User dan ramras - MathOverflow

Are there general position results in singular algebraic sets?

2013-06-02T17:13:57Z

Let $X$ be a real algebraic set, and let $Y \subset X$ be its singular set. In this question I'll focus on the analytic topology, so we can just imagine that $X$ is the zero set, in $\mathbb{R}^n$, of a certain polynomial $f(x_1, \ldots, x_n)$ with real coefficients. It's a classical result that $X$ can be triangulated with $Y$ as a subcomplex. Here is my question:

If $Y$ has codimension $d$ in $X$ (i.e. $X$ has dimension $n$ as a simplicial complex and $Y$ has dimension $n-d$) and $f: M^m\to X$ is a continuous map from a closed manifold of dimension $m\leq d-1$, is $f$ necessarily homotopic, inside $X$, to a map $g: M^m\to X\setminus Y$?

I rather doubt that this is true in full generality. I'd be quite interested in counterexamples (the simpler the better!) or any partial results of this nature.

Note that if $X$ were a smooth manifold and $Y$ were a submanifold, a positive answer to the question is one of the standard consequences of Thom's transversality theorem.

The question could be asked more generally for simplicial complexes $Y\subset X$. In this generality it's clearly false: let $X = Z \vee Z$, where $\pi_1 Z \neq 0$, and let $Y$ be the wedge point. Then $\pi_1 X = \pi_1 Z * \pi_1 Z$ and if $\gamma\in \pi_1 Z$ is non-trivial, every loop representing $\gamma*\gamma$ must pass through the wedge point $Y$. Here's a second question: is there an example of this form in which $X$ is actually a real algebraic set?

Finally, I'll mention that Lemma 2.5 of this paper gives a result somewhat along the lines I'm looking for, but just deals with loops in simplicial complexes (under somewhat strong hypotheses).

Zariski tangent spaces to representation varieties

2010-06-03T21:57:24Z

In Bill Goldman's paper "The Symplectic Nature of the Fundamental Groups of Surfaces" (Advances, 54, 200-225, '84) it is stated that the "Zariski tangent space" to a representation space Hom $(\pi, G)/G$ at a representation $\rho$ is the cohomology group $H^1 (\pi; Ad\ \rho)$ . Here G is a Lie group, and Ad $\rho$ is the representation of $\pi$ on the Lie algebra of G induced by the adjoint action of G. In the context of Goldman's paper, maybe this is just meant to refer to the case when $\pi = \pi_1 S$ with $S$ a Riemann surface.

My question is: is there some sense in which this is true for other discrete groups $\Gamma$ ? In general, $Hom(\pi, G)/G$ is only a semi-algebraic set, and not a variety, so maybe the question is not really meaningful. But I'd like to know whether these first cohomology groups behave like tangent spaces in some useful way.

Here are two specific questions. I'm most interested in the case when G = U(n).

Edit: Let me emphasize that I'm really talking about the topological quotient Hom(G, U(n))/U(n), which is a reasonably nice space. Since U(n) is compact, general nonsense implies that this space is Hausdorff, and even better, it's a semi-algebraic set. So in particular, it's homeomorphic to a simplicial complex.

Say $[\rho]\in$ Hom $\left(\Gamma, U(n)\right)/U(n)$ has an open neighborhood homeomorphic to $\mathbb{R}^m$ for some $m$. Is it then true that dim $H^1 (\Gamma; Ad\ \rho) = m$ ? In other words, does the cohomology group give the "topological" dimension at "smooth" points in Hom $\left(\Gamma, U(n)\right)/U(n)$ ? When $H^1 (\Gamma; Ad\ \rho) = 0$ , a theorem of Weil (Ann. of Math. (2) 80 1964 149--157) says that $[\rho]$ is an isolated point in Hom $\left(\Gamma, U(n)\right)/U(n)$ . This is the converse statement for m=0.
If $[\rho]$ is not a smooth point in the above sense, then in any triangulation of Hom $\left(\Gamma, U(n)\right)/U(n)$ , we see that $[\rho]$ must not lie in the interior of a maximal simplex. If $\sigma$ is a maximal simplex of dimension m containing $[\rho]$ (in its boundary), is it true that dim $H^1 (\Gamma; Ad\ \rho) > m$ ? In other words, does the dimension of the "tangent space" jump up at non-smooth points?

Any ideas, references, examples, or counterexamples would be welcomed!

Answer by Dan Ramras for Where does the splitting principle come from and does it generalize

2010-04-14T20:24:18Z

I'm not sure if this is really an answer to your question, but I like to think about the splitting principle as the statement that if you want to check a formula for all bundles, it usually suffices to check it for sums of line bundles.

As Anatoly explained, this "principle" works because you can always pull back your bundle $E\to X$ so that it becomes a sum of line bundles, and moreover you can do so using a map $f: Y\to X$ that's injective on cohomology. So to check some (cohomological) formula involving $E$ in the ring $H^*(X)$ , it's enough to check it in the larger cohomology ring $H^*(Y)$ , and back in $Y$ you get to work with the sum of line bundles $f^* (E)$ .

