## January 10, 2020

### Quotienting Out The Degenerate

#### Posted by Tom Leinster

This is a quick, off-the-cuff, conceptual question. Hopefully, it has an easy answer.

Often in algebra, we want to quotient out by a set of elements that we regard as trivial or degenerate. That’s almost a tautology: any time we take a quotient, the elements quotiented out are by definition treated as negligible. And often the situation is mathematically trivial too, as when we quotient by the kernel of a homomorphism.

But some examples of quotienting by degenerates are slightly more subtle. The two I have in mind are:

• the definition of exterior power;

• the definition of normalized chain complex.

I’d like to know whether there’s a thread connecting the two.

Let me now explain those two examples in a way that makes them look somewhat similar. I’ll start with exterior powers.

Take a vector space $X$ over some field, and take an integer $r \geq 0$. A multilinear map

$f: X^r \to Y$

to another vector space $Y$ is said to be alternating if

$(x_1, \ldots, x_r) \ \text{ linearly dependent} \implies f(x_1, \ldots, x_r) = 0.$

The $r$th exterior power $\bigwedge^r X$ is the codomain of the universal alternating map out of $X^r$.

That’s a characterization of $\bigwedge^r X$ by a universal property, but it’s not an actual construction. It’s constructed like this: the tensor power $X^{\otimes r}$ has a linear subspace $D_r$ generated by

$\{ x_1 \otimes \cdots \otimes x_r \mid (x_1, \ldots, x_r) \ \text{ linearly dependent} \},$

and then we put $\bigwedge^r X = X^{\otimes r}/D_r$. The letter $D$ is chosen to stand for either “degenerate” or “dependent”, linearly dependence being a kind of degeneracy condition. So, $\bigwedge^r X$ is $X^{\otimes r}$ quotiented out by its degenerate part.

Incidentally, the universal property and construction of the exterior power aren’t usually phrased this way. More often, a multilinear map $f: X^r \to Y$ is defined to be “alternating” if $f(x_1, \ldots, x_n) = 0$ whenever $x_i = x_j$ for some $i \neq j$. But this is equivalent. Similarly, $D_r$ can equivalently be defined as the subspace of $X^{\otimes r}$ generated by the elements $x_1 \otimes \cdots \otimes x_r$ where $x_i = x_j$ for some $i \neq j$.

Personally, my comfort with exterior algebra took a leap forward when I learned that the definition of exterior power could be expressed in terms of linear dependence, as opposed to the standard presentation via repeated arguments. I never felt entirely motivated by that standard approach, despite the volume interpretation and the nearly-equivalent condition that swapping arguments changes the sign. But the condition that degenerate terms get sent to zero feels more natural to me, whatever “natural” means.

Now let’s do normalized chains. I guess I could say this in the context of an arbitrary abelian category, but I’ll just say it for modules over some commutative ring.

Let $X$ be a simplicial module. It gives rise to a chain complex $C(X)$ of modules (the unnormalized complex of $X$). The $r$th module $C_r(X)$ is just $X_r$, and the boundary maps in $C(X)$ are the alternating sums $\sum (-1)^i d_i$ of face maps in $C(X)$, as usual.

Now, some of the elements of $X_r$ are “degenerate”, in the sense that they can be obtained from lower-dimensional elements. Specifically, let $D_r(X)$ be the submodule of $X_r$ generated by

$\{ (X f)(y) \mid f: [r] \to [q], \ q \lt r, \ y \in X_q\}.$

A little calculation shows that $D(X) = (D_r(X))_{r \geq 0}$ is a subcomplex of the unnormalized complex $C(X)$. So we can form the quotient complex $N(X) = C(X)/D(X)$, which by definition is the normalized complex of the simplicial module $X$.

Again, this isn’t quite the usual presentation. An important fact is that $N(X)$ is not just a quotient of $C(X)$, but a direct summand. In particular, it can also be viewed as a subcomplex of $C(X)$. There are two dual ways to view it thus: $N_r(X)$ can be seen as the intersection $\bigcap_{i = 0}^{r - 1} ker(d_i)$ of the kernels of all but the last face map, or dually as the intersection $\bigcap_{i = 1}^r ker(d_i)$ of the kernels of all but the first face map. Most often, $N(X)$ is defined to be one of these two subcomplexes. But to me it seems more natural to view it primarily as $C(X)/D(X)$: the quotient of $C(X)$ by its degenerate part.

So, those are the two situations I wanted to describe. They seem moderately similar to me: both involve a quotient by a subobject generated by degenerate elements; in both, the degenerate elements don’t actually form a subobject themselves (the words “generated by” are crucial); both involve an indexing over the natural numbers $r$. So I’m wondering whether the two situations are related. Maybe one is a special case of the other, or maybe there’s a common generalization. Even if not, perhaps there’s some good point of view on “quotient out the degenerate” constructions of this kind, including my two examples and maybe others. Can anyone shed any light?

