User jeff strom - MathOverflow

uniqueness of $f$-localization

2012-04-28T19:13:35Z

The $f$-localization I mean is the one described and studied in detail in the book by E. D. Farjoun; $L_f$ is a homotopy idempotent functor which associates to each space $X$ an $f$-equivalence $X\to L_f(X)$ where $L_f(X)$ is $f$-local.

$f$-localization has a kind of uniqueness: if $F$ is some other coaugmented functor with the property that $F(X)$ is $f$-local for every $X$ (I'm happy to assume that $F = L_g$ for some map $g$), then there is a commutative square of functors and natural transformations, which I don't know how to draw here. The square would show that the composites $$ id \xrightarrow{\iota} L_f \xrightarrow{L_f(j)} L_f\circ F \qquad \mathrm{and} \qquad id \xrightarrow{j} F \xrightarrow{\iota_F} L_f \circ F $$ are equal. And $\iota_F$ is a weak equivalence for every space $X$; thus $F$ factors through $L_f$ `up to weak equivalence'.

My Question: Suppose $X\to Y$ is an $f$-equivalence; then $L_f(X) \to L_f(Y)$ is a weak equivalence; does it follow that $(L_f\circ F)(X) \to (L_f\circ F)(Y)$ is a weak equivalence?

Answer by Jeff Strom for about relative homotopy group

2012-08-26T15:46:17Z

To define the relative homotopy groups of a pair $(X, A)$, let $i:A\to X$ be the inclusion, and write $F_i$ for its homotopy fiber. Then $$ \pi_n(X, A) = \pi_{n-1}(F_i). $$ In your examples, the inclusion maps are nullhomotopic, so the homotopy fibers are $$ \Omega \Sigma \mathbb{R}P^2 \times \mathbb{R}P^2 \qquad \mathrm{and} \qquad \Omega \Sigma \mathbb{C}P^2 \times \mathbb{C}P^2, $$ respectively. Since these spaces are path-connected, the relative homotopy "groups" in question are trivial.

Answer by Jeff Strom for Blackbox Theorems

2012-06-15T00:20:02Z

How about Haynes Miller's theorem resolving the Sullivan conjecture?

Characterizing the rationalization of spaces.

2010-07-08T16:48:14Z

In the category of rational spaces, loop spaces split as products of Eilenberg-Mac Lane spaces and SUSPENSIONS split as wedges of (rational) spheres. I wonder if anything of the following form is true:

(*) Any functor $F$ from spaces to spaces which splits suspensions and loop spaces as above must factor through the rationalization.

EDIT 1: Greg raises some fine questions, but I stand by my wording. This is a question that arises from curiosity, not because I need it for anything, so I'd be happy with "anything like" the given statement.

EDIT 2: At least for simply-connected spaces, rationalization commutes with loop and suspension. But, it seems to me that the power of the property is that the suspension of any F-space splits and the loops of any F-space splits. So I would go with:
the suspension of any rational space splits as a wedge of rational spheres and the loops of any rational space splits as a product of rational Eilenberg-Mac Lanes spaces.

Thus, we'd be looking for functors to some model-esque category with some relatively manageable list of objects whose products exhaust the homotopy types of loop spaces and whose wedges exhaust the homotopy types of suspensions.

Answer by Jeff Strom for Characterizing the rationalization of spaces.

2012-05-11T01:41:01Z

I have an answer.

Look at $f$-localization functors $L_f$. The restriction of $L_f$ to simply-connected spaces is rationalization if and only if the following three conditions hold:

$L_f(S^2)$ is nontrivial and simply-connected
$L_f$ commutes with cofiber sequences of simply-connected finite complexes
if $X$ is a simply-connected finite complex, then for large enough $k$, $\Sigma^k L_f(X)$ splits as a wedge of copies of $L_f(S^n)$ for various values of $n$.

Details can be found here: http://arxiv.org/abs/1205.2140

Reference request: splittings in rational homotopy theory

2012-05-04T02:00:59Z

It is well known that for simply-connected rational spaces, every suspension splits as a wedge of rational spheres and every loop space splits as a product of rational Eilenberg-Mac Lane spaces.

What are the best/original references for this?

