User robin chapman - MathOverflow

Answer by Robin Chapman for Relationships between the roots of an entire function and the roots of its derivative

2010-12-09T19:42:46Z

I don't know if there are any general results about these, but when $f$ is a polynomial, these must be in essence results about symmetric functions. If $f(z)=z^n+a_{n-1} z^{n-1}+\cdots+a_1z+a_0$ then $Z(1)=-a_1/a_0$, $Z(2)=Z(1)^2-2a_2/a_0$ etc. In this case your results surely specialize to polynomial identities in the $a_j$.

Answer by Robin Chapman for Rank of a free module without the axiom of choice

2010-12-09T17:47:14Z

Let $A$ and $B$ be infinite sets. Let $M$ be a rank $|B|$ module with basis $e_b$ for $b\in B$. If we take $|A|$ elements $m_a$ of $M$, then each can be expressed in terms of finitely many of the $e_b$. If $B$ is not the union of $|A|$ finite sets, then there is some $e_b$ not expressible in terms of the $m_a$, so $M$ cannot be free on $|A|$ generators.

I'm not expert on set theory without AC, but those here who are will surely tell us if there are models of ZF with non-equinumerous infinite sets such that $A$ is the union of $|B|$ finite sets and vice versa.

Answer by Robin Chapman for Proof that the factors of sigma(p^e) have two forms.

2010-12-08T07:57:10Z

It's well-known that if $a$ is an integer then a prime factor of the number $\Phi_n(a)$ is either a factor of $n$ or congruent to $1$ modulo $n$. Here $\Phi_n$ is the $n$-th cyclotomic polynomial. The reason is that if $p$ divdes $\Phi_n(a)$ but not $n$ then $a$ has order exactly $n$ in the multiplicative group $(\mathbb Z/p\mathbb Z)^*$. By Lagrange's theorem then $n\mid(p-1)$.

When $e+1$ is prime, then $\Phi_{e+1}(X)=X^e+X^{e-1}+\cdots+X+1$, so in this case a prime factor of any $\Phi_{e+1}(a)$ is either $e+1$ or congruent to $1$ modulo $e+1$. In general though $X^e+X^{e-1}+\cdots+X+1$ is the product of the $\Phi_m(X)$ over the factors $m>1$ of $e+1$, so a prime divisor of $a^e+a^{e-1}+\cdots+a+1$ is either a divisor of $e+1$ of congruent to $1$ modulo some prime divisor of $e+1$.

Answer by Robin Chapman for crookedness of convex curves (milnor)

2010-12-05T10:25:16Z

If you have a polygon with say a horizontal side, each point is a maxmimum (or minimum) of the projection onto the $y$-axis. So we must admit the possibility of an infinite number of maxima.

Answer by Robin Chapman for Hopf Algebras/Rings, A Question of Terminology

2010-12-02T17:11:33Z

The group operation corresponds to the multiplication map $\mu:A\otimes A\to A$ and the identity should be the natural map $\iota:k\to A$. Both these should be coalgebra maps. The inverse should correspond to a map $S:A\to A$ with $\mu\circ(\rm{id}\otimes S)\circ\Delta=\iota\circ\epsilon =\mu\circ(S\otimes\rm{id})\circ\Delta$, so $S$ is the antipode.

Answer by Robin Chapman for Suggestions for sonifying math

2010-12-01T11:45:44Z

There is Per Norgard's "infinity series" which he used in his Symphony no. 2.

Answer by Robin Chapman for Centers of Semidirect Products

2010-11-29T16:50:02Z

Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$.

As $y$ commutes with all elements of $N$ then $y$ lies in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$ is trivial.

Answer by Robin Chapman for mertens-function in the light of divergent summation - what summation method were best adapted

2010-11-27T08:13:02Z

Well, $$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$ for $s>1$, so setting $s=0$ should give $$\sum_{n=1}^\infty\mu(n)=\frac1{\zeta(0)}=-2$$ as $\zeta(0)=-1/2$. :-)

I should add that this is a trick often used in analytic number theory (for instance in Eisenstein series). More generally given a divergent sum $$S=\sum_{i\in I}a_i$$ then consider, for an appropriate choice of weights $b_i>0$ the series $$f(s)=\sum_{i\in I}\frac{a_i}{b_i^s}.$$ We hope this converges in a suitable half-plane and can be analytically continued to $0$. Then we "define" $S=f(0)$.

Answer by Robin Chapman for open but not affine subscheme?example?

2010-11-26T14:28:43Z

The standard example is to let $X$ be the affine plane over a field, and $U=X-\{(0,0)\}$.

Answer by Robin Chapman for From chain complex to simplicial abelian group

2010-11-24T21:34:23Z

This is the Dold-Kan correspondence.

