at.algebraic topology - a naive question about homogeneous polynomials

Assume that $p$ is non-zero. If the form $dp/p$ were exact, then locally a primitive would be $log(p)+const$; this is easily seen not to work as soon as you can "loop around" $S$ (e.g. restrict everything to a line intersecting $S$ and see what happens there). Thus the form $dp/p$ is exact if and only if $S$ is empty, and hence if and only if $p$ is constant.

abelian varieties - Distribution of dimensions of factors of the Jacobian of X_0(p)

Let me begin with what was formerly a comment above: the feeling among most experts is probably* that for each fixed $d$, the number of isogeny factors of $J_0(p)$ of dimension $d$ should be small compared to the dimension of $J_0(p)$, i.e. $o_d(p)$.

In what follows, I will cite some results which point in this direction. My overall response is not as logically coherent as I might have liked: perhaps a true expert will do better.

When $p$ is prime, it follows from a 1975 theorem of Ribet (reference below) that the $mathbb{Q}$-rational endomorphism algebra of $J_0(p)$ is the same as the geometric (i.e., $mathbb{C}$-rational) endomorphism algebra, and that this algebra is a product of formally real fields, each being the subfield of $mathbb{Q}$ obtained by adjoining the Fourier coefficients of the various weight $2$ cuspforms of level $p$.

Thus the problem can be viewed as a special case of a popular one in pure algebraic geometry: for which genera $g$ do there exist complex algebraic curves of genus $g$ with, e.g., Jacobians isogenous to a product of elliptic curves? (Or, more generally, with endomorphism algebra containing at least $N$ semisimple factors?) If you count codimensions in the Siegel moduli space corresponding to (say) principally polarized abelian varieties with certain nontrivial endomorphism algebras and the Torelli locus (i.e., of Jacobians), then you find that (at least in many special cases) the sum of these codimensions adds up to more than $frac{g(g+1)}{2}$, the dimension of the Siegel moduli space. Thus unless there is excess intersection between these two loci, for sufficiently large $g$ one expects Jacobians not to be highly decomposed. [This is a cue for Prof. JSE to weigh in on the matter.]

For instance, I think it is at least conjectured that for sufficiently large $g$, no $g$-dimensional Jacobian splits completely into a product of elliptic curves. This is known to be true over finite fields, but maybe not over $mathbb{C}$.

Serre has worked on both the general geometric question and on this special case. In his 1997 JAMS paper, Serre showed that the maximal dimension of a simple isogeny factor of $J_0(N)$ approaches infinity with $N$. I think the result is quantitative, so if you look there you may get the most information currently known on the question you asked.

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Ribet, Kenneth A. Endomorphisms of semi-stable abelian varieties over number fields. Ann. Math. (2) 101 (1975), 555--562.

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Serre, Jean-Pierre
Répartition asymptotique des valeurs propres de l'opérateur de Hecke $T_p$. (French) [Asymptotic distribution of the eigenvalues of the Hecke operator $T_p$]
J. Amer. Math. Soc. 10 (1997), no. 1, 75--102.

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*: I think it's true, and the small number of people I've spoken to about this explicitly (several years ago) thought it was true. We'll see what others have to say.

ct.category theory - What should the definition of "Yoneda property" be?

Well, I'll take a stab at it. I think you're going to want properties $P$ of functors $F$ of the form

$F$ has property $P$ if and only if $forall X$, $F(X)$ has property $P'$

as you'd suggested above. I think we can avoid $P'$ being a natural isomorphism with $h_X$ by making sure it's a property of all of these sets, rather than something we say about it as a functor (on the other hand, if we want to define Yoneda and CoYoneda properties of morphisms, this becomes a problem, and so this is really bothering me philosophically...)

