ag.algebraic geometry - A Galois Theory Computation

Excuse me for the specificity of this question, but this is a silly computation that's been giving me trouble for some time.

I want to explicitly realize the order 21 Frobenius group over ℂ(x), as ℂ(x,y,z) where y³=g(x) and z⁷=h(x,y). The order 21 Frobenius group is C₇⋊C₃, where the generator of C₃ acts by taking the generator of C₇ to its square. Or in other words < a,b|a³=1, b⁷=1, ba=a(b²) >. Furthermore, I want it to branch at exactly three points, two of which will have 7 preimages, each with ramification 3, and the third will have 3 points over it with ramification 7 each.

This can easily be shown to exist: Take ℙ¹ minus three points (say x=6,5,2); look at its algebraic fundamental group (=the profinite completion of < c, d, t| cdt = 1 >), and map it to the Frobenius group surjectively by c goes to a, d goes to b, and t goes to b^-1a^-1. This gives you a Galois cover of ℙ¹, with said ramification (because order(a)=3, order(b)=7, and order(b^-1a^-1)=3), and group the 21 order Frobenius group.

Of course, this construction is extremely difficult to track because of the topology involved. It would be much easier to deduce the field extension from the ramification behavior.

So: this can be broken down to two cyclic Galois extensions. The first, a ℂ(x,y), of the form y³=(x-2)²(x-6) is pretty easy to deduce (I need it to ramify at x=2 and x=6 and nowhere else -- this must be the equation up to change of variables). The second, a z⁷=h(x,y) is tricky. I want it to ramify only above x=5. There's some Abhyankar's lemma things going on here, and that makes the guesswork difficult, and my life much harder.

I should note that the distinction of the Frobenius group of order 21, and the reason that I'm at all interested in this example, is that it is the only order 21 group which isn't cyclic. Geometrically, it means that h is a function of both x and y, and not just x.

Thanks in advance.

at.algebraic topology - What is 'formal' ?

I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of Deligne-Griffiths-Morgan-Sullivan on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers.

I am not totally sure, but I believe that the definitions are as follows:

• A differential graded algebra $(A,d)$ is called formal if it is quasi-isomorphic (in general, if we work in the category of dg algebras and not, say, the category of A-infinity algebras, we need a "zig-zag" of quasi-isomorphisms) to $H^ast(A,d)$ considered as a dg algebra with zero differential.




• A space X is called formal (over the rationals resp. the reals) if its cochain dg algebra $C^ast(X)$ (with rational resp. real coefficients) with the standard differential is a formal dg algebra.

One of the things I'm not sure about is whether in the definition we should require $H^ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^ast(X)$ is always (graded-)commutative.

The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^infty$) differential forms on X with the standard de Rham differential is a formal dg algebra.

The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal.

Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally.

In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of A-infinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasi-isomorphic to $H^ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (Kontsevich-Soibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in this paper), and you can find it in the operads literature as the "transfer theorem") that you can put an A-infinity structure on $H^ast(A,d)$ which makes $A$ and $H^ast(A,d)$ quasi-isomorphic as A-infinity algebras. The A-infinity structure manifests itself as a series of $n$-ary products satisfying various compatibilities. Intuitively at least, these $n$-ary products should be thought of as being analogous to Massey products in topology. So $H^ast(A,d)$ with this A-infinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasi-isomorphic, as an A-infinity algebra, to $H^ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^ast(A,d)$.

---

*Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the A-infinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the A-infinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of A-infinity structure. From the paper:

... a minimal model is a formal consequence of its cohomology ring if, and only if, all the higher order products vanish in a uniform way.

and also

[Choosing a quasi-isomorphism from a minimal dg algebra to its cohomology] is a way of saying that one may make uniform choices so that the forms representing all Massey products and higher order Massey products are exact. This is stronger than requiring each individual Massey product or higher order Massey product to vanish. The latter means that, given one such product, choices may be made to make the form representing it exact, and there may be no way to do this uniformly.

(Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.)

nt.number theory - Special bases of number fields

Let K be a number field of degree n with a fixed embedding in the complex numbers. Let | . | be the normalized absolute value given by that embedding. (The square of the ordinary absolute value if the embedding is non-real.) Does there exist a basis x_1 , ... , x_n for K over the rationals with the following property:

| r_1 x_1 | + ... + | r_n x_n | = the maximum of | r_1 x_1 + ... + r_n x_n |_v

where | . |_v runs through the normalized archimedean absolute values of K ?

