User johannes ebert - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:08:50Z http://mathoverflow.net/feeds/user/9928 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128222/how-we-do-actually-compute-the-topological-index-in-atiyah-singer/128236#128236 Answer by Johannes Ebert for How we do actually compute the topological index in Atiyah-Singer? Johannes Ebert 2013-04-21T11:18:06Z 2013-04-21T11:30:35Z <p>''If we give certain boundary conditions on the domain (for example, the unit circle with a point removed, a triangle, a square, etc), can we extend the K-theory proof to this case?''</p> <p>At least not in general. The first problem is to formulate the appropriate boundary value problem to have the Fredholm property, which is already a highly nontrivial task, especially if you wish to consider pseudodifferential operators. First, you would like to consider a local boundary condition, i.e. one of the form $P(s|_{\partial W})=0$, where $P$ is a vector bundle map. There is an early paper by Atiyah and Bott on an extension of the $K$-theoretic index formula for differential operators with such local boundary conditions. It turns out that there is a topological obstruction to the existence of a (local) boundary condition. For geometrically interesting operators as the signature and the Atiyah-Singer-Dirac operator, this obstruction is nonzero, and so a more sophisticated type of boundary conditions is required. The index theorem for operators with these boundary conditions was proven by Atiyah-Patodi-Singer in a sequence of three magnificient papers. The theorem is in terms of cohomology; and there is no formulation in terms of $K$-theory alone (at least not without Kasparov theory). I recommend you to ignore manifolds with boundary for the time being, and also all questions of singularities etc.</p> <p>"It does not appear obvious to me how one may compute the Todd class or the Chern character in practical cases."</p> <p>It is indeed not obvious at all how to compute these things. As a general rule, the analytical index is more difficult to compute than the topological one, because the amount of data is reduced. And both indices become harder to compute the less symmetric the manifold is. On a generic manifold, it is very difficult to compute characteristic classes (just as it is very difficult to compute anything in mathematics in a generic situation).</p> <p>The strategy for calcluations is to start from a very symmetric situation, such as $CP^n$ or spheres, where the characteristic classes (and the analytic indices of the standard operators) are computable to other manifolds. Hirzebruch was a true master in computing these things. In his book ''Topological methods in algebraic geometry'' there is, for example, a formula how to calculate the characteristic numbers of complete intersection varieties in $CP^n$ and in a series of papers with Borel, he calculates characteristic classes of homogeneous spaces. </p> <p>There is one exception, where the analytical index is easier to compute, namely if some reason (like the Weitzenböck formula) forces the operator to be invertible and so to have index zero.</p> http://mathoverflow.net/questions/55365/counterexamples-in-algebraic-topology Counterexamples in algebraic topology? Johannes Ebert 2011-02-14T00:14:46Z 2013-03-29T02:19:02Z <p>In this thread</p> <p><a href="http://mathoverflow.net/questions/53036/books-you-would-like-to-read-if-somebody-would-just-write-them/53059#53059" rel="nofollow">http://mathoverflow.net/questions/53036/books-you-would-like-to-read-if-somebody-would-just-write-them/53059#53059</a>,</p> <p>I expressed my desire for a book with the title "(Counter)examples in Algebraic Topology". My reason for doing so was that while the abstract formalism of algebraic topology is very well-explained in many textbooks and while most graduate students are fond of the general machinery, the study of examples is somehow neglected. I am looking for examples that explain why certain hypotheses are necessary for theorems to hold. The books by Hatcher and Bredon contain some interesting stuff in this direction, and there is Neil Strickland's bestiary, which is mainly focussed on positive knowledge.