User ying zhang - MathOverflow

The etale cohomology``ring" structure of torsion sheaves on varieties

2013-04-02T20:52:21Z

For a topological manifold $M$, one can speak of the cohomology ring structure $H^*(M, k)$ where $k$ is a ring. If one replace $M$ by an arithmetic schemes $X$ over a base ring $S$, and replace $k$ by a torsion sheaf $\mu_n$, then one can define the "cup product" $H^i(X, \mu_n)\times H^j(X, \mu_n)\to H^{i+j}(X, \mu_n^2)$. However, if $S$ contains the n-th root of unity, then one can "untwist" the torsions by the canonical isomorphism $\mathbb{Z}/n\cong \mu_n^r$. (E.g., One can view the isomorphism of etale sheaves $\mu_n\cong \mathbb{Z}/n$ as induced by the isomprhism of the corresponding group schemes that represent them. ) Then one indeed has a cohomological "ring" structure for $H^*(X, \mathbb{Z}/n)$. My question was, people don't seem to be using this information a lot to talk about properties of arithmetic schemes. Maybe I am wrong. To my knowledge, even when $X$ is an elliptic curve over a local field $\mathbb{Q}_p$, with coefficients $\mathbb{Z}/2$, not much is obvious to me. In particular, does the congruence condition on $p$ make a difference? (My real question was, how does "arithmetic" play with "geometry" in this sense?)

I computed this "ring" structure for elliptic curves, but I am also worried that this may be trivial in the eye of the experts. I am not sure if MO will be a good place to post this, anyway.

Looking for reference on Serre's talk "linear rep and number of points mod p"

2010-10-26T04:30:25Z

Actually I am not sure this is a legitimate question on MO. In April and June of this year Serre gave two talks on the same title "linear representations and the number of points mod p", one in ETH Number theory Days Zurich, another during Prof. Gross's birthday conference in Boston. Unfortunately I was in neither of those, nor could I find another reference about this talk online.

In the proof of Weil conjecture for curves, say an elliptic curve $E$ over $\mathbf{Q}$ has good reduction mod $p$, then the characteristic polynomial of the Frobenius operator on the Tate module for $E$ mod $p$ will essentially give us the number of points on the curve in $\mathbf{F}_{p^n}$. So I would say this is an obvious example of relations between two-dimensional representation and number of points. But I wonder if Serre has more. E.g. (tangentially related) Here Mazur was interested in the Chebyshev bias (which is usually a quite analytic phenomenon) among the number of points corresponding to different $p$ (which, by the way, has few references also), and I'd like to know if the representation side could shed some light on this.

Of course since I didn't attend the talk, Serre could be talking about totally different things. I would greatly appreciate it if anyone attended the talk/have seen such notes/heard about this circle of ideas share some comments on this. Thanks!

Lemmas on etale cohomology with compact support from the book 'Arithmetic Duality Theorems'

2012-05-23T20:25:40Z

I was reading Milne's book "Arithmetic Duality Theorems". On page 166 there are a lot of useful lemmas on the etale cohomology with compact support on S-integers. However, I get confused when I tried to look at Prop 2.3 (a) and (d) at once.

Suppose $X$ is the ring of integers of a totally complex number field $K$. $U\hookrightarrow X$ is an open subscheme, $S=X\setminus U={p}$ has only one finite prime. Consider the sheaf $\mathbb{G}_m$ and its pull-back on the various schemes involved, by (d) of Prop 2.3, we have

$$ H^0(\mathbb{F}_p, \mathbb{G}_m)\to H^1_c(U, \mathbb{G}_m)\to H^1_c(X, \mathbb{G}_m) $$

Here $H^1_c(X, \mathbb{G}_m)=H^1(X, \mathbb{G}_m)$ since $K$ is a totally complex field, but anyway it is finite. Also $H^0(\mathbb{F}_p, \mathbb{G}_m)=\mathbb{F}_p^*$ is finite, therefore $H^1_c(U, \mathbb{G}_m)$ is finite.

