So what is topological complexity? To explain this I will paraphrase the motivation given in Farbers original paper. Let \(X\) be the state space of a mechanical system. Assume that \(X\) is path-connected. In order to plan motions of the system one needs a method to exhibit for every pair of states \((A,B) \in X \times X\) a continuous path connecting \(A\) and \(B\). Let \(PX\) be the space of continuous paths \([0,1] \to X\) and let \(\pi \colon PX \to X \times X\) denote the maps with maps a path to its start and endpoint. A great solution to the motion planning problem would be to come up with a continuous section \[ s \colon X \times X \to PX \] to the function \(\pi\colon PX \to X \times X\). However, such a section exists if and only if \(X\) is contractible. Indeed, one can use a contraction to move every state to one fixed state.

From a algorithmic point of view it is sufficient to ask for a weaker form of solution: Write \(X\times X\) as finite union of open sets \[ X\times X = U_1 \cup U_2 \cup U_3 \cup …\cup U_k\] and find a section \(s_i \colon U_i \to PX \) to \(\pi\) on each of these sets. Given a pair \((A,B)\) one can take the first \(U_i\) which contains the pair \((A,B)\) and use \(s_i(A,B)\) to move state \(A\) to state \(B\). Et voilá, this is the notion of topological complexity:

**Definition:** The *topological complexity* \(TC(X)\) of a path-connected topological space \(X\) is the smallest number \(k\) such that \(X \times X\) admits an open cover \[ X \times X = U_1 \cup U_2 \cup U_3 \cup …\cup U_k \] consisting of \(k\) sets such that there is a continuous section \(s_i \colon U_i \to PX\) to \(\pi\colon PX \to X \times X\) on each set \(U_i\).

Farber in particular notes that the topological complexity is a homotopy invariant and that \(TC(X) \leq 2\dim(X)+1\) if \(X\) is (for instance) a finite dimensional CW-complex. In fact, this bound is sharp, for instace, the topological complexity of the Klein bottle is 5; see this article (warning: they normalize topological complexity differently).

There is a considerable amount of results on topological complexity in the literature. Let me mention two that are related to the fundemental group of the topological space. (1) The arxiv preprint which made me fall into this rabit hole studies closed connected manifolds \(X\) with abelian fundamental group such that \(TC(X) < 2\dim(X) +1 \). In particular every non-orientable closed manifold with abelian \(\pi_1\) has this property. (2) A result of Farber and Mescher states that the topological complexity \(TC(X)\) of an aspherical finite CW-complex \(X\) with hyperbolic fundamental group \(\Gamma\) either equals the cohomological dimension \(c\) of \(\Gamma \times \Gamma\) or equals \(c+1\).

]]>Though being coarsely embeddable is an extremely general notion, it is not preserved under extensions as Arzhantseva and Tessera show (arXiv:1605.01192). So the question arises whether either the strong Novikov conjecture or the coarse Baum-Connes conjecture hold for such extensions. The case of the strong Novikov conjecture was affirmatively settled by Jintao Deng (arXiv:1910.05381), so let me focus on the coarse Baum-Connes conjecture.

In the following we need the notion of exactness, which is a bit stronger than being coarsely embeddable. We abbreviate being exact by A (this comes from the notion of ‘Property A’ which is equivalent to exactness) and being coarsely embeddable by CE, and if \(1 \to N \to G \to Q \to 1\) is a short exact sequence we say that it is, e.g., A-by-CE if N is exact and Q is coarsely embeddable.

