User adam hughes - MathOverflow

Is it known if the absolute Galois group is "divisible"?

2012-11-28T17:55:55Z

The definitions of a divisible group that I have seen all seem to assume abelian is an a priori property of the group. My question is as to whether or not it is known that--given a non-torsion element $\tau\in\mathbf{G}$* and $n\in\mathbb{N}$, do we know if $\exists \tau'\in\mathbf{G}_\mathbb{Q}$ such that $\tau=\tau'^n$?

The motivation is that--if this never happens--then the fields $\overline{\mathbb{Q}}^\tau$ should be sort of "maximal" subfields of $\overline{\mathbb{Q}}$ since fixing by the generator would be equivalent to fixed by any power of it.

Edit: anon has noted that quotients of divisible groups are divisible, so we can lay to rest that $\mathbf{G}_\mathbb{Q}$ is not divisible. Can we tell if the opposite is true? I.e. the answer to "can I find such a $\tau '$ given $n$" is "no", but how about the somewhat interesting and related question: If I have a cyclic subgroup of the absolute Galois group, $C=\langle\tau\rangle$ can we find a maximal (with respect to inclusion), cyclic group containing $C$? Of course this is equivalent to a minimal field in a chain, so it seems like if that is so something interesting must be going on.

The other related question which deals with the original spirit of the problem is: are there $\tau$ such that $\langle \tau^n\rangle$ fixes $k\subseteq\overline{\mathbb{Q}}$ then $\langle\tau\rangle$ fixes $k$? i.e. is $k=\overline{\mathbb{Q}}^\tau$ equivalent to $k=\overline{\mathbb{Q}}^{\tau^n}$ possible for some $\tau$ non-torsion? If so can such elements be characterized?

The immediate observation is that, by the FTGT, if we identify $\langle\tau\rangle\cong\mathbb{Z}$ and write $[n]$ for the index $n$ subgroup, that--in the topology of $\mathbf{G}_\mathbb{Q}$--necessarily $\overline{[n]}=\overline{[1]}$. I.e. the monothetic group $\overline{[1]}$ has every element is a generator. If $\overline{[1]}$ were connected, of course this would be trivial to check since the set of generators has Haar measure 1 in this case, and a finite index subgroup has positive Haar measure, and so contains a generator, but the topology here is totally disconnected so it is not easy to see one way or the other.

*My TeX was not rendering properly when I wrote $\mathbf{G}_\mathbb{Q}$ here, so I just included it as a sidenote rather than have it look like a mess.

Answer by Adam Hughes for Slick ways to make annoying verifications

2010-12-12T06:25:45Z

My example comes from algebraic geometry, specifically in the theory of $\lambda$-rings, which are used in K-theory and representation theory (and possibly other places of which I am unaware). If you look in Lecture Notes in Mathematics 308 $\lambda$-rings and the Reprsentation Theory of the Symmetric Group by Donald Knutson, on pp 27, there is a theorem which is called--quite appropriately--the "Verification Principle", and it is used to do exactly the kind of thing you're asking about. The statement is:

If $\mu$ is a $\lambda$-ring operation, then $\mu$ is uniquely a polynomial in the $\lambda$-operations and for any particular polynomial $f(\lambda^1,\lambda^2,\ldots , \lambda^n,\ldots)$ it is sufficient to check that $\mu = f$, operating on a sum $\xi_1+\ldots + \xi_r$ of elements of degree 1, for all $r>0$. If you read the first part of the book (i.e. up to page 27 where you read this) you learn that this is a very handy thing indeed, especially given the generating function polynomials that show up in the study of these objects and how much easier it is to check for things which are polynomials in the $\lambda$-operations (as they are called). It is this that so nicely ties this kind of ring in with the representation theory of the symmetric group, in particular the elementary symmetric functions, as Knutson mentions on p. 10 of the same publication.

Another useful technique is one that comes up in algebra, when you check something for general elements by checking on a basis. Eg. testing something for all the elements of your tensor or exterior algebra by testing it on the "pure" or "completely decomposable" tensors which form a generating set.

Place stabilizers for the absolute Galois Group

2012-09-13T02:37:06Z

Fix an algebraic closure, $\overline{\mathbb{Q}}$ for the rationals and consider the set, $B_p$, of all places of $\overline{\mathbb{Q}}$ over a fixed (possibly infinite) prime, $p$, of $\mathbb{Q}$. Let $G_\mathbb{Q}=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

1: Let $G_\mathbb{Q}$ act on $B_p$ in the natural way. Then what is the structure of the stabilizer, $H_v$ of $v$, for $v|p$ an element of $B_v$?