Often the real work will come in translating a formula that works for sums of line bundles into a formula that makes sense for arbitrary bundles. In the case of the Chern Character, for example, one has to introduce the Newton polynomials for precisely this purpose.

Answer by Dan Ramras for Constructing a simplicial set homology-equivalent to a given CW complex

2012-07-03T03:26:17Z

For a 2-dimensional complex, I think the following works, but I may not have checked carefully enough. Given a finite 2-dimensional CW complex, I'll describe a finite simplicial set with the same homology. (Note: this procedure uses the full attaching maps, not just their degrees.)

First, the 1 skeleton is a graph, maybe with multiple edges and loops, but in any case it's a simplicial set once you make choices of orientations on the edges. Let's assume that all the attaching maps for 2-cells are regular edge paths (i.e. ones that linearly traces out a sequence of n edges in equal time).

First, "double" every edge in the 1-skeleton by adding a new edge with the opposite orientation, and then add a "fake" 2-simplex whose faces are these paired edges along with (the degeneration of) one of their common vertices. (There's a choice involved in adding these 2-simplices, but I don't think it matters.)

Each 2-cell in your complex is attached along a regular edge-path of length n in the original graph. In some cases, it may traverse an edge in the wrong direction (according to the orientations chosen above) and in this case we'll use the double of that edge instead. (Note that the double still has the same endpoints just in the opposite order.)

To be precise: subdivide the disk into an n-gon, and attach it according to the given regular edge path, using the double edges as explained. This really amounts to attaching n 2-simplices which all have a common (new) vertex.

If you collapse the "fake" 2-simplices down to edges, you recover the original complex. I think this projection map should at least be a homology equivalence, if not a homotopy equivalence. Actually, if there are finitely many cells then the homology isomorphism follows immediately from the Vietoris-Begle mapping theorem, because the equivalence classes we're collapsing are either points or intervals: http://en.wikipedia.org/wiki/Vietoris%E2%80%93Begle_mapping_theorem

What is known about the Yang-Mills stratification over 3-manifolds?

2012-06-19T18:41:38Z

Rade proved in his thesis (Crelle's Journal, 1992, available here: digizeitschriften.de/dms/toc/?PPN=PPN243919689) that if $E\rightarrow M$ is a $U(n)$ -bundle over a 3-manifold, then the gradient flow of the Yang-Mills functional $L$ on the space $A(E)$ of (Sobolev) connections on $E$ is well-defined and, for any starting connection $A$ , the flow line converges to a critical point (i.e. a Yang-Mills connection) that I'll call $A_\infty$ . In particular, there are no finite time singularities (i.e. no bubbling) as there would be in 4 dimensions. This means that there is a well-defined "stratification" of $A(E)$ given by partitioning $A(E)$ according to the relation $A\sim B$ if $L(A_\infty) = L(B_\infty)$ . If $t\in \mathbb{R}$ is a critical value of $L$ , I'll write $C_t$ for the associated stratum. Rade showed that the gradient flow defines a deformation retraction from $C_t$ to its subset $L^{-1} (t)$ of Yang-Mills connections.

I'm curious if anything at all is known about this stratification. For instance:

Are the subsets $C_t$ actually submanifolds? I believe this would follow from general principals if $L$ satisfied the Palais-Smale Condition C, but it does not. (However, $L$ does satisfy a version of Condition C after modding out gauge transformations; I'm not sure how helpful that is.)
If the $C_t$ are submanifolds, is anything known about their codimensions?

In 2-d Yang-Mills theory, Daskalopoulos provided detailed answers to both questions (building on ideas of Atiyah-Bott). But his arguments make heavy use of complex analytic methods, using the equivalence between Hermitian connections and complex structures in the 2-d case.

Surely one needs to be a little careful about the exact Sobolev regularity of the connections used, but I'm not terribly concerned about that (i.e. I'm happy to assume extra regularity if it helps anything).

Is the singular homology of a real algebraic set always finitely generated?

2012-05-28T05:10:38Z

Here is a precise statement of my question:

Let $p\in \mathbb{R}[x_1, \ldots, x_n]$ be a polynomial, and let $Z(p)\subset \mathbb{R}^n$ be the set of zeros of $p$ . Must the singular homology $H_i (Z(p); \mathbb{Z})$ ( $i\geq 0$ ) be finitely generated as an abelian group?

Here I really just mean the singular homology groups of this set as a topological space with the Euclidean topology.

It's an old theorem of Whitney that $Z(p)$ has finitely many connected components, so $H_0 (Z(p); \mathbb{Z})$ is finitely generated. Note that it is possible to triangulate $\mathbb{R}^n$ with $Z(p)$ as a subcomplex, so $H_i (Z(p)) = 0$ for $i>n$.

