From: zare@cco.caltech.edu (Douglas J. Zare) Newsgroups: rec.puzzles,sci.math Subject: Re: Tetrahedron into a cube [Spoiler] Date: 13 May 1998 11:28:48 GMT Nils Kr. Rossing wrote: >[...] >I have a tetrahedron. Is it possible to divide this tetrahedron in a >number of pieces in such way that it is possible to put these pieces >together in a new way and get a cube with of course the same volume? >[...] Hilbert's 3rd problem was to show that this could not be done for all tetrahedra if the pieces must be polyhedra. Dehn solved this problem shortly afterwards by constructing an invariant of geometric decomposition. (Sum over the edges length(x)dihedral angle --> R(x)R/pi where (x) means tensor product over the field of rational numbers.) This invariant is 0 for any cube and nonzero for any regular tetrahedron, so no regular tetrahedron can be decomposed to produce a cube. Any polyhedron which tiles space has Dehn invariant 0, and since the Dehn invariant is an exact invariant (I believe this was proved by Snyder(?) in the 1950's) all of these are equidecomposable with cubes. See http://www.its.caltech.edu/~zare/dehninvstuff.html for a more complete discussion which does not require one to be familiar with the tensor product. Douglas Zare ============================================================================== Date: Wed, 14 Oct 1998 10:04:56 -0400 From: ruberman@BINAH.CC.BRANDEIS.EDU (Daniel Ruberman) Subject: Re: Cube & tetrahedron: was: Russian 53 To: rusin@vesuvius.math.niu.edu (Dave Rusin) > Nils Kr. Rossing wrote: > >[...] > >I have a tetrahedron. Is it possible to divide this tetrahedron in a > >number of pieces in such way that it is possible to put these pieces > >together in a new way and get a cube with of course the same volume? > >[...] > > Hilbert's 3rd problem was to show that this could not be done for all > tetrahedra if the pieces must be polyhedra. Dehn solved this problem > shortly afterwards by constructing an invariant of geometric > decomposition. (Sum over the edges length(x)dihedral angle --> R(x)R/pi > where (x) means tensor product over the field of rational numbers.) This > invariant is 0 for any cube and nonzero for any regular tetrahedron, so no > regular tetrahedron can be decomposed to produce a cube. > > Any polyhedron which tiles space has Dehn invariant 0, and since the Dehn > invariant is an exact invariant (I believe this was proved by Snyder(?) in > the 1950's) all of these are equidecomposable with cubes. > > See http://www.its.caltech.edu/~zare/dehninvstuff.html for a more complete > discussion which does not require one to be familiar with the tensor > product. > > Douglas Zare The complete invariant is the volume, together with the Dehn invariant. See Chi-Han Sah's book, "Hilbert's third problem : scissors congruence", San Francisco : Pitman Advanced Publishing Program, c1979. ============================================================================== From: Dave Rusin Date: Wed, 14 Oct 1998 09:36:02 -0500 (CDT) To: ruberman@BINAH.CC.BRANDEIS.EDU Subject: Re: Cube & tetrahedron: was: Russian 53 Thanks. Is this then the correct statement: "Let X and Y be two (finite) polyhedra in R^3. Then: there exist finite partitions X = Union A_i and Y = Union B_i with A_i congruent to B_i for each i, IFF vol(X)=vol(Y) and Dehn(X)=Dehn(Y)." ? If so, to whom is this due? Is there a constructive proof of the existence of the decompositions? (It's going to be nontrivial, e.g. splitting a sqrt(2) x sqrt(2) x 1 box into parts which reassemble into a 2 x 1 x 1 box cannot be done with cuts parallel to box sides). dave ============================================================================== Date: Wed, 14 Oct 1998 11:01:46 -0400 From: Daniel Ruberman Subject: Re: Cube & tetrahedron: was: Russian 53 To: Dave Rusin Dave: I think you've got the statement correct. I think the original reference is J.-P. Sydler's solution [Comment. Math. Helv. 40 (1965), 43--80; MR 33 #632] , although you might find the book of Sah easier going. (This is by reputation--I've never looked at Sydler's paper.) There is also the name of Jessen associated to the problem; he algebraicized Sydler's solution, and reputedly simplified it. If you search math reviews, looking for Dehn invariant', you'll get some of the key references, including an inkling of the link between this innocent geometric problem and very deep issues in homological algebra and algebraic K-theory. I don't know if anyone has made an algorithm from Jessen/Sydler's solution; I would guess it would be complicated in practice (as your example suggests), but exists in principle. Have fun, Danny Ruberman [previous letter was quoted -- djr] ============================================================================== Date: Wed, 14 Oct 1998 11:07:47 -0400 From: Daniel Ruberman