But this definition handles "is a group", "is initial", "is terminal" by "every $F(X)$ is a group", and statements that for a contravariant or covariant (as appropriate) we have a single element. We're going to want to allow products as well to be Yoneda, presumably. Now, $Xtimes Y$ is a product if for any $Zto X, Zto Y$ we get $Zto Xtimes Y$ commuting with projection. So I guess here we might want to have the property "There exist functors $G,H$ such that for all $X$, we have $F(X)=G(X)times H(X)$" perhaps? It avoids quantifying over the category in question, but instead does so over the category of functors, which I haven't thought through terribly well.

Mostly I'm just thinking aloud here, but no one else had much, so I thought I'd throw my two cents in.

graph theory - Proof of Bondy and Chvátal Theorem

Let me restate the theorem for variables.

Notation Read $pbowtie q$ as "$p$ is adjacent to $q$", and $pnotbowtie q$ as "$p$ is not adjacent to $q$".

Theorem Let $G$ be a graph whose order is greater than $2$. Let $v_1$ and $v_2$ be vertices such that $v_1neq v_2 land v_1notbowtie v_2$ and $deg v_1 + deg v_2 ge ninmathbb{N}$. Then $G$ is Hamiltonian iff $G' = G + v_1 v_2$ is Hamiltonian.

Proof Assume $G'$ is Hamiltonian but not $G$. Therefore, by definition, there exists a cycle $(p_1,ldots,p_n)$ in $G$ connecting $v_1$ (at $p_1$) to $v_2$ (at $p_n$) visiting all of $G$'s vertices. If $p_kbowtie p_1$ then $p_{k-1} notbowtie p_n$, since $(p_1,p_k,p_{k+1},ldots,p_n,p_{k-1},p_{k-2},ldots,p_1)$ is a Hamiltonian cycle of $G$. As such, $deg p_n le n - (1 + deg p_1)$, a contradiction.

dg.differential geometry - Existence of Fermi coordinates on a Riemannian manifold

I would answer this question this way:

First, without any further assumptions about the Riemannian metric on $M$, you don't even have a lower bound on the time $t_f$ for which the geodesic exists and does not intersect itself. The geodesic may fail to exist either simply because the metric is incomplete.

Second, if for some reason you know that the geodesic exists for time $t_f$ and does not hit any boundary or singular set of $M$, then you can definitely find Fermi co-ordinates on a neighborhood of the geodesic, but you have no control over the "thickness" of the neighborhood along the geodesic, so it may approach zero.

Third, the answers above hold regardless of whether you know anything about the metric at a single point $p$. That is far too little information, and you can make the metric do anything you want at a point arbitrarily close to the point $p$. As others have indicated, you must assume some information on an neighborhood of $p$, and the conclusion holds only for that neighborhood.

ADDED: Your question is analogous to the following one: Suppose I have a real-valued function $f$ on an interval, and for some (large) value of $k$, I know the $k$-th order Taylor polynomial of $f$ at a point $p$ in the interval. Is there a bound on the first derivative of $f$ on an open interval containing $p$, where the bound and the size of the interval depend only on the coefficients of the Taylor polynomial?

It is easy to show that the answer is no, and essentially the same proof can be used to show that the answer to your question is also no.

RESPONSE TO REVISED VERSION:
1) Yes, if your manifold is open, and you start with a geodesic segment, then there exist Fermi co-ordinates in a neighborhood of the geodesic. But you have no control over the thickness of the neighborhood around the geodesic on which the Fermi co-ordinates exist. It will vary and may approach zero as you approach one of the endpoints of the geodesic segment.

2) I'm a little baffled by why you would want to do everything relative to a background flat metric. Is this metric a natural part of what you need this for? This makes the question a bit contrived for me.

linear algebra - "Transient" of the discrete-time Riccati equation

It is a well-known result that, if the pair $(A,Q^{1/2})$ is stabilizable and the pair $(A, C)$ is detectable, the solution to the discrete-time Riccati recursion

$P(t+1) = A P(t) A^T - A P(t) C^T (CP(t)C^T + R)^{-1} CP(t) A^T + Q$

converges to the unique stabilizing solution of the Algebraic Riccati Equation

$P = A P A^T - A P C^T(CPC^T + R)^{-1}CP A^T + Q$

from any initial positive semidefinite matrix $P(0)=P_0$.