Infinite primes are just embeddings of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, so the case $p=\infty$ is really asking for the stabilizer of the action of left multiplication in $G_\mathbb{Q}$, and for finite $p$ it should be the inverse limit--over finite, Galois extensions of $\mathbb{Q}$--of the decomposition groups of primes $\mathfrak{p}|p$. However, actually getting ones hands on the structure of the stabilizers in question seems to be a hard question, at least no one I've talked to so far seems to know much more than the obvious things stated here.

2: Is is perhaps easier to describe the cosets $G/H_v$? If so, what should they look like?

When is $\mathbb{G}_m(R)$ enough to determine $R$?

2011-11-01T00:04:57Z

Say I have a ring, $R$, with 1 which I consider my universe, and I know its group of units $G=\mathbb{G}_m(R)$. Then given a subgroup, $H\le G$, can I determine if there is there a subring $S_H$ such that $\mathbb{G}_m(S)=H$? If so, is $S_H$ unique with this group of units? If so, is there in fact a--canonical in the sense above--1-1 correspondence between subgroups of $G$ and subrings of $R$ with 1? Preliminary attempts at a solution don't indicate any problems with the truth of the statement, but naturally one should be skeptical of limited data especially in a subject with so many intricacies as groups and rings.

The motivating example is $\overline{\mathbb{Q}}/\mathbb{Q}$, due to some interesting number theory that could come out of such a correspondence.

In the case of fields the question is supposed to collapse into the question "Can I add 0 to a subgroup of the group of units of some big field and get a subfield without doing anything else?"

There is no possibility for general rings, but are there assumptions on $R$ or $G$ which can ensure existence or uniqueness? And it is also fine to induce assumptions on what kind of $S$ we are allowed to have as well, fields instead of just rings for example.

Answer by Adam Hughes for Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions?

2011-04-27T18:23:46Z

Well it seems to me the Cauchy-Riemann equations say quite a lot. After all once you know the real and imaginary parts are harmonic, then you know their derivatives are as well, and $$f'(z)=u_x(z)+iv_x(z)$$ so you can do it again. I don't think this is boot-strapping it, but if anyone greater on the complex food-chain cares to disagree, I'm likely retreat quickly at their word.

Answer by Adam Hughes for Representation theory of $S_n$

2011-04-19T16:29:41Z

I recommend $\lambda$-Rings and the Representation Theory of the Symmetric Group by Donald Knutson. It helped me a lot. It's #308 in the Springer Lecture Notes series.

What is the smallest $n$ such that $G\le S_n$?

2011-04-04T17:17:31Z

Possible Duplicates:
Smallest permutation representation of a finite group?
Smallest n for which G embeds in $S_n$?

Cayley's theorem says that every finite group, $G$ can be thought of as a subgroup of some symmetric group, $S_n$, but just how small an $n$ can we take in understanding $G$? It is known that at worst, $n=|G|$ will work, but in the case of say $S_n$ it would be quite silly to embed $S_n$ in $S_{n!}$ when it fits just fine in the MUCH smaller $S_n$. :)

Other cases where it is smaller are $\mathbb{Z}/6$ which narrowly avoids a full embedding in $S_6$ and fits into $S_5$ where it can be modeled as $\langle (1\;2),(3\;4\;5)\rangle$

Indeed, this example lights the way to the general case for abelian groups, since we have the characterization of $G\cong P_1\times P_2\times\ldots\times P_\ell$ where each $P_i$ is the sylow $p_i$ subgroup of $G$ associated to the prime $p_i$.

Starting with the $P_i$, we know $P_i\cong \mathbb{Z}/p_i^{e_{i_1}}\times\mathbb{Z}/p_i^{e_{i_2}}\times\ldots\times \mathbb{Z}/p_i^{e_{i_k}}$, then it should fit into $S_n$ for $n=\sum\limits_{j=1}^k\; p_i^{e_{i_j}}$ and the more general case comes from summing over all the $1\le i\le \ell$, which I don't write out, because three subscripts for an exponent doesn't typeset well.