Solid Rings and Tor

2012-04-25T13:57:48Z

A solid ring is a ring $R$ such that the multiplication $R\otimes_{\mathbb{Z}} R \to R$ is an isomorphism.
These were classified by Bousfield and Kan; they are

subrings of $R\subseteq\mathbb{Q}$,
$\mathbb{Z}/n$,
products $R\times \mathbb{Z}/n$ with $R\subseteq \mathbb{Q}$ and every divisor of $n$ invertible in $R$
colimits of these.

I wonder how small the list gets if I put the additional constraint that $\mathrm{Tor}_{\mathbb{Z}}(R,R) = 0$.

REFERENCE: Bousfield, A. K.; Kan, D. M. The core of a ring. J. Pure Appl. Algebra 2 (1972), 73–81.

Answer by Jeff Strom for uniqueness of $f$-localization

2012-04-30T13:26:05Z

An $f$-equivalence is a map $\alpha:X\to Y$ which induces a weak equivalence $\alpha^\ast:map_\ast(Y,Q)\to map_\ast (X,Q)$ for all $f$-local spaces $Q$; equivalently, $L_f(\alpha)$ is a weak equivalence. Therefore the assumption implies that every $g$-local space is $f$-local, so an $f$-equivalence $\alpha$ is also a $g$-equivalence; so $L_g(\alpha)$ is a weak equivalence.

Answer by Jeff Strom for The most general context of Mather's Cube Theorems

2012-03-02T14:23:51Z

The second cube theorem (if base and sides are ok, then so is the top) is a straightforward consequence of the formally crazy fact that if $p:E\to B$ is a fibration, $i:A\to B$ is a cofibration, and $p_A:E_A\to A$ is the pullback of $p$ along $i$, then $i:E_A\to E$ is also a cofibration.

Answer by Jeff Strom for Moore path space.

2012-02-29T23:24:23Z

You can show the evaluation map is a weak fibration (I think this is the term: I mean a map homotopy equivalent in the category of spaces over $X\times X$ to a fibration -- namely $X^I\to X\times X$), which is good enough for many purposes, including the formation of homotopy pullbacks.

Answer by Jeff Strom for [v_4,l_4]=?, v_4 is the Hopf map in \pi_7(S^4) and l_4 is the generator of \pi_4(S^4)

2012-02-26T17:25:06Z

This is how far I can get without checking a book.

Since $v_4 = \pm {1\over 2} [i_4, i_4]$, you are interested in the triple Whitehead product $\alpha = {1\over 2}[ [i_4, i_4],i_4]$. Since it is a Whitehead product, its suspension is trivial. The Jacobi identity for Whitehead products shows that $[ [i_4, i_4],i_4]$ has order $3$; so $\alpha$ has order either $3$ or $6$.

Answer by Jeff Strom for Limits in category theory and analysis

2012-02-23T14:52:59Z

I think this doesn't quite work:

Let $\mathcal{C}$ be the category whose objects are the point of $X$, and define $$ \mathrm{mor}_\mathcal{C}(x,y) = \{ \mbox{closed sets containing both $x$ and $y$} \}. $$ Composition is union.

Now (for example) a sequence $\{ x_n\}$ in $X$ defines a functor $F: \mathbb{N} \to \mathcal{C}$ and a cone from $F$ to $y$ is essentially a single closed set containing the entire sequence and $y$. Since this set must contain the topological limit $x$ of the sequence, this means that the cone factors through the same closed set viewed as a morphism $x\to y$, so $x$ is the categorical colimit of $F$.

And since the morphism sets are symmetrical, the sequence $\{ x_n\}$ can be viewed as a contravariant functor $G: \mathbb{N}\to \mathcal{C}$, and the topological limit $x$ is the categorical limit of $G$.

PROBLEM: the factorization is not unique!

Answer by Jeff Strom for The Higman group

2012-02-03T02:15:11Z

Reference:

Higman, Graham

A finitely generated infinite simple group. J. London Math. Soc. 26, (1951). 61--64

It is shown that G is infinite and has no proper normal subgroups of finite index, except G.

It is easy to see that this group is perfect: it has trivial abelianization.

