The n-Category Café

In algebra the standard examples of this are called radicals (e.g., the Jacobson radical). One wants, in that case, that quotienting out by the radical should produce an object whose radical is trivial. That’s also what happens in your examples, but it doesn’t happen for quotients in general: quotienting a group by its center may well produce a group that has a nontrivial center, for example.

I don’t know whether this is an instance of what you are saying, but the situation reminds me of sigma ideals, sets which “can have measure zero and be safely ignored”.

More in general, there is always a notion of “ideal-like things” in mathematics, which can be seen as “things that can be safely quotiented out without breaking everything”. For groups, these are normal subgroups. For rings, these are ideals. Sometimes, such as in vector spaces, these coincide with subspaces. In other categories they only coincide in some cases (retracts).

But a general phenomenon seems to be: at least historically (before algebraic topology and category theory), people preferred to talk about subspaces than about quotients. This is why the circle is usually introduced as the subset of $\mathbb{R}^2$ that satisfies $x^2+y^2=0$, and only later seen as the quotient of $[0,1]$ obtained by identifying the extrema. I’d suspect that in many cases, whenever something can be written both as a subspace and as a quotient, the traditional definition is as a subspace, while for category theorists the more natural approach would be taking a quotient. It’s the difference between talking about equivalence classes, versus picking a representative for each class. The latter is more concrete, the former is more natural for us.

Another one favorite of mine is: why are probability measures defined as the subset of measures which have total normalization 1? The way I would have defined them is as the quotient of the positive measures under normalization (since in the end, only the relative ratios matter).

Maybe what I’m saying is obvious - but maybe it can help the brainstorming.

What’s your definition of ‘linearly dependent’? In particular is $(1)$ linearly dependent in $\mathbb{Z}/2\mathbb{Z}$ seen as a $\mathbb{Z}$-module? I’d usually say it was because $2\times 1 = 0$.

If so then every element of $\mathbb{Z}/2\mathbb{Z}$ is linearly dependent, so a map is alternating if and only if it sends everything to $0$. So then we would have $\bigwedge^1 \mathbb{Z}/2\mathbb{Z}=0$. This isn’t what I’d have expected since usually $\bigwedge^1$ is just the identity functor and $\bigwedge^1 M=M$ for all modules $M$

I have also wondered if there is a connection between these ideas. I’m afraid I never reached a firm resolution, so I’m just going to toss out a bunch of ideas and see if some of them resonate. Let’s work over a ground field $k$.

Let $\Delta$ be the simplex category, whose objects are finite totally ordered sets and whose morphisms are weakly order preserving maps. I’ll use the convention that $[n] = \{ 0,1,2,\ldots, n \}$, which seems confusing a first but is standard and useful eventually.

Warm up example Let $X$ be a finite simplical complex with vertex set $I$. Then there are two versions of the chain complex of $X$. Let an $n$-cell of $X$ be a map $f: [n] \to I$ whose image lies in a single face of $X$. Let $\tilde{C}_n$ be the free $k$-vector space on $n$-cells and let $\tilde{C}^n$ be the dual group. The $\tilde{C}_n$ (respectively, $\tilde{C}^n$) form a contravariant (respectively, covariant) functor from $\Delta$ to abelian groups and the homology (respectively, cohomology) of the resulting chain complex is the homology (respectively, cohomology) of $X$.

However, this isn’t what you see in a topology textbook. Rather, we define the chain group, $C_n$, to be the quotient of $\tilde{C}_n$ by the relations that (1) if $f: [n] \to I$ is not injective, then $f=0$ and (2) if $\sigma$ is a permutation of $[n]$, then $f \circ \sigma = (-1)^{\sigma} f$, where $(-1)^{\sigma}$ is the sign of the permutation $\sigma$. The co-chains, $C^n$, are then the subcomplex defined by the dual condition. These are no longer functors from $\Delta$, but they compute the same cohomology.

Suppose that $X$ is the complete simplicial complex – every subset of $I$ is in $X$. Then $\tilde{C}_n = (k I)^{\otimes n}$ and $C_n = \bigwedge^n (k I)$. So there is some connection!

The actual example I cared about Let $X$ be a topological space and let $U_i$ be an open cover of $X$ indexed by some set $I$. For example, we could take $X$ a simplicial complex with vertex set $I$ as above, and $U_i$ the open star around $i \in I$ (the union of the interiors of all faces $\sigma$ such that $\sigma \ni i$). Write $U_{i_0 i_1 \cdots i_n} := \bigcap_{k=0}^n U_{i_k}$. So, in our motivating example, $U_{i_0 i_1 \cdots i_n}$ is contractible if $(i_0 i_1 \cdots i_n)$ lies in a face of $X$ and empty otherwise.

Let $\mathcal{E}$ be a sheaf on $X$.

Then there are two ways to compute the Cech cohomology of $\mathcal{E}$ with respect to the cover $U_i$. One way is to define the Cech cochains $\tilde{C}^n$ to be $\prod_{f : [n] \to I} \mathcal{E}(U_{f(0) f(1) \cdots f(n)})$. So, if $\mathcal{E}$ is the sheaf of locally constant $\mathbb{Z}$-valued functions, this recovers the old $\tilde{C}^n$. The other way is a subcomplex of $C^n$, analogous to the one used before.

These give the same cohomology theory, but it isn’t obvious! See https://mathoverflow.net/questions/10056 . I was trying to find a conceptual way to think about this.

Trying to get to tensor products in a more canonical way

We saw that there is a natural way to make $(k I)^{\otimes n}$ into a simplicial vector space, for a set $I$. If $V$ is a vector space where we have not been given a basis, can we hope to likewise make $V^{\otimes n}$ into a simplicial set?