Posted at January 10, 2020 1:11 PM UTC

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### Re: Quotienting Out The Degenerate

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.

Posted by: Fernando on January 10, 2020 2:21 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

Thank you, Fernando. Can you say a bit more about how what you’re saying is connected to my two examples?

I know about the Jacobson radical of a module (or ring) and the nilradical of a ring, and I see a Wikipedia page on the general notion of a “radical of a ring”. But I don’t see how stuff about radicals of rings would be relevant to my question, given that the first example was about a quotient of a module and the second about a quotient of a chain complex of modules. What do you have in mind?

Posted by: Tom Leinster on January 10, 2020 2:51 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

I took the suggestion to be that “radical” is informally synonymous with “the collection of degenerate elements” (of whatever structure); so that, for the exterior algebra, the ideal generated by “linearly dependent tensors” (to coin an awful term) could be called a radical of the tensor power, and, for chain complexes, the ideal generated by non-primitive chains could be called a radical of the full chain complex.

Posted by: L Spice on January 11, 2020 12:21 AM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

In the context of algebraic groups, the quotient of a connected linear algebraic group by its unipotent radical has trivial unipotent radical. That’s just an easy consequence of the definition, not a miracle; but it has always seemed slightly miraculous to me that this step gets rid of the sort of counterexample you describe: once we get rid of the unipotent radical (getting a connected reductive group), further getting rid of the centre yields a centreless group, called the adjoint quotient.

Posted by: L Spice on January 11, 2020 12:16 AM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

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.

Posted by: Paolo Perrone on January 10, 2020 2:50 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

But a general phenomenon seems to be: at least historically […], people preferred to talk about subspaces than about quotients.

I think that’s true. On pages 132–134 of Basic Category Theory, I waffle on about limits vs. colimits in the history of geometry, using nearly the same example as you: a sphere rather than a circle. To view the sphere as

$\{ (x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$

is to construct it as an equalizer (indeed, an equalizer of two maps whose common domain is a product). To view it as the gluing-together of two patches each homeomorphic to a 2-disc is to construct it as a coequalizer (indeed, a coequalizer of two maps whose common codomain is a coproduct). As the text points out, both constructions involve arbitrary choices.

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.

I’m not sure I agree, though, that quotients are any more natural than subspaces. As I see it, they’re just two dual approaches.

Posted by: Tom Leinster on January 10, 2020 5:38 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

Oh sure, I don’t mean that the quotients are always the better way.

Posted by: Paolo Perrone on January 10, 2020 11:37 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

OK, sorry, I misunderstood.

Posted by: Tom Leinster on January 11, 2020 1:44 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

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 $M$

Posted by: Oscar Cunningham on January 10, 2020 3:47 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

Huh, yes. OK, I’ll retreat to the setting where the base ring is a field. When I was figuring stuff out on paper prior to posting, I assumed I was working with vector spaces over a field. Then when I was typing the post, I unwisely said to myself “might as well have modules over a ring”. Thanks for pointing out that piece of folly.

I’m going to correct the post now, so anyone reading this in the future should know that it originally did the exterior power construction over an arbitrary commutative ring (otherwise Oscar’s comment will look bizarre).

Posted by: Tom Leinster on January 10, 2020 5:22 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

If you every figure out a nice way of doing this, please answer this question!

Posted by: Oscar Cunningham on January 10, 2020 6:12 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

OK, I think I’ve got it now.

To begin by stating the obvious: in a vector space over a field, a finite list of elements is said to be linearly dependent if some nontrivial linear combination of them is zero.

Here “nontrivial” means that at least one of the coefficients is nonzero. But of course, in a field, being nonzero is equivalent to being invertible (i.e. a unit). So we could just as well say “at least one of the coefficients is invertible”.

When extending the definition of linear dependence to modules over an arbitrary commutative ring, we therefore have two options. In both, we define a list of elements to be linearly dependent if some nontrivial linear combination is zero. But in the first option, we take “nontrivial” to mean “at least one coefficient is nonzero”, and in the second, we take it to mean “at least one coefficient is invertible”.

You used the first definition (which I agree seems more obvious somehow). And as you point out, it makes many lists linearly dependent, including some that might seem like they shouldn’t be. For example (and I’m sure you’ve considered this), if $X$ is any finite abelian group regarded as a $\mathbb{Z}$-module, then every nonempty list of elements of $X$ is linearly dependent. (Consider multiplication by the order of the group.) Your example is the simplest instance of this. So in the language of my post, all nonempty lists of elements of $X$ are “degenerate”, which seems rather sweeping! And under this definition, the quotient of $X^{\otimes r}$ by the linearly dependent lists is trivial, and in particular isn’t the exterior power $\bigwedge^r X$.

