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.

  • $\begingroup$ I think you are overlooking Grothendieck Riemann Roch without denominators. From each graded piece $F^iK/F^{i+1}\to A^i$, you have the chern class map and the composite with $A^i\to F^iK/F^{i+1}K is multiplication by $(i-1)!$. Of course, $\phi$ is not a ring homomorphism and Todd class plays a role in GRR. $\endgroup$
    – Mohan
    Sep 19, 2015 at 0:23
  • $\begingroup$ I have to admit I'm not familiar with GRR without denominators. I'll consult the section on it in Fulton. $\endgroup$
    – peterx
    Sep 19, 2015 at 1:03
  • $\begingroup$ Okay, I see - it doesn't have that much to do with GRR w/o denoms per se, but there are morphisms $F^pK/F^{p+1}\to A^p$ given by $c_p$. This is just an explicit by-grading description of the induced Chern character ch' above since the $c_p$-coefficients in the series is $(-1)^{p-1}/(p-1)!$. Fine. And the Todd class plays a role I guess because $\tau_X$ also induces an isomorphism $K(X)_\mathbb{Q}\to A(X)_\mathbb{Q}$, since it also has the property that $\mathcal{O}_V$ gets mapped to $[V]$+lower-dim stuff, so on passing to gradings we can use the same argument as above. $\endgroup$
    – peterx
    Sep 19, 2015 at 2:52
  • $\begingroup$ But I don't see how this directly shows that $\phi$ is not a ring homomorphism, or answers any of my other questions. In fact I am even more confused now because I realized that I don't actually see why $K(X)_\mathbb{Q}$ is not already a graded ring - all the exact sequence relations $[B]=[A]+[C]$ when $0\to A \to B \to C \to 0$ will have the same support on both sides since $B$ is "morally the direct sum" of $A$ and $C$. But clearly $K(X)$ can't already have the grading which we impose on it via filtration, otherwise it would be already isomorphic to $\text{Gr }K(X)$ $\endgroup$
    – peterx
    Sep 19, 2015 at 2:57
  • 3
    $\begingroup$ As for why $K(X)$ is not graded, take the short exact sequence $0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O} \rightarrow k_x \rightarrow 0$ on $\mathbb{P}^1$. Also, this reference might be useful: link.springer.com/article/10.1007%2FBF02684300 $\endgroup$ Sep 19, 2015 at 5:00

2 Answers 2


$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\cO{\mathcal{O}}\def\FF{\mathbb{F}}$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 $\mathbb{Q}$.

I now have some concrete examples of algebraic varieties where $K(M, \mathbb{Z})$ is not isomorphic to $A(M, \mathbb{Z})$, and where $K(M, \mathbb{Z})$ does not support a grading with the correct Betti numbers.

The first example, due to Matt Larson, is to take $Q$ a quadratic $3$-fold in $\mathbb{P}^4$. Then $Q$ is a homogenous space for $SO(5)$, so it has a cellular stratification, as follows: Consider a chain $P \subset L \subset H \subset Q$, where $P$ is a point, $L$ is an isotropic line, $H$ is a hyperplane section and $Q$ is the whole space. Then $L\setminus P \cong \mathbb{A}^1$, $H \setminus L \cong \mathbb{A}^2$ and $Q \setminus H \cong \mathbb{A}^3$.

Thus, the Chow groups of $Q$ are $(\ZZ [Q], \ZZ [H], \ZZ [L], \ZZ [P])$. The non-obvious products are $$[H]^2 = 2[L],\ [H][L] = [P].$$

$K^0(Q)$ also has $\ZZ$-basis the structure sheaves of $Q$, $H$, $L$ and $P$. This time, the products are $$[\cO_H]^2 = 2[\cO_L]-[\cO_P],\ [\cO_H][\cO_L] = [\cO_P].$$ The latter one is because, if we think of $H$ as $\mathbb{P}^1 \times \mathbb{P}^1$, then the normal bundle is $\mathcal{O}(1,1)$, so the self intersection is two lines meeting at a point, and we need to subtract that point out.

