## Direct Image Sheaves Under Proper Maps

Today we continue our review/introduction of background material en route to stating and proving the Riemann Roch theorem. This theorem involves the relationship between proper maps and sheaves on the domain and target, so we need to understand how they are related. Really, we need to understand how to move from sheaves on the domain to sheaves on the target – so we’ll explain the (higher) direct image sheaves. Recall that all varieties are (quasi) projective and defined over algebraically closed fields.

Suppose that $f:X \rightarrow Y$ is a morphism and that F is a (coherent) sheaf on X. There are two possible ways to define the direct image sheaves (both of which agree). For each non-negative integer q and each open $U \subset Y$, define $R^qf(F) = H^q(f^{-1}(U), F)$. Note – often there’s a “lower star” in the notation but I will omit that for type-setting ease. Then we take the associated sheaf. In fact when q is zero, we call this the direct image sheaf (and no sheafification is necessary). Taking the direct image sheaf is a left exact functor from the category of sheaves on X to the category of sheaves on Y. It turns out that the higher direct image sheaves (that is, with q positive) are the derived functors of the direct image functor $F \mapsto R^0f(F)$. (I won’t prove this though it isn’t too hard – see Harthorne, EGA III for more). See this post for more.

Example 1: If f is a closed immersion, then $R^0f(F)$ is the sheaf extended by zero away from X and the sheaves $R^qf(F)$ are all zero. To see this, note that if U is open affine in Y, then $f^{-1}(U) = U \cap X$ is also affine and $H^q(U\cap X, F) = 0$ for positive q. From this description it is easy to see that the stalks of the direct image sheaf are zero at points of Y not contained in X.

Example 2: When Y is simply a point (which won’t be an uninteresting case for the future statement of the theorem!) then the $R^qf(F)$ are simply the cohomology groups $H^q(X, F)$. In this sense, taking the direct images is a generalization of taking cohomology.

Example 3: Suppose that f is a birational morphism (and say, that X and Y are smooth). Then we have that $R^0f(\mathcal{O}_X) = \mathcal{O}_Y$. If for example f is a blowup along a smooth subvariety, then all the higher direct image sheaves are zero.

Remark that in some sense the higher direct image sheaves are recording a relative cohomology. We could ask when is this true fiberwise: the Cohomology and Base Change theorem says that, roughly, if the sheaf F is flat over Y (which we assume integral) and the p-th cohomology groups of each fiber have the same dimension, then the natural map $R^pf(F) \otimes k(y) \rightarrow H^p(X_y, F_y)$ is an isomorphism for each point y of Y. Since we won’t need this in the following, you can read about it in Mumford’s Abelian Varieties, or in Hartshorne, or of course in EGA.

Further, we can ask how is the cohomology of the higher direct image sheaves on Y relate to the cohomology of F back on X. Here the relationship (purely topological, in the sense that it holds for topological spaces and continuous maps) is that there is a spectral sequence whose term $E_2^{p,q} = H^p(Y, R^qf(F))$ and which abuts to the group $H^*(X, F)$.

In general, the higher direct image of a (quasi)coherent sheaf is quasicoherent. However, when the map f is proper we have today’s main theorem, which some would call the fundamental theorem of proper maps:

Theorem: Suppose that $f:X \rightarrow Y$ is proper and that F is a coherent sheaf on X. Then the proper pushforward sheaves $R^qf(F)$ are coherent on Y.

The proof will proceed in several steps, (none of which is too hard).

Step 1: We may reduce to the case where X is projective space over Y, namely $X = P \times Y$. and the map f is the projection. Further we may assume that Y is affine.

To see this, let $\Gamma_f$ be the graph of f inside $P \times Y$. (Recall that we can do this for maps between projective varieties). By definition of properness, it is closed in the product. Let F’ be the extension of F by zero on $P \times Y$. Immediately we see that $R^q\pi(F') = R^qf(F)$ so of course one is coherent if and only if the other is. Since the property of being coherent is a local one, we may also assume that Y is affine.

