Auslander Conjecture for Complete Affine Manifolds

Auslander Conjecture for Complete Affine Manifolds

Definitions and mathematical statement

An affine transformation of ${\mathbb{R}}^n$ has the form $x\mapsto Ax+b$ with $A\in\operatorname{GL}_n({\mathbb{R}})$. An affine crystallographic group is a group of such transformations acting properly discontinuously and cocompactly on ${\mathbb{R}}^n$. Virtually solvable means having a solvable subgroup of finite index. Auslander's conjecture asserts that every affine crystallographic group is virtually solvable.

In the manifold formulation, a complete affine manifold is a quotient by a free properly discontinuous affine action; completeness identifies its developing space with all of ${\mathbb{R}}^n$. For compact such manifolds, the equivalent target in the record is that the fundamental group be virtually polycyclic, meaning that a finite-index subgroup admits a finite subnormal series with cyclic factors. No preserved positive-definite metric is assumed: restricting to Euclidean isometries would discard the essential affine difficulty.

Formulation and concept references: [S1].

Short English statement

Are groups acting properly and cocompactly by affine transformations always virtually solvable, with the corresponding compact complete manifold groups virtually polycyclic?

Research attempt

Scope and established results (27 September 2026).

Both properness and cocompactness are essential hypotheses. Here properness means that for every compact $K\subset\mathbb R^n$ only finitely many $\gamma$ satisfy $\gamma K\cap K\ne\varnothing$. This permits finite stabilizers in the group formulation. The manifold formulation additionally requires a free action. We never infer freeness from properness without a torsion-free hypothesis.

Goldman's text [S1] distinguishes the compact complete affine problem from the existence of complete noncompact manifolds with nonsolvable fundamental groups. Abels--Margulis--Soifer [E1] prove the conjecture for $n<7$; their result is much stronger than the elementary subcases proved here. This audit does not claim that a proof in unrestricted dimension has been found. The arguments below explore the translation subgroup, triangular linear holonomy, and compact quotient topology.

Linear and translational parts.

Write an affine transformation as $(A,b)$ with action $x\mapsto Ax+b$. Multiplication and inversion are \[ (A,b)(B,c)=(AB,b+Ac),\qquad (A,b)^{-1}=(A^{-1},-A^{-1}b). \] The linear-part map $\ell:\Gamma\to\operatorname{GL}_n(\mathbb R)$ is a homomorphism. Its kernel $T$ consists of translations and is abelian. Identifying $T$ with its translation vectors gives \[ 1\longrightarrow T\longrightarrow\Gamma \stackrel{\ell}{\longrightarrow}\ell(\Gamma)\longrightarrow1. \] Conjugation sends translation by $t$ to translation by $At$, so $T$ is normal and its real span $V$ is $\ell(\Gamma)$-invariant.

Properness implies that $T$ is a discrete subgroup of $\mathbb R^n$: infinitely many distinct bounded translation vectors would violate properness for a sufficiently large closed ball. A discrete subgroup of a finite-dimensional real vector space is a lattice in its span, and hence isomorphic to $\mathbb Z^r$ for some $r\leq n$. One elementary justification for finite generation is to choose a real basis $t_1,\ldots,t_r$ from $T$. Subtract integral combinations of these vectors to move every element of $T$ into their bounded parallelepiped. Its intersection with $T$ is finite by discreteness. These finitely many representatives, together with the $t_i$, generate $T$. A finitely generated torsion-free abelian group is free, and its discreteness forces its rank to equal the dimension of its real span.

A full-rank translation subgroup settles a genuine subcase.

Proposition 472.1. If $\Gamma$ acts properly on $\mathbb R^n$ and its translation subgroup has rank $n$, then $T$ has finite index in $\Gamma$. In particular $\Gamma$ is virtually abelian and hence virtually polycyclic.

