# $K_0$ of a non-separated scheme

This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($m\geq 1$). For any (noetherian) scheme $X$, let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$.

Firstly, let me sketch that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.

Let $X=\mathbf{P}^n$ be the projective $n$-space.

(I omit writing the base scheme in the subscript. In fact, you can take any noetherian scheme as a base scheme in the following, I think.)

Let $H\cong \mathbf{P}^{n-1}$ be a hyperplane with complement $U\cong \mathbf{A}^n$. By a well-known theorem on Grothendieck groups, we have a short exact sequence of abelian groups $$K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Now, let $i:H\longrightarrow X$ be the closed immersion. Then the first map in the above sequence is given by the "extension by zero", which in this case is just the K-theoretic push-forward $i_!$, or even better, just the direct image functor $i_\ast$. Now, there is a projection map $\pi:X\longrightarrow H$ such that $\pi\circ i = \textrm{id}_{H}$.

By functoriality of the push-forward, we conclude that $\pi_! \circ i_\ast = \pi_! \circ i_! = \textrm{id}_{K_0(H)}$.

Therefore, we may conclude that $i_\ast$ is injective and that we have a split exact sequence $$0 \rightarrow K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Thus, we have that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.

Q1: Let $\mathbf{P}^{n,m}$ be the projective $n$-space with $m$ origins ($m\geq 1$). For example, $\mathbf{P}^{n,1} = \mathbf{P}^n$. (Again the base scheme can be anything, I think.) Now, is it true that $$K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1,m}) \oplus K_0(\mathbf{A}^n)?$$

Idea1: Take a hyperplane $H$ in $\mathbf{P}^{n,m}$. Is it true that $H\cong \mathbf{P}^{n-1,m}$ and that its complement is $\mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:H\longrightarrow \mathbf{P}^{n,m}$ is proper, right? Also, is the projection $\pi:\mathbf{P}^{n,m}\rightarrow H$ proper? If yes, the above reasoning applies. If no, how can one "fix" the above reasoning? I think that in this case one could still make sense out of $i_!$ and $\pi_!$ (even if they are not proper maps.)

Idea2: Maybe it is easier to show that $K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^{n,m})$, where $\mathbf{A}^{n,m}$ is the affine $n$-space with $m$ origins. Then one reduces to computing $K_0(\mathbf{A}^{n,m})$...

Idea3: One could also take $m=2$ as a starting case and look at the complement of one of the origins. Then we get a similar exact sequence as above and one could reason from there.

Which of these ideas do not apply and which do?

Note: Suppose that the base scheme is a field. Since $K_0(\mathbf{A}^n) \cong \mathbf{Z}$ and $K_0(\mathbf{P}^n) \cong \mathbf{Z}^{n+1}$, this would show that $$K_0(\mathbf{P}^{n,m}) \cong \mathbf{Z}^{n+m}.$$ More generally, if $S$ is the base scheme, $K_0(\mathbf{P}^{n,m}) \cong K_0(S)^{n+m}$.

-

This is exactly the sort of example where it is relevant which version of K-theory you are employing. (In particular, which theorems you can employ here depends on this!) The issue here is a tad subtle.

To illustrate, let's restrict attention to the case of affine $n$-space $X$ with a doubled origin ($n\geq 2$). (An iteration of the discussion below, combined with your localization argument in Idea2, will let you address your general multiply-origined projective spaces.) For convenience, let me assume that the base is a regular noetherian scheme $S$.

Let me first answer your question as it was asked. Quillen's localization sequence gives fiber sequences of spectra or spaces

$$G(S)\to G(\mathbf{A}_S^n)\to G(\mathbf{A}_S^n-\{0\})$$

and

$$G(S)\to G(X)\to G(\mathbf{A}_S^n)$$

Here I'm using $G$-theory is the $K$-theory spectrum (or space) of coherent sheaves. This has the property that $G(S)\simeq G(\mathbf{A}_S^n)$. Putting this together, we see that

$$G(X)\simeq G(\mathbf{A}_S^n)\times G(S)\simeq G(S)\times G(S)$$

This answers your question for $G$-theory; in particular $\pi_0$ of $G$ is the Grothendieck group you seek, and so we conclude that $G_0(X)=G_0(S)\oplus G_0(S)$. Now use your localization arguments to get that $G_0(\mathbf{P}^{n,2})$ is $n+2$ copies of $G_0(S)$.

However: for $K$-theory, the way this computation gets done depends critically on which model of $K$ you use.

(A) If we define $K$-theory as in Thomason-Trobaugh (as the Waldhausen $K$-theory spectrum of perfect complexes), then in this case $K(X)\simeq G(X)$ and $K(S)\simeq G(S)$. (This follows from "Poincaré Duality;" see the end of section 3 of Thomason-Trobaugh.)

(B) If, however, we define $K^{\mathrm{naive}}(X)$ as the Quillen $K$-theory of the category of algebraic vector bundles on $X$, then things look different. [N.B. that this is the name given by Grothendieck, Illusie, et al. in SGA 6. It is not meant to be insulting!] Indeed, the inclusion $\mathbf{A}_S^n\to X$ induces an equivalence between the catgories of algebraic vector bundles on $X$ and those on $\mathbf{A}_S^n$, since the origin has codimension at least $2$. So then

$$K^{\mathrm{naive}}(X)\simeq K^{\mathrm{naive}}(\mathbf{A}_S^n)\simeq K(\mathbf{A}_S^n)\simeq K(S).$$

The second equivalence here follows from the fact that "naive" $K$-theory agrees with $K$-theory for schemes that admit an ample family of line bundles (see section 3 of Thomason-Trobaugh). The difference between (A) and (B) here reflects the failure of our $X$ to admit such a family.

An example like this is discussed is Thomason-Trobaugh I think at the end of section 8 (wherever they talk about Mayer-Vietoris). [Sorry, this is from memory; I don't have a copy of the Festschrift handy.]

Hope this helps!

-
Great! Just a couple of questions. You said, "the inclusion $\mathbf{A}^n \rightarrow X$ induces an equivalence between...". Are you supposing $n\geq 2$ here? What happens when $n=1$? I'm not good with vector bundles so details are much appreciated here. My question was originally just for $K_0(X)$. (I'm guessing this is old notation only used in the 60's.) Thanks for the enlightment on $K^0(X)$, i.e., the Grothendieck group of vector bundles on $X$. This clearly shows that the Cartan morphism need not be an isomorphism when $X$ is not separated. –  Ariyan Javanpeykar Mar 7 '10 at 8:05
Yes, I'm using $n\geq 2$, but only in (B) at the line you quote. When $n=1$, everything is the same, except the naive K-theory happens to agree with the "true" K-theory. You are, of course, correct that $K_0$ is older notation. Nowadays $K_0$ usually refers to the thing I describe in (A), and $G_0$ refers to the object you were interested in. (Of course you can use any notation you like, as long as you're clear about what you mean, just as you were here! The only risk is that it's sometimes tricky to keep track of which theorems apply to which objects...) –  Clark Barwick Mar 8 '10 at 21:43