I'm guessing the answer is well-known (either a theorem or a counterexample), but I couldn't find an answer on Google or MathSciNet...

Reference request: 2-dimensional Schonflies theorem

2012-05-24T22:53:21Z

Does anyone know a reference for the 2-dimensional version of the Schoenflies theorem? To be precise, I'd like a reference for the fact that every continuous, 1-1 map $S^1\rightarrow \mathbb{R}^2$ extends to a homeomorphism $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ . The discussions of the Jordan Curve Theorem that I can remember don't prove this stronger statement.

This statement is mentioned on the Wikipedia page for the Schoenflies problem . I looked through several papers on the generalized Schoenflies problem (which requires extra hypotheses in higher dimensions to rule out things like the Alexander Horned Sphere), but no luck...

Answer by Dan Ramras for Yang-Mills and Chern-Simons functionals as Morse functions

2012-04-15T21:50:23Z

This got too long for a comment.

Atiyah and Bott showed that the Yang-Mills functional on a Riemann surface is equivariantly perfect, i.e. it's perfect for gauge-equivariant (integral) cohomology. To be a little more precise, they showed that a certain stratification (the Harder-Narasimhan stratification) of the space of connections is perfect in this sense, and Daskalopoulos showed (using Uhlenbeck compactness among other things) that this stratification does in fact agree with stable manifolds of the Yang-Mills functional. (Atiyah-Bott had conjectured this, but did not prove it in their paper. Note that Uhlenbeck's compactness theorem came just after Atiyah-Bott.)

For non-orientable surfaces, the situation is different: in some cases the YM functional is "anti-perfect" in a certain sense, and in some cases it's neither perfect nor anti-perfect. These ideas are discussed in recent work of Melissa Liu and Nan-Kuo Ho, and also in recent work of Tom Baird.

Answer by Dan Ramras for Why is the fundamental group of a compact Riemann surface not free ?

2012-04-07T06:57:59Z

I just wanted to point out that Vitali's comment, and Daniel Litt's elaboration, can be explained without any hyperbolic geometry. As they point out, once we know that an orientable surface $M^g$ of genus $g>1$ has contractible universal cover, the desired result follow by computing $H_2 (M^g)$ (or $H^2 (M^g)$ ). So, here are 3 other proofs that the universal cover $X$ of $M^g$ is contractible:

$X$ is a 2-dimensional manifold, and it is non-compact because the fiber of the universal covering is the fundamental group, which is infinite (it has infinite abelianization). Any non-compact n-manifold has $H^i (M) = 0$ for $i>n-1$ (this is Proposition 3.29 in Hatcher). So in our case, $H_i (X) = 0$ for $i>1$ and since $X$ is simply connected, $H_1 (X)=0$ also. By the Hurewicz Theorem, all the homotopy groups of $X$ must be trivial as well. Since $M^g$ is a CW complex, so is its universal cover, so Whitehead's theorem says $X$ is contractible.
Hatcher Example B.14 proves this by describing $M^g$ in terms of a graph of groups in which the maps on the edges are injective.
Topologically, the only non-compact simply connected surface is $R^2$ by the classification of (non-compact) surfaces.

Answer by Dan Ramras for Classifying space commutes with geometric realization - reference request

2012-02-20T01:19:37Z

I think you should be able to prove this roughly as follows: first consider the loop space of your construction. For nice simplicial spaces, the loop space can be calculated level-wise (see May's Geometry of Iterated Loop Spaces, for instance), and hence the loop space of your construction is homotopy equivalent to the (realization of the) simplicial space $[n] \mapsto \Omega |\overline{W} Sing_n (G)|$ . But this space is level-wise weakly equivalent to $[n] \mapsto Sing_n (G)$ , whose realization is homotopy equivalent to $G$ . So the loop space of your construction yields $G$ , and now you need to apply some form of the statement: $B\Omega X \simeq X$ .

Also, note that it's a standard fact (due to Segal's "Classifying spaces and spectral sequences" and/or May's "Classifying spaces and fibrations") that the simplicial bar construction applied directly to your topological group $G$ gives a model for $BG$ (this requires the inclusion of the identity of $G$ to be a cofibration, which is of course true for Lie groups).

Computing `$\pi_1 S^1$` using groupoids

2010-10-29T05:24:48Z

I believe it is possible to compute $\pi_1 S^1$ by applying the groupoid version of the Seifert-Van Kampen Theorem (in the version presented in May's Concise Course) to a covering of the circle by three arcs. Is there an account like this somewhere in the literature? Ideally I'd like a discussion that a student familiar with May's book would be able to read. (May doesn't take a 2-categorical approach to groupoids, and so he does not discuss the fact that a diagram of groupoids that is a point-wise equivalence induces an equivalence of colimits. This is rather important for computations.)

Edit: this last statement is false in general! I was thinking of homotopy colimits. The relevant (correct) fact appears in Ronnie Brown's book: retracts of pushouts are pushouts. This is the means by which one compares the Van Kampen theorem for the full fundamental groupoid - as in May's book - with the Van Kampen theorem for the fundamental groupoid on a set of basepoints.)