- If the extension is A-by-A, then G is again exact and hence satisfies the coarse Baum-Connes conjecture. A proof of this may be found in ‘Anantharaman-Delaroche, Renault – Amenable groupoids’ or in a paper of Kirchberg and Wassemann (https://www.emis.de/journals/DMJDMV/vol-04/16.html).
- Similar to the previous point, Dadarlat and Guentner proved that CE-by-A implies CE (https://www.jstor.org/stable/1194929) and hence again the coarse Baum-Connes conjecture.
- The case of A-by-CE extensions is now much more interesting: examples of Arzhantseva-Tessera and of Delabie-Khukhro show that coarse embeddability might fail for groups that are A-by-CE extensions, but on the other hand Deng-Wang-Yu recently showed that such groups still satisfy the coarse Baum-Connes conjecture (arXiv:2102.10617).
- The remaining case of CE-by-CE is currently open, i.e., it is not known if groups defined by such extensions must satisfy the coarse Baum-Connes conjecture.

There are also other coarse geometric questions that one might ask about groups defined by extensions as discussed above. To mention one that interests me, consider the problem whether for a group G the maximal and reduced (uniform) Roe algebras coincide, resp. have isomorphic K-theories:

- Špakula and Willett proved that if G is exact, then its maximal and reduced (uniform) Roe algebras coincide, and if G is coarsely embeddable, then they have isomorphic K-theories (arXiv:1110.1531).
- Since an A-by-A group is again exact and a CE-by-A group is again CE, these cases are directly treated by the result of Špakula and Willett.
- So the remaining question is what happens in the case of A-by-CE, or even of CE-by-CE groups – do their maximal and reduced (uniform) Roe algebras have isomorphic K-theories?

edit (April 20th, 2021): Jintao Deng told me that their proof (the one in arXiv:2102.10617) *almost* also shows what I ask, i.e., that for A-by-CE groups the maximal and reduced (uniform) Roe algebras have isomorphic K-theories (they are showing it for the *twisted* Roe algebras). Hence it should be possible to adapt that part of their proof to really show what is asked.

The usual way of doing this requires \(n^3\) multiplications (and some additions) for multiplying two \((n\times n)\)-matrices. But there is actually a way of doing it with less than this: the current record by Alman-Williams from last October is roughly at \(\mathcal{O}(n^{2.37286})\)-multiplications (arXiv:2010.05846).

There is a nice, short article on the QuantaMagazine about this: link.

]]>… their foundational contributions to theoretical computer science and discrete mathematics, and their leading role in shaping them into central fields of modern mathematics.

]]>By definition, which goes back to arXiv:math/0208205, a *Hantzsche-Wendt manifold* is an orientable, n-dimensional flat manifold whose holonomy group is an elementary abelian 2-group of rank n-1, i.e., isomorphic to \((\mathbb{Z}/2\mathbb{Z})^{n-1}\).

Every n-dimensional flat manifold \(X\) is a quotient \(\mathbb{R}^n / \Gamma\), where \(\Gamma\) is a Bieberbach group, i.e., a torsion-free, co-compact and discrete subgroup of \(\mathrm{Isom}(\mathbb{R}^n) \cong \mathbb{R}^n \rtimes \mathrm{O}(n)\). The group \(\Gamma\) fits into a short exact sequence \[0 \to \mathbb{Z}^n \to \Gamma \to G \to 0\,,\] where the image of \(\mathbb{Z}^n\) in \(\Gamma\) is its maximal abelian normal subgroup and \(G\) is finite and coincides with the holonomy group of \(X\).

Hantzsche-Wendt manifolds exist in every odd dimension at least three, and since they are flat they are aspherical (their universal cover is contractible since it is \(\mathbb{R}^n\)). They also have some more interesting properties:

- They are rational homology spheres https://doi.org/10.1112/S0025579300010561.
- They are cohomologically rigid over \(\mathbb{F}_2\), i.e., two Hantzsche-Wendt manifolds are homeomorphic if and only if their graded cohomology rings with coefficients in \(\mathbb{F}_2\) are isomorphic: https://doi.org/10.1016/j.aim.2016.08.004.
- There are pairs of Hantzsche-Wendt manifolds which are isospectral but not homeomorphic to each other: https://doi.org/10.1515/crll.1999.077.
- No Hantzsche-Wendt manifold of dimension at least five admits a spin\({}^c\)-structure: https://arxiv.org/abs/2103.01051. (But the only one in dimension three is actually even spin.)

edit (April 14th, 2021): The (unique) three-dimensional Hantzsche-Wendt manifold has another interesting property. Its associated Bieberbach group \(\Gamma\) is the first counterexample to the unit conjecture (https://blog.spp2026.de/unit-conjecture-disproved/)!