Answer by Robin Chapman for Is every field the field of fractions of an integral domain?

2010-11-23T15:23:39Z

Every field $F$ of characteristic zero or of prime characteristic but not algebraic over its prime field is the field of fractions of a proper subring of $F$. But no algebraic extension of $\mathbb F_p$ is, since its only subrings are fields.

If $F$ is not an algebraic extension of some $\mathbb F_p$ then $F$ contains a subring $A$ isomorphic to $\mathbb Z$ or $\mathbb F_p[X]$. Each of these rings $A$ has a nontrivial valuation $v$. The valuation $v$ can be prolonged to $F$. Its valuation ring is a proper subring of $F$ whose quotient field is $F$.

Answer by Robin Chapman for Infinite direct product of the integers not a free module over the integers

2010-11-18T13:33:05Z

Another reference to a proof of Specker's theorem is Zagier's St Andrews problems.

Added Also rings such as $\mathbb{Z}$ with this property are called slender rings.

Answer by Robin Chapman for Parametrization of O(3)

2010-11-18T15:04:58Z

The general element is $\pm\exp(A)$ where $A$ is skew-symmetric. (This gives each element infinitely often). This trick essentially works for all compact Lie groups.

There is also the Cayley parameterization: $(I+A)(I-A)^{-1}$ for skew-symmetric $A$ is the general element of $SO(3)$ which lacks an eigenvalue $-1$ (so isn't a half-turn.) This parameterizes all such matrices once each.

Answer by Robin Chapman for Between mu- and primitive recursion

2010-11-17T13:47:36Z

You might look up fast-growing hierarchies.

Answer by Robin Chapman for Generalizations of Belyi's theorem

2010-11-16T17:08:50Z

The compactification is the usual one coming up in the theory of modular forms, with the cusps being orbits of $\Gamma$ on $\mathbb{Q}\cup\{\infty\}$.

As for the proof, I like this paper by Bernhard Koeck.

Answer by Robin Chapman for $A_5$-extension of number fields unramified everywhere

2010-11-04T11:23:45Z

Here's the standard example. I found it in Lang's Algebraic Number Theory where he attributes it to Artin. Let $K$ be the splitting field of $X^5-X+1$ over $\mathbb{Q}$. Then $K$ has Galois group $S_5$ over $\mathbb{Q}$ and $A_5$ over $L=\mathbb{Q}(\sqrt{2869})$. Also $K$ is unramified over $L$.

Answer by Robin Chapman for Does the Hausdorff dimension depend on the L^p-norm?

2010-11-03T11:34:43Z

Let $B_p$ denote the 1-ball with centre 0 with respect to the $l^p$ norm. For any $p$ and $q$ there is a number $N$ such that $B_p$ is covered by $N$ translates of $B_q$. Then any $\epsilon$-ball in the $l^p$ norm is covered by $N$ $\epsilon$-balls in the $l^q$ norm. Thus within a constant factor, the number of $\epsilon$-balls required to cover a set in the $l^p$ and $l^q$ norms is the same. This constant factor won't affect the asymptotic power in the number of $\epsilon$-balls required to cover a given set, so the Hausdorff dimension in both cases is the same.

Answer by Robin Chapman for Analytic continuation via square of absolute value

2010-10-29T13:00:22Z

Rather obviously not: if $f(z)=\sqrt{z}$ on $U$, the plane slit along the negative real axis, then $|f(z)|^2=|z|$ is real analytic on $V$ the plane with the origin removed but $f$ does not analytically continue from $U$ to $V$.

Answer by Robin Chapman for Which elements in SL2(Q) are conjugated to an element in SL2(Z)

2010-10-26T10:01:36Z

You can do this if and only if the trace of $M$ is an integer. By the theory of the rational canonical form if matrices $A$ and $B$ over $\mathbb{Q}$ have the same characteristic polynomial and neither has a repeated eigenvalue they are conjugate by a matrix over $\mathbb{Q}$. This almost does it, save for some fiddling about when the eignvalue of $M$ is repeated.

Answer by Robin Chapman for Analysis of a quadratic diophantine equation

2010-10-25T06:59:51Z

One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_1=6a-1$ etc. Then the variables $a_1$ etc are constrained to be congruent to $5$ modulo $6$.

Homogenizing these gives $$X^2+Y^2=Z^2+T^2$$ and $$X^2-Y^2=Z^2-T^2.$$ Searching for rational solutions of your equation is essentially looking for rational points on the intersection of these two quadrics in $\mathbf{P}^3$. In general the intersection of two quadrics in $\mathbf{P}^3$ is an elliptic curve, so it looks like your problem will boil down to something like finding the integer points on an elliptic curve.