Answer by Dan Ramras for Cohomology of representation varieties

2011-10-10T23:41:54Z

If $G=U(n)$ then I know a fair amount. Letting $n$ tend to infinity, Hom$(\pi_1 M^g, U)$ is homotopy equivalent to $U^{2g} \times BU$, so stably one can write down the cohomology using standard facts about the infinite unitary group. There is also a stability range for the inclusions Hom$(\pi_1 M^g, U(n))\to$ Hom$(\pi_1 M^g, U(n+1))$ (they are $(2n-2)$-connected maps). Roughly this range is the stability range for the unitary groups minus 2, if I remember correctly.

The way to prove these things is to follow Atiyah-Bott and think about the representation space as the space of flat connections modulo the based gauge group. If this is the sort of information you're looking for, and you'd like to know more details, I can say more. I think I wrote some notes a few years ago that go through this carefully. But since you mentioned Atiyah-Bott already, you may be looking in a different direction.

If you puncture the surface, the fundamental group becomes free. Then you might be interested in Tyler Lawson's paper about simultaneous similarity of unitary matrices (Math. Proc. Camb. Phil. Soc. 2008 or http://arxiv.org/abs/0809.0466 ). Or you might be interested in keeping track of some additional structure related to the punctures, in which case some of Tom Baird's work may be of interest to you.

Answer by Dan Ramras for When is a bijective map between bundles a homeomorphism?

2011-09-30T06:57:12Z

It has been pointed out in the comments that this sort of thing cannot hold for arbitrary fiber bundles.

To follow up on euklid345's comment regarding vector bundles, there is a statement of this type for arbitrary principal bundles, assuming the appropriate definitions. There's a detailed discussion of this in my course notes, available at http://sofia.nmsu.edu/~ramras/601.html (see p. 15 of Lectures 3-5). I believe it's also somewhere in Husemoller's book.

Answer by Dan Ramras for Are evaluation maps for sections of a fiber bundle weak homotopy equivalences?

2011-06-28T16:42:10Z

In your situation, the base space (a disk) is contractible, so your smooth fiber bundle is actually diffeomorphic to a product bundle, i.e. $E = B\times p^{-1} (x)$ . (This follows, for example, by passing to the associated principal bundles - with structure group the diffeomorphism group of the fiber, say.) Now the space of sections is just the mapping space $F(B, p^{-1}(F))$ , which is homotopy equivalent to $F(*, p^{-1} (x)) = p^{-1} (x)$ since $B$ is homotopy equivalent to a point. (Here $F$ is the unbased mapping space.)

In general, as you point out, there need not be any global sections at all.

Answer by Dan Ramras for On delta complex structures of complex quasi-projective varieties

2011-06-25T15:45:20Z

There are very general triangulability results for real (semi)algebraic sets (sets cut out by inequalities of real polynomials), and even for semianalytic and subanalytic sets. Lojasiewicz has some papers from the 60s on semianalytic sets; Hironaka and Hardt also have papers on the subject; and the book Real Algebraic Geometry by Bochnak, Coste, and Roy treats at least the semialgebraic case.

Answer by Dan Ramras for nontrivial theorems with trivial proofs

2010-06-20T03:25:39Z

I think Akhil may be right. I believe Grothendieck did say something along the lines of this quote, specifically in reference to Belyi's Theorem . My recollection is that Belyi proved this theorem without knowing that Grothendieck was interested in it, and in working out his theory of Dessin D'Enfants, Grothendieck found he needed this result, but couldn't prove it. He then discovered that Belyi had given a rather elementary proof (I'll hesitate to call it trivial myself, since I recall finding it pretty clever).

If anyone has a copy of Grothendieck's Esquisse D'un Programme, maybe the specific quote is in there? I don't seem to have an English copy on my laptop, and all of Grothendieck's writing has been removed from the Grothendieck Circle's webpage per Grothendieck's request. (Interestingly, Wikipedia says this request was made in a letter to Illusie in January 2010.) I don't immediately see such a quote in the French version.

Edit: Here is the English translation of a relevant passage from Esquisse d'un Programme due to Leila Schneps and Pierre Lochak, as it appears in London Math. Soc. Lecture Notes Series vol. 242 (pp. 254-255; around page 15 on Grothendieck's typewritten manuscript):

Every finite oriented map gives rise to a projective non-singular algebraic curve defined over $\overline{\mathbb{Q}}$, and one immediately asks the question: which are the algebraic curves over $\overline{\mathbb{Q}}$ obtained in this way -- do we obtain them all, who knows? In more erudite terms, could it be true that every projective non-singular algebraic curve defined over a number field occurs as a possible "modular curve" parametrising elliptic curves equipped with a suitable rigidification? Such a supposition seemed so crazy that I was almost embarrassed to submit it to the competent people in the domain. Deligne when I consulted him found it crazy indeed, but didn't have any counterexample up his sleeve. Less than a year later, at the International Congress in Helsinki, the Soviet mathematician Bielyi announced exactly that result, with a proof of disconcerting simplicity which fit into two little pages of a letter of Deligne -- never, without a doubt, was such a deep and disconcerting result proved in so few lines!