If we use (a) of Prop 2.3, we have

$$ H^0(U, \mathbb{G}_m)\to H^0(K_p, \mathbb{G}_m)\to H^1_c(U,\mathbb{G}_m)\to H^1(U,\mathbb{G}_m)$$

Here $K_p$ is the Henselization of $K$ at $p$. Since we have shown $H^1_c(U,\mathbb{G}_m)$ is finite, and since we know $H^0(U, \mathbb{G}_m)$, which is the $S$-units on $X$, is a finite rank abelian group, we should have $H^0(K_p, \mathbb{G}_m)=K_p^*$ is also a finite rank abelian group. This seems absurd to me, since the subfields of $K_p$ which are number fields has infinite rank multiplicative groups.

So I wonder, did I miss anything? Thanks very much in advance!

Collection of sub line bundles of a vector bundle

2011-10-18T04:04:18Z

Let X/k be a projective integral variety over a field, with fixed $\mathcal{O}(1)$, $\mathcal{E}$ be a rank two vector bundle, for simplicity assume it is stable. Regarding the collection of sub line bundles $\mathcal{L}\subset \mathcal{E}$ up to isomorphism,

EDIT: and moreover $\mathcal{L}$ is the line bundle given by an effective divisor,

denoted this collection by $Sub(\mathcal{E})$, I have the following questions:

(1) When is this collection finite?

(2) Take another scheme $T/k$, and base change $X_T/T$, when is $Sub(\mathcal{E}_T)=Sub(\mathcal{E})_T$ ?

(3) If the above equality doesn't hold, when does $Sub(\mathcal{E}_T)=Sub(\mathcal{E})_T$ up to tensoring pull back of a line bundle on T?

(4) If the answer to (2) and (3) are not very trivial, then consider the functor $Sub(\mathcal{E}_T\otimes f_T^*(\mathcal{N}))$ , where $f:X\rightarrow \mathcal{S}pec\ k$ is the structure morphism, $\mathcal{N}$ is a line bundle on T, and we identify two sub line bundles of $\mathcal{E}$ up to tensoring with $f^*_T(\mathcal{N})$. When is this functor representable?

Edit: Here by a sub line bundle I just mean a rank one subsheaf which is itself a line bundle; i.e. in an exact sequence like $0\rightarrow \mathcal{L}\rightarrow \mathcal{E}\rightarrow \mathcal{Q}\rightarrow 0$, I didn't mean both $\mathcal{L}$ and the "quotient" $\mathcal{Q}$ are line bundles.

Toric automorphism of P1 times P1 blown up at four pts

2011-09-15T15:55:49Z

A toric morphism between toric varieties is a morphism that is equivariant w.r.t. to the toric action, see e.g. section 3.2, Notes by H.Verrill and D.Joyner for definitions. In particular, any toric morphism comes from a morphism of the corresponding fans. For example, the toric automorphism group of $P^2$ is $D_3$. If we denote the fan for $P^2$ as generated by the rays (1,0), (0,1), (-1,-1), then other than the obvious reflection which has order 2, the element of order 3 in $D_3$ is given by the unimodular matrix

As another example, it is also easy to see that the toric automorphisms of $P^1\times P^1$ is $D_4$. Now if we blow up 4 points of $P^1\times P^1$ at such places so the fan after the blow-up has rays $(0,\pm 1), (\pm 1,0), (\pm 1, \pm 1)$, then the note I mentioned above says the toric automorphism group is $D_8$. My problem is that I can not find the element of order 8 in this group. The obvious rotation by $\frac{\pi}{4}$ wouldn't work, since we have to keep the lattice !

In other words, I want a unimodular matrix (i.e. integer matrix with determinant $\pm 1$)

to keep the lattice, and I need the ratio $(\frac{a+b}{c+d},\frac{a-b}{c-d})$=any of $\pm(1,-1),(0,\infty),(\infty,0), (0, \pm 1), (\pm 1, 0), (\infty, \pm 1), (\pm 1, \infty)$ to keep the fan. I couldn't find any order eight unimodular matrix satisfying these relations. Did I miss something here or I misunderstood any definitions?