</p> <p>To convey an idea of what I am after, here are a few examples from my private ''counterexamples in algebraic topology'' list. Some are surprising, some less so.</p> <ol> <li><p>The abelianization of $SL_2 (Z)$ is $Z/12$, the map $BSL_2(Z) \to BZ/12$ is a homology equivalence to a simple space. But it is not a Quillen plus construction, since the the homotopy fibre is $BF_2$ (free group on $2$ generators), hence not acyclic. See <a href="http://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz/43741#43741" rel="nofollow">http://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz/43741#43741</a>.</p></li> <li><p>Maps $f:X \to Y$ which are homology equivalences, the homotopy groups are abstractly isomorphic, but though, $f$ is not a homotopy equivalence (a number of examples has been given in the answers to these questions: <a href="http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409" rel="nofollow">http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409</a>, <a href="http://mathoverflow.net/questions/4665/are-there-pairs-of-highly-connected-finite-cw-complexes-with-the-same-homotopy-gr" rel="nofollow">http://mathoverflow.net/questions/4665/are-there-pairs-of-highly-connected-finite-cw-complexes-with-the-same-homotopy-gr</a>).</p></li> <li><p>Self-maps of simply-connected spaces $X$ which are the identity on homotopy, but not on homology (let $X=K(Z;2) \times K(Z;4)$, $u:K(Z;2) \to K(Z;4)$ be the cup square, and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). There are also self-maps of finite simply connected complexes that are the identity on homology, but not on homotopy, see Diarmuid Crowleys answer to <a href="http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/55609#55609" rel="nofollow">http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/55609#55609</a></p></li> <li><p>The stabilization map $B \Sigma_{\infty} \to B \Sigma_{\infty}$ induces a bijection on free homotopy classes $[X, B \Sigma_{\infty}]$ for each finite CW space $X$. However, it is not a homotopy equivalence (not a $\pi_1$-isomorphism).</p></li> <li><p>The fibration $S^1 \to B \mathbb{Q} \to B \mathbb{Q}/\mathbb{Z}$ is classified by a map $f:B \mathbb{Q}/\mathbb{Z} \to CP^{\infty}$, which can be assumed to be a fibration with fibre $B \mathbb{Q}$. Now let $X_n$ be the preimage of the n-skeleton of $CP^{\infty}$. Using the Leray-Serre spectral sequence, we can compute the integral homology of $X_n$ and, by the universal coefficient theorem, the homology of field coefficients. It turns out that this is finitely generated for any field, and so we can define the Euler characteristic in dependence of the field. It is not independent of the field in this case (the reason is of course that the integral homology of $X_n$ is not finitely generated).</p></li> <li><p>The compact Lie groups $U(n)$ and $S^1 \times SU(n)$ are diffeomorphic, their classifying spaces have isomorphic cohomology rings and homotopy groups, but the classifying spaces are not homotopy equivalent (look at Steenrod operations). </p></li> </ol> <p>Question: Which examples of spaces and maps of a similar flavour do you know and want to share with the other MO users?</p> <p>To focus this question, I suggest to stay in the realm of algebraic topology proper. In other words:</p> <ol> <li><p>The properties in question should be homotopy invariant properties of spaces/maps. This includes of course fibre bundles, viewed as maps to certain classifying spaces.</p></li> <li><p>Let us talk about spaces of the homotopy type of CW complexes, to avoid that a certain property fails for point-set topological reasons.</p></li> <li><p>This excludes the kind of examples from the famous book "Counterexamples in Topology".</p></li> <li><p>The examples should not be "counterexamples in group theory" in disguise. Any ugly example of a discrete group $G$ gives an equally ugly example of a space $BG$. Same applies to rings via Eilenberg Mac-Lane spectra.