In the form in which Bielyi states it, his result essentially says that every algebraic curve defined over a number field can be obtained as a covering of the projective line ramified over the points $0, 1$ and $\infty$. This result seems to have remained more or less unobserved. Yet it appears to me to have considerable importance. To me, its essential message is that there is a profound identity between the combinatorics of finite maps on the one hand, and the geometry of algebraic curves defined over number fields on the other. This deep result, together with the algebraic-geometric interpretation of maps, opens the door onto a new, unexplored world -- within reach of all, who pass by without seeing it.

Answer by Dan Ramras for Ideas on how to prevent a department from being shut down.

2011-04-27T21:04:19Z

Many applied mathematicians (based on no empirical evidence, I'd guess a vast majority) feel that pure mathematics is absolutely necessary because they apply pure mathematics to the real world. A university that is hostile to pure mathematics may thus find it difficult to maintain a strong status in the applied mathematics community. So I suggest that you try to find some high profile applied mathematicians who will support your cause. Hopefully some such people are reading MO these days...

Answer by Dan Ramras for On a proof of the existence of tubular neighborhoods.

2011-03-11T02:14:01Z

In the finite-dimensional setting, it's possible to construct tubular neighborhoods without anything like Godement's lemma. Many sources simply rely on a point-set topology argument that's based on the same idea as Godement's lemma (to be precise, I'm talking about the argument on p. 109 of Lang's book Differential and Riemannian Manifolds, which he says follows Godement). I'll explain another approach.

The idea is to use a Riemannian metric on the manifold $M$ , which also induces a Riemannian metric on $TM$ (viewed as a manifold in its own right). The geodesic distance then gives a (topological) metric on $TM$ . If $Y\subset M$ is a (not necessarily closed) submanifold, then a simple metric geometry argument can then be used to find a neighborhood of the zero section of $N(Y)$ (thought of as the perpendicular complement of $TY$ inside $TM$ ) on which the exponential map is injective. The key fact about the exponential map $f$ is that every point in the zero section of $N(Y)$ has a neighborhood on which $f$ is a diffeomorphism onto an open subset of $M$ . (Edit: Note that in the finite dimensional setting, $N(Y)$ is automatically a locally trivial vector bundle. This does not seem to be the case for arbitrary infinite dimensional Riemannian manifolds, as discussed here: http://mathoverflow.net/questions/17549/orthogonal-complements-in-hilbert-bundles. Hence the discussion that follows does not work in as great generality as arguments based on Godement's lemma.)

The general metric geometry fact is this:

Consider a metric space $T$ and subspaces $X, Y$ , and $D$ such that $Y \subset X$ and $Y\subset D$ . (Think: $T = TM$ , $X$ is the zero section of $TM$ , $Y$ is a submanifold of $M$ , and $D$ is the domain of the exponential map, lying inside $NY$ .) Let $f: D\to X$ be a continuous map that restricts to the identity on $Y$ (think: $f$ is the exponential map). Assume further that for each $y\in Y$ there exists $\epsilon_y >0$ such that $f$ restricted to $B_{\epsilon (y)} (y, D) = \{z\in D \,:\, d(z,y) < \epsilon(y)\}$ is a homeomorphism onto an open subset of $X$ . Then there exists a subspace $D'$ , open in $D$ , on which $f$ is injective.

Proof. For each $y\in Y$ , $f(B_{\epsilon (y)/2} (y, D))$ is open in $X$ , hence contains $B_{\epsilon'(y)} (y, X)$ , for some $\epsilon'_y < \epsilon_y/4$ (remember that $f(y) = y$ ). Now consider the inverse image $Z_y$ of $B_{\epsilon'_y} (y, X)$ under the restriction of $f$ to $B_{\epsilon (y)/2} (y, D)$ . Since $f$ is a homeomorphism when restricted to this ball, $Z_y$ is open as a subset of $D$ . Now I claim that $f$ is injective on $D' = \bigcup_{y\in Y} Z_y$ . Say $f(z_1) = f(z_2) = y_0$ with $z_1 \in Z_{y_1}$ and $z_2\in Z_{y_2}$ , and assume $\epsilon_{y_1} \geq \epsilon_{y_2}$ . Then we have \[d(z_2, y_1) \leq d(z_2, y_2) + d(y_2, y_0) + d(y_0, y_1) < \epsilon_{y_2}/2 + \epsilon'_{y_2} + \epsilon'_{y_1} \] \[< \epsilon_{y_1}/2 + \epsilon_{y_2}/4 + \epsilon_{y_1}/4 \leq \epsilon_{y_1}\] (for the second inequality, note that by definition, $y_0 = f(z_i) \in f(Z_{y_i}) \subset B_{\epsilon'_{y_i}} (y_i, X)$ for $i=1, 2$ ). So $z_2$ and $z_1$ both lie in $B_{\epsilon_{y_1}} (y_1, D)$ , and since $f$ is injective on this ball we have $z_1 = z_2$ .