I have heard through the grapevine that the space $X$ with four one-cells and four two-cells (corresponding, respectively, to generators and relations) is the classifying space of the group (I don't have a reference).

Fitting desired weak equivalences and cofibrations into a model category

2012-01-31T15:51:03Z

Suppose I have a category $\mathbf{C}$ and classes of morphisms $\mathcal{W}$ and $\mathcal{C}$, and I would like to know that $\mathcal{W}$ and $\mathcal{C}$ are the weak equivalences and the cofibrations of a model category structure.

I can certainly write down what the fibrations $\mathcal{F}$ would have to be, and I'm wondering if there are any theorems to provide shortcuts in verifying that the classes $\mathcal{W}$, $\mathcal{C}$ and $\mathcal{F}$ actually satisfy all the conditions of a model category.

Answer by Jeff Strom for Examples where it's useful to know that a mathematical object belongs to some family of objects

2011-12-15T15:24:28Z

I've written a paper (or two) about collection $\mathcal{R}$ of all pointed topological spaces $Y$ satisfying the property $\mathrm{map}_*(X,Y) \sim *$ (for fixed $X$). The interesting fact is that if $\mathcal{R}$ contains $S^{2n+1}$ for all sufficiently large $n$, then $\mathcal{R}$ contains all finite-dimensional simply-connected CW complexes. My proof works by induction on the cone length of $Y$ with respect to the collection of all wedges of spheres, and the passage from spaces of cone length $n$ to those of cone length $n+1$ requires information about a huge array of spaces with cone length $n$ (or less), and not just the ones in the given cone decomposition.

So: knowing that $Y$ has a finite cone decomposition leads to the apparently unrelated conclusion that $\mathrm{map}_*(X, Y)\sim *$; and the proof requires the collection approach.

Answer by Jeff Strom for looking for a book on banach manifolds

2011-11-21T16:17:03Z

I happen to know this

Abraham, Ralph; Robbin, Joel Transversal mappings and flows. An appendix by Al Kelley W. A. Benjamin, Inc., New York-Amsterdam 1967 x+161 pp.

exists.

Answer by Jeff Strom for Homotopic maps out of cofibration sequences

2011-11-10T01:27:58Z

No.

For example, if $X$ is a CW complex with skeleta $X_n$ and $f_n\simeq*$, then $f$ is a phantom map. Their homotopy classes are in bijective correspondence with $\lim^1 [\Sigma X_n, Y]$, and are frequently nonzero. For example, $\mathbb{C}P^\infty$ is the domain of nontrivial phantom maps.

It is true that there are no nontrivial phantom maps between (simply-connected at least) rational spaces.

It goes back to Milnor in the early 1960s, but in full generality I think it came a little later (perhaps Bousfield and Kan) that if $X$ is the colimit of your telescope diagram, then there is a natural short exact sequence of pointed sets $$ * \to {\lim}^1 [\Sigma X_n, Y] \to [X, Y]\to \lim [ X_n, Y]\to * . $$ For phantom maps, we take the telescope to be the skeleta of a CW decomposition of $X$, and we get the identification $\mathrm{Ph}(X, Y) \cong \lim^1 [\Sigma X_n, Y]$. This comes from just looking at the long cofiber sequence of the big fold/inclusion map $\bigvee X_n \to X$, whose cofiber $\Theta_X : X\to \bigvee \Sigma X_n$ is known as the universal phantom map (it is phantom and every other phantom factors (nonuniquely) through it; see the paper "Universal phantom maps" by Gray and McGibbon).

It is not too hard to show that if $\Sigma X$ is not a retract of a wedge of finite-dimensional spaces, then $\Theta_X\not\simeq *$, so there are nontrivial phantoms out of $X$. This is the case, for example, when the Steenrod algebra action takes elements to arbitrarily high dimension (as for $\mathbb{C}P^\infty$ or $\mathbb{R}P^\infty$ or $B\mathbb{Z}/p$, etc.)

Difference between represented and singular cohomology?

2010-06-24T20:28:27Z

Ordinary cohomology on CW complexes is determined by the coefficients. There are (more than) two nice ways to define cohomology for non-CW-complexes: either by singular cohomology or by defining $\widetilde H^n(X;G) = [X, K(G,n)]$. Are there standard/easy examples where these two theories differ?