But the second definition does what we want. Here we take “linearly dependent” to mean “some linear combination, at least one of the coefficients of which is a unit, vanishes”. This is equivalent to the familiar statement that at least one member of the list can be expressed as a linear combination of the others. Apart from being the way that we often first understand linear dependence and explain it to our students, this is also exactly what we need to get the exterior power thing to work. In other words, if we define linear dependence in this second way and write $D_r$ for the submodule of $X^{\otimes r}$ generated by the elements $x_1 \otimes \cdots \otimes x_r$ for which $(x_1, \ldots, x_r)$ are linearly dependent, then

$\bigwedge^r X = X^{\otimes r}/D_r.$

This is true over an arbitrary commutative ring.

(This doesn’t answer your MSE question, though: consider $\bigwedge^2 \mathbb{Z}$, viewing $\mathbb{Z}$ as a $\mathbb{Z}$-module.)

Posted by: Tom Leinster on January 10, 2020 8:13 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

Here’s a third possible definition of “linearly dependent”: let $X$ be an $R$-module, and say that $(x_1,\dots,x_r) \in X^r$ are “linearly dependent” if there exists $r_1,\dots, r_r \in R^r$ such that $\sum_i r_i x_i = 0$ and the ideal $(r_1,\dots, r_r) \subseteq R$ generated by the $r_i$’s is the unit ideal. This definition feels more natural to me, somehow, than requiring that some $r_i$ be a unit.

In fact, with this definition, I think we can put the two cases into a common framework, though I haven’t checked all the details. Let $A$ be a Cauchy complete additive category (i.e. an $Ab$-enriched category with finite sums and split idempotents), and let $Mod(A) = Fun^\oplus(A, Ab)$ be the category of additive copresheaves on $A$.

• In the case of $R$-modules, we let $A$ be the Cauchy completion of $R$ considered as a one-object category, i.e. the category of finitely-generated projective $R$-modules. Then $Mod(A) \simeq Mod(R)$, where $Mod(R)$ is the usual category of $R$-modules. Under the equivalence, a module $M$ corresponds to the restricted Yoneda functor sending $P \mapsto Hom(P,M)$

• In the case of chain complexes, we let $A$ be the Cauchy completion of the free additive category on the dual simplex category $\Delta^{op}$. Then $Mod(A)$ is the category of chain complexes of abelian groups.

Now for $M\in \Mod(A)$, $a \in A$, and $m \in M(a)$, say that $m$ is degenerate if there is a proper summand $a'$ of $a$ such that $m \in M(a')$. I claim that this recovers the desired definition in both cases.

• In the case of $R$-modules, note that if $\sum_i s_i r_i = 1$, then the map $[r_1,\dots,r_k]^T: R^k \to R$ is an epimorphism split by the map $[s_1,\dots,s_k]$, with a complementary summand $P$. This induces a decomposition $M^k = Hom(R^k,M) \cong Hom(R,M) \oplus Hom(P,M)$. If $\sum_i r_i m_i = 0$ (i.e. $(m_1,\dots,m_k)$ is “linearly dependent” in the sense I defined above, as witnessed by $(r_1,\dots,r_k)$), then $(m_1,\dots, m_k) \in M^k$ lies in the summand $Hom(P,M)$, so that $(m_1,\dots,m_k)$ is “degenerate”.

• In the case of chain complexes, the connection is even more immediate, since if $m \in M([n])$ is in the image of a lower-dimensional simplex $[n] \to [m]$, then we may factor $[n] \to [p] \to [m]$ where $[n] \to [p]$ is a split epi in $\Delta$, so that $m$ lies in the summand $M([p])$.

I haven’t verified the converse implications, i.e. it’s possible that the notion of “degenerate” given above is too strong, making too many things “degenerate”.

Posted by: Tim Campion on January 21, 2020 5:30 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

Er – I think I garbled a bit of that. In the third bullet point overall, It should say that the map $[r_1,\dots,r_k]^T: R \to R^k$ is a mono split by $[s_1,\dots,s_k]: R^k \to R$.

Posted by: Tim Campion on January 21, 2020 5:34 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

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?

Posted by: David E Speyer on January 13, 2020 6:25 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

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.

Posted by: David E Speyer on January 13, 2020 6:52 PM | Permalink | Reply to this

### Re: Quotienting Out The Degenerate

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.

Posted by: Maxime Lucas on January 14, 2020 10:52 AM | Permalink | Reply to this

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