What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes?

Question

More specifically, I was wondering if there are well-known conditions to put on $X$ in order to make $K_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?

Answer 1

Imposing that you can resolve by a length $2$ sequence of vector bundles is too strong. What you want is that there is some $N$ so that you can resolve by a length $N$ sequence of vector bundles. By Hilbert's syzygy theorem, this follows from requiring that the scheme be regular. (Specifically, if the scheme is regular of dimension $d$, then every coherent sheaf has a resolution by projectives of length $d+1$.)

Here is a simple example of what goes wrong on singular schemes. Let $X = \mathrm{Spec} \ A$ where $A$ is the ring $k[x,y,z]/(xz-y^2)$. Let $k$ be the $A$-module on which $x$, $y$ and $z$ act by $0$. I claim that $k$ has no finite free resolution. I will actually only show that $A$ has no graded finite free resolution. Proof: The hilbert series of $A$ is $(1-t^2)/(1-t)^3 = (1+t)/(1-t)^2$. So every graded free $A$-module has hilbert series of the form $p(t) (1+t)/(1-t)^2$ for some polynomial $p$; and the hilbert series of anything which has a finite resolution by such modules also has hilbert series of the form $p(t) (1+t)/(1-t)^2$. In particular, it must vanish at $t=-1$. But $k$ has hilbert series $1$, which does not.

There is, of course, a resolution of $k$ which is not finite. If I am not mistaken, it looks like

$$\cdots \to a[-4]^4 \to A[-3]^4 \to A[-2]^4 \to A[-1]^3 \to A \to k$$

Answer 2

You want coherent sheaves to have finite global resolutions by locally free sheaves. So definitely you need the regularity of $X$ to ensure that a locally free resolution stops at a finite stage. You also need a global condition such as quasiprojectivity over an affine base to guarantee that you can start the process. (The last condition is not optimal.)

Edit: In reading the follow up comments, I realize my answer was a bit cryptic. The inverse map $K_0(X)\to K^0(X)$ would send the class of a coherent sheaf to the alternating sum of the classes in a resolution. In general, these groups behave quite differently. $K^0(X)$ is contravariant like cohomology and $K_0(X)$ is covariant for proper maps like (Borel-Moore) homology. That they coincide for regular schemes is reminiscent of Poincaré duality.

Answer 3

Asking $K^0(X)$ to be isomorphic to $K_0(X)$ is not always "good enough". Of course, it will allow you to carry over constructions for $K_0(X)$ to $K^0(X)$, but not canonically. And it can happen that $K^0(X)\cong K_0(X)$ without $X$ being regular. For example, take $X= \textrm{Spec} A$, $A=k[x]/(x^n)$ with $n\geq 2$. Then you have an infinite resolution as given in David's answer for $k$. Computing $Tor^A_i(k,k)$ shows that $k$ has no finite resolution. (In fact, $Tor_i^A(k,k) = k^2$ for all $i>0$.) Now, although the above "existence of finite resolution" fails, it is not hard to see that $K^0(X)\cong \mathbf{Z}\cong K_0(X)$ in this case. (Use that $A$ is a local ring and the length map on $A$.) Of course, the natural map $K^0(X) \longrightarrow K_0(X)$ is not an isomorphism. (It is given by $1\mapsto n$.)

[Edit: I added another example]

[Edit 2: There was something wrong with the example below as noted by Michael. I fixed the problem]

Let me also add to my answer the following "snake in the grass". If you work with general schemes, even if regular, one requires the extra assumption of "finite-dimensionality". For example, take the scheme $X=\textrm{Spec} (k \times k[t_1]\times k[t_1,t_2] \times \ldots)$. Now, even though $A = k\times k[t_1]\times\ldots$ is regular, there is an infinite resolution for $k$ of the form $$\ldots \longrightarrow A\longrightarrow A\longrightarrow A \longrightarrow k \longrightarrow 0$$ which corresponds geometrically to taking a point, then adding a line, then adding a plane, etc. Again, take the Tor's to see that $k$ has no finite resolution. Do note that $X$ is not noetherian.

[Edit 3: I added the following for completeness]

Let $X$ be a regular finite-dimensional scheme. Assume that $X$ has enough locally frees. (This notion also arose in Are schemes that "have enough locally frees" necessarily separated ). Then the canonical morphism $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. In the second example, $X=\textrm{Spec} \ A$ is regular, but not finite-dimensional. Does $X$ have enough locally frees?