]]>Details of the above constructions can be found in my paper ‘Burghelea conjecture and asymptotic dimension of groups’ jointly written with Marcinkowski (arXiv:1610.10076) and in the paper ‘Operator ideals and assembly maps in K-theory’ by Cortiñas-Tartaglia (arXiv:1202.4999). The idea of using the Schatten-class operators as coefficients to define an algebraic version of the analytic assembly map is due to Yu (arXiv:1106.3796).

In a recent preprint (arXiv:2012.12359) Rouse-Wang-Wang constructed a Chern assembly map \[ch^{top}_\mu\colon K_{top}^*(G) \to H\!P_*(\mathbb{C}G)\,.\] Here the left-hand side is the domain of the assembly map \[\mu_*^{BC}\colon K_{top}^*(G) \to K^{top}_*(C^*_r G)\] as originally defined by Baum-Connes (employing heavily shriek maps in equivariant K-theory). In arXiv:1402.3456 Rouse-Wang defined a comparison map \[\lambda_G\colon K_{top}^*(G) \to RK_*^G(\underline{EG})\] and showed that it relates the assembly maps \(\mu_*^{BC}\) and \(\mu_*^{K\!K}\). When reading the introduction of their first mentioned preprint, the following question immediately came to my mind:

- Does the comparison map \[\lambda_G\colon K_{top}^*(G) \to RK_*^G(\underline{EG})\] relate the map \[\mu_*^{ch}\colon RK_*^G(\underline{EG}) \to H\!P_*(\mathbb{C}G)\] to the one defined by Rouse-Wang-Wang: \[ch^{top}_\mu\colon K_{top}^*(G) \to H\!P_*(\mathbb{C}G)\,?\] I strongly assume that this is indeed the case, so the question basically is ‘How hard (or easy) is it to prove this?’

Interestingly, in the paper by Rouse-Wang where they define the comparison map \(\lambda_G\), they do not prove that it is an isomorphism and instead leave this as an open problem. Again, I assume that it is actually an isomorphism for any group G, and so we end with the question:

- How hard (or easy) is it to prove that the comparison map \(\lambda_G\) is an isomorphism?

- The unit conjecture states that every unit in \(K[G]\) is of the form \(kg\) for \(k \in K\setminus\{0\}\) and \(g \in G\).
- The zero divisor conjecture states that the only zero divisor in \(K[G]\) is \(0\).
- The idempotent conjecture states that \(0\) and \(1\) are the only idempotents in \(K[G]\).

Torsion-freeness of \(G\) is a necessary condition in the conjecture since otherwise one gets obvious counter-examples:

- Let \(g\) be an element of order \(2\) in \(G\). Then for any \(k \in K\) we have \[((1-k)+kg)\cdot((1-k)-kg) = 1-2k\,.\] Hence, if \(1-2k\) is a unit in \(K\), then the element \((1-k)+kg\) is a unit in \(K[G]\) which is non-trivial for \(k \notin \{0,1\}\).
- If \(g\) has order \(n\) in \(G\), then \[(1-g)(1+g+g^2+ \cdots + g^{n-1}) = 1-g^n = 0\] in \(K[G]\) and consequently \(1-g\) is a non-trivial zero divisor in \(K[G]\).
- If the order \(n\) of \(g\) is invertible in \(K\), then \(n^{-1}\cdot\sum_{k=0}^{n-1} g^k\) is a non-trivial idempotent in \(K[G]\).

It is known that the unit conjecture implies the zero divisor conjecture, and the zero divisor conjecture implies the idempotent conjecture.

