## The Grothendieck Group of Coherent Sheaves on a Variety

Ok, to continue our quest toward the statement of the theorem, we need to explain a few more ideas.  Today, we’ll talk about the structure of the Grothendieck group K(X) of coherent sheaves on an algebraic variety X.  (Remember the ongoing convention that varieties are all (quasi) projective, defined over an algebraically closed field).  In fact, this group has a ring structure and, given a map $f: X \rightarrow Y$, there are maps between K(X) and K(Y) (in both directions).  It turns out that the most general statement of the Riemann Roch Theorem is an equality inside this group, so we had better have an idea of how it works!  There are some homological methods today, a topic which seems to me particularly ill-suited to this forum given my difficulty creating complicated commutative diagrams.  At a few instances I will sketch how certain facts are to be proven without going into all the details.

So let X be a variety, and let F(X) be the free abelian group generated by the coherent sheaves on X.  An element of F(X) looks like $\Sigma n_i F_i$ where this is read as integer times sheaf.  Given a short exact sequence (of coherent sheaves on X) $0 \rightarrow F \rightarrow G\rightarrow H\rightarrow 0$ from which we will from now on refer to as $(S)$, we form the element in F(X), Q(S) = G – F – H.  Then we define the Grothendieck group of coherent sheaves on X to be the quotient of F(X) by the sub group generated by all elements of the form Q(S) where S is a short exact sequence as above.  Clearly the group K(X) is generated by the images of sheaves on X, and given any map of abelian groups $a: F(X) \rightarrow G$ which satisfies a(G) = a(F) + a(H) for any short exact sequence S, we have a unique compatible map from K(X) to G.  All this seems very complicated and simple at the same time – the former because we don’t have a great understanding of all coherent sheaves on an algebraic variety, the latter because we haven’t really done anything.

However, we can undo the complication a bit by restricting our attention to vector bundles.  Let V be the set of vector bundles on X and let V(X) be the free abelian group generated by elements of V (as in the definition of F(X) above).  In an analogous way, quotient out by short exact sequences of vector bundles on X to get the group $K_1(X)$.  To every vector bundle we can associate a locally free sheaf – we have an injection of groups V(X) into F(X) and a canonical group homomorphism $e: K_1(X) \rightarrow K(X)$.  Today’s first goal is to outline the proof of:

Theorem: When X is a non-singular, quasi-projective variety (irreducible is implied), the homomorphism “e” above is a bijection.

Step 1: Given an exact sequence S as above with G,H in V, then F is also in V.

At each point p of X, the module F_p (over the local ring) is a submodule of G_p, which is free, so that F_p is a projective and thus free module.  Then a fairly easy exercise shows that a coherent sheaf all of whose stalks are free is itself locally free.

Step 2:  Suppose that X has dimension n, and that $0 \rightarrow Z \rightarrow L_q \rightarrow \ldots \rightarrow L_0 \rightarrow F \rightarrow 0$ is an exact sequence with each L_i in V.  If $q \geq n - 1$, then Z is also in V.

Again this is a local question, so let p be a (closed) point of X.  Since the local ring of X at p is regular of dimension n (note the use of the non-singular hypothesis), we can apply the syzygy theorem (which I don’t think has been written about here, but you can find it in many commutative algebra texts) to conclude that Z_p is also free.

Step 3: Any coherent sheaf F on X admits a resolution $0 \rightarrow L_n \rightarrow \ldots \rightarrow L_0 \rightarrow F \rightarrow 0$ with each L_i a vector bundle.

Embed X in projective space and let Y be its closure.  The sheaf F extends to a coherent sheaf F’ on Y (this is a nice fact, we’ll leave it as an exercise) and this sheaf is a quotient of a direct sum of sheaves of the form $\mathcal{O}_Y(n)$ (proof similar to one given last time, or see Hartshorne), so by restriction F is also.  Repeatedly using this fact and then Step 2 gives the conclusion.

As a consequence of this step, we have the following formal consequence: given coherent sheaves A, B, and C and surjective maps $u: A \rightarrow B, v: C \rightarrow B$, we may find a vector bundle L with surjective maps to A and C so that the compositions to B are equal.  To see this, take the fiber product sheaf (interpreted appropriately) and apply the step above.

Definition:  Given a coherent sheaf F on X, choose a resolution L by vector bundles as in Step 3, and set $g(L) = \Sigma (-1)^p L_p$.  This is an element inside $K_1(X)$.

Step 4: The element g(L) depends only on F, that is, not on the resolution.

This step in particular is a bit tricky to explain without a lot of diagrams, but isn’t too hard in and of itself.  The idea is show that for any other resolution M of F, we have g(L) = g(M).  To do this, we argue that there is a third resolution N, with surjective maps to L and M inducing the identity map on F.  Calling W the kernel of the map from N to L, we have that this is also a complex of vector bundles by step 1.  By construction/exercise, the homology of this complex W is trivial at each step.  Then we immediately have that g(W) = 0 but also that g(N) = g(L) + g(W), after which we are done.  The proof of the existence of N proceeds degree by degree using a repeated application of the consequence to step 3 above.  We leave it as another exercise.

Step 5: The map g is an additive function.

Given an exact sequence of sheaves like S above, we must construct a system of compatible resolutions (that is, of F, G, H so that they fit into a commutative diagram of resolutions).  This is sort of a standard fact in diagram chasing using again the consequence of Step 3 above.

