at.algebraic topology - Why does one think to Steenrod squares and powers?

Steenrod operations are an example of what's known as a power operation. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.

A cohomology class on $X$ amounts to a map $a: Xto R$, where $R = prod_{ngeq0} K(F_2,n)$. So the cup product of $a$ and $b$ is given by
$$Xtimes X to Rtimes R xrightarrow{mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes addition of cohomology classes.)

You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/Sigma_n$, where $Sigma_n$ is the symmetric group, i.e.,
$$X^n xrightarrow{a^n} R^n rightarrow R$$
should factor through the quotient $X^n/Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $Sigma_n$, and a product map:
$$mu_n' : E(n)times R^nto R$$
which is $Sigma_n$ invariant, so it factors through $(E(n)times R^n)/Sigma_n$. Thus, given $a: Xto R$, you get
$$P'(a): (E(n)times X^n)/Sigma_n to (E(n)times R^n)/Sigma_n to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/Sigma_n times Xto R.$$

If $n=2$, then $E(2)/Sigma_2$ is what Hatcher seems to call $L^infty$; it is the infinite real proj. space $RP^infty$. So $P(a)$ represents an element
in $H^* RP^infty times X approx H^*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$.

Other cohomology theories have power operations (for K-theory, these are the Adams operations).

You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.

mg.metric geometry - Is there a neat formula for the volume of a tetrahedron on the surface of $S^3$?

On the volume of a hyperbolic and spherical tetrahedron, by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, Roberts, Woodward, Frohman, Kania-Bartoszynska, etc.)

Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.

stochastic calculus - Relation between regularities of the trajectory of a mean zero gaussian process and its covariance operator

Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $mathbb P$. The covariance operator $C : X^* to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $mathbb P$ is then the closure of the image of $X^*$ under $C$: $$operatorname{supp} mathbb P = overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$langle f, g rangle = f(Cg)$$ for $f, g in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $iota : X^* hookrightarrow H$ be the inclusion map. Define a map $iota^* : H hookrightarrow X$ first on the dense subspace $iota X^* subseteq H$ by $$iota^*(iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = iota^* circ iota$.

The Hilbert space $iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $mathbb P$, we have $$operatorname{supp} mathbb P = overline{iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $mu$ is a Radon measure on $[0,1]$, then $$(Cmu)(s) = int_0^1 c(s,t) , dmu(t).$$

If we write $c_s(t) = c(s,t)$, then the support of $mathbb P$ is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $overline{operatorname{span}{c_s}}$ is a space with regularity property $P$, then the process $xi_t$ satisfies property $P$ with probability one.

homotopy theory - Do h-coequalizers and coproducts give all h-colimits?

It is well known that if a category has all coequalizers and all (small) coproducts then in fact it has all (small) colimits. More important is the proof which shows that every colimit can be built by using coproducts and coequalizers. This implies that if a functor commutes with coproducts and coequalizers, then it must commute with all (small) colimits as well.

Is there homotopical analog of this? If I have a functor which commutes with all small (homotopy) coproducts and all homotopy coequalizers, does it necessarily commute with all homotopy colimits in general?

This question makes sense for general model categories, but I am particularly interested in the usual model structure on spaces.

foliations - Foliated bundles and suspensions

Hi,

Is there any criteria, except for the existence of a flat connection, for a foliated bundle $E$ to be a suspension ( a foliated flat bundle)? For example, the Kronecker foliation on the torus is a suspension $mathbb{R}times_{mathbb{Z}} mathbb{S}^1$ , i.e. its of the form $Mtimes_Gamma F$, where $Gamma$ acts freely and transitively on the manifold $M$ and there is a free action $
rho: Gamma rightarrow Diff(F)$.

Note: The Kronecker foliation is induced by the vector field , $a frac{partial}{partial x} + b frac{partial}{partial y}$ on $mathbb{R}^2$ with $a,b$ constants.
The action of $mathbb{Z}$ on $mathbb{R}times mathbb{S}^1$ is given by:

$(r, exp{iz}).m= (r+m, exp{(iz+malpha)})$ for some $alpha in mathbb{R}$.