Later: As posed, the answer is no. Suppose I started with the field generated by a cube root and was unlucky enough to start with the real embedding. Then I'd be in effect trying to show ( x + y + z )^2 > x^3 + y^3 + z^3 on the positive octant. But if I was lucky enough to start with the complex embedding ...

ag.algebraic geometry - Parametric polynomial solution of a single polynomial equation

Let $P$ be a polynomial in $n$ variables with rational coefficients,
$P in {mathbb Q}[Z_1,Z_2, ldots ,Z_n]$, and consider the algebraic
set
$Z=lbrace (z_1,z_2,z_3, ldots ,z_n) in {mathbb Q}^n |
P(z_1,z_2, ldots ,z_n)=0 rbrace$

If $r$ is a nonnegative
integer, $x_1,x_2, ldots ,x_r$ are variables, and $Q_1,Q_2, ldots ,Q_n$
are polynomials in $x_1,x_2, ldots ,x_r$ such that
$(Q_1(x_1, ldots ,x_r),Q_2(x_1, ldots ,x_r),ldots,Q_n(x_1, ldots ,x_r)) in Z$
for all $(x_1, ldots ,x_r) in {mathbb Q}^r$, we call $(Q_1,Q_2, ldots ,Q_n)$
a $r$-dimensional parametric solution
of the equation $P(z_1,z_2, ldots ,z_n)=0$. It is also
natural to define a maximal parametric solution as one with the largest possible $r$
(to avoid trivialties, we also impose
that there is no variable upon which none of the $Q_i$ depends. I'm not sure
that this last condition avoids all degenerate cases, but I'd like to avoid
definitions that involve advanced notions such as the dimension of an algebraic
variety ).

My questions : is the problem of computing the largest $r$ known to be undecidable in general ? What are the most general cases in which algebraic geometry allows us to compute the largest $r$ (and the associated parametric solutions) ?

ag.algebraic geometry - Intuition/Heuristic behind I/I^2 definition of Kähler differentials

Hello,

this one has always been mysterious to me. The Kähler differentials $Omega_{A/k}$ are definined, by the universal property
$$Der_k(A,M)=A-Mod(Omega_{A/k},M)$$
so for $M=A$ we get that $Omega_{A/k}$ is the cotangent space of $spec(A)$.
(or a relative version of it if k is no field).

There are two constructions of Kähler Differentials I know.
The first one is $$Omega_{A/k}=langle df : text{relations satisfied by any derivation} rangle$$
I think I sort of understand this one, it says that the differential of a function just contains enough information to extract the derivation of the function out of it.
And this is what a section into cotangent space should be. Something that contains just enough information to pair it with a vector-field into a function.
The other construction is
$$Omega_{A/k}=I/I^2$$
Where $I$ is the Ideal of functions vanishing on the diagonal in $spec(A)times_{spec(k)} spec(A)$.

More geometrically it says
sections into cotangent space=functions vanishing on the diagonal mod higher order.
But still I don't think I understand this equality on an intuitive level. Can someone explain the heuristic behind this equality? Or maybe explain $Omega_{A/k}=I/I^2$ from another intuitive viewpoint?

big list - Interesting applications of the Pigeon-hole Principle

I think the solutions of these questions are very interesting (by using pigeon-hole principle), first question is easy, but second question is more advanced:

1) For any integer $n$, There are infinite integer numbers with digits only $0$ and $1$ where
they are divisible to $n$.

2) For any sequence $s=a_1a_2cdots a_n$, there is at least one $k$, such that $2^k$ begin with $s$.

Graphs where every two vertices have odd number of mutual neighbours

Maybe I'm missing something, but I'm not sure that the third condition actually generates what I'll call odd graphs. For example, if I let $A$ be the graph consisting of a single vertex and $B$ be the graph consisting of two isolated vertices, then clearly both $A$ and $B$ are even. However, if I form the $A-B-C$ construction with this choice of $A$ and $B$ I get a graph with an even number of vertices, which can't be odd by the proof of the cute question.

We can fix this by further insisting that each vertex of $A$ and $C$ have odd degree. I'll call such graphs oddly even. Note that a disjoint union of two oddly even graphs is still oddly even, so it really isn't necessary to have a $A-B-C$ construction, we only need a $A-B$ construction. Furthermore, it is not necessary that $B$ is an odd clique; $B$ can in fact be any odd graph.

We thus have the following theorem.

Theorem. Let $A$ be an oddly even graph and $B$ be an odd graph. Then the graph $A-B$ formed by taking $A$ and $B$ and adding all edges between $A$ and $B$ is odd.

So, I guess the answer to the question is no.