</p></li> <li><p>I prefer examples from unstable homotopy theory.</p></li> </ol> <p>To get started, here are some questions whose answer I do not know:</p> <ol> <li><p>Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. EDIT: Appropriate Moore spaces do the job, see Eric Wofseys answer.</p></li> <li><p>Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? EDIT: the answer has been given by Tyler Lawson, see below.</p></li> <li><p>Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs? EDIT: without the finiteness assumptions, this question was ridiculously simple.</p></li> <li><p>Do you know examples of fibrations $F \to E \to X$, such that the integral homology of all three spaces is finitely generated (so that the Euler numbers are defined) and such that the Euler number is not multiplicative, i.e. $\chi(E) \neq \chi(F) \chi(X)$? Remark: is $X$ is assumed to be simply-connected, then the Euler number is multiplicative (absolutely standard). Likewise, if $X$ is a finite CW complex and $F$ is of finite homological type (less standard, but a not so hard exercise). So any counterexample would have to be of infinite type. The above fibration $BSL_2 (Z) \to BZ/12$ is a counterexample away from the primes $2,3$, but do you know one that does the job in all characteristics?. Of course, the ordinary Euler number is the wrong concept here.</p></li> </ol> <p>I am looking forward for your answers.</p> <p>EDIT: so far, I have gotten great answers, but mostly for the specific questions I asked. My intention was to create a larger list of counterexamples. So, feel free to mention your favorite strange spaces and maps.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/125437#125437 Answer by Johannes Ebert for What are your favorite instructional counterexamples? Johannes Ebert 2013-03-24T05:44:05Z 2013-03-24T05:44:05Z <p>The Warsaw circle $W$ <a href="http://en.wikipedia.org/wiki/Continuum_%28topology%29" rel="nofollow">http://en.wikipedia.org/wiki/Continuum_%28topology%29</a> is a counterexample for quite a number of too naive statements.</p> <p>Some observations: $W$ is weakly contractible (because a map from a locally path connected space cannot ''go over the bad point''). There is a projection map $g:W \to S^1$ onto the usual circle. The point-preimages of $g$ are either points or, for a single point on $S^1$, a closed interval. </p> <p>Thus the assumptions of the Vietoris-Begle mapping theorem hold for $g$, proving that $g$ induces an isomorphism in Cech cohomology. Thus the Cech cohomology of $W$ is that of $S^1$, but it has the singular homology of a point, by Hurewicz. These observations imply:</p> <ol> <li><p>A map with contractible point-inverses does not need to be a weak homotopy equivalence, even if both, source and target, are compact metric spaces. Assuming that the base and the preimages are finite CW complexes does not help.</p></li> <li><p>The Vietoris-Begle Theorem is false for singular cohomology (in particular, the wikipedia version of that Theorem is not quite correct).</p></li> <li><p>$W$ does not have the homotopy type of a CW complex (since it is not contractible).</p></li> <li><p>Even though the map $g$ is trivial on fundamental groups, it does not lift to the universal cover $p: \mathbb{R} \to S^1$, because $g$ cannot be nullhomotopic. Thus the assumption of local path connectivity in the lifting theorem is necessary.</p></li> </ol> http://mathoverflow.net/questions/125412/is-there-any-algebraic-function-that-has-a-specific-relation-to-transcendental-on/125429#125429 Answer by Johannes Ebert for is there any algebraic function that has a specific relation to transcendental one? Johannes Ebert 2013-03-24T04:55:55Z 2013-03-24T04:55:55Z <p>Yes: $g(x):= \sum_{k \geq 0} X^{k!}$ is transcendental; and so is $F(x)=g(x) + \frac{1}{1-x}$. Put $M=2$ and $A(x)= \frac{2}{1-x}$.