Subject: Re: Cube & tetrahedron: was: Russian 53 To: Dave Rusin Dave: By the way, the example you give is easy: take a sqrt(2) x sqrt(2) square, and split it into 4 triangles along the two diagonals. These reassemble to give a 1 x 2 rectangle. The fact that you can do this in the plane (and hence for all rectangular prisms) was presumably part of Hilbert's motivation for the problem in the first place. Is the 2d result constructive? Danny >? If so, to whom is this due? Is there a constructive proof of the existence >of the decompositions? (It's going to be nontrivial, e.g. splitting a >sqrt(2) x sqrt(2) x 1 box into parts which reassemble into a 2 x 1 x 1 box >cannot be done with cuts parallel to box sides). > >dave ============================================================================== From: Dave Rusin Date: Wed, 14 Oct 1998 11:28:58 -0500 (CDT) To: ruberman@BINAH.CC.BRANDEIS.EDU Subject: Re: Cube & tetrahedron: was: Russian 53 >By the way, the example you give is easy: Yes, I know; it's just a little non-intuitive. I guess a better question is how to decompose a 2 x 1 x 1 block into a cube? >Is the 2d result constructive? I don't know; I guess I should've started with that... dave ============================================================================== From: Robin Chapman Newsgroups: sci.math Subject: Re: Cube & tetrahedron: was: Russian 53 Date: Wed, 14 Oct 1998 08:13:13 GMT In article <7015ns$11m@staff.cs.usyd.edu.au>, ftww@cs.usyd.edu.au (Fred the Wonder Worm) wrote: > > In article , > Gerry Myerson wrote: > >In article <6vvk1v$624$1@gannett.math.niu.edu>, > >rusin@vesuvius.math.niu.edu (Dave Rusin) wrote: > >=> In article <6vvee8$4n0$2@cantuc.canterbury.ac.nz>, > >=> Bill Taylor wrote: > >=>> So is it possible to carve up a cube into a finite number of regions and > >=>> reassemble them into a single tetrahedron? > >=> > >=> [ Quoting Douglas Zare: ] > >=> This invariant is 0 for any cube and nonzero for any regular tetrahedron, > >=> so no regular tetrahedron can be decomposed to produce a cube. > > > > But Bill didn't ask about a *regular* tetrahedron. On the face of it, > > the Dehn result says nothing about the general case. > > From part of that post by Douglas Zare: > > > Any polyhedron which tiles space has Dehn invariant 0, and since the Dehn > > invariant is an exact invariant (I believe this was proved by Snyder(?) in > > the 1950's) all of these are equidecomposable with cubes. Does Snyder's proof give a construction for decomposing polyhedra with equal volumes and Dehn invariants into each other? > So if there is a tetrahedron which tiles space then there is such a > decomposition. I'm not sure what restriction is placed on the tiling, > however - I assume that rotations are allowed, but what about enantiomorphs? > The tetrahedron formed by joining (0,0,0), (1,0,0), (1,1,0) and (1,1,1) > tiles space together with its enantiomorph (it is trivial to construct a > cube out of these), but not by itself. Is its Dehn invariant still 0? > > Actually, after a little more thought (using the rhombic dodecahedron as > a basis rather than the cube) I think that the tetrahedron with vertices > (0,-1,0) (0,1,0) (-1,0,1) (1,0,1) tiles space without requiring an > enantiomorph. So this should be equidecomposable with the cube. I can > make 6 of them out of one cube, and hence 6 of them out of 6 cubes, but > not 1 out of 1 cube yet. You can reduce to decomposing a 6 x 1 x 1 block into a cube. One can't do this by Euclidean methods though! In general, accepting Synder's results, one should be able to decompose any rectangular block into a cube. I must confess, I haven't a clue how. Robin Chapman + "They did not have proper Room 811, Laver Building - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20 -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: ftww@staff.cs.usyd.edu.au (Geoff Bailey) Newsgroups: sci.math Subject: Re: Cube & tetrahedron: was: Russian 53 Date: 15 Oct 1998 15:30:13 +1000 In article <701meo$b08$1@nnrp1.dejanews.com>, Robin Chapman wrote: >In article <7015ns$11m@staff.cs.usyd.edu.au>, > ftww@cs.usyd.edu.au (Fred the Wonder Worm) wrote: > >>> Any polyhedron which tiles space has Dehn invariant 0, and since the Dehn >>> invariant is an exact invariant (I believe this was proved by Snyder(?) in >>> the 1950's) all of these are equidecomposable with cubes. > > Does Snyder's proof give a construction for decomposing polyhedra with > equal volumes and Dehn invariants into each other? No idea - I was just quoting. >> Actually, after a little more thought (using the rhombic dodecahedron as >> a basis rather than the cube) I think that the tetrahedron with vertices >> (0,-1,0) (0,1,0) (-1,0,1) (1,0,1) tiles space without requiring an >> enantiomorph. So this should be equidecomposable with the cube. I can >> make 