Definition of multiplication in Grothendieck ring

Let $X$ be a smooth variety over an algebraically closed field $k$ of dimension $n$. Consider the Grothendieck Group $K(X)$ of coherent sheaves on $X$, i.e. the free abelian group generated by expressions $[\mathscr F]$ with $\mathscr F$ a coherent sheaf on $X$, modulo the relation $[\mathscr F]=[\mathscr K]+[\mathscr Q]$ whenever there is an exact sequence $$0\to\mathscr K\to\mathscr F\to\mathscr Q\to 0.$$

You can turn $K(X)$ into a ring with multiplication induced by the tensor product, i.e. $[\mathscr F]\cdot[\mathscr F]:=[\mathscr F\otimes \mathscr G]$. However, I saw the product defined as $$[\mathscr F]\cdot[\mathscr G] := \sum_{i=0}^n (-1)^i [\operatorname{Tor}_i(\mathscr F,\mathscr G)]$$ in Positivity in the Grothendieck group of complex flag varieties by Brion, it's on page 5 in the arXiv version.

Now this should come out of the very definition of $\operatorname{Tor}_i$ as the left derived of the tensor product. However, it does not work out that way for me, so I am looking for the mistake in my calculation.

Choose a projective resolution of $\mathscr G$, say $0\to\mathscr R_n\to\cdots\to\mathscr R_1\to \mathscr R_0 := \mathscr G\to 0$.

Apply $\mathscr F\otimes (-)$ to this sequence and with $\mathscr P_i := \mathscr F\otimes \mathscr R_i$, we get the exact sequence $$0\xrightarrow{d_{n+1}=0} \mathscr P_n \xrightarrow{d_n} \cdots \xrightarrow{d_2} \mathscr P_1 \xrightarrow{d_1} \mathscr P_0 = \mathscr F\otimes \mathscr G\xrightarrow{d_0=0} 0$$ Now set $\mathscr K_i := \ker(d_i)$, $\mathscr I_i := \operatorname{im}(d_i)$ and $\mathscr T_i := \operatorname{Tor}_i(\mathscr F,\mathscr G)= \mathscr K_i / \mathscr I_{i+1}$. We then have two exact sequences, \begin{align*} 0&\to \mathscr I_{i+1} \to \mathscr K_i \to \mathscr T_i \to 0 &&\text{and}& 0&\to \mathscr K_{i} \to \mathscr P_i \to \mathscr I_i \to 0 \end{align*} Note that $\mathscr I_{n+1}=0$ and $\mathscr K_0=\mathscr P_0 = \mathscr F\otimes\mathscr G$, so we have \begin{align*} \sum_{i=0}^n (-1)^i [\mathscr T_i] &= \sum_{i=0}^n (-1)^i ([\mathscr K_i]-[\mathscr I_{i+1}]) = [\mathscr K_0] + \sum_{i=1}^{n} (-1)^i ([\mathscr K_i]+[\mathscr I_i]) \\ &= [\mathscr P_0] + \sum_{i=1}^n (-1)^i [\mathscr P_i] = \sum_{i=0}^n (-1)^i [\mathscr P_i] = \sum_{i=0}^n (-1)^i [\mathscr F\otimes \mathscr R_i] \\ &= \sum_{i=0}^n (-1)^i [\mathscr F]\cdot[\mathscr R_i] = [\mathscr F]\cdot \sum_{i=0}^n (-1)^i [\mathscr R_i] \end{align*} But by basically the same calculation, by exactness of the complex $\mathscr R_\bullet$, we have $\sum_{i=0}^n (-1)^i [\mathscr R_i] = 0$.

What am I doing wrong?

• A resolution of a module doesn't include the module itself. Put it another way, a resolution is an exact sequence if and only if it is a resolution of $0$. – Zhen Lin Aug 28 '14 at 23:50
• @ZhenLin: Yea, I think I see now. I should replace $\mathscr P_0$ and $d_1$ by $0$ in my notation. I will get roughly the same, but will end up with $[\mathscr F]\cdot\sum_{i=1}^n (-1)^{i+1} [\mathscr R_i]$ which is precisely $[\mathscr F]\cdot[\mathscr G]$. – Jesko Hüttenhain Aug 29 '14 at 0:11

I think it's only a matter of indexing. For example, according to the notation in Weibel's book we would have $\cdots\to \mathscr R_1\to \mathscr R_0\to \mathscr G\to 0$ in the projective resolution, rather than defining $\mathscr R_0 = \mathscr G$. That is, in a projective (or free) resolution, the final map $\mathscr R_1\to\mathscr R_0$ is typically defined so that it has cokernel isomorphic to the resolved module/sheaf, rather than incorporating the module/sheaf directly into the complex as a term.
Alternatively, you can just start your sum at $i=1$. (This statement is a little misleading, since we would also need to re-index; see comments below.)
• Hm. I actually thought about this? But starting the sum at $i=1$ doesn't seem to change anything, since $[\mathscr T_0]=[\mathscr K_0]-[\mathscr I_1]=[\mathscr P_0]-[\mathscr P_0]=0$ in my notation. – Jesko Hüttenhain Aug 28 '14 at 23:57
• You should add Zhen Lin's comment to your answer: The problem is really that I treated $\mathscr G$ as part of the resolution. It is not enough to just start the sum at $1$, one needs to look at a different complext $\mathscr P_\bullet$ than I did. – Jesko Hüttenhain Aug 29 '14 at 0:13
• Dear @JeskoHüttenhain, it seems to me that starting at $i=1$ doesn't change the Tor-zero term, but it changes the final product so you get $[\mathscr F]\cdot[\mathscr G]$. (We also might want to change $(-1)^i$ to $(-1)^{i-1}$....) – Andrew Aug 29 '14 at 0:30
• Well, changing that $(-1)^i$ to $(-1)^{i-1}$ is pretty much the same as losing the Tor-zero term. – Jesko Hüttenhain Aug 29 '14 at 0:34
• Ah, I see, perhaps saying we can "start at $i=1$" is a little sloppy. Sorry about that. – Andrew Aug 29 '14 at 0:41