Deforming ample line bundles vs cohomology group

2011-09-08T18:20:14Z

Let X be a smooth projective variety over the complex numbers, of dimension at least two. $D$ is an ample divisor on X. Then we know for $m>>0$, $H^i(mD)=0$. Now suppose $E$ is another divisor that is algebraiclly equivalent to $mD$, i.e. the line bundle $\mathcal{L}(E-mD)\in Pic^0(X)$ lies in the identity component of of Picard variety of X. Then is it true that even if they are not linearly equivalent, we still have $dim H^0(\mathcal{L}(E))=dim H^0(\mathcal{L}(mD))$, for $m>>0$ ?

Since the Euler character is a topological invariant, we know $\chi(\mathcal{L}(E))=\chi(\mathcal{L}(mD))$. Therefore if we know $H^i(\mathcal{L}(E))=0$ for $i>0$, we are done. However it is not obvious to me if that is true or not.

Some of my mumbling which may or may not be related:

For a general line bundle $\mathcal{L}\in Pic^0(X)$, if $\mathcal{L}\neq \mathcal{O}_X$, then $H^0(\mathcal{L})=0$ since for effective divisor in a projective variety we have a notion of degree, see Hartchorne chap II Exer 6.2. But I don't know if $H^i(\mathcal{L})=0$ or not.

In a series of paper by Green and Lazarsfeld, they looked at the case where X is compact Kahler, not necessarily projective, the behavior where $\mathcal{L}\in Pic^0(X)$ but $H^i(\mathcal{L})\neq 0$. see paper. But I don't know how to use that to construct an example where $E\sim_{alg}mD$ but $H^i(\mathcal{L}(E))\neq 0$ for $i>0$, or $H^0(\mathcal{L}(E))\neq H^0(\mathcal{L}(mD))$.

Is the tensor product of regular rings still regular

2010-12-21T17:08:03Z

An imprecise version of the question is that when A and B are regular rings, is $A \otimes B$ also regular? Please allow me to put more restrictions, here I am only interested in the case when A and B are f.g. k algebras. When k is perfect, the answer is yes, see http://arxiv.org/abs/math/0210359. In fact, we can view A and B as coordinate rings of some affine k varieties (correspondingly X and Y). Since k is perfect, regular is equivalent to smooth and it can be shown easily $X \times_k Y$ is smooth. Hence the conclusion. Now when k is not perfect, and assume in addition X and Y are both geometrically (absolutely) integral, moreover they contain some (regular but) not smooth point (so the above method doesn't apply), then is $X \times_k Y$ still regular?

Example, $k=\mathbb{F}_p(t)$, $p>2$, $A=k[x,y]/(x^p-x^{p-1}y-t)$, $X=Spec A$. Then it is easy to show X is geometrically integral, the maximal ideal generated by $(Y)$ in A is regular but not smooth, and that A is regular. The question is that is $X\times_k X$ regular?

Note, it is easy to produce a counter example when X is not assumed to be geometrically integral, e.g. $A=k[x]/(x^p-t)$, then it is easy to show $X\times_k X$ is a 0 dimension local ring but not a domain, therefore it can not be regular.

Some trivial questions on Tate's conjectures

2011-04-28T21:03:02Z

X is a smooth projective variety over a number field k. Write $\bar{X}$ for $X_{\bar{k}}$. Consider the cycle class map on Neron-Severi group of divisors: $cl_l: NS(\bar{X})\hookrightarrow H^2_{et}(\bar{X}, Q_l(1))$. Then Tate's conjecture 1 predicts ~~$cl_l(NS(\bar{X}))\otimes_Q Q_l=H^2_{et}(\bar{X}, Q_l(1)$~~ . Edit: *should be $cl_l(NS(\bar{X}))\otimes_Q Q_l=H^2_{et}(\bar{X}, Q_l(1)$.* (See e.g. Tate, "conjectures on algebraic cycles in l-adic cohomology", Motives, Jannsen et, proceeding of symposia vol55.)

