Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$

Consider the semiring $$\mathbb{N}[H,H^{-1}]/(H^p+H^q = H^{p+q}+1)_{p,q \in \mathbb{Z}}.$$ Is it finitely presentable? Is there any simplification of the relations (except for $p \geq q \geq 0$)?

Topological background. Whereas the semiring of vector bundles on the scheme $\mathbb{P}^1_{\mathbb{C}}$ is just $\mathbb{N}[H,H^{-1}]$ (old result by Dedekind-Weber, 1892, which has been rediscovered many times), the semiring of vector bundles on the topological space $\mathbb{C}\mathbb{P}^1$ seems to be the semiring above, where $H$ represents the tautological bundle. The relations follow from Example 1.13 in Hatcher's VBKT. The reason is that $\begin{pmatrix} z^p & 0 \\ 0 & z^q \end{pmatrix}$ is homotopic to $\begin{pmatrix} z^{p+q} & 0 \\ 0 & 1 \end{pmatrix}$ in $\mathrm{Map}(S^1,\mathrm{GL}_2(\mathbb{R}))$. The $K$-theory ring is just $K(\mathbb{C}\mathbb{P}^1)=\mathbb{Z}[H]/(H^2+1=2H)$.

• Very good question. Any element can be uniquely presented in the form $N$ or $N+MH^p$ with $p>0$. Multiplication and addition in this basis seems tricky though, e.g. $(2+3H^3)(4+5H^2) = 44 + H^{131}$ or something like that. – Piotr Achinger Apr 16 '14 at 7:09
• @Piotr: $H^{-1}$ cannot be written like that. I think the normal form is rather $\mathbb{N} \sqcup \{H^p + n : p \in \mathbb{Z} \setminus \{0\}, n \in \mathbb{N}\}$. – Martin Brandenburg Apr 16 '14 at 8:12
• right, I forgot I cannot subtract ;) – Piotr Achinger Apr 16 '14 at 8:20
• Regarding history, in 1886 del Pezzo classified surfaces of minimal degree, and this was extended to minimal varieties of all dimensions by Bertini. So the "Birkhoff-Grothendieck splitting lemma" goes back further than Dedekind-Weber. – Jason Starr Apr 18 '14 at 12:11

Some observations: Consider the monoid $P = \{ (N, p)\in\mathbb{N}\times\mathbb{Z} \,:\, N = 0 \Rightarrow p = 0\}$ with the semiring structure $(N, p)(M, q) = (NM, Mp+Nq)$. I claim that the map $(N, p)\mapsto N-1+H^p$ is an isomorphism between $P$ and your semiring. In particular, the additive monoid underlying your semiring is not finitely generated.
• Isn't part of your conclusion correct - that $H^p$ for $p \in \mathbb Z$ must be monomials in the generators? So the generators are $H^r$ for some set of $r$ that generate the semigroup $\mathbb Z$. Now infinitely many distinct relations involve a sum of at least $2$ of these things, and two such relations cannot combine to form a third, so there is no finite presentation. – Will Sawin Apr 16 '14 at 16:35
• The condition in the def. of $P$ should be $p = 0 \Rightarrow N \geq 1$, right? – Martin Brandenburg Apr 16 '14 at 17:25