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I'm not much good at even basic number theory, so this is really mysterious to me. I think the actual question might want to restrict to finite Laurent series, but I don't think it matters: I'm pretty sure what I want to do is figure out what each homogeneous component is allowed to look like. Moreover, I'm pretty sure that by plugging in $h=k=1$ I get that all the coefficients in nonnegative degrees need to be integral, so the first place to look is in degree $-1$.

So suppose we have \begin{equation*} f(u,v)_{-1}=\frac{a_1}{a_2 u}+\frac{b_1}{b_2 v}. \end{equation*} Then \begin{equation*}f(ht,kt)_{-1}=\frac{a_1}{a_2 ht}+\frac{b_1}{b_2 kt} = \frac{a_1 b_2 k+b_1 a_2 h}{a_2b_2hkt};\end{equation*} for this to lie in $\mathbb{Z}[t^{\pm},1/h,1/k]$ we need $a_2b_2|a_1b_2k+b_1a_2 h$. I believe it's a basic fact that since this has to be true for all nonzero $h$ and $k$, we need that $a_2b_2|gcd(a_1b_2,b_1a_2)$. This implies that $a_2b_2|a_1b_2 \Rightarrow a_2|a_1$ and similarly $b_2|b_1$. But assuming we started with our fractions in reduced form, this implies that $a_2=b_2=1$; that is, $f(u,v)_{-1}\in \mathbb{Z}[t^{\pm}]$.

Since I don't think I gained much insight from that calculation, I just tried to find restrictions on \begin{equation*} f(u,v)_{-2}=\frac{a_1}{a_2u^2}+\frac{b_1}{b_2uv}+\frac{c_1}{c_2v^2}. \end{equation*} Now we have \begin{equation*} f(ht,kt)_{-2} = \frac{a_1}{a_2 h^2t^2}+\frac{b_1}{b_2 hkt^2} + \frac{c_1}{c_2 k^2t^2} \end{equation*} \begin{equation*} = \frac{a_1b_2hkc_2k^2 + b_1a_2h^2c_2k^2 + c_1a_2h^2 b_2hk}{a_2b_2c_2h^3k^3t^2} \end{equation*} \begin{equation*} = \frac{a_1b_2c_2k^2 + b_1a_2c_2hk + c_1a_2b_2h^2}{a_2b_2c_2h^2k^2t^2},\end{equation*} and what we need is for $a_2b_2c_2|a_1b_2c_2k^2+b_1a_2c_2hk+c_1a_2b_2h^2$ (for all $h$ and $k$, of course). This is where I'm stuck. Reducing mod $a_2$ and recalling that $gcd(a_1,a_2)=1$ we should get that $a_2|b_2c_2k^2$, and similarly $b_2|a_2c_2hk$ and $c_2|a_2b_2h^2$. Putting $h=k=1$ this means that $a_2|b_2c_2$, $b_2|a_2c_2$, and $c_2|a_2b_2$. I tried breaking down these conditions into conditions on the prime factorizations but didn't get very far. And in any case, numbers that satisfy these need not satisfy the original relation. Help please????


Motivation: In case you're still reading and you're wondering why I care about this specific calculation, it's because it's closely related to the "$K$-theoretic homology of $K$-theory"; precisely, it's the image of $K_*K\rightarrow K_*K\otimes \mathbb{Q}$. This is important because for any (co)homology theory $E$, the $E$-cohomology of $E$ (written $E^*E$) is exactly the algebra of stable cohomology operations, and with some additional assumtions on $E$ (which are satisfied by $K$-theory), the $E$-homology of $E$ is the coalgebra of stable homology cooperations. Furthermore, the $e$-invariant is a homomorphism from the stable homotopy groups of spheres to a subquotient of $K_*K$!

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I just came across a very pleasant algebro-geometric description of this ring $K_*K$, so I figure I'll record it here (a mere year and a half after I asked the question). My sense is that there isn't a better "elementwise" description, anyways.

Here are a few background facts, each of which builds on the previous. $\underline{\mbox{Iso}}(G,H)$ denotes the scheme of isomorphisms between two group-schemes $G \rightarrow X$ and $H \rightarrow Y$; a $Z$-point is a pair of maps $(X \xleftarrow{\varphi} Z \xrightarrow{\psi} Y)$ along with an isomorphism $\alpha:\varphi^*G \xrightarrow{\sim} \psi^*H$ of group-schemes over $Z$.

