### 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?

## 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.