Frankl, F.; Pontrjagin, L. (1930). "Ein Knotensatz mit Anwendung auf die Dimensionstheorie". Math. Annalen 102 (1): 785–789. doi:10.1007/BF01782377.
Translated by Google Translate, no further editing.
----------------------------------------------------
The following is proved that any (and every aueh knotted) simple closed
polygon of R3 edge of a singularity-free orientable Manifold is. For
example, the trefoil knot edge of a surface of genus 1 With the help of
this theorem is then shown that fo: r compact parts R3 of the Alexandro.
ffsche Dimensionsbegri: tP) by Brouwer MengerUryoohnschen match 2). We
first prove the theorem on polygons. So be a p simple closed polygon of
R3, so we project it from a outlying point n, creating a cone is formed.
The point n we choose in general position to the edges and vertices of
the Polygon. Then occur at most double generatrix. we triangulate the
cone so that each double generatrix (ie the distance between the Point
and the first point of intersection with the node) as an edge de: r
occurs triangulation. The triangles are consistent with a particular-
voted oriented sense of circulation of the node. We now want the area
change at the double generatrix and at the top so that it to a is
singularity-free two-sided variety. Along a double generatrix a bump
together exactly four triangles.
We number them according to their cyclical succession in space with L1l,
L1: p. L18, L14 in such a way that a in Al. oppositely oriented and L12
appear (and ebensQ course in L13 and L14). Lf1 L1 and close the angle
space A L14 A3 and the angle B is a space. We now connect the center of
a of a with the opposite corner of Ai (i = 1, 2, 3, 4). thereby LJI
decays into two triangles and AI LJ: '. We are now in A in near "a
point" A and B as in a point "B and 'LJ' by triangles LJ, LJ, .1,, LJ,
who, the triangles LJ, LJ2, LJ replace instead of "have the corner" A,
and similarly the triangle .. 1, LI, LJ, 'LJ' by triangles J3, LJ, 13,
14 If aA and "B sufficiently close to ge« · selects and we run this
process on all double generatrix Thus from iJ> creates a surface
oriented iJ> ', in which each of J; Cante the Triangulation occurs more
than twice. Only at the point 7t remains still a singularity, which is
to resolve as follows: We set · n to a small ball K. These cuts iP 'in
finitely many foreign simple closed curves, in which one resident inside
of K, simply can clamp contiguous areas. We replace it with these areas
in the interior of K lying part of iJ> 'get so we are an area P, which
all: demands are met. The accompanying figure represents one by above
method for the trefoil knot constructed Area represents We now prove the
following Lemma: Let M one polyedrale orientable bounded two-dimensional
Manifold in R3 and r its edge, then there are just as many such
manifolds M1, M2, ..., Mn of the rim R, which but otherwise either among
themselves or with M points common.
We first show that such arbitrarily close to M Manifold M1 to the edge r
indicates which otherwise with M no. Point in common. For this purpose
we decompose the R8 simplicial so that M consists of triangles of this
triangulation, and each tetrahedron a maximum of one (nu1l, one or
zweidi.Jp.ensionales) edge element with M common with. We consider all
tetrahedra, which border element M is a common, which is not on r, and
under all these, based on a given side of M lie. Let T be such a
tetrahedron, and the x common border with M element; y x is the opposite
edge · element of T. We consider all links of points of x with points of
y and consider the set of their centers, one is two-dimensional element.
All resulting elements together form a manifold M, the M not
schneidet3). Its edge is r; he herandet with r a complex of N, the group
consisting of: a plurality of cylindrical strip (each corresponding to a
boundary curve of M). M1 = M + N then has the required property. -
Repetition this configuration, we reach as many areas MP M3, • • •, Mn.
We conclude another lemma, namely: Z is a cylinder (ie, the closed area,
the shell of the and the two Basisfiächen) consists. «1 and« 2 are the
endpoints of its Axis. These endpoints were to Z by several lines 11,
12, • • •, ln , said li = ril + egg + ri2, ei a generatrix of the
cylinder and rli ri and radii of the two base circles. The numbering is
the cyclic sequence of radii in the base areas correspond. Also were "1
and a, p in the interior of Z by a regular polygon connected. Then there
is the interior of Z two-sided manifolds Wed p with the edge + li, which
are up to foreign points of p to each other. B e i we see we first
construct, according to previous records Areas N1, N2, • • •, Nn with
the edge p + l1, which is not otherwise intersect, they are arranged
around the axis in the same sense of rotation as li; as is easily seen,
it can be chosen so that they in Interior of Z lie. We now consider a Z
lying in the interior of (For instance in relation to the center of
homothetic) slightly smaller Cylinder Z '. We set li = Ni · Z 'and
describe the inside of Z' downstream part of Ni with Ni. From Z 'project
two routes p1 and p2 of p out. In the closed curves p1 + p2 + li + li,
you can now pairwise disjoint, between Z and Z 'preferred orientable
surfaces Oi Clamp 4). The surfaces Mi = Nf + Oi then fulfill our
demands. We now proceed to prove the theoretical dimension set. For this
purpose, we rely on Alexandro: ffs "doctrine of justification" 5), the
question for us coming into special case reads as follows: In a compact
subset F of R3 itself is then and only then in Meaning of Alexa.ndro. Ff
most one-dimensional, if there has been no the R decomposed third There.
the Alexa.ndro. ffsche dimension is never greater than the
Brouwer-Menger-Urysohn 6), we need only show the following: if F
decomposed no area of R3, then F in the sense of Brouwer, Menger and
Drysohn) at most one-dimensional seventh
To prove this we decompose the simplicial R 3 such that no vertex this
triangulation in F contain jst. There. F be nowhere dense in R 3 needs,
there are arbitrary fine decompositions of this kind Let k an edge this
Decomposition and S the union of all tetrahedra with the edge we k now
construct a cylinder Z, whose axis is a part of k, in the Interior of S
is the average, and k · F contains inside, and this we do for each edge
in a way that all the resulting cylinder are alien to each other.
According to the doctrine of justification, it is now possible the
endpoints of the axis Z in the interior of Z by a regular To combine
polygon p, which is alien to F. Now let T1, T2, • • •, Tn the
tetrahedron with the edge k, according to their cyclical arrangement
numbered in space, and Ai the triangles with the side k (where J1 = T1 •
T2, etc.). If we set li = Z · Ai, we can at Z and the P li lines apply
the lemma just proved. we get Areas Mi, which the interior of Z, Y (Z)
(in a manner analogous to the Ai) divide into n parts, wherein in place
of Y (Z) · T, occurs approximately Wed we change Ti from now so that J
(Z) · Ti is replaced by Mi, and lead this process for all edges and
tetrahedra by. Takes the place of the tetrahedron T then a
three-dimensional manifold with boundary U. Let F now as Sum of all F ·
U represents, so have at most two common points of these parts (ie, if
the corresponding U that share a face). because these parts, by a
sufficiently fine division of the space originates, can be made
arbitrarily small, F is after the patch set of dimension at most
one-dimensional theory in the sense of Brouwer, Menger and Urysohn.
Z usa tz both K or re cture (7th 10th 1929). That in the above Proof of
the curves p li generally not accepted unknotted can be shown by the
following example which Alexandro. ff and Gruzewski Have found (Warsaw)
independently. It starts from the cube 0