Proof. Take a compact parallelepiped $Q$ for the full lattice $T$. For every $\gamma\in\Gamma$ choose $t\in T$ so that $t\gamma(0)\in Q$. Thus each left coset $T\gamma$ has a representative sending $0$ into $Q$. Only finitely many such representatives exist: take a compact set $K$ containing $Q\cup\{0\}$ and apply properness to it. Therefore there are finitely many cosets.$\square$

Notice that cocompactness was not required once a full translation lattice was given. The missing step in applying this proposition to the general problem is not its proof; it is the false hope that every affine crystallographic group must have a full translation lattice. An explicit compact affine example below disproves that hope.

There is also a useful converse subcase. If $\ell(\Gamma)$ is finite, then $T$ has finite index. Cocompactness for $\Gamma$ gives cocompactness for $T$: enlarge a compact covering set by finitely many coset representatives. A translation lattice of rank less than $n$ leaves an unbounded quotient direction, as the projection to $\mathbb R^n/V$ is constant on its orbits. Thus finite linear holonomy forces rank $T=n$ and again yields the conclusion.

Dimension one, including finite stabilizers.

Proposition 472.2. Every affine crystallographic group of the line is virtually infinite cyclic.

Proof. Let $\Gamma^+$ be the subgroup with positive linear part. It has index at most two. An element $x\mapsto ax+b$ with $a>0$ and $a\ne1$ has the fixed point $b/(1-a)$ and infinite order, so its powers give an infinite stabilizer. This contradicts properness. Thus every element of $\Gamma^+$ is a translation. The discrete translation subgroup of the line is either zero or $c\mathbb Z$ for some $c>0$. The zero case would make $\Gamma$ finite, and a finite group cannot cover the unbounded line by translates of a compact set. Hence the second case occurs.$\square$

The same reasoning must not be copied to higher dimension by replacing $a\ne1$ with $A\ne I$: a nonidentity matrix can have eigenvalue one, and an affine map with that linear part can be fixed-point-free.

A solvable linear-holonomy criterion.

Proposition 472.3. If the linear image $\ell(\Gamma)$ is virtually solvable, then $\Gamma$ is virtually solvable. In particular this holds if a finite-index subgroup of the linear image is simultaneously upper triangular over $\mathbb C$.

Proof. Take the inverse image $\Gamma_0$ of a solvable subgroup of finite index in $\ell(\Gamma)$. If its image has derived length $d$, then $\Gamma_0^{(d)}\subseteq T$. Since $T$ is abelian, $\Gamma_0^{(d+1)}=1$. This proves the first assertion.

For the second, the diagonal map on the upper triangular group has abelian image and kernel the upper unitriangular group $U$. Let $U_j$ be the subgroup whose entries on the first $j-1$ superdiagonals vanish. Matrix multiplication gives $[U_i,U_j]\subseteq U_{i+j}$, and $U_n=1$. Consequently repeated commutators eventually vanish in $U$. The triangular group is an extension of this nilpotent group by an abelian group, hence is solvable.$\square$

This establishes virtual solvability directly. It does not by itself prove the virtually polycyclic conclusion for every finitely generated solvable linear group: that general assertion is false. For example, the group generated on the line by $x\mapsto2x$ and $x\mapsto x+1$ contains translations by every dyadic rational. It is a solvable affine group, but those arbitrarily small nonzero translations violate properness. Its translation subgroup $\mathbb Z[1/2]$ is not finitely generated, so the group is not polycyclic. Proper crystallographic actions supply additional restrictions in the established equivalence discussed in [S1]; we do not replace those restrictions by solvability alone.

A compact non-Euclidean example with too few translations.

For integers $a,b,c$, define \[ g_{a,b,c}(x,y,z)=(x+a,\ y+b,\ z+c+ay). \] Direct substitution yields \[ g_{a,b,c}g_{a',b',c'} =g_{a+a',\,b+b',\,c+c'+ab'},\qquad g_{a,b,c}^{-1}=g_{-a,-b,-c+ab}. \] Thus these maps form the integral Heisenberg group, here given by an explicit affine action on all of $\mathbb R^3$.