One idea that comes to mind is the paper by Milnor and Barratt (about Anomolous Singular Homology) which says that the $n$-dimensional Hawaiian earring $H^n$ has nontrivial singular homology in arbitrarily high dimensions. But I don't see an easy way to compute $[H^n, K(G, m)]$.

Answer by Jeff Strom for $H_2$ of a simply connected Lie group vanishes

2011-05-26T20:04:21Z

One way to do this is to use a very nice theorem of Weingram:
if $G$ is finitely generated, then no nontrivial map $\Omega S^{2n+1} \to K(G, 2n)$ can factor through a finite-dimensional CW complex.

Now if $X$ is a Lie group and $\pi_2(X) \neq 0$, then we get (using the James construction) a map $\alpha: \Omega S^3 \to X$ nonzero on $\pi_2$, and a cohomology class $u: X\to K(G,2)$ such that $u \circ \alpha \neq 0$. Since $X$ is a Lie group, this contradicts Weingram's theorem.

Reference:
Weingram, Stephen On the incompressibility of certain maps. Ann. of Math. (2) 93 (1971), 476–485.

Answer by Jeff Strom for Techniques for computing cup products in singular cohomology

2011-05-11T13:10:35Z

The cup product is ultimately the map induced by the diagonal $\Delta: X\to X\times X$. You can get lots of information by studying the actual map.

Reference for an automorphism in a paper of Toda

2011-04-07T02:31:52Z

In Selick's very pretty paper "Odd primary torsion in $\pi_*(S^3)$" he makes use of an automorphism which was established by Toda in his paper "On the double suspension $E^2$".

Unfortunately, Selick just refers to [10], but the paper is about 40 pages and dense reading; after skimming it several times, I fear that the automorphism is there either implicitly, or with different notation.

Can someone provide a more precise reference? A page or theorem number would be ideal.

Left and right eigenvalues

2010-07-17T17:13:43Z

A quaternionic matrix $A$ gives rise to a function $\mathbb{H}^n \to \mathbb{H}^n$ given by $x \mapsto A \cdot x$. This is real linear, but not complex- or quaternionic-linear (in general) if we consider $\mathbb{H}^n$ as a left $\mathbb{C}$ or $\mathbb{H}$ module, but is pretty good if we use right actions.

A right eigvenvalue of $A$ is a quaternion $q$ such that $A\cdot x = x \cdot q$ for some $x\in \mathbb{H}^n$; a left eigenvalue is quaterion $q$ such that $A \cdot x = q\cdot x$ for some $x\in \mathbb{H}^n$.

The algebra of right eigenvalues is pretty good, but the algebra of left eigenvalues is quite interesting. For example, it is not hard to see that there are matrices $A$ with infinitely many left eigenvalues, even for $2$-by-$2$ matrices.

Now let's assume that $A\in Sp(n)$, so that the left eigenvalues are all contained in $S^3\subseteq \mathbb{H}$. What sort of geometric properties must the set $L(A)$ of left eigenvalues have?

EDIT: An example is $$ \left[ \begin{array}{cc} 0 & 1 \cr -1 & 0 \end{array} \right] \cdot \left[ \begin{array}{c} 1 \cr q
\end{array} \right] = q \cdot \left[ \begin{array}{c} 1 \cr q
\end{array} \right] $$ for any $q\in S^3 \subseteq \mathbb{H}$ with zero real part, since then $q^2 = -1$.

EDIT 2: Examples like this show that for some symplectic matrices, the set of left eigenvalues is a union of copies of $S^2$.

Answer by Jeff Strom for Surprising and Useful Physical Intuition for Mathematical Objects

2011-03-30T00:21:15Z

Archimedes figured out the area of a parabolic section by slicing it into pieces and moving the pieces to the other side of an imaginary lever and observing that they balance a certain triangle.