Comment for edit: *Consider a product of two modular elliptic curves $E_1$ and $E_2$ over Q. Then it was proved in the following cited paper by Rosen and Silverman that their product satisfies both Tate's conjectures. However, we know the $NS(\bar{E_1}\times \bar{E_2})=2, 3 or 4$. On the other hand, by Kunneth formula $H^2(E_1\times E_2(C), C)=C^6$, so by the comparison theorem the algebraic cycles have no hope of generating the whole space in $H^2_{et}(\bar{E_1}\times \bar{E_2}, Q_l(1))$. One will have to look at $H^2_{et}(\bar{E_1}\times \bar{E_2}, Q_l(1))^{G_Q}$ instead.*

Moreover, there is a stronger conjecture let's call it Tate's 2:

consider the Hasse-Weil L function defined by the Galois representation of $G_k:=Gal(\bar{k}, k)$ on $H^2_{et}(\bar{X}, Q_l(1))$. For notational convenience we denote the latter as $V_l(1)$. Conjecture 2 says that: $-ord_{s=2}L_2(\bar{X}, s)=dim_{Q_l}Hom_{Gal(k/Q)}(1, V_l(1))$.

Question 1. (Edit: to keep track my original notation I'll leave question 1 and 2 still separated, but actually after editting I only have one question now.)

If X is a smooth projective variety over a finite field then conjecture 2 follows from conjecture 1, however over a number field I don't know if it still follows. In the article by Rosen and Silverman "on the rank of an elliptic surface" MR1626465, they proved conjecture 2 on rational surfaces over a number field. (Here conjecture 1 is known and I have no problem with that.) However I do not understand their proof.

Here is roughly how their argument goes:

Edit: *for the speacial case when $Q_l\otimes cl_l(NS(\bar{X}))=H^2_{et}(\bar{X}, Q_l(1))$ before taking the Galois invariants of the latter,*

Then $L_2(\bar(X),V_l(1), s)=L(G_k, NS(\bar{X})\otimes_Z Q, s)$. The latter is a classical (rather than motivic) Artin L function, and the order of poles at s=1 equals the Q-dimension of $G_k$-invariant of Neron-Severi space. My confusion is, if this line of argument would work, then ~~isn't it ture in general that conjecture 1 would imply~~ Edit: in this special case we would get conjecture 2 over a number field?

Question 2:

Note here we are only talking about divisors, i.e. 1-cocyles, so we don't need to worry about the difference between numerical equivelance and homological equivalence. Tate has shown $cl_l(NS(\bar{X}))\otimes Q_l\cong Q_lNS(\bar{X})$. (see prop 2.9, 2.10 from Tate's article in the above cited "motive" book. The following argument is mine, which might contain mistakes:) Therefore $dim_Q NS(\bar{X})=dim_{Q_l}V_l(1)$. Moreover, $dim_{Q_l} V_l(1)=dim_{Q_l}V_l$, by the compasison theorem, $dim_{Q_l}V_l=dim_C H^2_{singular}(\bar{X}(C), C)$. Therefore if I know $dim_Q NS(\bar{X})= H^2_{singular}(\bar{X}(C), C)$, then I would know conjecture 1 is true. (e.g. this happends when $H^1(X(C))=0$. Edit: and say $H^2(X, O_X)=0$.) This seems a very strong result, but I've never seen it anywhere, so my feeling is that there has to be something wrong with this.

Immerse an affine schemes into $A^n_S$

2011-02-14T22:36:03Z

Suppose $f: X\rightarrow S$ is of finite type, S is Noetherian. Now X=Spec B is affine, but the morphism f is not an affine morphism. S is not affine (or really f does not factor through any affine subscheme on S), nor is S of finite type over $\mathbb Z$. Now my question is, can we have an immersion of $X\rightarrow \mathbb{A}^n_S$ as S schemes?

Remark: If S is affine this is trivial. Moreover when f is an affine morphism I think it can also be done. When S is of finite type over $\mathbb Z$ by transitivity of finite type we get a map $X\rightarrow \mathbb{A}^n_{\mathbb Z}$ therefore by universal property of fiber product we get a map $X\rightarrow \mathbb{A}^n_S$.

