There is an algebraic K-theory seminar at my school and we are struggling to find applications from areas other than topology. We'd like a nice statement like "If X then Y" whose proof makes unexpected use of $K_0(R)$ or $K_1(R)$, where X and Y are purely algebraic statements about rings (or something like that).

An example from group cohomology (not K-theory) is the following Sylow-style theorem, which is elementary to state.

Problem A. Suppose $G$ is a finite group of order $mn$ where $\gcd(m,n)=1$. Suppose $A$ is a normal abelian subgroup of order $m$. Prove that $G$ has a subgroup of order $n$, and show that any two such subgroups are conjugate.

Surprisingly a quick proof can be given with group cohomology. We're looking for something similar, except maybe for rings/fields, and instead of group cohomology it must use K-theory.

I've also seen the following result as a bonus problem in commutative algebra, and my prof told me the only solution he knew used K-theory. I don't know the solution though.

Problem B. Suppose $R\subseteq S$ is an integral extension of commutative domains. Suppose that $S$ is a PID. Prove that $R$ is a PID.

Not sure if the above statement is true as it stands, but I'm definitely pretty sure it's true if you assume that $R$ is a finitely-generated free $R$-module.

Anything else like Problem B above?

  • $\begingroup$ For problem B, $\mathbb{R}[x,y]/(x^2+y^2-1)\subset\mathbb{C}[x,y]/(x^2+y^2-1)$ is a counterexample. I asked JB about this problem, and he said he wasn't sure what the right statement was supposed to be. $\endgroup$ – Julian Rosen Nov 1 '14 at 21:29
  • $\begingroup$ My research gives some combinatorial results using algebraic K-theory of Grassmannians, but it doesn't sound like that's what you want. $\endgroup$ – Matt Samuel Nov 1 '14 at 21:55
  • $\begingroup$ @Julian: in your example, is $S$ is f.g. free over $R$? $\endgroup$ – Ehsaan Nov 1 '14 at 22:10
  • $\begingroup$ Yes, a basis for $S$ as an $R$-module is $1,i$. $\endgroup$ – Julian Rosen Nov 1 '14 at 22:57

Let $R$ be the ring $\mathbb{Z}[T]$ after inverting $T$ and $T^m-1$ for all $m\geq 1$. One can prove that $R$ is a principal ideal domain [Gray2]. Is it a Euclidean domain? It's hard to prove that Euclidean domains are principal ideal domains, so maybe this a job for $K$-theory after all.

Recall that for a commutative ring $SK_1(R)$ is the quotient $SL(R)/E(R)$ or equivalently, the kernel of map $K_1(R)\to R^\times$ induced by the determinant. It is easy to prove [Ros, Prop. 2.3.2] that if $R$ is a Euclidean domain, then $SK_1(R) = 0$.

However, for this particular $R$, Grayson in [Gray2] notes a remark of Stein and Franks that the results in [Gray1] imply that $SK_1(R) = SSR$ where $SSR$ is a certain abelian group that functions as the receptacle for obstructions to a diffeomorphism (of a certain type, see [Gray2] for details) of a compact smooth manifold to be isotopic to a Morse-Small diffeomorphism. Apparently, these are important in the theory of dynamical systems. However, $SSR$ can be defined completely algebraically.

Anyways, H.W. Lenstra Jr. proved that $SSF\cong \oplus_{n\geq 1}{\rm Cl}(\mathbb{Z}[\zeta_n,1/n])$ where $\zeta_n$ is a primitive $n$-th root of unity and ${\rm Cl}(-)$ denotes the class group. One can see this is nonzero by looking at $n$ a prime power, so that the class group of $\mathbb{Z}[\zeta_n,1/n]$ is the same as that of $\mathbb{Z}[\zeta_n]$. There are infinitely many of these that are nontrivial. For example ${\rm Cl}(\mathbb{Z}[\zeta_{23}]) \cong \mathbb{Z}/3$. You could also consider this another aspect of $K$-theory since for a Dedekind domain the class group is the reduced $0$-th $K$-group of $R$.

Thus, $SK_1(R)$ is nonzero and hence $R$ cannot be a Euclidean domain. So we get an example of a principal ideal domain that is not Euclidean domain.

  • [Gray1] - Grayson, '$K$-Theory of Endomorphisms'
  • [Gray2] - Grayson, '$SK_1$ of an Interesting Principal Ideal Domain'
  • [Ros] - Rosenberg, 'Higher Algebraic $K$-Theory and Its Applications'

If you are prepared to consider $K_{0}$ of varieties (schemes) rather than rings, then the paper Algebraic $K$-theory and sums-of-squares formulas by Dugger and Isaksen fits the bill.

Let $F$ be a field. The sums-of-squares problems asks for which $r,s$ and $n$ there exist identities $$ (x_{1}^{2} + \cdots + x_{r})(y_{1}^{2} + \cdots + y_{s}^{2}) = z_{1}^{2} + \cdots + z_{n}^{2} $$

where the $z_{i}$ are bilinear in $X=(x_{1},\dots,x_{r})$ and $Y=(y_{1},\dots,y_{s})$.

Dugger and Isaksen establish bounds on possible triples $[r,s,n]$ by considering algebraic vector bundles on the space $\mathbb{P}^{s-1} - V_{q}$, where $V_{q}$ is the subvariety of $\mathbb{P}^{s-1}$ given by $\sum_{i} x_{i}^{2} = 0$. They show that the existence of an $[r,s,n]$ formula implies the existence of a certain vector bundle over $\mathbb{P}^{s-1} - V_{q}$, which then places some constraints on the structure of $K_{0}(\mathbb{P}^{s-1} - V_{q})$.

In section 3 of their paper, which is essentially independent of the rest, they calculate $K_{0}(\mathbb{P}^{s-1} - V_{q})$, using fairly classical methods. This result is the black box they use to obtain bounds on the possible triples $[r,s,n]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.