Slick Proof of Kudo Transgression Theorem

2011-03-11T01:27:27Z

The Kudo Trangression Theorem has to do with the transgression in the Leray-Serre spectral sequence for cohomology in $\mathbb{Z}/p$ ($p$ odd). It can be proved by the method of the universal example, once it is shown that in the path-loop fibration sequence $K(\mathbb{Z}/p,2n) \to P(K(\mathbb{Z}/p,2n+1)) \to K(\mathbb{Z}/p,2n+1)$

the fundamental class $v$ of the fiber transgresses to $u$, that of the base
this forces a zig-zag of cancellation, up to $v^{p-1}\mapsto u \otimes v^{p-2}$
also $v^p$ transgresses to $P^n(u)$, and
$u\otimes v^{p-1}$ "transgresses" to $\beta P^n(u)$.

Parts (1), (2) and (3) are easy, but part (4) seems difficult. There is a proof along these lines in a paper of Browder from the mid 1960's (he attributes the proof to Milgram), but the proof of (4) is actually quite hard and leans heavily on algebraic mucking around in the spectral sequence.

Does anyone know of a clever way to prove (4)?

Edit: Let's say we know by induction that the cohomology of the fiber is what it has to be. Then I think the behavior of the spectral sequence is forced in dimensions below that of $u\otimes v^{p-1}$. Does this show that $u\otimes v^{p-1}$ "transgresses"? Suppose it does; then its image is $Q(u)$, where $Q$ is a cohomology operation that vanishes when looped (since it is not the transgression of a class in the fiber). Perhaps we can argue that $Q$ must be $\beta P^n$, up to sign?

Answer by Jeff Strom for Is there the Whitehead theorem for cohomology theory?

2011-03-08T10:58:53Z

Sure.

The basic point is that for simply-connected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber.

In homology, you determine the connectivity of the cofiber by looking at $H_*(C;\mathbb{Z})$, because of the Hurewicz Theorem.

In cohomology, you appeal to: if $X$ is simply-connected, then $X$ is $n$-connected if and only if $[ X, K(G,m)] = *$ for all $m \leq n$ and all abelian groups $G$. This is because (by basic obstruction theory) if $X$ is $(n-1)$-connected, then $[X, K(G,n)] \cong \mathrm{Hom}(\pi_n(X), G )$.

This results in a long list of groups to check, admittedly; but it can be whittled down by Universal Coefficients theorems if you like.

The fiber of a Serre fibration

2011-01-29T17:41:51Z

If $p:E\to B$ is a Serre fibration (assume it is surjective), then for each $b\in B$ we get a comparison map $p^{-1}(b) \to F_b$, where $F_b$ is the homotopy fiber of $p$ over $b$.

It is easy to see that these maps induce isomorphisms on $\pi_n$ for $n\geq 1$, but I wonder about $\pi_0$.

Question: Is it true that $p^{-1}(b) \to F_b$ is a weak homotopy equivalence?

Answer by Jeff Strom for What does the adjective "natural" actually mean?

2011-03-01T01:02:56Z

According to Mac Lane (as I remember it from Categories for the Working Mathematician), Eilenberg and Mac Lane invented categories so they could talk about functors, and they wanted to talk about functors so they could define "natural."

Answer by Jeff Strom for The fundamental group of space which has both an H and a co-H structure

2011-02-26T21:22:46Z

If $X$ is co-H, then $\pi_1(X)$ must be a free group. If $X$ is H, then $\pi_1(X)$ must be abelian. The only free group that is abelian is $\mathbb{Z}$.

Argument for the first assertion: The co-H structure (and Van Kampen) gives a factorization $$ \pi_1(X)\xrightarrow{i_*} \pi_1(X)*\pi_1(X)\xrightarrow{j_*} \pi_1(X)\times\pi_1(X) $$ of the map $\Delta_*$ induced by the diagonal $\Delta:X\to X\times X$. This shows that $\pi_1(X)$ is isomorphic to a subgroup of $G = (j_*)^{-1}( \mathrm{im}(\Delta_{*}))$, which is free on the elements ${ x \bar x}$ ($x$ and $\bar x$ represent the same element $x\in\pi_1(X)$ in the two summands of $\pi_1(X)*\pi_1(X)$). Now we are done because a subgroup of a free group is free.

Answer by Jeff Strom for homotopy pushout of spaces homotopic to finite CW complexes

2011-02-27T11:44:00Z

I don't have a reference, but here is an easier argument, based, like John's, on the homotopy invariance of the homotopy pushout.