Does reduced+Noetherian space imply Noetherian scheme

2010-07-14T15:16:16Z

In the special case of affine schemes, there is an exercise on Hartshorne saying that when Spec A is a Noetherian topological space A may not be a Noetherian ring. While it is easy to find an example for that when A has nilpotent elements, e.g. $A=k[x_1,...,x_i,...]/(X_1, x_2^2,...,x_i^i,...).$ It is not clear to me that whether we could still find a counter example when A is a domain. In the general case, I am asking for a reduced scheme, if the underlying topological space is Noetherian, is the scheme necessarily a Noetherian scheme?

My guess would be no, consider the direct limit of the series of localizations,

$$\underline{lim}\ k[x^{1\over 2^n}]_{(x^{1\over 2^n})}$$

each process within the limit is a one dimension scheme, and I think the limit is also a one dimension scheme. (In general, will dimension necessarily be held constant, non increasing or non decreasing in a limit process? or none of the above?). However, it is not hard to see the limit ring is not a Noetherian ring.

Answer by Ying Zhang for Is the tensor product of regular rings still regular

2010-12-21T23:44:59Z

I will try to sketch something I know or collected from elsewhere. We know "regular$\Rightarrow$complete intersection$\Rightarrow$Gorestein$\Rightarrow$Cohen Macaulay". In thm 2 link text it is shown when A and B are E algerbas, A is flat over E, B is finitely generated over E, then:

If A is a complete intersection, B and E are regular, then $A\otimes_E B$ is a complete intersection.
If A, B, E are Gorestein, then $A\otimes_E B$ is Gorestein.
It is proved in EGA when A, B, (also E, but this may be superluous) are Cohen Macaulay then so is $A\otimes_E B$.

When E is a field k, B is finitely generated, the above hypothesis is satisfied, therefore our question is "almost correct". Now in my old notation as in the original question, suppose X and Y are geometrically integral, therefore the function field of Y $K_Y=Frac(B)$ is linearly disjoint from the perfect closure of $k$, i.e. $k^{1\over {p^{\infty}}}$. Now by analogy with the "genus drop" phenomenon, see e.g.link text , I suspect (caveat!) $A\otimes_k Frac(B)$ is regular, in particular normal. Similarly $Frac(A)\otimes_k B$ is also regular, now $A\otimes_k B=A\otimes_k Frac(B)\cap Frac(A)\otimes_k B$ is normal.(Edit: fiber product of normal varieties is normal just by universal property, this part is superfluous) However we will show it is not regular.

Take the example as suggest by Qing Liu's comment, i.e. $A=B=k[S,T]/(T^2-S^p-t)$, $X=Spec A$, Y=$Spec B$, $x_0\in X$ is the maximal ideal generated by $(T)$, then there is a (in this case unique) maximal ideal in $A\otimes_k B$ containing $(T)\otimes_k B$ and $A\otimes_k (T)$, call it the point $(x_0, x_0)$.

(Note in general given $x\in X$, $y\in Y$, I suspect we do not get the point $(x,y)\in X\times_k Y$ for free by the universal property of fiber product. The reason is that a "closed point" in a scheme corresponds to a morphism from a field to the scheme, rather than the other way around.) The residue field at $(x,y)$ will be the composite field (ambiguity of Galois conjugation) of the corresponding residue field $k_{X,x}$ and $k_{Y,y}$, which may not be linearly disjoint even when $Frac(A)$ and $Frac(B)$ are.

Localize $A\otimes_k B$ at $(x_0, x_0)$, call it C with maximal ideal $m$. We need a lemma from EGA IV 17.1.8:

Lemma: Suppose C is Noetherian local ring, with maximal ideal $m$, $t\in m$, the following is equivalent:

C/tC is regular, and t is not a zero divisor of C;
C is regular, $t\notin m^2$.

