points \(F_1(\underset{1}{\mathbf R})\) are situated in the normal plane on a straight line not passing through the point of the surface \(D_2\). Similarly, we obtain that the focal points \(F_2(\underset{2}{\mathbf R})\) lie on the straight line

\[ \sigma_2x+\tau_2y=1. \tag{5} \]

We shall call the straight lines (4) and (5) the axes of curvature of the surface \(D_2\). From (4) and (5) we find the geometric meaning of the invariants \(\chi_1=\sqrt{\sigma_1^2+\tau_1^2}\) and \(\chi_2=\sqrt{\sigma_2^2+\tau_2^2}\): their reciprocals \(\frac{1}{\chi_1}\) and \(\frac{1}{\chi_2}\) give the distances from the point of the surface \(D_2\) to the axes of curvature (4) and (5), respectively.

Let us pose the question: do there exist envelopes of the family of normal planes of the surface \(D_2\)?
Consider the points

\[ \mathbf R=\mathbf r(u^1,u^2)+a(u^1,u^2)\mathbf X+b(u^1,u^2)\mathbf Y \]

in the normal plane \(D_2\), and require that at these points it be tangent to some surface, i.e. that the derivatives \(\mathbf R_1,\mathbf R_2\) decompose with respect to \(\mathbf X\) and \(\mathbf Y\). We have

\[ \mathbf R_i=(\delta_i^s-ah_i^{\,s}-bk_i^{\,s})\mathbf r_s+a_i\mathbf X+b_i\mathbf Y, \]

whence we obtain the condition

\[ \delta_i^s-ah_i^{\,s}-bk_i^{\,s}=0. \tag{6} \]

Let \(a^i\) and \(\widetilde a^i\) be common principal directions of the tensors \(h_{ij}\) and \(k_{ij}\). Then from (6) we have

\[ a^i-a\sigma_1a^i-b\tau_1a^i=0,\qquad \widetilde a^i-a\sigma_2\widetilde a^i-b\tau_2\widetilde a^i=0 \]

or

\[ \begin{aligned} a\sigma_1+b\tau_1&=1,\\ a\sigma_2+b\tau_2&=1. \end{aligned} \tag{7} \]

In this case our requirements are satisfied and the envelope exists. Comparing (7) with (4) and (5), we see that \(a(u^1,u^2)\) and \(b(u^1,u^2)\) are the coordinates of the points of intersection of the axes of curvature, which, thus, lie on the envelope of the family of normal planes \(D_2\). Thus, we have proved that the normal planes \(D_2\), embedded in \(E_4\), admit an enveloping surface. We shall call this surface the evolute surface.

We shall now prove that if the normal planes of a two-dimensional surface embedded in \(E_4\) admit an envelope, then this surface is dually normalizable.

Let

\[ \mathbf R=\mathbf R(u^1,u^2) \]

be the parametric equation of the evolute surface. If \(\mathbf m,\mathbf n\) are unit and mutually perpendicular normal vectors of the evolute surface, then its basic derivative equations have the form

\[ \nabla_j\mathbf R_i=a_{ij}\mathbf m+b_{ij}\mathbf n, \tag{8} \]

where \(a_{ij}=-\partial_j\mathbf R\,\partial_i\mathbf m=\mathbf m\nabla_j\mathbf R_i\), \(b_{ij}=-\partial_j\mathbf R\,\partial_i\mathbf n=\mathbf n\nabla_j\mathbf R_i\), and \(\nabla\) is the symbol of covariant differentiation in the intrinsic connection of the evolute surface. If the radius vector of a point of the original surface is represented in the form

\[ \mathbf R^*=\mathbf R(u^1,u^2)+\lambda^k(u^1,u^2)\mathbf R_k, \tag{9} \]

then we must require that the derivatives \(\mathbf R^*_1,\mathbf R^*_2\) be resolved along \(\mathbf m\) and \(\mathbf n\). We have:

\[ \mathbf R_i^*=(\delta_i^m+\nabla_i\lambda^k)\mathbf R_k+\lambda^k a_{ik}\mathbf m+\lambda^k b_{ik}\mathbf n, \]

whence we obtain the condition

\[ g_{ij}+\nabla_i\lambda_j=0; \tag{10} \]

\(\lambda_i\) is a gradient, since \(g_{ij}\) is symmetric.

Forming the integrability condition for (10), we obtain \(R_{\cdot kij}^{\ \ \ l}\lambda_l=0\), or \({}^{(3)}\)

\[ K\varepsilon_{jk}\varepsilon_i^{\ l}\lambda_l=0, \tag{11} \]

where \(K\) is the Gaussian curvature of the intrinsic geometry of the evolute surface. From (11) it follows that \(K=0\), and this means that the intrinsic geometry of the evolute surface is Euclidean. The line element of this surface can be reduced to the form \(ds^2=du^2+dv^2\). Hence, in turn, it follows that

\[ \mathbf R_1^2=1,\qquad \mathbf R_2^2=1,\qquad \mathbf R_1\mathbf R_2=0. \]

From (10) we obtain

\[ \partial_1\lambda_1=-1,\qquad \partial_2\lambda_2=-1,\qquad \partial_1\lambda_2=\partial_2\lambda_1=0; \]

whence it follows that

\[ \lambda_1=-u+a,\qquad \lambda_2=-v+b, \]

where \(a,b\) are constants.

From what has been said it is clear that equation (9) can be rewritten in the form

\[ \mathbf R^*=\mathbf R(u,v)+(a-u)\mathbf R_1+(b-v)\mathbf R_2. \tag{12} \]

It is easy to see that equation (12) represents the surface \(D_2\). Indeed, the vectors \(\mathbf R_1, \mathbf R_2\) are unit vectors, and for them the conditions
\(\partial_i\mathbf R_j=a_{ij}\mathbf m+b_{ij}\mathbf n\) hold, where \(\mathbf m\) and \(\mathbf n\) are normal vectors. Hence

\[ \mathbf R_1\partial_j\mathbf R_2=0,\qquad \mathbf R_2\partial_j\mathbf R_1=0,\qquad j=1,2. \]

This means that the surface is doubly normalized.

Thus, we have proved the following theorems:

Theorem 1. In order that a surface be doubly normalizable, it is necessary and sufficient that its normal planes have an envelope of zero curvature.

Theorem 2. Every \(D_2\) in \(E_4\) cuts orthogonally the tangent planes of some surface of zero curvature.

Thus, the problem of finding \(D_2\) in \(E_4\) is equivalent to the problem of finding \(X_2\) of zero curvature. We shall call surfaces that cut orthogonally the tangent planes of a given surface its evolvent surfaces. Then this result can be formulated as follows:

Every \(D_2\) in \(E_4\) is an evolvent surface of an arbitrary surface of zero curvature.

In conclusion I express my gratitude to A. P. Norden, under whose guidance the present work was carried out.

Yerevan State
University

Received
2 XII 1961

REFERENCES CITED

  1. A. P. Norden, Spaces of Affine Connection, 1950.
  2. A. V. Chakmazyan, Dokl. Acad. Sci. ArmSSR, a) 28, No. 24, 151 (1959); b) 29, No. 1, 3 (1959); c) 31, 129 (1960); d) 30, No. 4, 187 (1960); e) 33, No. 3 (1961).
  3. A. P. Norden, Theory of Surfaces, 1956.

Submission history

Evolute Surfaces of a Two-Dimensional Dually Normalized $D_2$ in $E_4$