To see that these rings are not isomorphic, note that, in Chow, the square of any element in the maximal ideal is divisible by $2$ (since $(a[H]+b[L]+c[P])^2 = a^2 [H^2] + 2 ab [H][L] = 2a^2[L] + 2ab[P]). In $K$-theory, we have $[\cO_H]^2 = 2[\cO_L]-[\cO_P]$. So the rings are different.

Let's see further that the $K$-theory ring can't be graded as $K_0(Q) = \ZZ \oplus \ZZ \gamma_1 \oplus \ZZ \gamma_2 \oplus \ZZ \gamma_3$, with $\gamma_i$ in degree $i$. Degree considerations show that $\gamma_2$ and $\gamma_3$ square to $0$, so they must lie in $\ZZ [\cO_L] + \ZZ [\cO_P]$, and thus must span it. Since $\gamma_1$ is nilpotent, it must be of the form $a [\cO_H] + b[\cO_L] + c[\cO_P]$; since $\ZZ \gamma_1 \oplus \ZZ \gamma_2 \oplus \ZZ \gamma_3$ is the maximal ideal, we must have $a = \pm 1$ and, WLOG, we can take $a=1$. Then $\gamma_1^2$ should be $m \gamma_2$ for some integer $m$. But $\gamma_1^2 = ([\cO_H] + b[\cO_L] + c[\cO_P])^2 = 2 [\cO_L] + (2b-1) [\cO_P]$ and $GCD(2, 2b-1) = 1$ so we must have $m=1$ and $\gamma_2 = 2 [\cO_L] + (2b-1) [\cO_P]$. Similarly, $\gamma_3$ must be $\gamma_1^3/2 = [\cO_P]$. But $([\cO_H] + b[\cO_L] + c[\cO_P], 2 [\cO_L] + (2b-1) [\cO_P], [\cO_P])$ only span an index $2$ sublattice in $K^0$, a contradiction.

I have asked various people for toric counterexamples, and now I have one. I'll do this more quickly. Let $C$ be the toric variety whose fan is combinatorially the normal fan to the cube, but with rays in directions $$(1,0,0), (0,1,0), (0,0,1), (-1,0,0), (1,-1,0), (0,2,-1).$$ This is a so-called "Bott tower".

This space is a tower of $\mathbb{P}^1$ bundles, so I can compute $K$-theory and cohomology/Chow using the projective bundle theorem. (Cohomology) ($K$-theory).

For cohomology/Chow, I get $$H^{\ast}(C) = A^{\ast}(C) = \ZZ[X,Y,Z]/\langle X^2, Y(Y+X), Z(Z+2Y) \rangle$$ In other words, $$X^2=0, Y^2 = -XY,\ Z^2 = -2YZ.$$ Note that the square free monomials in $X$, $Y$, $Z$ are a basis.

In $K$-theory, I get $$K^{0}(C) = \ZZ[u,v,w]/\langle (u-1)^2,\ (v-1)(vu-1),\ (w-1)(wv^2-1) \rangle.$$

Put $u = 1+x$, $v = 1+y$ and $w = 1+z$. Since $u$, $v$, $w$ are classes of line bundles, $x$, $y$ and $z$ are in codimension $1$. We get the relations $$x^2 = 0,\ y(y+x+xy)=0,\ z(z+2y+y^2+2yz+y^2z)=0.$$ So $$x^2=0$$ $$y^2 = -xy-xy^2 = -xy+x(xy+xy^2) = -xy.$$ $$z^2 = -2yz - y^2 z - 2y z^2 = -2yz + xyz - 2y(-2yz +\cdots) = -2yz + xyz +4 y^2 z = -2yz - 3xyz.$$ Here the "$+\cdots$" are terms that I know will die at the next level for dimension reasons. To summarize $$x^2=0,\ y^2 = -xy,\ z^2 = -2yz-3xyz.$$

We can see that the Chow relations are the associated graded of the $K$-theory relations.