Do you know any result on the behavior of $P(t)$ during the transient, or more generally
for all $t geq 0$? More precisely, I need some result that ensures that $P(t)$ is
stabilizing for all $t geq 0$, under fair conditions on $A$, $C$, $Q$, $R$, and for arbitrary $P_0$.

Thank you very much in advance,

f.

EDIT. Some background: We consider a discrete-time, linear, time invariant system of the form

$x(t+1) = Ax(t) + v(t)$

$y(t) = Cx(t) + w(t)$

where $x(t) in R^n$, $Ain R^{ntimes n}$, $y(t)in R^p$, $Cin R^{ptimes n}$, and $v$ and $w$ are zero-mean, uncorrelated white noises (say, Gaussian) with appropriate dimensions, with variances $Q$ and $R$ respectively. The predictor $hat{x}(t+1|t)$, that is, the best linear estimator of $x(t+1)$ given $y(0), cdots, y(t)$, is given by the Kalman filter. It can be expressed recursively substituting the equations of the Kalman filter into each other. The variance $P(t+1)$ of the corresponding prediction error $tilde{x}(t+1|t) = hat{x}(t+1|t)-x(t+1)$ is then given (recursively) by the Riccati equation.

The "dual" of the linear estimation problem above is the "linear quadratic regulator" problem of optimal control, and the Riccati equation is fundamental also in this context.

Given matrices $Ain R^{ntimes n}$ and $B in R^{ntimes m}$ with $mleq n$, to say that the pair $(A, B)$ is reachable means that the matrix $[B AB cdots A^{n-1}B]$ has full rank ($=n$). The pair $(A, B)$ is stabilizable if with a suitable "change of base" $(A,B)mapsto(T^{-1} A T, T^{-1}B)$ it can be put in the form

$left(left[begin{matrix} A_{11} & A_{12} \\ 0 & A_{22} \\ end{matrix}right], left[begin{matrix} B_1 \\ 0 \\ end{matrix}right]right)$

where $(A_{11}, B_1)$ is reachable and $A_{22}$ is Hurwitz (all eigenvalues in the interior of the unit disk).

Dually, given matrices $Ain R^{ntimes n}$ and $C in R^{ptimes n}$ with $pleq n$, to say that the pair $(A, C)$ is observable means that the matrix

$left[begin{matrix} C \\ CA \\ vdots \\ CA^{n-1} end{matrix}right]$

has full rank. The pair $(A, C)$ is detectable if with a suitable "change of base" $(A,C)mapsto(T^{-1} A T, CT)$ it can be put in the form

$left(left[begin{matrix} A_{11} & 0 \\ A_{21} & A_{22} \\ end{matrix}right], left[begin{matrix} C_1 & 0 end{matrix}right]right)$

where $(A_{11}, C_1)$ is observable and $A_{22}$ is Hurwitz.

Finally, the matrix $P=P^Tgeq 0$ is stabilizing if the "closed loop" matrix $A - KC$ is Hurwitz, where $K = A P C^T(CPC^T + R)^{-1}$.

For more details, see for example the Wikipedia pages on Kalman filter, controllability, observability, and Kalman decomposition. For a full reference, see e.g. A. H. Jazwinski, Stochastic Processes and Filtering Theory.

reference request - Is every real vector bundle over the circle necessarily trivial?

In the spirit of Theo's comment, I'll say something about sufficient conditions.

A real vector bundle over the circle is trivial if and only if it is orientable. I discussed this a bit here.

The main point is that up to isomorphism, every real vector bundle over the circle is either trivial, or the Whitney sum of a trivial bundle with the Mobius bundle. The latter is not orientable.

The other answers at the question I linked to above may also be helpful to you.