13
votes

### Representation theory of $\mathrm{GL}_n(\mathbb{Z})$

For $n=2$, these groups are very close to free groups, so their representation theory is not very tractable. For $n \geq 3$, however, there is a beautiful story arising from Margulis superrigidity. ...

10
votes

Accepted

### Derived category of local systems of finite type on a $K(\pi,1)$ space: an explicit counterexample

I can give an example that is even a smooth complex algebraic variety that shows up naturally in algebraic geometry.
Let $X$ be the moduli space of abelian varieties of genus $g>1$ with full level $...

6
votes

Accepted

### Why should we study the total complex?

Chain complexes arise from simplicial abelian groups via the Dold–Kan correspondence: the normalized chain complex functor establishes an equivalence of categories from simplicial abelian groups to (...

5
votes

Accepted

### Hattori-Stallings trace

The answer to (1) is yes. In fact this is true more generally: for any $f: M\to N, g:N\to M$, you have $tr(fg) = tr(gf)$.
The point is that $\hom_R(M,N)\otimes \hom_R(N,M)\cong \hom_R(M,R)\otimes_R N \...

5
votes

Accepted

### Projective objects in chain complexes of an abelian category: Further question

In this case, there's no difference between the direct sum and the direct product, because in every degree $n$, we are only taking the direct sum of two things. With reference to the answer you linked,...

4
votes

Accepted

### Pontryagin product on the homology of cyclic groups

I recommend always looking at the canonical reference: Ken Brown's "Cohomology of Groups". Here Chapter V.5 is literally titled "The Pontraygin product" and then the very next ...

4
votes

Accepted

### On infinity-morphisms between algebras over algebraic operads

It is a typo. The map $f$ should only be assumed to be a morphism of the underlying graded $\mathbb S$-modules.

4
votes

Accepted

### Group homology for a metacyclic group

The name metacyclic is normally used for a group which is cyclic-by-cyclic (ie. a group $G$ with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic). I will therefore refer to a finite group $...

3
votes

### Hochschild cohomology of an Azumaya algebra

It is true that $HH^*(A) \cong HH^*(Z)$ when $Z$ is a commutative $k$-algebra (i.e. in the setting of affine varieties). The following argument is due to Vadim Vologodsky, although any errors are mine:...

3
votes

### Group homology for a metacyclic group

Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa ...

3
votes

### The Krull dimension of the tensor product of rings

You are looking at infinite tensor products so we really shouldn't expect the infinite product to have finite dimension in most cases. Your second example was simply an infinite tensor product of $\...

1
vote

### When do faithfully semiinjective complexes exist?

Turns out it isn't that hard. We can do better and always find faithfully injective modules (I'm certain this is written down somewhere but I'm a "noob" and (i) didn't already know it and (...

1
vote

### Rings of weak dimension ≤ 1 vs. semihereditary rings?

(NB: all rings mentioned below are commutative.)
The most natural example of a ring which has weak global dimension equal to 1, but is NOT semihereditary, is $B[[X]]$, where $B$ is $$\frac{\prod_{\...

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