Proposition 472.4. This action is free, proper and cocompact. The group is polycyclic and nilpotent of class two, is not virtually abelian, and its subgroup of translations has rank two rather than three.

Proof. A fixed point would force $a=b=0$ from the first two coordinates and then $c=0$ from the third. Hence the action is free.

If a compact set is contained in $[-M,M]^3$ and its image by $g_{a,b,c}$ intersects it, the first two coordinates give $|a|,|b|\leq2M$. The third gives $|c|\leq2M+|a|M$. There are only finitely many integral triples satisfying these bounds, proving properness.

Given any $(x,y,z)$, choose integers $a,b$ so that $x+a,y+b\in[0,1]$, and then choose $c$ so that $z+c+ay\in[0,1]$. Its orbit meets the unit cube, proving cocompactness.

The map $g_{a,b,c}\mapsto(a,b)$ has central kernel $\{g_{0,0,c}\}$ and quotient $\mathbb Z^2$. More explicitly, \[ [g_{a,b,c},g_{a',b',c'}]=g_{0,0,ab'-a'b}. \] This proves nilpotence of class two. The subgroups with $a=0$ and then $a=b=0$ give a subnormal series with three infinite cyclic factors, proving polycyclicity. Any finite-index subgroup contains positive powers of $g_{1,0,0}$ and of $g_{0,1,0}$. Their commutator is $g_{0,0,mn}\ne1$, so that subgroup is not abelian.

Finally the linear matrix is \[ \begin{pmatrix}1&0&0\\0&1&0\\0&a&1\end{pmatrix}. \] It equals the identity exactly when $a=0$. The translation subgroup therefore consists of vectors $(0,b,c)$ and has rank two. $\square$

This example satisfies the conjecture, but defeats the stronger proposed intermediate conclusions "finite linear holonomy" and "virtually abelian". It also prevents an argument that averages a positive-definite inner product over the linear holonomy without proving that averaging is legitimate. A nontrivial unipotent matrix cannot preserve a positive-definite form: a matrix preserving such a form is conjugate to an orthogonal matrix and is diagonalizable over $\mathbb C$, whereas the displayed nonidentity shear is not.

What fixed-point tests really imply.

For $\gamma=(A,b)$, the fixed-point equation is \[ (I-A)x=b. \] If $1$ is not an eigenvalue of $A$, the equation has a unique solution. Thus every infinite-order element of a proper affine group must have $1$ as an eigenvalue of its linear part. In a free action this holds for every nonidentity element, whether or not its order was specified.

Even the converse fixed-point test needs care: if $1$ is an eigenvalue, there is a fixed point exactly when $b\in\operatorname{im}(I-A)$. Writing $\lambda$ for a covector fixed by $A^T$, the necessary condition $\lambda(b)=0$ for all such $\lambda$ is also sufficient, by elementary duality between image and annihilator. Translation in a direction detected by one of these covectors can remove all fixed points. Freeness still does not imply properness of the whole group.

The spectral restriction alone does not force solvability. Consider the action of $\operatorname{SL}_2(\mathbb R)$ on the three-dimensional space of trace-zero matrices by $X\mapsto AXA^{-1}$. For every $A$, the matrix $A-\tfrac12\operatorname{tr}(A)I$ is a fixed trace-zero matrix when it is nonzero; if it vanishes, $A=\pm I$ and the whole space is fixed. Thus every matrix in this representation has eigenvalue one. Nevertheless its image contains a nonabelian free group.

