User tom bachmann - MathOverflow

Answer by Tom Bachmann for Hyper(co)homology of exact (acyclic) complexes

2012-09-07T08:11:27Z

EDIT: reading your question again, I'm a bit confused about both terminology and what you are actually asking. Below by an exact complex I mean one with vanishing cohomology (as a complex!), whereas by an acyclic complex I mean a complex of ($\Gamma$-)acyclic objects. In general, the hypercohomology of an acyclic complex is easy to compute (see first part of my answer), and the hypercohomology of an exact complex is zero (see second half).

To elaborate on Damian's comment (as Sándor Kovács points out, the second half is wishful thinking):

One way to define hypercohomology of a complex $K^\bullet$ is to take a Cartan-Eilenberg resolution $K^{\bullet, \bullet}$ and then take the cohomology of the total complex of $K^{\bullet, \bullet}$. Cartan-Eilenberg resolutions are pretty awesome (they have basically all nice properties you could ask for), but in particular every columns is an injective resolution of the respective term of $K^\bullet$.

Now given any left-bounded double complex $K^{\bullet, \bullet}$ there exist two spectral sequences converging to the cohomology of the total complex (see Weibel). The $E_2$ pages are obtained by taking first horizontal and then vertical cohomology, and vice versa.

In what follows, I shall write $h^p(K^\bullet)$ for the cohomology of the complex $K^\bullet$, and $H^p(K)$ for the derived functor of global sections on $K$.

In our case, taking vertical cohomology first, we get an $E_1$ page $H^q(K^p)$. By your acyclicity assumption, we have $H^q(K^p) = 0$ for $q>0$, i.e. the $E^1$ page just consists of $K^\bullet$ again in the lowest row, and zeros else. Thus the spectral sequence degenerates at the $E_2$ page, leaving you with $h^p(\Gamma(K^\bullet))$ for the cohomology of the total complex, i.e. hypercohomology (see Weibel again for how the separate pages relate to the cohomology of the total complex).

This is the statement you were asking for (to the extent that I am aware of a correct formulation): the hypercohomology of an acyclic complex is just the cohomology of its complex of global sections. In particular an exact acyclic complex has vanishing hypercohomology.

More generally, any exact complex has vanishing hypercohomology: this time look at the spectral sequence where we take horizontal cohomology first. Another awesome property of cartan-eilenberg resolutions: the horizontal boundaries and cycles are themselves injectives, and hence taking horizontal cohomology of the cartan-eilenberg resolution yields an injective resolution of the cohomology of $K^\bullet$. There is a small difficulty: we are not supposed to compute the hypercohomology of $K^{\bullet,\bullet}$ but of $\Gamma(K^{\bullet,\bullet})$. However, since injectives are acyclic, "taking horizontal cohomology" and "taking global sections" actually commutes in our case, so we are good. In summary: there exists a convergent spectral sequence with $E_2$ page $H^q(h^p K^\bullet)$, which is evidently zero if $K^\bullet$ is exact.

Even more generally (and by a very similar argument), hypercohomology factors through quasi-isomorphism.

Note: again as Damian points out, "hypercohomology on some abelian category" does not really make sense. You can in general define hyperderived functors, and then hypercohomology is usually the name for the hyperderived functor of global sections. My above statements hold for the hyperderived functors of an arbitrary left-exact additive functor $\Gamma$.

Cup products and hypercohomology

2012-08-31T17:34:08Z

This is a cross-post of the following math.stackexchange question: http://math.stackexchange.com/questions/188760/cup-product-and-hypercohomology

I always found the cup product slightly mysterious. Recently I discovered the following interesting theorem (in Voisin's book Hodge theory and complex algebraic geometry I, chapter 4.3):

For the setup, let $(X, \mathcal{O})$ be a ringed space, $\mathcal{F}$, $\mathcal{G}$ sheaves of $\mathcal{O}$-modules, $\mathcal{F}^\bullet, \mathcal{G}^\bullet$ acyclic resolutions of $\mathcal{F}, \mathcal{G}$, and $\mathcal{H}^\bullet$ an acyclic resolution of $\mathcal{F} \otimes \mathcal{G}$. Suppose given a morphism of complexes $$\phi^\bullet: Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet) \to \mathcal{H}^\bullet,$$ (where $Tot$ denotes the total (simple) complex associated to a double complex). This data naturally yields homomorphisms $$H^p(X, \mathcal{F}) \otimes H^q(X, \mathcal{G}) \to H^{p+q}(X, \mathcal{F} \otimes \mathcal{G}) \quad(*).$$

The theorem is this: if $\phi^\bullet$ is compatible with the resolutions (that is, the evident triangle involving $\mathcal{F}\otimes\mathcal{G}$, $Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet)$ and $\mathcal{H}^\bullet$ is commutative), then the induced morphism $(*)$ on cohomology is the cup product pairing.