6 of them out of one cube, and hence 6 of them out of 6 cubes, but >> not 1 out of 1 cube yet. Sorry, those 6s should have been 12s. I've looked at this further and I believe I can describe a decomposition, although it is not very good. The specified tetrahedron does have Dehn invariant 0, which is a good sign. (Assuming that I calculated correctly, using the definition in http://www.cco.caltech.edu/~zare/dehninvstuff.html. I make the dihedrals angles pi/2 and pi/3, hence the invariant is 0.) Description is below. > You can reduce to decomposing a 6 x 1 x 1 block into a cube. One can't > do this by Euclidean methods though! In general, accepting Synder's results, > one should be able to decompose any rectangular block into a cube. > I must confess, I haven't a clue how. I'm not sure what reduction you are talking about, but I will describe the procedure for decomposing rectangular blocks into equal volume blocks as part of the description for the decomposition anyway. Here goes: 1) A cube can be decomposed into 12 such tetrahedra as described above. Join the centre of the cube to each vertex, giving a decomposition into 6 square pyramids. Join the centre of each square to the vertices in that square and the centre of the cube, giving a decomposition into 24 tetrahedra. If the cube "wraps around" on each face then these tetrahedra are joined in pairs along the faces which were part of the cube. The resulting 12 such tetrahedra are all congruent to the one described. This is perhaps not very clear. If we take the cube with vertices at (+-1, +-1, +-1) then two of the original tetrahedra which become joined have vertices: (0, 0, 0) (0, 0, 1) (1, -1, 1) (-1, -1, 1) and (0, 0, 0) (0, 0, -1) (1, -1, -1) (-1, -1, -1) They are joined along the faces z = 1 and z = -1 to give a tetrahedron whose vertices are a translation of: (0, 0, 0) (0, 0, 2) (1, -1, 1) (-1, -1, 1) 2) n^3 congruent such tetrahedra can form a similar tetrahedron n times as big. (Handwavy a bit - construct some models to convince yourself of it or point out that I am wrong.) Take the n-times-as-big tetrahedron and n-sect every edge, making the obvious joins to turn it into the union of tetrahedra and octahedra. The tetrahedra are similar to the original and the octahedra can be cut into four such tetrahedra by cutting along the diagonals of the square. The net result is a dissection into n^3 congruent tetrahedra of the desired type. Reversing it gives the desired assembly. 3) A cube can be decomposed into any rectangular prism. This is a standard technique in dissection, although usually it is done with squares to rectangles (or vice versa). To change a square into a given rectangle with side ratio between 1/4 and 4 (inclusive), use the following dissection: +---------------+ |~-_ | | ~-_ | +------~-=------+-------+ | ~-_ | | | ~-_| | | T-_ | | | ~-_ | +---------------+-----~-+ For ratios outside that range, chop the rectangle in half and stack on itself until the ratio lies in that range. Perform the dissection and then unstack to get the rectangle. To get a cube into an arbitrary rectangular prism, perform the above dissection on one face to get one edge to the required length. Then operating on the orthogonal face to that edge change it to a square and then to the appropriate rectangle. 4) A cube can be decomposed into a tetrahedron similar to the given one. Using (3), the cube can be dissected into a rectangular prism with sides in the ratio 2:3:3. This can then be dissected into 18 smaller cubes, each of which can be dissected by (1) into 12 small tetrahedra. These 216 tetrahedra can be formed into the desired tetrahedron by (2). Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.usyd.edu.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ============================================================================== From: ruberman@binah.cc.brandeis.edu (Daniel Ruberman) Newsgroups: sci.math Subject: Re: Cube & tetrahedron: was: Russian 53 Date: Wed, 14 Oct 1998 10:04:04 -0400 > Nils Kr. Rossing wrote: > >[...] > >I have a tetrahedron. Is it possible to divide this tetrahedron in a > >number of pieces in such way that it is possible to put these pieces > >together in a new way and get a cube with of course the same volume? > >[...] > > Hilbert's 3rd problem was to show that this could not be done for all > tetrahedra if the pieces must be polyhedra. Dehn solved this problem > shortly afterwards by constructing an invariant of geometric > decomposition. (Sum over the edges length(x)dihedral angle --> R(x)R/pi > where (x) means tensor product over the field of rational numbers.) This > invariant is 0 for any cube and nonzero for any regular tetrahedron, so no > regular tetrahedron can be decomposed to produce a cube. > > Any polyhedron which tiles space has Dehn invariant 