Here is an explicit verification of the last assertion. The matrices \[ U=\begin{pmatrix}1&2\\0&1\end{pmatrix},\qquad V=\begin{pmatrix}1&0\\2&1\end{pmatrix} \] act on the real projective line by $t\mapsto t+2$ and $t\mapsto t/(2t+1)$. Put $X=\{|t|>1\}\cup\{\infty\}$ and $Y=\{|t|<1\}$. For every nonzero integer $m$, $U^m(Y)\subset X$ and $V^m(X)\subset Y$: for the latter, $|2mt+1|\geq2|t|-1>|t|$ when $|t|>1$, and the assertion at infinity follows from $V^m(\infty)=1/(2m)$. The ping-pong argument now shows that $U,V$ freely generate. To spell out the usual endpoint issue, a nontrivial reduced word may be conjugated to a pure power or to an alternating word beginning and ending in nonzero powers of the same generator; in the latter case it carries the other generator's domain strictly into that generator's domain. It is not the identity. Pure powers are visibly nonidentity as well. The projective action also excludes $-I$ from this free subgroup. The adjoint representation, whose kernel is $\{I,-I\}$, therefore remains faithful on it.

This is only a counterexample to a proposed algebraic implication from the eigenvalue condition. Its linear action fixes the origin and is not proper. It is not a counterexample to Auslander's conjecture.

A compactness obstruction from cohomological dimension.

Suppose now that the action is free, proper and cocompact. The quotient $M=\Gamma\backslash\mathbb R^n$ is a compact manifold without boundary with contractible universal cover. Its lifted cellular chain complex gives a free resolution of the trivial module over $\mathbb F_2[\Gamma]$ of length $n$, using a finite-dimensional triangulation of the smooth compact manifold. On the other hand, \[ H^n(\Gamma;\mathbb F_2)=H^n(M;\mathbb F_2)\ne0 \] by the mod-two fundamental class. Hence $\operatorname{cd}_{\mathbb F_2}\Gamma=n$.

A nontrivial free group has cohomological dimension one over $\mathbb F_2$: its tree gives the upper bound, and its nonzero first cohomology gives the lower bound. Therefore a nonabelian free group cannot act freely, properly and cocompactly on $\mathbb R^n$ for $n>1$. This explains why a proper free affine action in dimension three, of the kind treated in the established Margulis-spacetime theory, does not refute the compact conjecture. It does not exclude a much larger group containing a free subgroup: the cohomological dimension of a subgroup need not equal that of the ambient group.

Exact diagnostics and remaining gap.

The accompanying integer-arithmetic script checks the Heisenberg composition, inverse and commutator identities on a finite grid and checks the adjoint matrices of the two displayed generators. These checks detect transcription and convention errors; the universally quantified proofs are the symbolic calculations above. No finite matrix search certifies properness for an arbitrary infinite affine group.

The proved subcases are the full translation lattice case, finite linear holonomy, dimension one, and virtually solvable (in particular virtually triangular) linear holonomy. The explicit compact example shows why these do not exhaust the problem. The first unresolved step is to exclude a nonsolvable linear image using both the translational part and cocompactness; neither the eigenvalue-one restriction nor subgroup cohomological dimension achieves this. The general conjecture remains unresolved by this attempt.

Sources

• S1 William M. Goldman, "Geometric Structures on Manifolds," AMS Graduate Studies in Mathematics 227 $2022$, Conjecture 8.6.2; cf. H. Abels, G. Margulis, G. Soifer, arXiv:2011.12788 (Auslander conjecture for n < 7). https://www.math.umd.edu/~wmg/gstom.pdf. Formulation source.
• E1 Herbert Abels, Grigory A. Margulis and Gregory A. Soifer, The Auslander conjecture for dimension less than 7, arXiv:2011.12788, 24 November 2020. https://arxiv.org/abs/2011.12788; https://arxiv.org/html/2011.12788v1. Primary abstract and introduction inspected 27 September 2026. The dimension bound is $n<7$, not a theorem in arbitrary dimension.
• E2 William M. Goldman, author draft of Geometric structures on manifolds, dated 11 December 2021, Chapters 8 and 15. https://www.math.umd.edu/~wmg/gstom.pdf. Author-hosted text inspected 27 September 2026. The date of this accessible draft is distinguished from the 2022 book in [S1].