Proof: The proof is easy now that we’ve done (interpreted appropriately) all the work.  Via Step 4 and Step 5, we have a homomorphism $n: K(X) \rightarrow K_1(X)$ by setting n(F) = g(L).  Then we have $e(g(L)) = F$ since the choice of L is a resolution of F.  This finishes the proof and implies that we to understand the group we really need to know how to work with locally free sheaves, which are in general, easier to understand than arbitrary coherent ones.

There are four operations on these groups that I would like to point out.

1.  Ring Structure:

If F and G are coherent sheaves on X, the sheaves $Tor_p(F,G)$ are also.  Since X is smooth all local rings are regular and this implies the sheaves are zero for p > dimension of X.  Then we can define $\chi(F,G)$ as the alternating sum of the tors.  The exact tor sequence shows that this is bilinear in F and G so defines a map $K(X) \times K(X) \rightarrow K(X)$.  When F and G are locally free we have that $\chi(F,G) = F \otimes G$.  Since K(X) is generated by vector bundles (and the map is associative on such sheaves), we can show this implies that the map is associative in general.   It is pretty clearly commutative, implying that we have a ring structure on K(X).

2.  Exterior Product:

When E is a vector bundle, $\Lambda^pE$ is also one, and we denote this class in K(X) by $\lambda^p(E)$.  Given an exact sequence of vector bundles S (as above), we have the formula (exercise) $\lambda^p(G) = \Sigma \lambda^r(F) \lambda^s(H)$ where the sum is taken over r + s = p.  Introducing a formal variable t, we may write $\lambda_t(E) = \Sigma \lambda^p t^p$ which is an element of K(X)[[t]] (beginning with 1 by convention).  The formula above may be expressed as $\lambda_t(G) = \lambda_t(F) \lambda_t(H)$.  The association $E \mapsto \lambda_t(E)$ may be thought of as a map from K(X) to formal (invertible) power series in t with coefficients in K(X).

3.  Pullback:

Given a morphism of varieties $f: Y \rightarrow X$ and a vector bundle E on X, we have that $f^*E$ is a vector bundle on Y.  This operation is additive and so extends to a map $f^*: K(X) \rightarrow K(Y)$.  We’ll leave it as an (easy) exercise that this is compatible with the ring operation and the exterior product map in the obvious way.  Note that for any coherent sheaf F, the pullback may be defined directly as the alternating sums of $Tor_p(\mathcal{O}_Y, F)$.

4. Pushforward:

Let f be a proper map from Y to X.  When F is coherent on Y, we have that the higher direct image sheaves are coherent on X from the last post.  Their alternating sum is a well defined element of K(X).  From the long exact sequence formula we have a homomorphism $f_* K(Y) \rightarrow K(X)$.  This is not compatible with multiplication, but we have the projection formula : $f_* (y \cdot f^*(x)) = f_*(y) \cdot x$ where x is in K(X) and y is in K(Y).  Since this post is already a bit long, we’ll leave this as yet another exercise, noting that it suffices to prove it when F is coherent on Y and L is locally free on X and then check it locally as an equality of the higher direct image sheaves.  A spectral sequence argument shows that if g is a proper map from Z to Y then we have $(fg)_* = f_* g_*$.

Just so you can’t say there were no examples, here are three:

$K(\mathbb{A}^1) = \mathbb(Z)$

$K(X) = Pic(X) \oplus \mathbb{Z}$ for X a smooth curve.

$K(\mathbb{P}^r)$ is freely generated by the collection of $\mathcal{O}_{V_i}$ where $V_i$ is a linear subspace of dimension i.

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### 12 Responses to The Grothendieck Group of Coherent Sheaves on a Variety

1. Pingback: Worth Reading « Not Even Wrong

2. Yeah, homological algebra is a bit tricky because wordpress doesn’t support the tex packages xy or dcpic, as far as I know. I think John Armstrong at Unapologetic Mathematician creates the diagrams offline, makes images, and then uploads them.

3. Correct. Commutative diagrams are pretty much beyond WordPress’ LaTeX ken, so I have to make them by hand.

4. Gino says:

In step 3 do you need the base field k to be algebraically closed? I think it should work over any field but I am not so sure. Can someone explain me whether or not we can assume k to be any field? Thank you!

5. Matt DeLand says:

Yes, those steps do not use the assumption that k is algebraically closed. On a smooth projective variety over a field, any coherent sheaf admits a locally free resolution (in fact, you only need a resolution of length dim(X)).

6. Gino says:

Thank you Matt! As you suggest in your notes I have found the explanation in Hartshorne bible! Thank you very much!

7. Roland says:

Hi,

I wondered if the projection formula you wrote in 4 is correct. If x and y are “only” coherent sheaves (and none of them is assumed to be locally free) shoudn’t you replace the pull back by the derived pullback ? I haven’t been able to find any refererences for a projection formula which would hold at this level of generality (even in Verdier’s paper). If you have a “serious” reference I would be extremely thankful if you could mention it.

Best,

8. Matt DeLand says:

Hi,

I’m confused by your question. In (3) it explains that the pullback map on a coherent sheaf first requires resolving by locally free sheaves and then applying the pullback “additively”. Is this what you are worried about? In any case, you can read the proof of this projection formula in the original Borel-Serre paper.

Matt

9. Matei Toma says:

Hello,

I am interested in computing the pullback in the Grothendieck group of a (torsion free) sheaf $F$ by, say, a desingularization map. Maybe someone can enlighten me on the following questions: Is there anything that can be said about the codimensions of the supports of the sheaves $Tor_i(O_Y,F)$? Is there any relation between the torsion of the usual sheaf pullback $f^*F$ and $Tor_1(O_Y,F)$?

Matei

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