0
$\begingroup$

This is just a nomenclature question. Let $T$ be a commutative monoid, and let $T^*$ be its Grothendieck group. That is, $T^* \cong T \times T \ / \sim$, where $(s,s') \sim (t, t')$ if $s+t'+e = s'+t+e$ for some $e \in T$.

Let $i : T \to T^*$ be the natural inclusion map, where $i(t) = [(t,0)]$, and let $K$ denote the kernel of this map.

Does the kernel $K \subseteq T$ have a common name in the literature? What are elements $k \in K$ called?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ I assume by the kernel you mean the inverse image of 0. It is not usual to use the term kernel for the inverse image of the identity in semigroup theory since in general it is not interesting. The fibers of a semigroup homomorphism can all have radically different cardinality. I know of no name for $K$ in semigroup theory although perhaps in K-theory or some place else it has a name. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 26, 2013 at 3:27
  • 2
    $\begingroup$ Here's a negative result: no name given in Ogus's log geometry book. $\endgroup$
    – S. Carnahan ♦
    Commented Jan 26, 2013 at 10:35

1 Answer 1

Reset to default
3
$\begingroup$

I don't know a name for $K$ either, but here is a suggestion.

The correct definition of $\sim$ is $(a,b) \sim (c,d) \Leftrightarrow \exists e \in T ~ (a+d+e=b+c+e)$. In particular $(a,b) \sim (0,0) \Leftrightarrow \exists e \in T : a+e=b+e$. I would suggest to call such elements $a,b$ stably equal. Remark that this coincides with the terminology stably isomorphic from topological $K$-theory (where $T$ is the commutative monoid of $\cong$-classes of vector bundles on a space $X$ and one usually takes w.l.o.g. $e$ to be the class of a trivial vector bundle when $X$ is paracompact). In particular, $a \in K \Leftrightarrow \exists e \in T : a+e=e$, in which case one may say that $a$ is stably zero.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Similarily one could say that for a function $f : X \longrightarrow X$ two elements $x,y \in X$ are stably equal if $f^n(x)=f^n(y)$ for some $n \in \mathbb{N}$ (which means that they are equal in the localization $X_f$). $\endgroup$
    – Martin Brandenburg
    Commented Jan 27, 2013 at 10:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .