Abstract
Full Text
MATHEMATICS
Academician A. D. ALEKSANDROV
MAPPINGS OF FAMILIES OF SETS
1. Let \(A_n, A_{m'}\) be affine spaces \((n \geq 1)\), let \(M\) be a bounded set in \(A_n\), and let \(f\) be a one-to-one mapping of \(A_n\) onto \(A_{m'}\) having the following property. The image of every set \(t(M)\), obtained from \(M\) by some parallel translation, is \(t'(M')\), obtained by translating the set \(M' = f(M)\), and conversely: every \(t'(M')\) is the image of some \(t(M)\). In short, the equality holds
\[ ft(M)=t'f(M), \tag{1} \]
where to each \(t\) there corresponds \(t'\), and conversely, so that \(f\) realizes a mapping of the family \(\{t(M)\}\) onto \(\{t'(M')\}\). The question consists in investigating mappings \(f\) of this kind. Obviously, every affine mapping of \(A_n\) onto \(A_{m'}\) will be a mapping \(f\) for any \(M\); what other mappings \(f\) exist for a given \(M\)?
2. To formulate our results, we introduce some definitions. Let \(P\) be any \((n-1)\)-dimensional plane in \(A_n\), parallel to some given \(P_0\), and let \(l\) be a vector not parallel to \(P\).
We shall call a shift of the planes \(P\), preserving \(l\), such a homeomorphism \(d\) of the space \(A_n\) onto itself that: 1) \(d\) takes every directed segment equal to \(l\) into a segment equal to \(l\); 2) on every plane \(P\), \(d\) is its parallel translation. This is equivalent to the following. If in \(A_n\) we introduce coordinates \(x_1,\ldots,x_n\) so that the axis \(x_1\) is parallel to \(l\), and the axes \(x_2,\ldots,x_n\) are parallel to \(P\), then the mapping \(d_{Pl}\) is represented by the formulas:
\[ x_1'=\varphi(x_1),\quad x_2'=x_2+a_2,\ldots,\ x_n'=x_n+a_n, \tag{2} \]
where the \(a_i\) are constants, and \(\varphi\) is such a homeomorphism of the axis \(x_1\) onto itself that \(\varphi(x_1+|l|)=\varphi(x_1)+|l|\).
Obviously, every shift \(d=d_{Pl}\) takes any cylinder \(C\) with generator \(l\) and base plane \(P\) into the same kind of cylinder, i.e., there exists a translation \(t\) such that \(d(C)=t(C)\). We shall call a quasicylinder any (nonempty) bounded set \(Q\) possessing the same property. In this case \(P\) is called the base plane of the quasicylinder \(Q\), and \(l\) its generator, under the additional condition that for no vector \(kl \ne \pm l\) (\(k\) an integer) does \(Q\) possess the same property. (That is, shifts \(d=d_{Pkl}\) are found such that \(d(Q)\ne t(Q)\). Otherwise we could regard as the “generator,” for example, half the generator of an ordinary cylinder.)
The case \(l=0\) is not excluded, when the quasicylinder \(Q\) is a plane set: \(Q\subset P\), and the shift \(d_{Pl}\) is any homeomorphism of \(A_n\) onto itself that is a translation on every plane \(P\). At the same time it is not difficult to see that a set \(N\) will be a quasicylinder with base plane \(P\) and generator \(l\ne0\) if and only if it admits the following representation. There are planes \(P_1,\ldots,P_p\) parallel to \(P\), the distances between which in the direction \(l\) have common measure \(|l|\), and \(N\) is composed of sets lying in the planes \(P_i\), and of open segments parallel to \(l\) with endpoints on the planes \(P_i\). In this case it is not necessary that such segments occur between every pair \(P_i, P_j\)—there may be none at all, just as it is not necessary that on every \(P_i\) there be points of \(N\); but no plane \(P_i\) can-
be excluded without the indicated representation of the set \(N_\varphi\), with \(l\) replaced by \(kl \ne \pm l\), ceasing to be possible.
One and the same quasicylinder may have several generating and base planes, as, for example, a parallelepiped: it has \(n\) generating and \(n\) base planes.
We say that a quasicylinder \(Q\) is degenerate if there exists a base plane \(P\) such that \(Q\) is contained in a finite number of planes parallel to \(P\); otherwise \(Q\) is nondegenerate. Note that if a nondegenerate quasicylinder has two generators \(l_1, l_2\) and base planes \(P_1, P_2\), then, as can be shown, \(l_1 \parallel P_2\) and \(l_2 \parallel P_1\). Hence it follows, in particular, that a nondegenerate quasicylinder can have at most \(n\) generators and base planes.