In order to do this, we would want to have:

(1) A map $V^{\otimes n+1} \to V^{\otimes n}$, corresponding to the injection $[n-1] \to [n]$ which misses the last element. The only natural choice seems to be $v_0 \otimes v_1 \otimes \cdots \otimes v_{n-1} \otimes v_{n} \mapsto v_0 \otimes v_1 \otimes \cdots \otimes v_{n-1} \otimes \lambda(v_{n})$. So our vector space $V$ would need a fixed linear map $\lambda : V \to k$.

(2) A map $V \to V \otimes V$, corresponding to $[1] \to [0]$. The absence of such a natural map is sometimes referred to as the “no cloning theorem”, since it is what prevents us from cloning a quantum state.

But, if $V$ is a co-algebra, then it does come with natural maps $V \to k$ (the co-unit) and $V \to V \otimes V$ (the co-product). So perhaps something could be done in that case?

While I’m brain dumping, I’ll mention how I think about the two chain complex constructions. John Wiltshire-Gordon and I were going to write an expository note about this, but we never got to it, and it is quite possible that, to the experts, all of this is obvious anyway.

Let $\Delta$ be the simplex category as above. Let $k \Delta$ be the category-rng whose elements are formal $k$-linear combinations of morphisms of $\Delta$, and where multiplication is composition when composition is defined and $0$ otherwise. The fact that this is a rng rather than a ring will introduce only minor inconveniences at the level of exposition here, but is definitely one of the reasons we didn’t get to writing the note.

Let $V$ be a functor from $\Delta$ to $k$-vector spaces (so a co-simplical vector space); we can think of $V$ as a $k \Delta$-module. A standard way to study modules over a noncommutative ring $R$ is to find a projective generator $\mathcal{P}$ of the category of $R$-modules and turn $R$-modules into $\mathrm{End}(\mathcal{P})$ modules by sending $M$ to $\mathrm{Hom}(\mathcal{P}, M)$. We will do this or two different projective generators, and get the two chain complexes.

The “free” modules Let $F_j$ be the co-simplical vector space where $(F_j)_n = k \mathrm{Hom}_{\Delta}([j],[n])$. This has the universal property that $\mathrm{Hom}(F_j, V) \cong V_j$ for any co-simplical vector space $V$. It is therefore immediate that $F_j$ is projective, and $F_j$ acts like a “free module generated in degree $j$”. Then $\mathcal{F}:=\bigoplus F_j$ is a projective generator, and $\mathrm{Hom}(\mathcal{F}, M) = \prod M_j$. However, $\mathrm{End}(\mathcal{F})$ is quite large, and one has to be a little clever to see the particular elements of $\mathrm{End}(\mathcal{F})$ which give the maps in the co-chain complex.

The indecomposable projectives In non-commutative ring theory, one usually wants to work with indecomposable projectives, those which do not split as a non-trivial direct sum. $F_j$ is not indecomposable. For example, $(F_1)_n$ has basis $\{ (i, j) : 0 \leq i \leq j \leq n \}$, where $(i,j)$ is short for the map $0 \mapsto 1$, $1 \mapsto j$. This is the direct sum of the spans of $\{ (k,k) : 0 \leq k\leq n \}$ and $\{ (i,j) - (j,j) : 0 \leq i \lt j \leq n \}$. This splitting can be shown to decompose $F_1$ as a direct sum of two smaller modules, $F_1 = P_0 \oplus P_1$.

In general, let $P_j$ be the quotient of $F_j$ where we set all non-injective maps to $0$. We found a splitting of the surjection $F_j \to P_j$, so $P_j$ is a summand of $F_j$ and hence projective. Moreover, the other summand is isomorphic to $P_{j-1}$. (As in our example, $F_1 \cong P_1 \oplus P_0$ above.) So $\mathcal{P} = \bigoplus P_j$ is also a projective generator.

Applying $\mathrm{Hom}(P_j, V)$ to a co-simplical vector space $V$ picks out the $j$-th term of the reduced co-chain complex, and now $\mathrm{End}(\mathcal{P})$ is very small – essentially, the only non-obvious maps are the ones in the chain complex, so no cleverness is needed to find the chain complex structure.

I don’t know if this is what you have in mind, but I think rewriting theory has something to say here.

Let me look at an other example. One often learns that $\mathbb{C}$ is the quotient of $\mathbb{R}[X]$ by $X^2 + 1$. But this is a slight abuse of notations: we are really quotienting by the ideal generated by $X^2 + 1$.

From the point of view of rewriting, it is useful to see $X^2 + 1$ as the rewriting rule $X^2 \to -1$. Starting from a polynomial in $\mathbb{R} [X]$, you can then “normalize it” until you reach a normal form, which is something of the form $a + Xb$. In this case, rewriting theory tells you that there is an isomorphism (as $\mathbb{R}$-vector spaces) between the quotient $\mathbb {R}[X]/(X^2 +1)$ and the space of normal forms, which I will denote by $NF(\to)$.

The general formulation of this result tells you than whenever you have a confluent and terminating rewriting relation $\to$ on a free R-module, then you get an isomorphism between $R/\to$ and $NF(\to)$. This is more or less the Church-Rosser property.

I believe you get the cases described in your post by looking at the quotients you take as rewriting systems. For example if you choose a basis $E = \{e_1,\ldots,e_n\}$ of $X$, you get a basis of $X^{\otimes n}$, and you can look at the rewriting system generated by $e_{i_1} \otimes \ldots \otimes e_{i_n} \to 0$ whenever $i_k = i_{k'}$ for $k \neq k'$. Then you get that quotienting by the “degenerate generators” is the same as looking at the subspace spanned by non-degenerate generators. For the normalized chain complex example, the same thing happens: quotienting by “degenerate cells” is the same as restricting oneself to the “non-degenerate” ones.