linear algebra - Extremum under variations of a traceless matrix

Sorry for my precedent tentative, I was a little hasty:

Ok, I think I'd better put the original problem:

I have an action of three fields: $A$ which is the spin-connection, $B$ an skew-symmetric 2-form and $Phi$ which is traceless and skew-symmetric scalar field. These fields take their values on some algebra, index their components in this algebra by $i,j,k,... = 1,2,3$

I want to implement a certain condition on B by using equations of motion of $Phi$, the action is:

$S=int (B_i wedge F^i + Lambda B_i wedge B^i + Phi_{ij} B^i wedge B^j) $

Now for me equations of motions are simply:

$B^i wedge B^j=0$

perhaps with the condition that all diagonal elements are equal (as jc showed) but this is automatically satisfied for a skew-symmetric matrix (here $B^i wedge B^j$).

But in all papers I find:

$B^i wedge B^j - frac{1}{3}delta^{ij}B_kwedge B^k = 0$

So I see that they all took the traceless part of the matrix representing equations of motion, necessarily it has a relation with the traceless character of $Phi$ but I do not see which one.

In addition, this expression is not antisymmetric in $i,j$.

Would anyone have an idea?

ca.analysis and odes - Notation for eventually less than

I agree with Joel Hamkins's answer, but I don't entirely agree with his comment on that answer. I generally use asterisks to mean "with finitely many exceptions" or "modulo finite sets", so I'd use $fleq^*g$ and $Asubseteq^*B$ as Joel says. But when working modulo some ideal $I$ other than the ideal of finite sets, I'd ordinarily avoid asterisks and instead write $fleq_Ig$ and $Asubseteq_IB$.

I'd like to protest vigorously against the use of $ll$ in this situation. To me, $fll g$ means that $f$ is a lot smaller than $g$ (at least eventually), whereas here you might have $f(n)=g(n)-1$ for all $n$.

terminology - Is there a name for this property of a topology?

In spaces where singleton points are closed, your property is equivalent to saying that the space has no isolated points. Or in other words, that it is perfect.

Clearly, no space with an isolated point can have your property. Conversely, when singletons are closed, then you can subtract one point from any open set and thereby have a proper open subset. So if U has at least 2 points x,y, then U = U-{x} union U-{y}, giving an instance with I of size 2.

However, your property does not imply that points are closed, since the space on reals R, where open sets have the form (-infty, a), has your property, but points are not closed in this space.

big list - One-step problems in geometry

Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:

There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?

To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)

P.S. When I was asked the question, I failed miserably!

st.statistics - Uniqueness of the function used in the Khinchine criterion for characteristic function

Khinchine criterion for $varphi$ being a characteristic functions states that it is a ch.f. if and only if $varphi(0)=1$ and

$varphi(t) = int_{-infty}^infty g(t+theta) overline{g(theta)} dtheta$

Is function $g$ uniquely defined for a given $varphi$?

vector bundles - Uniqueness of Chern/Stiefel-Whitney Classes

This question is closely related to this previous question.

Chern and Stiefel-Whitney classes can be defined on bundles over arbitrary base spaces. (In Hatcher's Vector Bundles notes, he uses the Leray-Hirsch Theorem, which appears to require paracompactness of the base space. The construction in Milnor-Stasheff works in general, as does the argument given by Charles Resk in answer to the above question. A posteriori, this actually shows that Hatcher's construction works in general too, since he really just needs $w_1$ and $c_1$ to be defined everywhere.)

The proof of uniqueness (as discussed in Milnor and Stasheff, or in Hatcher's Vector Bundles notes, or in the answers to the above question) relies on the splitting principle, and hence (it seems to me) requires the existence of a metric on the bundle in question. More precisely, if we have two sequences of characteristic classes satisfying the axioms for, say, Chern classes, and we want to check that they agree agree on some bundle $Eto B$, the method is to pull back $E$ along some map $f: B'to B$ (with $f^*$ injective on cohomology) so that $f^*E$ splits as a sum of lines. Producing the splitting seems to require a metric on $E$ (or at least on $f^*E$).

If $B$ is not paracompact, bundles over $B$ may not admit a metric (and may admit a classifying map into the universal bundle over the Grassmannian), so my question is:

Are Chern and/or Stiefel-Whitney classes unique for arbitrary bundles? If not, do $w_1$ and $c_1$ at least determine the higher-dimensional classes?

mg.metric geometry - Hausdorff measure question

Say we have some compact metrisable topological space $X$ with a measure $mu$ defined on the Borel sets of $X$. Then is there some way to determine whether $mu$ is the Hausdorff measure associated to some metric $d$ compatible with the topology of $X$? And if so, is there some process to recover a metric from the measure? I'd imagine that there would have to be some conditions placed on the space $X$, eg. that it's connected, and it might even be necessary to assume that it's some nice space such as a manifold, with a "gauge" metric $d_{0}$ relative to whose Hausdorff measure $mu$ is absolutely continuous, but I'd like to ask the question in the greatest generality possible, in the hope that there is an answer out there.