0, and since the Dehn > invariant is an exact invariant (I believe this was proved by Snyder(?) in > the 1950's) all of these are equidecomposable with cubes. > > See http://www.its.caltech.edu/~zare/dehninvstuff.html for a more complete > discussion which does not require one to be familiar with the tensor > product. > > Douglas Zare The complete invariant is the volume, together with the Dehn invariant. See Chi-Han Sah's book, "Hilbert's third problem : scissors congruence", San Francisco : Pitman Advanced Publishing Program, c1979. ============================================================================== Date: Thu, 15 Oct 1998 09:32:58 +0100 (BST) From: "Dr R.J Chapman" To: rusin@math.niu.edu Subject: Re: Cube & tetrahedron: was: Russian 53 > How about the 2D version -- is it clear how to decompose polygons of equal > area into congruent parts? No it isn't. I vaguel recall seeing a proof many years ago, but please don't challenge me to recreate it! At least in two dimensions it is not a priori impossible to contruct the pieces within the Euclidean constraints. Do you know whether one can construct a dissection in general within these constraints? Robin ============================================================================== Subject: scissors congruences From: AMS ! Date: Nov 24 1998 00:23 CDT (from timestamp) To: rusin 91b:22016 22E41 (19D55 22E40 52B45 57Q99) Dupont, Johan L.(DK-ARHS); Sah, Chih-Han(1-SUNYS) Homology of Euclidean groups of motions made discrete and Euclidean scissors congruences. Acta Math. 164 (1990), no. 1-2, 1--27. The authors pursue their work on scissors-congruence problems. The main purpose of their paper is to give a new proof of the classical theorem of J. P. Sydler about the completeness of volume and Dehn invariants, for the scissors congruence of polyhedra in Euclidean 3-space [Comment. Math. Helv. 40 (1965), 43--80; MR 33 #632; see also P. Cartier, Asterisque No. 133-134 (1986), 261--288; MR 87i:52010]. Here are the main features of this work. The proof of Sydler used complicated geometrical arguments: here the geometrical arguments are reduced to very simple and actually very elementary ones. The proof is homological: this is a line of investigation initiated in the subject by B. Jessen, who gave a first striking simplification of the proof of Sydler [Math. Scand. 22 (1968), 241--256; MR 40 #4860]; by earlier work of Dupont [Osaka J. Math. 19 (1982), no. 3, 599--641; MR 85f:52014], Sydler's theorem was known to be equivalent to a fact in homology of $\roman{SO}(3,\bold R)$ considered as a discrete group, namely $H\sb 2(\roman{SO}(3,\bold R),\bold R\sp 3)=0$, but there was no direct algebraic proof of this fact: the paper under review fills the gap, so that Sydler's theorem is now definitely integrated into the current mathematical trend. There is a clever use of quaternions and classical isomorphisms of classical groups. Hochschild homology enters the scene, and particularly an involution on it which already appeared, as the dihedral involution, in the study of cyclic homology. If it were still necessary after previous work of the two authors, all this reinforces the idea that scissors-congruence problems have to be studied through the homology of classical and algebraic groups: this point of view provides interesting perspectives towards algebraic $K$-theory and cyclic homology, where such homological problems are also considered. Reviewed by Jean-Louis Cathelineau _________________________________________________________________ Next Review 90h:57054 57T10 (17B55 22E41 22E99) Cathelineau, J. L.(F-NICE) Sur l'homologie de ${\rm SL}\sb 2$ à coefficients dans l'action adjointe. (French) [${\rm SL}\sb 2$-homology with coefficients in the adjoint action] Math. Scand. 63 (1988), no. 1, 51--86. _________________________________________________________________ The main result: Let $k$ be an algebraically closed field of characteristic zero; then $H\sb 1(\roman{SL}(2,k),\roman{sl} (2,k))\cong\Omega\sp 1\sb k$, $H\sb 2(\roman{SL}(2,k),\roman{sl} (2,k))=0$. Here $\roman{sl}(2,k)$ is taken with its $\roman{SL}(2,k)$-module structure with respect to the adjoint representation; $\Omega\sp 1\sb k$ is the $k$-vector space of Kahler differentials, i.e. the space generated by the symbols $da$, $a\in k$, subject to the relations $d(ab)=a\,db+b\,da$, $d(a+b)=da+db$. As the author states the known related results $H\sb 1(\roman{SO}(3), \roman{so}(3))\cong\Omega\sp 1\sb R$, $H\sb 2(\roman{SO}(3), \roman{so}(3))=0$, where SO(3) and so(3) are taken with their discrete structure [cf. J. L. Dupont, Osaka J. Math. 19 (1982), no. 3, 599--641; MR 85f:52014] cannot be obtained by descent from $\bold C$ to $\bold R$ from the above main result. However, the above result is obtained purely by algebraic methods, whereas the corresponding result for discrete SO(3) needs (up to now) the Sydler scissors congruence result [J. P. Sydler, Comment. Math. Helv. 40 (1965), 43--80; MR 32 #1098]. Nevertheless the method employed by the author is in many respects strongly reminiscent of the proof of Sydler's result by B. Jessen [Math. Scand. 