- The strongest of these conjectures, the unit conjecture, is currently only proven in those cases where one can first show a stronger, purely group-theoretic property: having unique products. The group \(G\) is said to have unique products if for every choice of finite, non-empty subsets \(A,B \subset G\) there are elements \(a \in A\) and \(b \in B\) such that the product \(ab\) can not be expressed as any other product of elements from \(A\) and \(B\). Standard examples of groups with unique products are orderable groups.
- The zero divisor conjecture is proven for all elementary amenable groups and holds over \(\mathbb{C}\) for groups satisfying the Atiyah conjecture.
- The idempotent conjecture is known to hold for many groups since it follows from the isomorphism conjectures: for \(K[G]\) it follows from the corresponding Farrell-Jones conjecture and for \(\mathbb{C}[G]\) from both the Baum-Connes and the Burghelea conjecture.

Today I want to report on recent progress on the unit conjecture. Namely, if was disproven! You can find the corresponding preprint of Giles Gardam on the arXiv:2102.11818. In his counter-example the group is virtually abelian (i.e., from the general point of view of geometric group theory an extremely ‘simple’ group) and the field is \(\mathbb{F}_2\). Note that since the group is virtually abelian, the zero divisor conjecture holds for it.

I am very interested in the follow-up question whether one can also construct counter-examples to the unit conjecture over the field \(\mathbb{C}\) (or any other field of characteristic \(0\))? The motivation for asking this stems from the fact that the other two conjectures (the zero divisor and the idempotent conjecture) are known in the case of \(\mathbb{C}\) for many more groups than in the general case (i.e., for arbitrary fields \(K\)) since in this case they follow from other, more analytic conjectures (i.e., the Atiyah and the Baum-Connes conjecture). But there is currently no analytic conjecture known which implies the unit conjecture.

edit (April 14th, 2021): There is a connection of this counterexample to Hantzsche-Wendt manifolds about which I blogged shortly after this post (https://blog.spp2026.de/hantzsche-wendt-manifolds/). In fact, if I understand it correctly, the group in Giles counterexample is exactly the Bieberbach group of the (unique) three-dimensional Hantzsche-Wendt manifold.

]]>- Does a given tetrahedron tile space?
- Which tetrahedra are scissors-congruent to a cube?
- Can one describe the tetrahedra all of whose six dihedral angles are a rational number of degrees?

The first question goes back to Aristotle, the second is from Hilbert’s list of problems, and the third one was asked by Conway and Jones.

These questions are related by the following implications: any tetrahedron that tiles space is scissors-congruent to a cube, and also any tetrahedron with all six dihedral angles a rational number of degrees is scissors-congruent to a cube.

In a recent paper (arXiv:2011.14232) Kedlaya, Kolpakov, Poonen and Rubinstein solve the third question completely!

I recommend reading the press release of MIT regarding their result (link). It’s only three pages and contains some nice pictures!

]]>Today appeared a nice article by Andrew Putman on the arXiv which describes surprisingly simple free generating sets for the commutator subgroups of free and surface groups. Let me briefly state the result for surface groups. Let \(\Sigma_g\) be a closed surface of genus \(g \geq 1\) with fundamental group

$$\pi_1(\Sigma_g) = \langle x_1, x_2,\dots,x_{2g} \mid [x_1,x_2][x_3,x_4]\cdots[x_{2g-1},x_{2g}]\rangle.$$

Then the following is a free generating set for the commutator group of \(\pi_1(\Sigma_g)\):

$$ \{[x_i,x_j]^{x_i^{k_i}x_{i+1}^{k_{i+1}} \cdots x_{2g}^{k_{2g}}} \mid 1\leq i < j \leq 2g, (i,j) \neq (1,2), k_i,\dots,k_{2g} \in \mathbb{Z} \}.$$

The very readable article of Putman provides a nice geometric proof for this result (and a similar result for free groups).

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