It's easy to see that this version of the classification gives the smaller of the two candidate values for $n$ because if we do the version where $G\cong A_1\times A_2\times\ldots\times A_k$ where each $A_i$ is cyclic and $|A_{i+1}|$ divides $|A_i|$ for $0< i< k$, since this more or less amounts to $x+y <xy$ for $x,y\ge 2$, since we can, WLOG, assume $y=\max\{ x,y\}$ so that $x+y\le 2y\le xy$.

I'm fairly certain that the $n$ above is the minimal such $n$ for abelian groups, but then some questions which remain are: "Is there a simple proof of this fact?" (my guess is yes), "What can be said for nonabelian groups?", "Can nonabelian groups always fit into $S_n$ with $n<|G|$?", and combining the last two a little, "If so, is there an upper bound for this $n$ in terms of just |G|, which always gives a better estimate than $n\le |G|$?"

Axiom of Choice in a weaker system

2011-01-29T11:04:37Z

Is it known whether or not there is a consistent system of logic where two or all of the axiom of choice, well-ordering principle, and Zorn's lemma have no (known) proof of equivalence?

I was thinking about the old adage "The Axiom of Choice is clearly true, the well-ordering principle is clearly false, and nobody really knows about Zorn's lemma" and the content of http://mathoverflow.net/questions/21367/proof-that-pi-is-transcendental-that-doesnt-use-the-infinitude-of-primes, particularly François G. Dorais' beautiful answer, where he explains the existence of systems where it isn't known whether infinite primes is necessary for the transcendence of $\pi$, and it occurred to my rather logic-ignorant self that perhaps a sufficiently weak system might allow for this adage to be true, at least in the form $X$ is (clearly or otherwise) true and $Y$ is (again, clearly or otherwise) false for suitable $X$ and $Y$ from the above list. I think it would be particularly poetic to find one in which $X$ is the Axiom of Choice and $Y$ is the well-ordering principle.

Answer by Adam Hughes for Why need the morphisms to form a set ?

2010-12-09T19:42:13Z

If you look at Steve Awodey's book Category Theory (Oxford Logic Guides * 49) you'll see that on p. 22, Definition 1.12 is that a category in which $\hom_\mathbf{C}(A,B)$ is a set for every pair of objects $A$ and $B$ is called locally small.

Functions whose antiderivative behaves like xf(x)

2010-11-19T01:19:39Z

I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by

$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$

I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.

It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on n and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.

A preliminary investigation yields some interesting first starts:

1) A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$

2) If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.

3) A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$

Answer by Adam Hughes for Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$?

2010-11-13T20:04:53Z

That should only be true for things which have $\mathbb{R}^n$ as their universal covering space. In particular I think it fails for things like Lens spaces.

Is it possible to recover the degree of a field extension from a list of elements and the ground field?

2010-11-07T21:24:16Z

I'm interested to know if there is anything known about recovering the degree of a field extension, $E/k$, given $E=k(\alpha_1,\ldots, \alpha_n)$ (here I'm assuming that the extension is of finite degree. Obviously there are some silly examples anyone could eyeball, like $\sqrt[m]{d}$ for $n=1$ and $\alpha_1=d\in k$ or the case $E=\mathbb{F}_q$ and $k=\mathbb{F}_p$ where we can just divide orders (intermediate fields are clearly equally trivial). Where, to give the necessary bit of care, we assume this is a nontrivial extension. If $E/k$ is Galois and we can appeal to other bits of theory, we might also get the degree by calculation of the Galois group('s order). Is there anything known about more general extensions? It is conceivable given that $E/k$ is an extension of algebraic number fields that the theory of ideals might give an insight, especially given the (IMO) rather fascinating fact that $\mathcal{O}_E$ is finitely generated as a $\mathbb{Z}$-module is an equivalent statement, and using the machinery of algebraic number theory, or some other extra structure, but I'm principally concerned with the more general theory if any exists or if just anything is known about this problem.

Generalizations of divided-power algebras over finite fields

2010-11-08T06:48:36Z

In Andrews, Askey, and Roy's Special Functions, the authors state that Gauß sums are finite field analogs of the $\Gamma$-function as Jacobi sums are to B-function. The $\Gamma$-function is well-known as the analytic continuation for the factorial function and if $x$ is in a divided power algebra then $$x^{[m]}\cdot x^{[n]}={(m+n)!\over (n!)(m!)}x^{[m+n]}.$$ where $x^{[n]}$ is the nth power of x under the divided power meaning of $x^n$. For characteristic 0, this combinatorial coefficient can be written as $${\Gamma(1+m+n)\over \Gamma(1+m)\Gamma(1+n)}=B(1+x,1+y)^{-1}\cdot[(1+x)+(1+y)].\quad (*)$$ What I mean to ask is: "Is there value in considering these objects--i.e. divided power algebras and these coefficients-- from this point of view--i.e. use of Gauß and Jacobi sums? Is this formula just a coincidence or a pattern that I want to see, or is there more to it?"