However, we can also see that the rings are not isomorphic. Tensor with $\FF_2 = \ZZ/2 \ZZ$ to get $$\FF_2[X,Y,Z]/(X^2=Z^2=0,\ Y^2 = YZ)$$ and $$\FF_2[x,y,z]/(x^2=0, y^2=yz, z^2 = xyz)$$ respectively. The two rings are not isomorphic, because there squares form a $2$-dimensional $\FF_2$ vector space in the first case (basis $\{ 1, YZ \}$) and $3$-dimensional vector space in the second (basis $\{ 1,yz, xyz \}$).

  • $\begingroup$ Why is the class $\log [L]$ well defined in $K(X)$? Does the series defining it stop at some point? $\endgroup$
    – Qfwfq
    Sep 21, 2015 at 19:17
  • $\begingroup$ Yes, because $[L]-1$ is equivalent to something supported in codimension $1$. If, for simplicity, $L$ has a section (other than $0$), then we have $0 \to \mathcal{O} \to L \to L\otimes \mathcal{O}_Z \to 0$, where $Z$ is the zero locus of the section. $\endgroup$ Sep 21, 2015 at 19:25
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    $\begingroup$ The filtration on $K(X)$ is the one coming from the motivic version of the Atiyah-Hirzebruch spectral sequence for algebraic K-theory (also known as the coniveau spectral sequence), whose $E_2$ term can be expressed with motivic cohomology (whence Chow groups in the appropriate degrees). This spectral sequence degenerates with rational coefficients, which is why the Chern character becomes an isomorphism. But, with integral coefficients, it does not degenrate, and thus the filtration on $K$-theory does not come from a grading. The same thing happens with topological K-theory. $\endgroup$ Sep 21, 2015 at 23:14
  • $\begingroup$ Yes, this is exactly the kind of thing I was looking for! Thank you. $\endgroup$
    – peterx
    Sep 21, 2015 at 23:45

The product in $\mathrm{Gr}K(X)$ is defined precisely out of the product of $K(X)$. The definition is not straightforward because one has to prove that the product of $K(X)$ is compatible with the filtration. It is a standar result. Find a proof here in Theorem, page 322. (I'm sure Fulton's also has this but I am not used to his book).

Now for your questions:

1.- I don't know what you mean by change the multiplicative structure .

2.- by the Grothendieck-Riemann-Roch the Chern character $\mathrm{ch}\colon K(X)_\mathbb{Q}\to \mathrm{Gr}K(X)_\mathbb{Q}$ is an isomorphism of rings, so in particular it is also compatible with products. The fact that $\mathrm{ch}$ is an morphism of rings is not particularly elaborated. Recall that the Chern character is the multiplicative extension of the series $e^x$ and that the Chern classes are additive so that $$ \mathrm{ch}([L]\cdot [L'])=e^{c_1(L\otimes L')}=e^{c_1(L)+c_1(L')}=e^{c_1(L)}\cdot e^{c_1(L')}=\mathrm{ch}([L])\cdot \mathrm{ch}([L']). $$ See it done with detail once again here in page 323.

  • $\begingroup$ Thanks for the answers, but I think you were answering slightly different questions. 1 - I understand the concept of getting a new product out of the old one by the graded construction. My question was more along the lines of the following, though: $K(X)_\mathbb{Q}$ is abstractly isomorphic to $\text{Gr }K(X)_\mathbb{Q}$ as groups (though it obviously can't respect the grading). I'm really asking more about what the product on $K(X)$ looks like when the graded product is "transported back" by the isomorphism. Sorry for not being clear. $\endgroup$
    – peterx
    Sep 21, 2015 at 16:53
  • $\begingroup$ 2 - I was asking about what I referred to as $\text{ch}'$, which I realize now is just the functor $\text{Gr}$ applied to $\text{ch}$, and hence is also a ring homomorphism - and hence so is its inverse $\phi$ once we pass to rational coefficients. So I believe I already figured this one out. $\endgroup$
    – peterx
    Sep 21, 2015 at 16:56
  • $\begingroup$ I don't understand your first question. For the second: note that the Chern character is an isomorphism of rings, not just of groups. $\endgroup$
    – Tintin
    Sep 21, 2015 at 21:27

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