Let $X$ be a nonsingular variety. (Perhaps some/all of this works over more general smooth schemes, but let's stick to the simple case.)
In, e.g., Fulton's Intersection Theory chapter 15, and Soule's Lectures on Arakelov Geometry chapter 2, there are brief expositions on the associated graded $\mathbb{Z}$-algebra $\text{Gr }K(X)$ of the Grothendieck group under its filtration by dimension of support, and its relationship to the Chow ring $A(X)$. These treatments are pretty much just glancing mentions, so I'm left somewhat confused about the situation. Specifically:
We have a surjective graded-group homomorphism $\phi:A(X)\to \text{Gr }K(X)$ given by $[V]\mapsto [\mathcal{O}_V]$. Further, some geometry shows that the Chern character takes $[\mathcal{O}_V]$ to $[V] +\alpha$ for $\alpha$ of lower-dimensional support, hence induces a graded-group homomorphism in the opposite direction $\text{ch}':\text{Gr }K(X)\to \text{Gr }A(X)\otimes \mathbb{Q}$, so that $\text{ch}'\circ \phi$ is the natural inclusion. So when we pass to rational coefficients everywhere, everything becomes an isomorphism, and hence $\text{ch}$ is also an isomorphism of graded groups since the associated graded module is a faithful functor, and hence it is also an isomorphism of rings. All of this is fine. (Unless I made a mistake!)
But now since $A(X)$ is already a graded algebra, it is naturally isomorphic to $\text{Gr }A(X)$. So we have that $K(X)\otimes \mathbb{Q}$ is also isomorphic to $\text{Gr }K(X)\otimes \mathbb{Q}$ as graded groups, which makes sense – after killing torsion, $K(X)$ already seems to be a graded group since direct sum of sheaves preserves dimension so long as we don't get killed off by one of the exactness relations. I am pretty sure this is all right.
So my clarifying questions are:
1) How exactly does taking the associated graded algebra of $K(X)$ by dimension of support change the multiplicative structure? And – does it? $K(X)\otimes \mathbb{Q}$ is already a graded $\mathbb{Q}$-algebra by the Chern character isomorphism, since the Chow ring is. But it seems that the induced gradation by this isomorphism is very weird since, e.g. $[\mathcal{O}_V]$ corresponds not to $[V]$ but $[V]+\alpha$ as mentioned above, so it can't be the same gradation.
2) On a related note, $\phi,\text{ch'}$ can't be ring homomorphisms, right? Since then that would imply the impossible thing from #1. But this seems odd, because in the case of $\phi$ at least, $\phi([V]\cdot [W])=\phi([V])\cdot\phi([W])$ seems to be exactly the (true) Serre intersection multiplicity formula – or would be, if all the nonvanishing Tors had support on $V\cap W$, which I'm not sure is the case. What exactly is going on here? I'm particularly interested in whether there is a quick way to prove the Serre intersection multiplicity formula here quickly in full rigor.
Best Answer
Does this help? $K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the Adams operators $\psi^k$ act by $k^j$ on the $j$-th graded piece. The corresponding descending filtration is the filtration by codimension.
For example, let $L$ be a line bundle with $c_1(L) = D$. Then $ch(L) = e^D$. So $ch(\log L) =D$ and $\log [L]$, defined as the class $([L]-1) - ([L]-1)^2/2 + ([L]-1)^3/3 - \cdots$ in $K(X)$, is pure of degree $1$. Indeed, $\psi^k [L] = [L^k]$, so $\psi^k \log [L] = \log [L^k] = k \log [L]$. Let's abbreviate $\log [L]$ by $z$. The structure sheaf of the vanishing locus of a section of $L$ has $K$-class $1-L^{-1} = 1-e^z = z-z^2/2+z^3/6-z^4/24+\cdots$. So this class is in the filtered part that has degree $\geq 1$, but is not of pure degree.
A natural question, to which I don't know the answer, is whether the integer $K$-ring has a natural grading $K(X) \cong \bigoplus K_i(X)$, which turns into this grading when we tensor by $\mathbb{Q}$.