Take t to be $1\otimes_k T\in m\setminus m^2$, then $C/tC=$ some localization of $A\otimes \mathbb{F}_p(t^{1\over p})$ which is not regular at the maximal ideal $m/tm$. Q.E.D

However, I don't know if there is a more direct way to see why $(x_0, x_0)$ is not a regular point of $X\times_k Y$, I would appreciate any comment.

Which covers of Lie groups will I get

2010-10-01T02:17:27Z

Here is a question I get from sitting in my Lie algebra class: Fix a Lie algebra $\mathfrak{h}$, we know there is a unique simply connected Lie group $H$ which serves as the universal cover of other connected Lie groups with the same Lie algebra. Now assuming $H$ is a $n_1$ sheeted cover of $H_1$, and a $n_2$ sheeted cover of $H_2$, (we are not restricting the Deck groups yet). we know $\phi: \mathfrak{h_1}\cong \mathfrak{h_2} (\cong \mathfrak{h})$ but we only cares about the isomorphism between the first two factors. Now as we travel along all $a\in \mathfrak{h_1}$, $(a, \phi (a))$ will form a Lie subalgebra in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, denoted by $\mathfrak{h'}$. By the existence theorem of Lie subgroups we have a uniqueLie subgroup $G\subset H_1\times H_2$ corresponding to $\mathfrak{h'}$. Note $\mathfrak{h_1},\mathfrak{h_2}, \mathfrak{h'}$ are isomorphic Lie subalgebras in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, however since they sort of "sit in different directions" in the ambient Lie algebra when we form the exponential map we are producing non isomorphic Lie subgroups. Anyhow, Project down from $G$ to $H_1$ and $H_2$ induce an iso on the Lie algebra therefore we know $G$ actually covers $H_i$. That's the set up of the question.

Now if I know $n_1$ and $n_2$ are coprime to each other, then automatically $G$ has to be the universal cover $H$. However, if say the maximal common divisor of $n_1$ and $n_2$ is 2, and that there are two non isomorphic Lie groups that could cover both $H_1$ and $H_2$, namely H and the one doubly covered by H donoted $\tilde{H}$, (here I need some compatible condition on the Deck group, but let assume $\tilde{H}$ covers $H_i$). So my question is which one is isomorphic to $G$ when I draw the graph? I am worried a little bit that the there may not a canonical way of choosing this isomorphism $\phi$ to make this question make sense, (or is there)? My guess is that if there is a "canonical choice", then the $G$ we get from the graph should be the smaller $\tilde{H}$, while my colleage thinks that by choosing different $\phi$, you can get both. When $H_1$ and $H_2$ have a lot of non isomorphic common covers, then by choosing different $\phi$ you can produce all of them as subgroups of $H_1\times H_2$? (fixing $H_1$ and $H_2$.) I really hope someone ccan shed lights on this, thanks very much!

Answer by Ying Zhang for How to determine the homotopy groups of the suspension of a space ?

2010-04-13T22:09:28Z

I can't help but mention a cute application of the suspension isomorphism of (co)homology combined with Pioncare duality to show that for a compact closed manifold $M^n$, if the suspension $\sum M$ is homotopic to a closed orientable manifold, then $M$ is a homology sphere. This actually concerns the homology of the suspension, rather than homotopy groups. I apologize if it doesn't help your question.

In general, for a CW complex $M$, we have the suspension isomorphism $\tilde H_i(M)\cong \tilde H_{i+1}(\sum M)$, the reduced homology. And $\tilde H^i(M)\cong \tilde H^{i+1} (\sum M)$ for reduced cohomology. Now if we have Poincare duality on both sides, we would have $\tilde H^i(M)\cong \tilde H^{i+1}(M)$. passing from reduced and non reduced (co)homology tells us $M$ is a homology sphere.

There is a relevent exercise in "Elements of homology theory" by Viktor Vasilʹevich Prasolov on P45.

Edit: $\tilde H^i(M)\cong \tilde H^{i+1}(M)$ for suitable dimensions. e.g. you can start the induction from $H^{n-1}(M)=0$ and then $H^{k-1}(M)=H^{k-2}(M)$ etc.