- From the definitions of the displacement \(d_{Pl}\) and of a quasicylinder it follows immediately that if the set \(M \subset A_n\) is a quasicylinder with base plane \(P\) and generator \(l\), then for every displacement \(d=d_{Pl}\) and every translation \(t\) there exists a translation \(\bar t\) such that
\[ dt(M)=\bar t(M). \tag{3} \]
Therefore, if \(M\) has generators \(l_1,\ldots,l_k\) and base planes \(P_1,\ldots,P_k\), then for any \(d_i=d_{P_i l_i}\) and any translation \(t\) there also exists a translation \(\bar t\) such that
\[ d_1 \ldots d_k t(M)=\bar t(M). \tag{4} \]
This is nothing other than formula (1) with \(f=d_1\ldots d_k\), \(f(M)=M\), \(t'=\bar t\). Thus the product of the displacements \(d_i\) is a mapping \(f\) of the space \(A_n\) onto itself, as defined in item 1. If, in addition, we perform any affine mapping \(f_0\) of \(A_n\) onto \(A_n'\), then we obtain a mapping \(f\) of \(A_n\) onto \(A_n'\):
\[ f=f_0 d_1\ldots d_k. \tag{5} \]
At the same time the following holds.
Theorem 1. Let the mapping \(f\), defined in item 1, be continuous (so that \(m=n\)). Then, if the set \(M\) is not a quasicylinder, \(f\) is affine. If \(M\) is a nondegenerate quasicylinder having exactly \(k\) generators \(l_i\) and base planes \(P_i\), then \(f\) is representable in the form (5), where \(f_0\) is an affine mapping of \(A_n\) onto \(A_n'\) and \(d_i\) are the displacements \(d_{P_i l_i}\). Moreover, one may take \(f_0(M)=f(M)\), so that also in this case the set \(M'=f(M)\) is an affine image of \(M\).
Finally, if \(M\) is a degenerate quasicylinder, then continuous mappings \(f\) may be quite arbitrary.
However, they are easy to characterize, relying on the preceding assertions of the theorem. For example, if for a base plane \(P\) there is a \(P_1 \parallel P\) such that \(M\cap P_1\) is not a quasicylinder in \(P_1\) (and is nonempty), then \(f\) is affine on every plane \(P'\parallel P\).
Note that for a nondegenerate quasicylinder the displacements \(d_i\) commute, so that their order in (5) is immaterial, and repeated application of them leads to the same formula (5). This follows from the fact that, for a nondegenerate quasicylinder, when \(i\ne j\), \(l_i \parallel P_j\) and \(l_j \parallel P_i\).
- Theorem 1 reduces the question of the nature of the mappings \(f\) to the question of the conditions for their continuity. To formulate such, possibly more general, conditions, we introduce the following construction.
Fix a point \(O\in A_n\) and, putting \(M=M_0\), define by induction the sets \(M_i\): \(M_i\) is the union of all \(t(M_{i-1})\) that contain \(O\):
\[ M_i=\bigcup t(M_{i-1}); \qquad O\in t(M_{i-1}), \quad M_0=M. \tag{6} \]
Theorem 2. If \(M'=f(M)\) is bounded, then \(f\) is continuous under one of the following conditions: (I) among the sets \(M_i\) there are open ones; (II) among the \(M_i\) there are closed sets with interior points and \(f^{-1}\) is bounded.
On the other hand, for any given \(M\) there exist discontinuous mappings \(f\) of the space \(A_n\) onto any \(A_m\).
Let us note that condition (I) of Theorem 2 is certainly satisfied if \(M\) itself is open, or even \(M=G\setminus N\), where \(G\) is open and \(\overline N\subset G\). Likewise, condition (II) on \(M_i\) is satisfied if \(M\) itself is closed with interior points, or \(M=F\setminus N\), where \(F\) is closed with interior points and \(N\) is contained inside \(F\). Therefore, in particular, \(f\) is continuous if \(M\) is the boundary of a domain and \(M'\) is bounded, and \(f^{-1}\) is bounded. At the same time, if \(M\) is closed but none of the sets \(M_i\) has interior points, then a discontinuous mapping \(f\) is certainly possible, even if \(M'\) is bounded and \(f^{-1}\) is bounded.
If among the \(M_i\) there are closed sets with interior points, then the requirement that \(f^{-1}\) be bounded is probably superfluous, but we can prove this only in special cases, for example when some \(M_i\) is a polyhedron or a strictly convex body. In general this requirement can be replaced, for example, by one of the two following requirements: (a) the \(f(M_i)\) are measurable; (b) the interior measure of one of the \(f(M_i)\) is positive.
Theorem 2 can also be supplemented by conditions ensuring the continuity of \(f\) also when none of the sets \(M_i\) is either open or closed. However, the question of necessary and sufficient conditions on \(M\) under which \(f\) must be continuous (for bounded \(M'\)) remains open.
- The construction (6) of the sets \(M_i\) can be represented as follows.
Take the point \(O\) as the origin of the vectors. Then, as is easy to see, the set \(M_i\) is formed by the endpoints of all vectors \(y-x\), where the endpoints of the vectors \(y\) and \(x\) independently run through the whole set \(M_{i-1}\). Thus, \(M_i\) is the set of endpoints of the vectors
\[ \sum_{1}^{i}(y_j-x_j), \]
where the endpoints \(y_j\) and \(x_j\) run through the set \(M\).