  1. For any line bundle $M$ over a scheme $S$, there is a canonical isomorphism $(\mathbb{G}_m)_S \cong \underline{\mbox{Aut}}(M)$: given an $S$-scheme $\iota:\mbox{spec }R \rightarrow S$ and a point $(\mathbb{G}_m)_S(R) = \mbox{Hom}_S(\mbox{spec }R,(\mathbb{G}_m)_S)$ given by $\lambda \in R^\times$, we obtain the automorphism $\cdot \lambda: \iota^*M \xrightarrow{\sim} \iota^*M$.
  2. The natural map $$\underline{\mbox{Aut}}((\hat{\mathbb{G}}_a)_\mathbb{Q}) \rightarrow \underline{\mbox{Aut}}(\mbox{Lie}((\hat{\mathbb{G}}_a)_\mathbb{Q}) \cong (\mathbb{G}_m)_\mathbb{Q} $$ given by $f \mapsto (df)_0$ is an isomorphism. If $R$ is a $\mathbb{Q}$-algebra, then an $R$-point of the source is an element $f(x) \in x\cdot R[[x]]$ with $f'(0)\in R^\times$ and $f(x +_{(\hat{\mathbb{G}}_a)_R} y) = f(x) +_{(\hat{\mathbb{G}}_a)_R} f(y)$, but when you unwind this, having all denominators means that this must take the form $f(x)=\lambda x$. Then $(df)_0$ is multiplication by $\lambda$, which of course corresponds to the $R$-point of $(\hat{\mathbb{G}}_m)_\mathbb{Q}$ picked out by $\lambda$.
  3. There is an isomorphism $\exp : (\hat{\mathbb{G}}_a)_\mathbb{Q} \xrightarrow{\sim} (\hat{\mathbb{G}}_m)_\mathbb{Q}$, which induces an isomorphism $\underline{\mbox{Aut}}((\hat{\mathbb{G}}_m)_\mathbb{Q}) \cong \underline{\mbox{Aut}}((\hat{\mathbb{G}}_a)_\mathbb{Q}) \cong (\mathbb{G}_m)_\mathbb{Q}$.
  4. For each $k \in \mathbb{Z} \backslash \{0\}$, the endomorphism $[k]:\hat{\mathbb{G}}_m \rightarrow \hat{\mathbb{G}}_m$ pulls back to an automorphism over $\mathbb{Z}[1/k]$, which yields a commutative diagram $$ \begin{array}{ccccccc} \coprod_{k \not= 0} \mbox{spec }\mathbb{Q} & \xrightarrow{\coprod [k]} & \underline{\mbox{Aut}}((\hat{\mathbb{G}}_m)_\mathbb{Q}) & \cong & \underline{\mbox{Aut}}((\hat{\mathbb{G}}_a)_\mathbb{Q}) & \cong & (\hat{\mathbb{G}}_m)_\mathbb{Q} \\ \downarrow & & \downarrow \\ \coprod_{k \not= 0} \mbox{spec }\mathbb{Z}[1/k] & \xrightarrow{\coprod [k]} & \underline{\mbox{Aut}}(\hat{\mathbb{G}}_m). \end{array} $$ (The right-hand vertical map is hard to define diagramatically, but easy to define on functors of points once we observe that $|(\mbox{spec }\mathbb{Z})(R)|=1$ and $|(\mbox{spec }\mathbb{Q})(R)| \leq 1$.)

Theorem [Adams-Harris-Switzer '71, reinterpreted]: This diagram is a pushout in the category of affine schemes.

That is, the natural map $\mathcal{O}(\underline{\mbox{Aut}}(\hat{\mathbb{G}}_m)) \rightarrow \mathbb{Q}[w,w^{-1}]$ (coming from the right-hand vertical map composed with the two isomorphisms) is injective, and its image consists of those Laurent polynomials $f(w)$ such that for all nonzero integers $k$, $f(k) \in \mathbb{Z}[1/k]$.

The connection with K-theory is the following. Let $E$ and $F$ be two Landweber-exact spectra with formal group laws $f_E$ and $f_F$ (so $E_0X \cong MUP_0X \otimes_{\eta_R,MUP_0,f_E} E_0 \cong E_0 \otimes_{f_E,MUP_0,\eta_L} MUP_0 X$, naturally in X, and similarly for $F_0X$; here, $MUP$ is the periodified complex cobordism spectrum). Then, \begin{eqnarray} E_0 F &=& MUP_0 F \otimes_{\eta_R,MUP_0,f_E} E_0 \\ &=& F_0 MUP \otimes_{\eta_R,MUP_0,f_E} E_0 \\ &=& F_0 \otimes_{f_F,MUP_0,\eta_L} MUP_0 MUP \otimes_{\eta_R,MUP_0,f_E} E_0. \end{eqnarray} Thus, by Quillen's theorem, $\mbox{spec }\pi_0(E\wedge F) = \underline{\mbox{Iso}}(G_E,G_F)$ (the scheme of isomorphisms between the associated formal groups). In the special case that $E=F=K$, we recover that $\mbox{spec }K_0 K = \underline{\mbox{Aut}}(\hat{\mathbb{G}}_m)$.

The two-variate version in the original question arises when you consider $K_* K$ (with its natural map to $(K_\mathbb{Q})_*K_\mathbb{Q}=H\mathbb{Q}P_* H\mathbb{Q}P \cong \mathbb{Q}[u^\pm,v^\pm]$) instead of just $K_0 K$ (with its natural map to $(K_\mathbb{Q})_0 K_\mathbb{Q} = H\mathbb{Q}_0 H\mathbb{Q}P \cong \mathbb{Q}[w^\pm]$).

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