An exercise in group cohomology

2010-04-06T14:21:02Z

Here is an exercise from Serre's "local fiels" when he starts to do cohomology: Let G act on an abelian group A, f be an inhomogenous n cochain, i.e. $f\in C^n(G,A).$ Define an operator T on f, $Tf(g_1,g_2,\cdots,g_n)=g_1g_2\ldots g_n f(g_n^{-1},g_{n-1}^{-1},\ldots,g_1^{-1})$. It is clear that $T^2f=f$. It is also not too hard to show $T(df)=(-1)^{n+1}d(Tf)$. Thus f is a cocycle iff Tf is, and f is a coboundary iff Tf is. When n=1, it is straightforward to see -f is cohomologous to Tf. Then the exercise wants us to show when n= 0,3 mod 4, f is cohomologous to Tf, while when n=1,2 mod 4, Tf is cohomologous to -f.
Any idea will be appreciated.

Maps inducing zero on homotopy groups but are not null-homotopic

2010-04-04T02:46:25Z

Today my fellow grad student asked me a question, given a map f from X to Y, assume $f_*(\pi_i(X))=0$ in Y, when is f null-homotopic?

I search the literature a little bit, D.W.Kahn

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102995805

And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces.

http://www.jstor.org/stable/pdfplus/2037939.pdf

However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint?

or, maybe even some examples arising from manifolds?

Why the rank of a locally free sheaves is well defined?

2009-11-18T02:01:43Z

In Hartshorne P109 he defines a sheaf $\mathcal{F}$ of $O_X$-modules to be locally free if there is an open cover of X, s.t. on each U, $\mathcal{F}|U$ is a free $O_X|U$ module of rank $I$. Then if X is connected, rank $I$ is globally well-defined. Here $(X,O_X)$ is any ringed topological space(e.g. not necessarily the structure sheaf of a ring). A similar definition is here link text

However, it didn't seem obvious to me that if $V$ is a smaller open set included in $U$(say $U$ connected), then the number of copies $J$ of $(O_X|V)^J=\mathcal{F}|V$ would remain the same, because in general the restriction map of the sheaf $O_X$ or $\mathcal{F}$ from $U$ to $V$ need be neither subjective nor injective, why would the index $J$ stay the same as $I$?

How many primes stay inert in a finite (non-cyclic) extension of number fields?

2010-01-20T02:20:02Z

In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic, there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8,

In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves. So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic.

I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?

Comment by Ying Zhang

2012-05-24T20:36:07Z

Thanks David, that helps. I might be missing out a "trend" here, but if people find it is better to stick with the old tags as in Arxiv rather than using more specific but less common tags, that's good for me :) Since I don't want to end up answering my own question, I've written up what I believe to be what's going on in a pdf [here] following Minhyong's suggestion to "unravel the definitions". [here]:hans.math.upenn.edu/~yinzhang/Work/…

Comment by Ying Zhang

2012-05-24T20:02:12Z

Thanks Minhyong, sometimes it helps me a lot to be pointed out by others

`you are wrong about this and this..." and I'll check again the assumptions I implicitly assumed to be true. I am definitely wrong in saying that the pull-back of a sheaf defined by a group scheme is still the sheaf defined by the same group scheme: we see counter examples here, and also Milne has it on page 68 of his

`etale cohomology'' book (It is a different example there).

Comment by Ying Zhang

2012-05-23T22:47:18Z

Sorry about the notation, $K_p$ here is the Henselization, therefore it is the algebraic elements in the usual p-adic field. But you are right, it is not a finitely generated abelian group, which is why I am confused. I thought either I misinterpret something or one of the lemmas is wrong?

Comment by Ying Zhang

2012-05-23T22:44:20Z

That could the case. But since $\mathbb{G}_m$ is represented by $Hom(-,Spec \mathbb{Z}(t,\frac{1}{t}))$, I thought the pull-back of $\mathbb{G}_m$ on any scheme is always $\mathbb{G}_m$ on that scheme, isn't it? The push-forward, on the other hand, could be something else.