22 (1968), 241--256; MR 40 \#4068]. In any case this applies among others to the so-called $d\log$ exact sequence $0\to{\scr B}\sb k\overset D\to \to k\sp \times\otimes k\sp +\overset L\to\to \Omega\sp 1\sb k\to 0$ established by the author as a first crucial step toward the proof of the main result. Here ${\scr B}\sb k$ denotes the $k$-vector space generated by the symbols $[a]$, $a\in k\sbs\{0,1\}$ subject to the relations $[a]=[1-a],[1/a]=-[a]/a$, $[a]-[b]+(1-a)[(1-b)/(1-a)]+ a[b/a]=0$, $a,b\in k\sbs\{0,1\}$, $a\neq b$, and furthermore $D\:[a]\mapsto a\otimes a+(1-a)\otimes (1-a),L\:a\otimes b\mapsto b(da/a)$. By a series of somewhat delicate technicalities the $d\log$ exact sequence is shown to "map" into an exact sequence, similar to the one due to Bloch-Dupont-Sah, viz. $0\to H\sb 2(G,\germ g)\to(E\sb 1)\sb G\overset{D'}\to\to k\sp \times\otimes k\sp +\to H\sb 1(G, \germ g)\to 0$, in the following fashion: $$\matrix 0\to H\sb 2(G,\germ g) & \to & (E\sb 1)\sb G & \overset{D'}\to\to & k\sp \times\otimes k\sp + & \to & H\sb 1(G,\germ g)\to 0\\ & & {\ssize{\Lambda}}\uparrow & & \Vert \\ 0 & \to & {\scr B}\sb k & \overset D\to \to & k\sp \times\otimes k\sp + & \to & \Omega\sp 1\sb k\to 0 \endmatrix$$ in which $\Lambda$ turns out to be a bijection, and where $G=\roman{SL}(2,k)$, $\germ g=\roman{sl}(2,k)$ (both discrete). The commutativity of the diagram then automatically entails the main result. Apart from the main result the paper contains various interesting observations. Reviewed by W. T. van Est Cited in reviews: 91i:57023 _________________________________________________________________ 85f:52014 52A25 (20J05 57Q99) Dupont, Johan L. Algebra of polytopes and homology of flag complexes. Osaka J. Math. 19 (1982), no. 3, 599--641. Hilbert's third problem (H3P) began as a problem in the foundation of geometry: Show that there is no "elementary" geometric theory of volume for Euclidean polyhedra, as against the existence of such a theory for area. The immediate and difficult (for that time) solution by Dehn probably caused H3P to be overshadowed by H$n$P, $n\neq3$. Nevertheless, interest in H3P was kept alive by H. Hopf, Hadwiger, Jessen, Sydler, and others. In particular, Jessen's simplification of the tour de force converse to Dehn's result found by Sydler suggested connections with homological algebra. Recent investigations in differential geometry, algebraic $K$-theory (in the form of Eilenberg-Mac Lane homology theory of groups), etc., caused a revival of interest in variations of H3P. The present work puts forth a number of basic ideas, techniques, reformulations, new results, new problems, etc., connected with H3P. Section 1 gives a rapid and concise summary of the main results. A number of these have been extended. Section 2 gives an algebraic topological setting that enables one to study various extended forms of H3P. One of the main results is a resolution of some confusions (or "controversies") connected with H3P. Dehn's solution consisted of a set of necessary conditions that were identical to the set of necessary conditions found by R. Bricard in 1896 for a problem that appeared to be identical to H3P. Unfortunately, combinatorial topology was not yet established on firm grounds. The difficulty with equating the problem solved by Bricard and H3P was clearly exposed by H. Hopf in his 1946 lectures [see Differential geometry in the large, Lecture Notes in Math., 1000, Springer, Berlin, 1983; MR 85b:53001]. For Euclidean 3-space, the equivalence of these two problems may be deduced as a corollary of Sydler's result. For Euclidean $n$-space, the equivalence of obvious generalizations is supposed to be "well known" and a direct argument is contained in an unpublished Danish manuscript of A. Thorup. Theorem 2.3 in the present work gives a uniform proof for Euclidean, spherical and hyperbolic $n$-spaces. In particular, all of these settle (after the fact) that Bricard's problem was in fact equivalent to H3P. Section 3 deals with the translational version of the scissors congruence problem ($=$ cutting and pasting problem for polytopes $=$ extended H3P) in Euclidean $n$-space. The problem was first resolved by Jessen-Thorup through the use of Hadwiger invariants. In theory, the Euclidean version can be solved by describing the orthogonal invariants on the space of invariants "generated" by the Hadwiger invariants. Since there are "relations" among the Hadwiger invariants, the reviewer posed a "syzygy problem" in the hope of connecting