Let $X$ be a smooth hypersurface in $\mathbb{P}^n$ and let $\mathfrak{P}$ be the completion of $\mathbb{P}^n$ along $X$.

Let $\mathcal{R}$ denote the category of reflexive sheaves on $\mathbb{P}^n$ which are bundles outside a finite set of points in $X$ (the finite set is not fixed, it could/will be different for different sheaves). The motivation for this is Grothendieck's effective Lefschetz condition.

Let $\mathcal{V}$ denote the category of (formal) locally free sheaves on $\mathfrak{P}$ and consider the functor

$$F: \mathcal{R} \to \mathcal{V}.$$ Then by Grothendieck's Lefschetz conditions, this functor is an equivalence of categories. This therefore implies that

$$K_0(\mathbb{P}^n) \cong K_0(\mathfrak{P}).$$

On the other hand, the subgroup $CH^{n}(\mathbb{P}^n) \subset K_0(\mathbb{P}^n)$ maps to zero in $K_0(\mathfrak{P})$. This can be seen in the following way:

take any point $x \in \mathbb{P}^n \setminus X$ and consider the resolution of the skyscraper sheaf $k_x$ $$0 \to F_{\bullet} \to k_x.$$ On restricting this resolution to $\mathfrak{P}$ (restrict to any thickening $X_m$ otherwise), we get an exact sequence $$ 0 \to F_{\bullet}\otimes O_{\mathfrak{P}} \to 0.$$

Therefore we see that $\sum(-1)^iF_i \mapsto 0$ under the map $K_0(\mathbb{P}^n) \to K_0(\mathfrak{P})$.

Clearly I am making some (stupid) mistake. I would be glad if someone points out what it is. In particular which of the arguments is wrong.



I will first assume that, by $K_0$, you mean the quotient of the free abelian group generated by isomorphism classes of objects in your category, modulo the relation $$\sum_i(-1)^i [F_i] = 0 \quad (A)$$ for every exact sequence $0 \to F_0 \to F_1 \to \dots \to F_{n-1} \to F_n \to 0$.

Well, since the functor does not preserve exactness of such long exact sequences, there is no reason that it should induce an isomorphism of $K_0$ using this definition.

OTOH, if you use as definition of the K-group the quotient generated by relations for short exact sequences, then I do not see why the relation (A) should hold. Of course, in an abelian category, you can decompose a long exact sequence into short exact sequences, and deduce it from the relation . Then the above relation for long exact sequences follows from the relation coming from the short exact sequences. But in an exact category, I do not see how to deduce (A).

  • $\begingroup$ @ABayer: Thanks. I realise the mistake I am making. I was indeed considering relations by short exact sequences. $\endgroup$ – MOfan Apr 11 '11 at 15:47

Dear all,

First, I apologize for writing this as an "answer" -- it is just a comment on ABayer's answer (but I don't have permission to comment on answers). ABayer's point is right on: the kernels and cokernels in the standard resolution of a skyscraper sheaf of a point are not reflexive. So you cannot break things up into short exact sequences. One case where you do get a short exact sequence is for P^2. But Grothendieck rules out the equivalence in this case (because he requires depth at least 3 or 4, can't remember which at the moment).

Best regards,


  • 1
    $\begingroup$ The text inside <> is supposed to be "not". I am not sure what is wrong with the parser. $\endgroup$ – Jason Starr Apr 11 '11 at 11:15
  • $\begingroup$ Hi Jason. It seems the computer thinks your method of emphasis looks a lot like an HTML tag. I changed it to a boldface. $\endgroup$ – S. Carnahan Apr 11 '11 at 16:12

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