Comment by Ying Zhang

2011-10-18T15:09:42Z

@Parsa: Thanks for your suggestion, but I think the term "sub line bundles" might be confusing, so I have edited my question. Here in exact sequence like $0\rightarrow \mathcal{L}\rightarrow \mathcal{E}\rightarrow \mathcal{Q}\rightarrow 0$, I didn't require both $\mathcal{L}$ and the "quotient" $\mathcal{Q}$ are line bundles. In my terminology, I just require $\mathcal{L}$ to be a rank one subsheaf which is locally free. Sorry about not clarifying my terminology in the beginning!

Comment by Ying Zhang

2011-10-18T14:56:00Z

@Sawin: Thanks for your example. As for the finiteness for "effective" line bundles, I think it is true when $Pic^0(X)=0$ i.e. $Pic(X)$ is discrete (and we know it has finite rank), then by similar arguments of "maximal degree" one can show it is finite. But I don't know how to do it in general. Actually, the situation that interests me most is when X is a threefold, $\mathcal{E}$ is rank two vector bundle.

Comment by Ying Zhang

2011-10-18T14:40:55Z

(continued) And yes by an effective line bundle I meant a line bundle given by an effective divisor, by equivalently $H^0(X, \mathcal{L})\neq 0$. Now if I give $H^0(X,\mathcal{E})/k^*$ the natural projective space structure, where we assume $H^0(X, \mathcal{O}_X)=k$, and when I rule out those (hopefully finite) closed subset $H^0(X,\mathcal{L})/k^*$, then I get an open set $U$. My core interest is whether this open set $U$ represents any reasonable functor. However, when the collection of "effective" sub line bundles is not finite, this $U$ is only constructible and less desirable.

Comment by Ying Zhang

2011-10-18T14:34:20Z

@Angelo: Thanks for the editing! I checked out "backticks" on wikipedia and figured out what is going on. However, I didn't see the link to "How to write math" as you mentioned, is there a wonderful page like that? As for the motivation, the idea is this: generically a zero locus of a section in $\mathcal{E}$ will cut out a codimension two subscheme of X, given $\mathcal{E}$ is a rank two vector bundle, and generically it is even smooth. But when the section comes from a sub line bundle, it will cut out a divisor. So I want to rule them out.(to be continued...)

Comment by Ying Zhang

2011-10-18T04:07:38Z

Could someone please tell me what is wrong with the formatting? It is horrible, but I don't know what went wrong...

Comment by Ying Zhang

2011-09-17T15:30:17Z

Nice. Looking back at this example, the strict transform of a nice $P^1$ is "strict" on the top since it passes through one of the points we blow up. It is a nice exercise to write down the total transform and see it on the nose that pulling back keeps the self-intersection number 0.

Comment by Ying Zhang

2011-09-09T18:14:33Z

Thanks, I realized this is a stupid question earlier today...

Comment by Ying Zhang

2011-04-29T02:14:13Z

@mephisto: Really? That will be great, and $dim_Q H^{2r}(X(C), Q)(r)=dim_Q H^{2r}(X(C), Q)$, right? The reason I ask question 2 is because whenever I check some reference on "cases where Tate's conjecture is known", I never see such comments on these trivial cases, so I assumed it might be not correct.

Comment by Ying Zhang

2011-02-15T01:59:12Z

@Qing: Thanks, this completes resolves the problem when S is separated.

Comment by Ying Zhang

2011-02-15T00:30:28Z

Dear Mattia, thanks for your response. I am sorry for the confusion, but I actually am not looking for a closed immersion. E.g $A^1\rightarrow P^1$ would be a counterexample such that this kind of closed immersion can not be realized.

Comment by Ying Zhang

2011-01-10T01:13:40Z

On p.140 Prop 1. line 7 of the proof, it is very confusing to write $\delta(r/n)\in\hat{H}^0(G,\mathbb{Z})$ and $\delta(r/n) =r$, since we know $\hat{H}^0(G,\mathbb{Z}) =\mathbb{Z}/n\mathbb{Z}$. I think this is a typo? Actually, I think the entire proof of Prop.1 is just restating the fact of line 6. It is tautological.