Thanks

gn.general topology - Non-homeomorphic spaces that have continuous bijections between them

Recycling an old (ca. 1998) sci.math post:

" Anyone know an example of two topological spaces $X$ and $Y$
with continuous bijections $f:Xto Y$ and $g:Yto X$ such that
$f$ and $g$ are not homeomorphisms?

Let $X = Y = Z times {0,1}$ as sets, where $Z$ is the set of integers.
We declare that the following subsets of $X$ are open for each $n>0$.
$${(-n,0)}, {(-n,1)}, {(0,0)}, {(0,0),(0,1)}, {(n,0),(n,1)}$$
This is a basis for a topology on $X$.

We declare that the following subsets of $Y$ are open for each $n>0$.
$${(-n,0)}, {(-n,1)}, {(0,0),(0,1)}, {(n,0),(n,1)}$$
This is a basis for a toplogy on $Y$.

Define $f:Xto Y$ and $g:Yto X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$
Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.

This example is due to G. Paseman.

David Radcliffe "

More generally, take a space X with three successively
finer topologies T, T' and T''. Form two spaces which have underlying
set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' ....
and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer
topology in one sequence to a rougher topology in the other. You can
make them bijective, and show that they are obviously non-homeomorphic
for a judicious choice of X, T, T', and T''.

Gerhard "Ask Me About System Design" Paseman, 2010.07.05

ca.analysis and odes - Closed forms for Monotonic polynomial recurrences?

A lot depends of what you mean by "fair accuracy" and on what exactly you are going to do with your formula. If a 30% upside error in each $d_n$ is tolerable, you can do the following.

We look at the recursion $d_{n+1}=qd_n(1-d_n)$ with $0<q=2p<1$ starting with some $d_1in[0,1]$. It'll be convenient to do the first step separately, so we have $d_2=q d_1(1-d_1)$. The reason is that $q^{-1}d_2in[0,1/4]$. Now, denote $b_n=q^{-(n-1)}d_n$ and rewrite the recurrence as $b_{n+1}^{-1}=b_n^{-1}+q^{n-1}frac{b_n}{b_{n+1}}$. Note now that the ratio of each term to the next is between $1$ and $4/3$ and tends to $1$, so, replacing it by $1$, we get $b_n^{-1}approx b_2^{-1}+frac{q-q^{n-1}}{1-q}$ with accuracy 30% and being sure that it is an underestimate. Putting all this stuff together, we get
$$
d_napprox q^{n-1}left[frac 1{d_1(1-d_1)}+frac{q-q^{n-1}}{1-q}right]^{-1}
$$
for $nge 2$.

Note also that 30% is a very rough estimate for the accuracy. In practice, I've never seen it being worse that 6% (though I haven't done too many simulations).

P.S. The estimate
$$
d_napprox q^{n-1}left[frac 1{d_1(1-d_1)}+frac{q-q^{n-1}}{1-q}
+logleft(1+d_1(1-d_1)frac{q^2-q^{2(n-1)}}{1-q^2}right)right]^{-1}
$$
is even better (3% accuracy for small $n$ and about 0.5% accuracy for large $n$) but noticeably uglier. As several people have already mentioned, there is no exact formula, so the higher precision you want, the longer and uglier the approximation gets.

P.P.S. The main idea behind the second approximation is that if $B_n=b_{n}^{-1}$, then we make an error $q^{n-1}frac{B_{n+1}-B_n}{B_n}approx frac{q^{2(n-1)}}{B_n}$ during each step that led us to the first approximation. To compensate, we would like to take the partial sums of that series. We also know that $B_napprox frac 1{d(1-d)}+q+q^2+dots+q^{n-2}$. Unfortunately, we cannot sum the resulting series nicely. But if we double all the exponents in the approximate formula for $B_n$ (which will lead only to a small error when $qapprox 1$), then we'll get the Riemann sum type expression, which we can replace by the corresponding integral. Clearly, it may overshoot on the long run but to my own surprise, it not only corrects the first few terms nicely, but also corrects the asymptotics giving an error below 0.5% for large $n$ in the entire range of parameters (well, at least that is so for all values I looked at and I tried quite a few). Why it works this way remains a mystery, but it does. It can be shown rigorously that the total relative overshot coming from all steps after $n$ is at most $n^{-1}$ and I've never seen the overshots of more than 0.1% up to $n=300$.