Step 2: Every coherent sheaf F on $X = P \times Y$ is the quotient of a direct sum of sheaves of the form $\mathcal{O}_X(n)$.

This Lemma is proved in Hartshorne (II.5) so I’m not going to repeat it. Basically you choose (finitely many) generators for F on each piece of the standard affine cover of projective space – then up to a power of the coordinate functions these sections will extend to all of projective space and will in fact generate F.

Step 3: For the line bundles $\mathcal{O}_X(n)$, we can directly calculate that $R^q\pi (\mathcal{O}_X(n))$ are coherent on Y.

If U is an affine open of Y we have that $R^q\pi (\mathcal{O}_X(n))(U) = H^q(P \times U, \mathcal{O}_X(n))$. Now, taking an affine cover of projective space $U' = {U_i}$, we have the following facts. Cech cohomology computes sheaf cohomology on this affine cover, and there is an equality of Cech complexes $C(U', \mathcal{O}_X(n)) = C(U', \mathcal{O}_P(n)) \otimes H^0(U, \mathcal{O}_U)$. This shows that the sheaf $R^q\pi(\mathcal{O}_X(n))$ is isomorphic to $V^q(n) \otimes \mathcal{O}_Y$ where $V^q(n)$ is the q-th cohomology of the line bundle $\mathcal{O}_P(n)$ on projective space. This is great, because we know it’s finite dimensional (again, see Hartshorne for the explicit computation and the other facts swept under the rug in this abbreviated argument), so that these sheaves are coherent on Y.

Step 4: We finish the proof of the theorem.

Let’s proceed by descending induction on q – note that the statemtent of the theorem is clear for $q > dim(X)$ because then the pushforwards are zero. Suppose that the theorem is true for (q + 1). By Step 2, we have an exact sequence $0 \rightarrow K \rightarrow L \rightarrow F \rightarrow 0$ where L is a direct sum of line bundles on projective space and K is coherent (as it is the kernel of a map of coherent sheaves). Now since the derived functor of the pushforward gives us the higher direct images, we can pushforward this sequence and take its long exact sequence “in cohomology”, that is, in higher direct images. Letting the reader write down this sequence on her/his own, we immediately see that $R^qf(F)$ is “trapped” in between two coherent sheaves (the induction hypothesis applies to K and Step 3 applies to L). Because the image of a map of coherent sheaves is coherent and the extension of two coherent sheaves is coherent (exercise), we have that $R^qf(F)$ must also be coherent. The proof is complete.

As a last remark, note that we don’t really need that the varieties are (quasi)-projective, we can apply Chow’s Lemma to prove the theorem in full generality, but we’ll leave that step to the ambitious reader!

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### 6 Responses to Direct Image Sheaves Under Proper Maps

1. Zygmund says:

Just curious- I’m currently a beginner in algebraic geometry, working through Hartshorne and I was curious: Where do you use the higher direct image sheaves R^i? In his proof of the Riemann-Roch theorem he only uses Serre duality.

2. Zygmund, Hartshorne only proves the Riemann-Roch theorem for the structure morphisms of smooth, complete curves and surfaces. In general, you can prove a theorem that can be rightly called the Riemann-Roch theorem for the structure map of any smooth, complete variety in a similar way (I’m not sure on the necessity of smoothness, but it’s the case that comes up most often.)

Now, here’s the however: what if you want to relate sheaves on X and Y when you have a proper map $f:X\to Y$? There’s a Riemann-Roch theorem there too! It just needs the higher direct images to even state it. The reason they don’t come up in the versions in Hartshorne is that the direct images of a sheaf in the map to a point are just the cohomology groups, which he manipulates directly and treats in terms of invariants of the variety, rather than handling things relatively. Look in the Appendix on intersection theory, there’s some sketching of the generalized stuff there, don’t recall how detailed it is offhand, though.

3. Zygmund says:

Thanks for the explanation.

4. jc says:

I have a question. When f is birational, as ex3, what is the definition of direct image? Thank you.

• Matt says:

What do you mean? The definition is the same.

• Anonymous says:

I mean, under this condition, f is actually not a morphism, then how can we define the direct image…I know little about birational thing