The construction therefore consists in the successive construction of the additive group of vectors generated by the set \(M\) as a set of vectors. Hence the condition that the sets \(M_i\) cover all of \(A_n\),
\[ \bigcup_{i=0}^{\infty} M_i = A_n, \tag{7} \]
is equivalent to saying that \(M\) is a generating set of the entire group \(A_n\).
If among the \(M_i\) there are sets having interior points, then (7) is obviously satisfied. Let us also note that if some \(M_k\) is closed or open, then the same is true for all \(M_i\), \(i>k\). Therefore, in particular, if \(M_k\) is closed and (7) holds, then the whole space \(A_n\) is represented as the sum of a countable number of closed sets \(M_k\subset M_{k+1}\subset\cdots\). Then, by known theorems of dimension theory (see, for example, \(({}^1)\)), among the \(M_{k+i}\) there is a set with interior points. Thus, the condition of Theorem 2 that among the \(M_i\) there is a closed set with interior points is equivalent to saying that among them there are closed sets and that (7) holds.
If (7) is not satisfied, then the mapping \(f\) certainly may fail to be continuous, even if \(M'\) is bounded and \(f^{-1}\) is bounded. Indeed, if \(\bigcup M_i=N\ne A_n\), then take in \(A_n\) a point \(O_1\) not belonging to \(N\). Construct around it the sets \(M_{i1}\) in the same way as the \(M_i\) are constructed around \(O\), and form their sum \(N_1\). In the sense of the group of vectors, \(N\) is a subgroup of the whole group \(A_n\), and \(N_1\) is its adjacent coset, so that \(N\cap N_1=\varnothing\). Interchanging \(N\) and \(N_1\) by parallel translations, we obtain a discontinuous mapping \(f\) of the space \(A_n\) onto itself.
It seems probable that, conversely, for the continuity of \(f\) it is sufficient that (7) be satisfied, provided only that \(M'\) is bounded.
- We give one consequence of our results. Let a metric \(\rho(XY)\) be given in the space \(A_n\), satisfying the condition of invariance under parallel translation. The sets
\[ \{Y:\rho(XY)\leq 1\},\qquad \{Y:\rho(XY)<1\},\qquad \{Y:\rho(XY)=1\} \]
will be, respectively, the unit “ball,” the open ball, and the “sphere” with center \(X\); a translation of the center \(X\to X_1\) induces the same translation of each of them.
It is known \((^2,^3)\) that if in the spaces \(A_n, A_n'\) there are translation-invariant metrics \(\rho,\rho'\) and there is an isometric mapping \(g\) of \(A_n\) onto \(A_n'\), then \(g\) is affine. From our Theorems 1 and 2, however, a much stronger result follows immediately.
Theorem 3. Let translation-invariant metrics be given in the spaces \(A_n, A_n'\), and suppose that the unit ball \(S\) in \(A_n\) is not a quasicylinder. Let \(g\) be a one-to-one mapping of \(A_n\) onto \(A_n'\) under which either unit balls, or open unit balls, or unit spheres in \(A_n\) are mapped onto the corresponding sets in \(A_n'\), and conversely. Then \(g\) is affine.
If, however, \(S\) is a quasicylinder, then \(g\) need not be affine, and the definitions are as in Theorem 1 (so that, in particular, the unit ball \(S'\) in \(A_n'\) is nevertheless an affine image of \(S\)). But for \(g\) to be affine it is sufficient, for example, that for each generator \(l_i\) of the ball \(S\) there be an irrational \(\alpha_i\) such that for every pair of points \(X,Y\) with \(\overline{XY}=\alpha_i l_i\) one has \(\rho(XY)=\rho'(X'Y')\), where \(X'=g(X),\ Y'=g(Y)\) and \(\rho,\rho'\) are metrics in \(A_n,A_n'\).
In view of the generality of Theorems 1 and 2, the symmetry of the metric, i.e. the condition \(\rho(XY)=\rho(YX)\), is not at all essential.
- The proof of Theorems 1 and 2, though elementary, is somewhat too complicated for us to outline it here; only the proof of the first part of Theorem 2 for open \(M\) turns out to be simple.
The construction (6) of the sets \(M_i,\ i\geq 1\), plays the main role. First, it is easy to see that the point \(O\) will be their center of symmetry. Further, from the definition of the mapping \(f\) it follows directly that the sets \(M_i'=f(M_i)\) are constructed from \(M'\) in exactly the same way about the point \(O'=f(O)\). In general, if \(M_{iX}\) is constructed about the point \(X\), then \(M'_{iX'}=f(M_{iX})\) is constructed in the same way about \(X'=f(X)\). At the same time \(M_{iX}\) is obtained from \(M_i\) by the translation \(t:O\to X\), and \(M'_{iX'}\) from \(M_i'\) by the translation \(t':O'\to X'\). Therefore for the sets \(M_i\) the same formula (1) holds:
\[ ft(M_i)=t'f(M_i) \]
with the important additional fact that the center \(t(M_i)\) is mapped to the center \(t'f(M_i)\). In view of this we may, instead of the given \(M\), consider any of the sets \(M_i\); and they are “better arranged.” This is the starting point of our arguments.
Received
27 X 1969
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