Added There's a blunder in the above: I must thank Fedor for noticing that the second equation should be $$X^2-Y^2=W^2-T^2.$$ So the variety is the intersection of two quadrics in $\mathbf{P}^4$. Hartshorne mentions in passing that in general this construction gives a del Pezzo surface. Del Pezzo surfaces are rational so there should be a birational parametrizion (in terms of two affine parameters) of the rational solutions to the original pair of equations.

Answer by Robin Chapman for Integration problem: $\int_{-\pi}^{\pi} | \log( | 1 + \exp(- I \nu ) | ) | \mathrm{d}\nu < \infty$

2010-10-22T14:35:46Z

You want to show that $$\int_{-\pi}^\pi|\log|1+e^{-it}||dt$$ is finite. Now $$|1+e^{-it}|=|e^{it/2}+e^{-it/2}|=2\cos(t/2)$$ so your integral is $$\int_{-\pi}^\pi|\log|2\cos(t/2)||dt =2\int_0^\pi|\log|2\cos(t/2)||dt.$$ Replacing $t$ by $\pi-2$ in the last integral gives $$2\int_0^\pi|\log|2\sin(t/2)||dt.$$ The integrand is nicely continuous away from $0$. Near $0$, $\sin (t/2)=tf(t)$ where $f(t)\to1/2$ as $t\to0$. Then the integrand is $|\log t+g(t)|$ where $g$ is continuous at $0$ and now finiteness follows since $$\int_0^1|\log t|dt$$ is finite (integration by parts).

Answer by Robin Chapman for How many Hecke operators span the level 1 Hecke algebra?

2010-10-19T17:45:55Z

The answer is yes when $k$ is a multiple of $4$. There is a unique form of weight $k$ of the form $f_k=1+a_dq^d+\cdots$. When $k$ is a multiple of $4$ this is the theta series for a putative extremal even unimodular lattice of rank $2k$. Theorem 20 in chapter 7 of Conway and Sloane's Sphere Packings, Lattices and Groups asserts that $a_d>0$. They give several references for the proof, including a 1969 paper of Siegel.

Answer by Robin Chapman for Algorithms for finding rational points on an elliptic curve?

2010-10-13T14:04:52Z

There is a whole industry devoted to this. The basic method is by descent, which is a formalized version of the infinite descent proofs of Fermat and Euler. It helps if there are rational 2-torsion points but it's not essential. Chapter X in Silverman's The Arithmetic of Elliptic Curves is called "Computing the Mordell-Weil group". It has lots of good information, but maybe isn't so easy for a beginner due to its heavy use of group cohomology.

Answer by Robin Chapman for Eigenvalues of sum of two anti-commuting matrices

2010-10-11T08:23:42Z

Suppose for simplicity's sak that $A$ and $B$ are diagonalizable over $\mathbb{R}$ and are non-singular.

Let $V_a$ be the $a$-eigenspace of $A$. Then by anti-commutativity, we find $BV_a\subseteq V_{-a}$ etc. As $A$ and $B^2$ commute then there is an eigenvector $v\in V_a$ with $B^2v=b^2 v$ for some $v$. If we let $w=bv+Bv$ then $Bw=bw$ so $b$ is real (assuming $B$ has real eigenvectors). On the space $W$ spanned by $v$ and $w=b^{-1}Bv$ the linear transformation $A+B$ has matrix $$\left(\begin{array}{rr} a&b\\ b&-a\\ \end{array}\right)$$ which has an eigenvectors with eigenvalues $\pm\sqrt{a^2+b^2}$.

Answer by Robin Chapman for Convex sets and projections

2010-10-09T16:13:34Z

I presume what you want to prove is the following. Let $S$ be a nonempty closed subset of $\mathbb{R}^n$. Then if there is a point $y\in\mathbb{R}^n$ and there are at least two points $p$ and $q$ in $S$ with Euclidean distance $d$ from $y$ (where $d$ is the distance of $y$ from $S$), then $S$ is not convex. To see this, note that the midpoint $r$ of the line segment $pq$ is closer to $y$ than $p$ of $q$ is, and so cannot lie in $S$. Hence $S$ isn't convex.

Answer by Robin Chapman for How to resolve an issue with Pranesachar et al.'s formula for the number of four-line Latin rectangles?

2010-10-07T06:40:14Z

Obviously, as you know, writing down something like $(-3)!$ is absurd and meaningless. But I would contend that absurd and meaningless as an expression like $$\frac{(-3)!}{(-6)!}$$ is, that it still equals $(-3)(-4)(-5)=-60$. If you have these negative factorials paired up into numerators and denominators, you can calculate like this. Essentially this is interpolating the factorial in the obvious way via the gamma function and noting that sometimes poles cancel to give removable singularities.

I don't know if using a convention like this will successfully resolve your particular problem though.