The proof says, somewhat mysteriously to me, that the result follows by defining cup products on hypercohomology, and then using commutativity. While I know about hypercohomology, it is unclear what cup products should even mean in this situation. Can you explain what Voisin means, or provide a reference?

Note: the theorem essentially says that all such $\phi^\bullet$ induce the same morphism on cohomology (independent of the resolutions even), so we need not acutally know here what the cup product pairing is.

Thanks in advance.

EDIT: Since this may not have been clear, my question is this: How do you prove the above theorem, potentially by defining a cup product on hypercohomology and exploiting its properties?

It is quite easy to construct, for arbitray left-bounded complexes $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$, a canonical product $\mathbb{H}^p(\mathcal{F}^\bullet) \otimes \mathbb{H}^q(\mathcal{G}^\bullet) \to \mathbb{H}^{p+q}(Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet))$ (by using "Godement double-resolutions" and the standard fact that godement resolutions remain resolutions after tensoring). It is not clear to me if this is a sensible construction, since $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet$ is not stable under replacing $\mathcal{F}^\bullet$, $\mathcal{G}^\bullet$ by quasi-isomorphis complexes.

EDIT 2: the above claim is false; I made a mistake in my computation. As Donu points out below, the natural target for cup product in hypercohomology is the "derived tensor" $\mathcal{F}^\bullet \otimes^L \mathcal{G}\bullet$. This is indeed what the construction I had in mind yields.

Ideal class groups and extension of number fields

2011-12-27T16:19:15Z

[I already posted this question on stackexchange a while ago, but did not get any response: http://math.stackexchange.com/questions/93437/ideal-class-groups-and-extension-of-number-fields]

Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be schemes and $ f:X \to Y$ be a morphism. Suppose $f^\#:\mathcal{O}_Y \to f_*\mathcal{O}_X $is injective. Then so is the restriction $\mathcal{O}_Y^* \to f_*\mathcal{O}_X^*$ and we can complete to an exact sequence $1 \to \mathcal{O}_Y^* \to f_*\mathcal{O}_X^* \to \mathcal{Q} \to 1$. From this we get an exact sequence in cohomology starting as $ 1 \to H^0(\mathcal{O}_Y^*) \to H^0(f_*\mathcal{O}_X^*) \to H^0(\mathcal{Q}) \to H^1(\mathcal{O}_Y^*) \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{Q}) $ .

Question(s)

Consider the case where $L/K$ is an extension of number fields with rings of integers $\mathcal{O}_L$, $\mathcal{O}_K$, $X = Spec(\mathcal{O}_L)$, $Y = Spec(\mathcal{O}_K)$ and $f$ induced from the inclusion $\mathcal{O}_K \to \mathcal{O}_L$.

How do $H^1(f_*\mathcal{O}_X^*)$ and $H^1(\mathcal{O}_X^*) = Pic(X)$ relate? (In particular, are they equal?)
Can we describe $\mathcal{Q}$ explicitely? In particular, is there a good expression for $Q = H^0(\mathcal{Q})$? Does $H^1(\mathcal{Q}) = 0$?.

Examples/Motivation

The motivation for considering the above is to end up with an interesting exact sequence relating the Picard and unit groups of $X$ and $Y$ via $Q$. Nice results are obtainable for example if $Y$ is the spectrum of a Dedekind domain and $X$ is an open subset or if $Y$ is the spectrum of a one-dimensional integral domain, $X$ its normalisation, and some finiteness conditions hold. These two cases are worked out "by hand" in Neukirch, algebraic number theory, sections 1.11 and 1.12.

A start on question 1

This is probably not the most elegant method, but the five-term exact sequence from the Leray spectral sequence starts as $ 1 \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{O}_X) \to \Gamma(Y, R^1f_*\mathcal{O}_X^*) $ . A sufficient condition for $ Pic(X) = H^1(f_*\mathcal{O}_X^*) $ is thus that the stalks of the first higher direct image are trivial, which is easily seen to be equivalent to $Pic((f_*\mathcal{O}_X)_p) = 0$ for all $p \in Y$. This does hold for an open immersion of number rings, but it is not clear to me if it applies to a normalisation or to an extension of number fields.

It is also easy to see that $H^1(\mathcal{F}) = H^1(f_* \mathcal{F})$ for all $\mathcal{F}$ if $f_*$ is exact, for example a closed immersion. This does not seem applicable here either, though.

conformally embedding complex tori into R^3

2010-04-06T19:23:34Z

Let $L$ be a lattice in $\mathbb{C}$ with two fundamental periods, so that $\mathbb{C}/L$ is topologically a torus. Let $p:\mathbb{C}/L \mapsto \mathbb{R}^3$ be an embedding ($C^1$, say). Call $p$ conformal if pulling back the standard metric on $\mathbb{R}^3$ along $p$ yields a metric in the equivalence class of metrics on $\mathbb{C}/L$ (i.e. a multiple of the identity matrix).