The invariance implies that you can replace the maps $i: A\to B$ and $j: A\to C$ with homotopic maps and get the same homotopy pushout, up to homotopy type. So assume they are cellular. It is easy to give the inclusions $A\hookrightarrow M_i$ and $A\hookrightarrow M_j$ finite CW structures so that $A$ is a subcomplex of each, and then the union $M_i \cup_A M_j$ inherits a finite CW structure.

Answer by Jeff Strom for Why does homotopy behave well with respect to fibrations and homology with respect to cofibrations?

2011-02-24T21:30:24Z

Fibrations are defined as target-type concepts -- they have good formal homotopy properties when you map into them, for example, if you apply the functor $\pi_n(-) = [S^n,-]$.

Now I'll subject you to my point of view on homology.

Dually, cofibrations are a domain-type concept, so they behave well when you map out of them, say if you apply (represented) cohomology: $\widetilde H^n(-;G) = [-, K(G,n)]$.

Homology is a bit of a monster: it is a covariant functor that works well with domain-type input. It is a small miracle that such functors exist at all; this is why they are hard to construct. But the answer for homology has to be: homology works well with cofibrations because we built it to work well with cofibrations, and a more informative answer to your question would depend on the construction you use.

Comment by Jeff Strom

2013-05-31T01:17:43Z

This sequence is best (from my point of view) derived from the "Mayer-Vietoris fiber sequence" $\Omega Z \to E\to X\cross Y$ associated to the (homotopy) pullback square. This is in my book.

Comment by Jeff Strom

2013-05-08T21:41:26Z

Can we express $X$ as a homotopy colimit of the various intersections and the map $f$ as a pointwise homotopy equivalence of diagrams?

Comment by Jeff Strom

2013-03-23T14:48:49Z

The analogs in higher dimensions have nonzero homology in arbitrarily high dimensions!

Comment by Jeff Strom

2013-02-19T13:57:07Z

I think this is just straightforward obstruction theory. If I'm right, this will be in Whitehead's account of obstruction theory. I'll check when I have the book and some time, if I remember to.

Comment by Jeff Strom

2012-10-24T16:20:46Z

Actually, $\pi_6(S^3)$ acts on $[S^3 \times S^3, S^3]$ and the orbit of one H-structure is all the H-structures.

Comment by Jeff Strom

2012-10-24T16:18:55Z

Isn't it conceivable that $S^3$ has several H-structures, one of which is homotopy commutative? Then you might be detecting the wrong structure.

Comment by Jeff Strom

2012-05-29T15:01:15Z

It would be nice to have a definition of "order type" here in this question.

Comment by Jeff Strom

2012-05-15T16:29:48Z

Have you tried to work out the homotopy limit of the Samelson product and see what happens?

Comment by Jeff Strom

2012-05-11T12:57:05Z

@Fernando: Since I had just posted my answer, I misunderstood your saying that $F$ was a counterexample to (*) as saying that it was a counterexample to my answer. Sorry for the confusion.

Comment by Jeff Strom

2012-05-11T12:55:23Z

My question was (explicitly) intended to be open-ended. Your example clearly shows that the stated splitting conditions are not enough to force $F$ to factor through rationalization.

Comment by Jeff Strom

2012-05-11T11:37:40Z

This is not a localization and it does not commute with cofiber sequences. For example, apply it to $S^1 \to * \to S^2$.

Comment by Jeff Strom

2012-05-09T19:32:27Z

I'd be tempted to look at the RPT approach to modeling loop spaces (developed by my advisor, S. Husseini, in the early 1960s).

Comment by Jeff Strom

2012-04-26T14:06:45Z

This is perfect!

Comment by Jeff Strom

2012-04-26T14:06:31Z

I apologize for the mangling of the classification of solid rings; fixed now, I think.

Comment by Jeff Strom

2012-03-31T00:39:43Z

And the answer to this question is that the map in question is $\alpha \wedge c$, where $\alpha$ is an admissible map for the fibration. An $H_*$-orientable fibration is one in which all admissible maps induce the same map, so it does induce $\id\wedge c$.