This argument is useful for the construction of equivariant tubular neighborhoods in certain infinite-dimensional settings. See http://arxiv.org/abs/1006.0063; I just updated it so I guess the new version will show up tomorrow. The equivariant version of the above argument is in Proposition 2.3 or 2.4, depending on the version.

Answer by Dan Ramras for Third differential in Atiyah Hirzebruch spectral sequence

2011-04-22T21:39:21Z

Okay, I can't pass up the chance to try and be more industrious than Tyler (this is really a comment on Tyler's answer).

I'll try to explain why there's a non-zero $d_3$ differential in the AHSS for $\mathbb{R}P^2 \times \mathbb{R}P^4$ .

The K-theory of $\mathbb{R}P^{k}$ is $\mathbb{Z} \oplus \mathbb{Z}/2^k$ in dimension 0 and is trivial in dimension 1 (at least for $k$ even). Applying the Kunneth Theorem tells you that $K^0 (\mathbb{R}P^2 \times \mathbb{R}P^4) = \mathbb{Z}\oplus \mathbb{Z}/2\oplus \mathbb{Z}/4\oplus \mathbb{Z}/2$ . Comparing with the cohomology of $\mathbb{R}P^2 \times \mathbb{R}P^4$ , one sees that one factor of $\mathbb{Z}/2$ appearing on the line $y-x = 0$ in the E_2 page of the AHSS has to be killed by a differential.

After the 3rd page, all differentials coming in (or out) of the line $y-x = 0$ in the AHSS start (or end) at trivial groups. So there must be a non-zero differential on the 3rd page. It's not clear to me which one it is, but I haven't thought about the multiplicative structure.

Answer by Dan Ramras for learning $\mathbf{A}^1$-homotopy theory

2011-04-19T19:33:39Z

Dan Dugger's paper on the subject is an extremely valuable reference, since he manages to set up the foundations in a natural manner. Some familiarity with model categories is certainly needed. A version is available on his web page:

http://pages.uoregon.edu/ddugger/univ.html

Of course the long Morel-Voevodsky paper is the original reference. It has a lot of good information, although I did not find it easy going.

Answer by Dan Ramras for Why are connective spectra called "connective"?

2011-04-18T21:26:10Z

Maybe just because (-1)-connected is a mouthful? And connected would not be the right term, since that would imply trivial $\pi_0$. Other than that, I don't know.

Answer by Dan Ramras for Is there an alternative characterisation of vector bundles with vanishing characteristic classes?

2011-04-01T00:02:16Z

Since no one has mentioned it yet, let me point out one possibly interesting observation. If the base manifold $M$ is compact and has no torsion in its integral cohomology, then a vector bundle $E$ with vanishing Chern classes is stably trivial. This was pointed out to me by Robert Lipshitz. The reason is as follows: from looking at the Atiyah-Hirzebruch spectral sequence, one can see that there can't be any torsion in the complex K-theory of $M$. Looking at the Chern character, one concludes that $[E]$ must be trivial in $\widetilde{K}^0(M)$, i.e. $E$ is stably trivial.

Answer by Dan Ramras for More open problems

2011-03-29T03:27:37Z

Mark Hovey maintains a list of open problems in algebraic topology (which, as he points out, hasn't been updated in a while): http://math.wesleyan.edu/~mhovey/problems/

Answer by Dan Ramras for What are your favorite instructional counterexamples?

2011-03-28T00:59:30Z

Homotopy groups do not, in general, commute with sequential colimits, even for nice maps between nice spaces.

I just learned this beautiful example from Bill Dwyer.
Take the sequence

$S^1\stackrel{2}{\longrightarrow}S^1\stackrel{3}{\longrightarrow}S^1\stackrel{4}{\longrightarrow}\cdots.$

Here $n$ denotes the $n$ th power map on $S^1$ . Thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$ , one finds that the colimit of this sequence (in the category of topological spaces) is the quotient group $\mathbb{R}/\mathbb{Q}$ . Note that this quotient group, topologized as a quotient space of $\mathbb{R}$ by the relation $x\sim y$ if $x-y\in \mathbb{Q}$ , has the indiscrete topology. In particular, the colimit of this sequence is a contractible topological space and has trivial homotopy groups.

On the other hand, the colimit of the corresponding sequence of fundamental groups is the group $\mathbb{Q}$ (checking this is a fun exercise).

(There's something sort of odd here, because one might have guessed that $\mathbb{R}/\mathbb{Q}$ would be a model for $K(\mathbb{Q}, 1)$ , since after all $\mathbb{R}$ is a free $\mathbb{Q}$ -space. But there are no interesting open sets in the quotient and hence no chance of local triviality.)