Is there an explicit formula for such a p in the case of L an oblique lattice?

Backgorund

The existence of such $C^1$ embeddings is implied by the nash embedding theorem (fix a metric on $\mathbb{C}/L$, pick any short embedding, apply nash iteration to make it isometric and hence conformal).

For orthogonal lattices, the solution is simple: Parametrise the standard torus of radii $r_1$, $r_2$ in the usual way. Make the ansatz $\pi(\theta, \phi) = (f(\theta), h(\phi))$, pull back the standard metric on $\mathbb{C}/L$ and solve the resulting system of ODEs. This relates $r_1/r_2$ to the ratio of the magnitudes of the periods. This shows that no standard torus can be the image of $p$ in the original question (oblique lattice), although that is geometrically clear anyway.

Comment by Tom Bachmann

2013-05-14T16:24:52Z

I don't think that makes a difference, just consider the same example in the plane and "close up" the two bits via a "connection far enough out" (i.e. a something like a thickened semi-circular arc in the upper half plane).

Comment by Tom Bachmann

2013-04-11T09:43:26Z

Let $k$ be a field and $P$ a prime of $k[X,Y].$ Then $p$ contains an irreducible element $f$, hence $(0) \subset (f) \subset P \subset M,$ where $M$ is a maximal ideal containing $P.$ The first inclusion is strict. Since $k[X,Y]$ has dimension two, $P=M$ or $P=(f).$ This reduces the problem to classifying irreducible polynomials and maximal ideals. Maximal ideals of $k[X_1, \dots, X_n]$ are well-known, by the Nullstellensatz.

Comment by Tom Bachmann

2013-04-11T09:40:01Z

Let $P$ be a prime ideal of $R := \mathbb{Z}[X,Y]$. Then $P' := P \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ If $P' = (p)$ for a rational prime $p,$ then $P/p$ is a prime of $\mathbb{F}_p[X,Y]$. Otherwise, let $S = \mathbb{Z} \setminus {0} \subset \R.$ It follows that $S^{-1}P$ is a prime of $S^{-1}R = \mathbb{Q}[X,Y].$ This reduces the problem to polynomial rings over fields.

Comment by Tom Bachmann

2013-04-08T13:42:18Z

Indeed regarding "sufficiently nice": it suffices that for each $n$, the complex $C^{n, \bullet}$ computes the cohomology of $\scr{F}^n$ (consider the spectral sequence taking "vertical cohomology" first).

Comment by Tom Bachmann

2012-09-07T18:53:17Z

I see; my use of the terminology may be wrong as well. In any case, just to have "almost all" cohomology of $K$ vanish is obviously not enough.

Comment by Tom Bachmann

2012-09-02T08:11:13Z

6.2.1 is indeed precisely what I was looking for. Thanks very much.

Comment by Tom Bachmann

2012-09-01T12:02:34Z

Thanks for the answer. I'll check out the reference as soon as I have time.

Comment by Tom Bachmann

2012-08-31T18:44:02Z

My question is not really "what is the cup product in hypercohomology" but "how do I prove this theorem". I'll check out cartan-eilenberg.

Comment by Tom Bachmann

2012-08-12T10:41:00Z

(I think) he's saying that $\pi_U : U \to \pi(U)$ is a closed immersion.

Comment by Tom Bachmann

2012-05-16T10:56:14Z

That's not James Martin's point (I believe). You are basically asking a question about distribution functions (the distribution function determines a measure on $\mathbb{R}$, and from this you get the expectation). So if you can find any random variables X, Y with the distributions $F_X$ and $F_Y$ such that $E(X) \le E(Y)$, then the statement must be true for all random variables with these distributions. And James Martin's comment shows how to construct such $X, Y$.

Comment by Tom Bachmann

2012-04-11T17:41:26Z

I didn't really think about it again until today g.

Comment by Tom Bachmann

2012-04-11T11:57:37Z

Huh. I just saw this question pop up on the front page and thought "This guy asked something quite similar to what I asked a long time ago.". Then I realised it's my own question :D. Thank you for providing the reference, I will look into it.

Comment by Tom Bachmann

2012-03-27T13:45:38Z

Thanks for your quick answer!

Comment by Tom Bachmann

2012-03-27T13:25:15Z

Sorry to hijack this beatiful answer with a (presumably) trivial question, but how do you obtain the cohomology ring of a quotient space as a subring of the cohomology ring of the original space (any kind of reference is more than enough... maybe I'm just incapable of searching Hatcher's book thoroughly).

Comment by Tom Bachmann

2011-12-30T16:18:29Z

In fact the exact sequence you quote is the five-term exact sequence of the Leray spectral sequence for the Etale topology. Reading Kleiman now; this material seems very interesting and a good way to learn about lots of stuff. I gladly accept your answer.