up the extended Euclidean version of H3P with the Eilenberg-Mac Lane (co-)homology of ${\rm O}(n)$ with coefficients in suitable modules. In the present section, the idea of Section 2 is extended. Using the Tits complex, extending a proposition of Lusztig, as well as adopting spectral sequence techniques, the translational problem is reformulated. In particular, the Hadwiger invariants are identified. Section 4 deals with the Euclidean version of the scissors congruence problem in all dimensions and uses results from Section 3. The main point is that the scissors congruence group for Euclidean $n$-space is the module of coinvariants of the orthogonal group acting on the translational scissors congruence group. The "syzygy problem" can be resolved in the dual formulation through the use of an exact sequence of Lusztig (Proposition 4.7). With the present approach, various higher Dehn invariants can be identified with suitable higher differntials in suitable spectral sequences. The connection with various Eilenberg-Mac Lane homology of orthogonal groups (trivial as well as nontrivial coefficient modules) is clearly and firmly established. Section 5 deals with the spherical version of the scissors congruence problem. Steinberg modules (related to Tits geometry and representation theory) now enter the picture. The principal result is an exact sequence (Proposition 5.33$=$Theorem 1.3). The Cheeger-Chern-Simons invariant (connected with differential geometric investigations) may be identified. In particular, the (yet to be resolved) spherical version of H3P (in dimension 3) is shown to be equivalent to the injectivity problem of the Cheeger-Chern-Simons invariants on $H\sb 3({\rm SU}(2))$. There is a little bit of ambiguity on 2-torsion hidden by the use of Serre classes. However, this has been resolved in later works. This section also reviews Schlafli's formula for the volume of an "orthoscheme". The volume of a general spherical 3-simplex would require a cutting and pasting procedure reducing it to a suitable integral combination of orthoschemes. This is a rather messy process that requires clarification. The present section ends with a discussion of the relation with the Euclidean case. Section 6 deals with the hyperbolic version of the scissors congruence problem. The results are quite similar to those of section 5. A large number of the results in the present section have been extended in later works. It appears that H3P has as many interesting possibilities as H$n$P for $n\neq 3$ as pointed out by the present work. Reviewed by Chih-Han Sah Cited in reviews: 91g:52010 91b:22016 90h:57054 _________________________________________________________________ 82e:52012 52A45 Debrunner, Hans E. Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln. (German) Arch. Math. (Basel) 35 (1980), no. 6, 583--587 (1981). H. Hadwiger hat im Jahre 1963 die Vermutung vorgelegt, dass bei einer Pflasterung (P{sub} alpha : alpha {in}J) des euklidischen Raumes E {sup}d, in der die Polyeder paarweise kongruent sind, jedes dieser Polyeder P {sub} alpha zu einem inhaltsgleichen Wurfel zerlegungsgleich sei. Der Autor der vorliegenden Arbeit beweist die Vermutung von Hadwiger fur d=3 und d=4 ohne der Art des Pflasterns eine Einschrankung aufzuerlegen. Der Beweis benutzt das Dehnsche Funktional und stutzt sich auf Ergebnisse von J. P. Sydler [Comment. Math. Helv. 40 (1965), 43 - 80; MR 33 #632] und B. Jessen [Math. Scand. 22 (1968), 241 - 256 (1969); MR 40 #4860]. Reviewed by H. Kramer _________________________________________________________________ 81g:51011 51M20 (20H15 51M25 57S30) Sah, C. H. Hilbert's third problem: scissors congruence. Research Notes in Mathematics, 33. Pitman (Advanced Publishing Program), Boston, Mass.-London, 1979. vi+188 pp. ISBN 0-273-08426-7 The third Hilbert problem was solved (in a negative sense) by Dehn a few months after its formulation but this result generates a new problem: to find a complete system of invariants of a polyhedron with respect to scissors congruence. The present book contains an account of some of the recent developments associated with this problem that are not covered by the book of V. G. Boltjanskii (Hilbert's third problem, English translation, Winston, New York, 1978; MR 58 #18074). Its main feature is an algebraic approach to the problem. In Chapters 2 - o5 the author gives an account of the complete solution of the problem in the translational case. This includes a determination of the relations between the Hadwiger invariants. In Chapters 6 - 8 the spherical, Euclidean and hyperbolic scissors congruences are discussed. In the spherical case the scissors congruence classes (in all dimensions) form a ring, and a certain quotient ring of this ring is provided with a structure of a Hopf algebra, say A. In all three cases the group of scissors congruence classes is a comodule over A. This comodule structure is actually an algebraic version of Dehn's invariants. Besides the trivial case n <= 2, the sufficiency of the Dehn invariants is known only for n=3, 4 in the Euclidean case (Sydler, Jessen). The proof is given in Chapter 7. A number of open problems are formulated. Reviewed by E. Vinberg Cited in reviews: 91a:52017 84b:53062c _________________________________________________________________ 57 #7346 50A30 {\cyr Boltyanski\u\i, V. G.