Which torsion classes in integral cohomology are Chern classes of flat bundles?

2011-03-24T21:14:31Z

Chern-Weil theory tells us that the integral Chern classes of a flat bundle over a compact manifold (i.e. a bundle admitting a flat connection) are all torsion. Given a compact manifold $M$ whose integral cohomology contains torsion, one can then ask which (even-dimensional) torsion classes appear as the Chern classes of flat bundles. What is known about this question? I would be interested both in statements about specific manifolds and about general (non)-realizability results.

One specific thing that I know: if $S$ is a non-orientable surface, then there is a flat bundle $E\to S$ whose first Chern class is the generator of $H^2 (S; \mathbb{Z}) = \mathbb{Z}/2$ . This shows up, for example, in papers of C.-C. Melissa Liu and Nan-Kuo Ho. As Johannes pointed out in the comments, this also shows that the fundamental class of a product of surfaces can be realized by a flat bundle.

However, I suspect that for a product of 3 Klein bottles, not all the 4-dimensional torsion classes can be realized as second Chern classes of flat bundles. In fact, I think I know a proof of this if one restricts to unitary flat connections: the space of unitary representations has too few connected components.

What is the history of the Y-combinator?

2010-07-14T19:08:03Z

Inspired by the comments to this question, I wonder if someone can explain the history of the fixed point combinator (often called the Y combinator) in lambda calculus.

Where did it first appear? Was it directly inspired by the Arithmetic Fixed Point Theorem? The two are very similar in spirit.

Based on the dates of Church's introduction of lambda calculus and Goedel's incompleteness theorem, it seems to me the Arithmetic Fixed Point Theorem must have come first.

Answer by Dan Ramras for Applications of the group completion theorem

2011-03-24T01:15:43Z

I use the group completion theorem quite a lot.

For example, I used it (along with Yang-Mills theory and work of Tyler Lawson) to study the spaces Hom $(\pi_1 S, U)$ and Hom $(\pi_1 S, U)/U$ when $S$ is an aspherical surface (or a product of aspherical surfaces) and $U =$ colim $U(n)$ is the infinite unitary group. These spaces turn out to be the homotopy group completions of the monoids $\coprod_n$ Hom $(\pi_1 S, U(n))$ and $\coprod_n$ Hom $(\pi_1 S, U(n))/U(n)$ , respectively. Here the monoid structure comes from block sum of unitary matrices. This means, in particular, that both of these representation spaces are infinite loop spaces (because block sum is commutative up to coherent isomorphisms). For orientable surfaces, the picture is quite nice:

Hom $(\pi_1 S, U)/U \simeq (S^1)^{2g} \times \mathbb{C} P^\infty$ .

The first factor can be seen entirely via the determinants of representations ($g$ is the genus). The second factor gives a (non-canonical) 2-dimensional cohomology class (or line bundle) over the moduli space of representations. This should be a reflection of Goldman's symplectic form, but I don't have any idea how to prove that.

The group completion story in this case is spelled out in quite a bit of detail in my paper "Excision for deformation K-theory..." (AGT, 2007). It goes back to the basic ideas of McDuff and Segal. The applications to surface groups show up in "The stable moduli space..." (Trans. AMS, 2011). You also can find these papers on my webpage or on the arXiv.

Answer by Dan Ramras for Examples of common false beliefs in mathematics.

2010-09-27T01:41:30Z

If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.

A better, but still false, version:

If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.

(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)

The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G={1}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.

Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.

Answer by Dan Ramras for First Chern class of a flat line bundle

2011-02-21T00:52:41Z

It's probably not exactly what you want (in particular, they're dealing with real bundles and the Stiefel Whitney classes), but something sort of close is discussed in the appendix to

MR2003827 (2004h:53116) Ho, Nan-Kuo(3-TRNT); Liu, Chiu-Chu Melissa(1-HRV) Connected components of the space of surface group representations. Int. Math. Res. Not. 2003, no. 44, 2359–2372. 53D30 (22F05 57N05)

Answer by Dan Ramras for Are infinite dimensional constructions needed to prove finite dimensional results?

2011-02-17T20:37:49Z

There are certain results about representation spaces of surface groups that I only know how to prove using Yang-Mills theory, i.e. by studying the infinite-dimensional space of connections (or rather a Sobolev space completion thereof) on a principal bundle. In particular, one can compute the homotopy groups of the real algebraic varieties Hom$(\pi_1 M^g, U(n))$ through a range (roughly the stable range for $\pi_* U(n)$) by identifying this space with the space of flat connections on $M\times U(n)$ modulo based gauge equivalence. Yang-Mills theory (in particular, work of Atiyah-Bott, Uhlenbeck, Daskalopoulos, and Rade) can be used to show that the space of flat connections is highly connected. Roughly speaking, it's the minimum critical set for the Yang-Mills functional, and the other critical sets have high Morse index. This means the the space of flat connections modulo based gauge equivalence is reasonably close to the classifying space of the gauge group. The latter is (a component of) Map$_* (M^g, BU(n))$, whose homotopy groups are computable in a range of dimensions.