Answer by Robin Chapman for Any sum of 2 dice with equal probability

2010-10-06T18:39:55Z

You can write this as a single polynomial equation $$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$ where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$. So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$ where the factors satisfy some extra conditions (coefficients positive, $p(1)=1$ etc.).

This is a standard method (generating functions).

Answer by Robin Chapman for Proof that bases etc. exist in early linear algebra course?

2010-10-06T09:59:08Z

If a vector space had bases of two different finite sizes $m < n$ say, then expressing one in terms of the other gives $m$ by $n$ and $n$ by $m$ matrices $A$ and $B$ such that $BA=I_n$. Now use Gaussian elimination (quote their first-year course) to "prove" (with an appropriate amount of hand-waving) that $Av=0$ has a nonzero solution, and thus derive a contradiction. Surely less than a lecture :-)

Answer by Robin Chapman for A coverage question

2010-10-05T18:14:22Z

These odd multiples are (save for $q=3$) the class numbers $h_{-q}$ of the fields $\mathbb{Q}(\sqrt{-q})$. (This follows from the analytic class number formula,)

By the Brauer-Siegel theorem, $\log h_{-q}\sim\log\sqrt{q}$. From this it looks plausible that all possible odd $h$ could occur. I'm not aware of any proof that this is the case (nor of any potential counterexample).

Answer by Robin Chapman for fibonacci series mod a number

2010-10-02T06:51:18Z

This is really just an expansion of Gerhard's comment. One has the matrix formula $$\begin{pmatrix} 1&1\\ 1&0 \end{pmatrix}^n= \begin{pmatrix} F_{n+1}&F_n\\ F_n&F_{n-1} \end{pmatrix} $$ so the problem reduces to computing $A^n$ modulo $k$ where $$A=\begin{pmatrix} 1&1\\ 1&0 \end{pmatrix}.$$ This can be done by the repeated squaring method often used in modular exponentiation. The idea is to compute $A^n$ recursively either as $(A^m)^2$ or $A(A^m)^2$ according to whether $n=2m$ or $n=2m+1$.

Comment by Robin Chapman

2010-12-17T14:05:22Z

What is "free/independent"?

Comment by Robin Chapman

2010-12-16T14:10:13Z

According to the authors of uk.arxiv.org/abs/1012.3235 "no explicit triangulation of $CP^3$ was known so far".

Comment by Robin Chapman

2010-12-15T08:07:26Z

An elliptic curve has (affine) equation $y^2=f(x)$ where $f$ is a cubic or a quartic. In either cases there is a map from $E$ to $P^1$ given by $(x,y)\mapsto x$. This is a double cover, ramified at the zeros of $f$, and also at $\infty$ when $f$ is cubic.

Comment by Robin Chapman

2010-12-14T16:06:13Z

What's the question?

Comment by Robin Chapman

2010-12-09T19:59:19Z

And also in Macdonald's Symmetric Functions and Hall Polynomials. I don't know off-hand if they have results like this though.

Comment by Robin Chapman

2010-12-09T19:44:39Z

Ans others think it's the usual proof in disguise :-)

Comment by Robin Chapman

2010-12-09T18:06:42Z

For finitely generated free modules over commutative rings, the result is elementary. For finitely generated free modules over non-commutative rings, the assertion can fail.

Comment by Robin Chapman

2010-12-09T16:50:26Z

So why does $p/q$ being in lowest terms entail that $p^2/q^2$ is? If "in lowest terms" means having no common factors save units, then this implication doesn't hold in all integral domains.

Comment by Robin Chapman

2010-12-09T16:28:01Z

Version 2 of the first paper has now appeared on ArXiv. In it, "Lemma 2" has been downgraded to "Conjecture 1", but a footnote claims that a cited reference has proved it. :-)

Comment by Robin Chapman

2010-12-08T10:40:54Z

No, Riemann gives an aysmptotic formula not for the number of zeroes on the critical line, but rather in the critical strip. A translation of Riemann's paper appears in an appendix to Harold Edwards's book Riemann's Zeta Function now avaliable as a Dover paperback. Edwards proves all the results in Riemann's paper and a lot more.

Comment by Robin Chapman

2010-12-06T10:22:42Z

Clockwise and anticlockwise don't mean anything in three dimensions.

Comment by Robin Chapman

2010-12-05T10:26:37Z

amazon.co.uk/…

Comment by Robin Chapman

2010-12-05T09:52:19Z

Could you please remind those of us without immediate access to Milnor's paper of what $\mu(P,b)$ denotes.

Comment by Robin Chapman

2010-12-04T11:17:53Z

This question is very difficult to read :-(

Comment by Robin Chapman

2010-12-02T18:35:51Z

Thanks Todd, but doesn't one also have group objects in monoidal categories?