} {\cyr Tret\cprime ya problema Gil\cprime berta.} (Russian) [Hilbert's third problem] Izdat. `Nauka'', Moscow, 1977. 207 pp. The third of Hilbert's famous 23 problems asks for two tetrahedra in Euclidean 3-space, with congruent bases and equal heights, which can neither be dissected into congruent tetrahedra nor enlarged, by adjoining congruent tetrahedra, to polyhedra which can be so dissected. In short, the problem is to find polyhedra with equal volumes which are not equi-decomposable. The question goes back to the use of a continuity axiom for stereometric theorems in Euclid's Book XI, whereas this can be avoided in the proof of the corresponding plane results. M. Dehn solved the problem in 1900 by finding necessary conditions for equi-decomposability which certain polyhedra of equal volumes do not satisfy. This began the general study of the connection between equi-decomposability (with respect to some group of motions) and equality of volumes for polyhedra in Euclidean $n$-space; it is the topic of the book under review. The book is a complete re-working and extensive amplification of the author's earlier one [Equivalent and equidecomposable figures (Russian), Gostehizdat, Moscow, 1956; RZMat 1957 \#5070K; English translation, D. C. Heath, Boston, 1963; Turkish translation, Turk Matematik Dernegi, Istanbul, 1964; MR 37 #5761], which presented the fundamental results of H. Hadwiger and his school on this topic. This new work, written with the author's usual expository skill and geometric feeling, includes J. P. Sydler's proof of the sufficiency of Dehn's conditions for equi-decomposability in $R\sp 3$ and the research of B. Jessen, both as it affects Sydler's result and as it applies to higher dimensional questions. A table on pp. 198--199, appropriately dedicated to the memory of Hilbert, summarizes the current state of all these problems. This monograph fills the gap in the recent survey of Hilbert's problems [Mathematical developments arising from Hilbert problems (Proc. Sympos. Pure Math., Northern Illinois Univ., DeKalb, Ill., 1974), Amer. Math. Soc., Providence, R.I., 1976; MR 54 #7158] which lacks only a discussion of the third problem, though it is written at a much more elementary and discursive level than those survey articles. Reviewed by W. J. Firey Cited in reviews: 81b:52006 81b:01002 80k:28001 58 #18074 58 #12668 _________________________________________________________________ 50 #8303 52A25 (50B30) Mürner, P. Zwei Beispiele zur Zerlegungsgleichheit $4$ dimensionaler Polytope. (German) Elem. Math. 29 (1974), 132--135. Die Zerlegungsgleichheit und die translative Zerlegungsgleichheit von Polyedern des $E\sp 4$ ist gleich wie die fur Polyeder des $E\sp 3$ auf Grund der Kriterien von Dehn, Sydler, Hadwiger und Jessen anhand der Kantenwinkel und -inhalte kontrollierbar, doch ergeben sich dabei keine konstruktiven Hinweise, z.B. uber die Anzahl und Art der Zerlegungsstucke. Der Autor stellt hier zuerst anhand der genannten Kriterien die Zerlegungsverhaltnisse zwischen den regularen Polytopen des $E\sp 4$ zusammen und zeigt dann durch effektive geometrische Konstruktion: die Vereinigung zweier translationsgleicher Wurfel der Kantenlange 1 des $E\sp 4$ ist einerseits durch Zerstuckelung in 9 Teilpolyeder mit einem passend liegenden regularen 24-Zell der Kantenlange 1, anderseits durch Zerstuckelung in 18 Teilpolytope mit der Vereinigung von drei passend liegenden regularen 16-Zellen (Hyperoktaedern) der Kantenlange $\surd 2$ translativ zerlegungsgleich. Reviewed by H. E. Debrunner _________________________________________________________________ 44 #1963 39.38 (13.00) Jessen, Børge; Karpf, Jørgen; Thorup, Anders Some functional equations in groups and rings. Math. Scand. 22 1968 257--265 (1969). In connection with the simplified proof of the Dehn-Sydler theorem given by the first author [Math. Scand. 