Nonuniqueness of maps of representing spaces

2010-05-06T20:37:46Z

In Rudyak's On Thom Spectra, Orientability, and Cobordism, two variants of Brown's representability theorem are presented: given a natural transformation $f^*: E^* \to F^*$ of cohomology theories, Brown's representability theorem asserts that we can lift $f^*$ uniquely to a map $f: E \to F$ of spectra, and we can also lift $f^*$ to map $f^i: \Omega^\infty \Sigma^i E \to \Omega^\infty \Sigma^i F$ of representing spaces -- but he does not assert that this second kind of lift is unique.

My feeling is that this second map becomes unique after stabilization. Is there an accessible example of such a natural transformation with two nonhomotopic representing maps on classifying spaces, illustrating nonuniqueness?

Comment by Adam Hughes

2012-11-28T18:48:58Z

That's a fantastically simple reason. Thank you, anon. I've edited the original question to be a bit stronger now that the easy part is revealed.

Comment by Adam Hughes

2012-10-29T21:42:05Z

Just so we're clear, the way to think of the big prime over p as an embedding is to look at a uniformizer and looking at the power series for the element of $\overline{\mathbb{Q}}$ in $\overline{\mathbb{Q}_p}$, or is there something I'm missing?

Comment by Adam Hughes

2012-10-29T21:33:38Z

Fixed, thanks darij.

Comment by Adam Hughes

2011-11-01T16:39:00Z

Mariano: Thanks for the feedback, I wasn't aware of this. Indeed I was trying to keep my question short for fear of rambling on, I'll keep that in mind in the future.

Comment by Adam Hughes

2011-11-01T06:39:32Z

Alex: you're allowed to hypothesize on S as well, as I said down with the question posed after Steven's answer. Implicitly I'm requiring S be a subfield.

Comment by Adam Hughes

2011-11-01T06:13:30Z

Alex: yes, sorry, I messed up my notation, I meant the field extension without the "Gal" bit at the beginning.

Comment by Adam Hughes

2011-11-01T02:07:54Z

(and of course S should be required to be a subfield)

Comment by Adam Hughes

2011-11-01T02:04:28Z

There is a very degenerate case which definitely works, the finite fields. It's immensely restrictive, but it's clear based on the fact that their unit groups are cyclic.

Comment by Adam Hughes

2011-11-01T01:45:29Z

M Turgeon: if R is required to be a field that circumvents the problem with $k[x]$, so there is hope in that direction.

Comment by Adam Hughes

2011-11-01T00:40:36Z

Alex: Yes, I was playing with torsion bits in $\overline{\mathbb{Q}}/\mathbb{Q}$ and thinking it would be hopeless if we have these finite little groups, one of the assumptions I think should be necessary, but which I didn't list in the original question to avoid bias in responses, is that we consider groups without finite subgroups like that which mess things up. The question doesn't have much hope in the most general setting, but I'm more interested in the "right" hypotheses on $G$ and $R$ so that it becomes true than I am on general truth.

Comment by Adam Hughes

2011-11-01T00:34:51Z

That's great for general rings, and I thought I said this, but I apparently did not: can I include assumptions on $R$ which will make it hold true. I'll edit the original question. Thanks!

Comment by Adam Hughes

2011-06-08T06:25:25Z

This might be more appropriate on math.stackexchange.

Comment by Adam Hughes

2011-05-03T22:32:48Z

I think this belongs on math.stackexchange.com, it's not a research-level question.

Comment by Adam Hughes

2011-04-27T21:21:15Z

And I should add that I think that perhaps the so-called "magic" is less mind-boggling as you get used to it. Just because you don't see some masterful underlying reason which trivializes things now, doesn't mean you won't later on. And if you don't, then I don't think it's that much of a loss, you just move on.

Comment by Adam Hughes

2011-04-27T21:17:48Z

Haha, I think the "magic" is just what makes it so nice. Classical complex analysis is a beautiful subject and has a lot to bring to the table in the ways of powerful tools. I'm certain you'll never find success at removing all the magic from things though, things are far too complicated for that.