I suppose there could be some finite-dimensional approach to these results, but I'm not aware of one. (I should say that Hom$(\pi_1 M^g, U(n))$ is connected, and this does have a finite-dimensional proof due to Nan-Kuo Ho and C.-C. Melissa Liu, using some Lie theory and facts about quasi-Hamiltonian moment maps.)

Comment by Dan Ramras

2013-06-07T04:40:33Z

Well, you could think of $2^T$ as the set, or space, of maps from $T$ to a discrete two point space, with the compact-open topology. Of course this is not very interesting if $T$ is connected, and maybe still not very interesting in general.

Comment by Dan Ramras

2013-06-06T20:03:50Z

You have to create a meta account separately, I think.

Comment by Dan Ramras

2013-06-03T01:12:58Z

Ben: Yeah, fair enough. Sean Lawton and I were able to generalize the Gomez-Pettet-Souto result to similar sort of statement about $\pi_2$ which seems to serve as a replacement for an honest general position result. I'll edit the question to mention this once I get our draft posted on my webpage.

Comment by Dan Ramras

2013-06-02T23:36:12Z

And I'm definitely interested in seeing irreducible examples.

Comment by Dan Ramras

2013-06-02T23:35:27Z

Sorry, that should say $f(x_1, \ldots, x_n)^2$ not $f(x_1, \ldots, x_n)2$.

Comment by Dan Ramras

2013-06-02T23:34:15Z

Sorry for being slow, but can you make your second paragraph more explicit? If $X=Z(f(x_1,\ldots,x_n))$ (the zero set of a polynomial $f$ in $n$ variables) and we assume that $0\in X$, then the wedge $X\vee X$ (with $0$ as the basepoint) can be realized as the zero set of the polynomial in $2n$ variables $(f(x_1,\ldots ,x_n)2+y^2_1+\cdots +y^2_n))(f(y_1,\ldots,y_n)^2+x^2_1+\cdots+x^2_n)$. What if $0$ is not in $X$? Then this construction just gives a disjoint union of two copies of $X$ instead of the wedge. Maybe there's some change of variables that fixes this?

Comment by Dan Ramras

2013-05-30T22:38:53Z

Thanks Sean! These examples are helpful.

Comment by Dan Ramras

2013-05-30T22:35:18Z

Thanks for the reference, Misha. Section 6 looks very interesting, and appears to be making explicit some issues I've been confused about for a long time... I'll have to look at it carefully. The question was really motivated by some work I did a couple of years ago, giving upper bounds on the dimension (as a CW complex) of the moduli space of $U(n)$ representations of crystallographic groups.

Comment by Dan Ramras

2013-05-10T02:29:55Z

Gottlieb's paper is also useful if you want to deal with the based gauge group, which one often does.

Comment by Dan Ramras

2013-05-10T02:28:53Z

The result is originally due to Daniel Gottlieb, but people often forget this or aren't aware. Anyway, all the details are in his paper "Applications of bundle map theory" which appears to be on his webpage: google.com/…

Comment by Dan Ramras

2013-05-08T06:19:31Z

My guess is that this is related to the Asterisque volume by May and Kriz that talks about motives, but that is just a guess. There's a version of the volume here: math.uiuc.edu/K-theory/0061

Comment by Dan Ramras

2013-05-05T19:58:13Z

Teena Gerhardt and her collaborators have done a number of computations for rings like Z[x]/(x^2) in which the answers are given (essentially) in terms of K(Z). So this would be in the direction of your last sentence.

Comment by Dan Ramras

2013-03-29T16:00:56Z

By saying $J$ is cofinal in $I$, I just meant that for each $i$ in $I$, there is an element $j$ in $J$ such that $i$ is contained in $j$. Here I'm thinking of $I$ and $J$ as posets.

Comment by Dan Ramras

2013-03-28T18:27:59Z

When $S$ is finite then $S$ itself is a terminal element of $I$ and you can just take $J$ to be the trivial category with this one element. If $S$ is countable, then we can assume $S =\mathbb{N}$ and let $J$ be the subcategory consisting of the sets $\{1, \ldots, n\}$. Then $J$ is cofinal (and directed), so the homotopy colimit restricted to $J$ is the same as the full homotopy colimit (up to homotopy); this is proven somewhere in Bousfield and Kan.

Comment by Dan Ramras

2013-03-28T18:27:43Z

The morphisms in your category $I$ are simply inclusions? By directed, do you mean isomorphic to the poset of natural numbers? Since the union of two finite sets is finite, your category $I$ is always a directed poset in the usual sense (for any two objects, there is a third that is greater than both) but if you want to talk about mapping telescopes I guess you need a poset isomorphic to the natural numbers under the usual ordering.