22 (1968), 241--256 (1969); MR 40 #4860] one is led to the study of certain functional equations of independent interest which are closely related to those appearing in homological algebra. Let $M$ be a module over a commutative integral domain $R$. For any function $f$ from $R$ to $M$, the functions $F$ and $G$ from $R\times R$ to $M$ defined by $F(a,b)=f(a+b)-f(a)-f(b)$, $G(a,b)=f(ab)-bf(a)-af(b)$ (which measure the deviation of $f$ from being a derivation) satisfy certain obvious functional equations. When $M$ is a torsion-free and divisible $R$-module, these functional equations are shown to characterize completely the functions $F$ and $G$ that arise from the above construction. Analogous results are proved for ordered integral domains and functions with positive arguments. For (ordered) abelian groups similar characterizations are obtained for functions measuring the deviation from linearity. Finally, the results are stated in the form in which they are actually needed in the proof of the Dehn-Sydler theorem, i.e., for vector spaces over the real field. Reviewed by Chr. U. Jensen Cited in reviews: 97a:39026 87h:39006 86m:39018 80k:39002 _________________________________________________________________ 40 #4860 50.90 Jessen, Børge The algebra of polyhedra and the Dehn-Sydler theorem. Math. Scand. 22 1968 241--256 (1969). Necessary conditions that two polyhedra of euclidean 3-space can be decomposed into pairwise congruent sub-polyhedra were given in 1901 by M. Dehn. Only recently J.-P. Sydler [Comment. Math. Helv. 40 (1965), 43--80; MR 33 #632] has succeeded in establishing that Dehn's conditions are also sufficient. The present paper gives a substantial simplification of Sydler's proof by turning geometric situations into algebraic ones and by operating on the latter. As a new byproduct it appears that the cardinality of linearly independent equivalence classes of polyhedra is $2\sp {\boldsymbol\aleph\sb 0}$. The algebraization of the problem proceeds as follows. The polyhedron group $\scr P$ is introduced as the free abelian group over the set of closed nondegenerate polyhedra of $E\sp 3$ as a base; $\scr E$ denotes the subgroup generated by all elements $A-A\sb 1-\cdots-A\sb n$ and $A-B$, where $A$ is geometrically composed of $A\sb 1,\cdots,A\sb n$ and $B$ is congruent to $A$, and $\scr F$ denotes the subgroup generated by $\scr E$ and all prisms. The natural projections $\kappa'\colon\scr P\rightarrow\scr P/\scr E$ and $\kappa\colon\scr P\rightarrow\scr P/\scr F$ then check equivalence and equivalence modulo prisms. Group homomorphisms on $\scr P$ are defined from their values on polyhedra, e.g., the volume, or "Dehn's function" $\Delta\colon\scr P\rightarrow R\otimes\sb {{z}}R/(\pi)$ by $\Delta(A)=\sum a\sb i\otimes\alpha\sb i$, where the summation runs over the edges of $A$, $a\sb i$ denoting the length and $\alpha\sb i$ the corresponding dihedral angle. One easily sees that $C,D\in\scr P$ are equivalent if and only if they are equivalent modulo prisms and in addition have equal volume. Dehn's result now appears in the form that $\Delta$ factors through $\scr P/\scr E$ and even through $\scr P/\scr F$; hence $\Delta=\delta\circ\kappa$ for some $\delta\colon\scr P/\scr F\rightarrow R\otimes R/(\pi)$. Sydler's result states that $\delta$ is injective. In a natural way, domain and range of $\delta$ can be considered as real vector spaces and $\delta$ as a linear map, so in order to prove injectivity the equivalent problem is considered of exhibiting, for each linear map $\tau\colon\scr P/\scr F\rightarrow V$ into any $R$-vector space $V$, a linear map $\Phi\colon R\otimes R/(\pi)\rightarrow V$ such that $\tau\circ\kappa=\Phi\circ\Delta$. To do this, a special two-parameter family of polyhedra $T(a,b)$ generating $\scr P/\scr F$ as a vector space is utilized geometrically (this key step is due to Sydler [loc. cit.]) to yield relations, which in their turn appear in the determination of the second cohomology of groups and rings. These problems are essentially known, but have to be solved under an additional complication arising from the fact that $a,b$ have restricted domain; this algebraic part is contained in a separate paper by B. Jessen, J. Karpf and A. Thorup [Math. Scand. 22 (1968), 257--265 (1969)]. In the final section the image of $\scr P/\scr F$ under $\delta$ is characterized by means of derivations. Reviewed by H. E. Debrunner Cited in reviews: 91b:22016 82e:52012 _________________________________________________________________ 15,550b 48.0X Sydler, J.-P. Sur l'équivalence des polyèdres à dièdres rationnels. (French) Elemente der Math. 8, (1953). 75--79. {The review of this item is not currently available in MathSciNet. See the (paper) Mathematical Reviews journal for the review.} Reviewed by B. Jessen _________________________________________________________________ 14,309f 52.0X Sydler, J.-P. Sur les conditions nécessaires pour l'équivalence des polyèdres euclidiens. (French) Elemente der Math. 7, (1952). 49--53. {The review of this item is not currently available in MathSciNet. See the (paper) Mathematical Reviews journal for the review.} Reviewed by B. Jessen Previous Review © Copyright American Mathematical Society 1998