</p> http://mathoverflow.net/questions/109604/is-every-group-completion-map-an-acyclic-map Is every ''group-completion'' map an acyclic map? Johannes Ebert 2012-10-14T11:51:18Z 2013-03-21T22:15:33Z <p>I start with a longer discussion which will result in a precise version of the question. A am puzzled about an issue with the Quillen plus construction. I have seen outstanding experts being confused about this point. There are the following different ways of calling a map $f:X \to Y$ a homology equivalence:</p> <ol> <li>$f_*:H_*(X;\mathbb{Z}) \to H_*(Y;\mathbb{Z})$ is an isomorphism ("weak homology equivalence").</li> <li>For each abelian system of local coefficients $A$ on $Y$ ($\pi_1 (Y)$ acts through an abelian group), the induced map $H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("strong homology equivalence").</li> <li>For each system of local coefficients $A$ on $Y$, the induced map $H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("acyclic map").</li> </ol> <p>The third condition is equivalent to each of</p> <p>3'. The homotopy fibres of $f$ are acyclic. 3'''. $f$ is can be identified with the Quillen plus construction.</p> <p>EDIT: Before, I included the statement ''3''. $f$ is weak homology equivalence, $\pi_1 (f)$ is epi and $ker(\pi_1 (f))$ is perfect.'' This is false (does not imply the other two conditions); in my answer to <a href="http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409" rel="nofollow">http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409</a> I gave an example of a weak homology equivalence that is even an isomorphism on $\pi_1$, but whose homotopy fibre is not acyclic. END EDIT</p> <p>The implications $(3)\Rightarrow (2)\Rightarrow ( 1)$ hold. If all components of $Y$ are simply connected, then all these notions coincide; if $\pi_1 (Y)$ is abelian (each component), then $(2)\Rightarrow(3)$. In that case, $\pi_1 (X)$ is quasiperfect (i.e., its commutator subgroup is perfect). If $\pi_1 (Y)$ is nonabelian, then $(2)$ does not imply $(3)$ (take the inclusion of the basepoint into a noncontractible acyclic space). Even if $Y$ is an infinite loop space, a weak homology equivalence does not have to be strong: Take $X=BSL_2 (Z)$, $Y=Z/12$. The abelianization of $SL_2 (Z)$ is $Z/12$, and the map $SL_2 (Z) \to Z/12$ is a weak homology equivalence. The kernel, however, is a free group on two generators.</p> <p>Now, many cases of such maps arise in the process of ''group completion''. Here are some examples</p> <ol> <li>$X=K_0 (R) \times BGL (R)$ for a ring and $Y=\Omega B (\coprod_{P} B Aut (P))$ ($P$ ranges over all finitely generated projective $R$-modules). The commutator subgroup is perfect due to the Whitehead lemma.</li> <li>$X=\mathbb{Z} \times B \Sigma_{\infty}$; $Y=QS^0$. The alternating groups are perfect.</li> <li>$X=\mathbb{Z} \times B \Gamma_{\infty}$ (the stable mapping class group); $Y$ the Madsen-Weiss infinite loop space. Here there is no problem, $\Gamma_g$ is perfect for large $g$.</li> <li>$X=\mathbb{Z} \times B Out(F_{\infty})$ (outer automorphisms of the free group), $Y=Q S^0$. Galatius proves that this is a weak homology equivalence and he states implicitly this map is a strong homology equivalence.</li> </ol> <p>I explain why I am interested: if you only look at the homology of $X$ and $Y$, this is only an aesthetical question. I want to take homotopy fibres and as explained above, the distinction is essential and a mistake here can ruin any argument.</p> <p>In all these cases, there is a topological monoid $M$ (for example $\coprod_{P} B Aut (P)$) and $X$ is the limit $M_{\infty}$ obtained by multiplying with a fixed element. There is an identification $\Omega BM$ with $Y$ that results from geometric arguments and does not play a role in this discussion.</p> <p>There is a map $\phi:M_{\infty} \to \Omega BM$, which is the subject of the ''group-completion theorem'', see the paper "Homology fibrations and the ''group completion'' theorem" by McDuff-Segal. The map arises from letting $M$ act on $M_{\infty}$ and forming the Borel construction $EM \times_M M_{\infty} \to BM$. The point-preimage is $M_{\infty}$, the space $EM \times_M M_{\infty}$ is contractible and so the homotopy fibre is $\Omega BM$. $\phi$ is the ''geometric-fibre-to-homotopy-fibre'' map.</p> <p>What Segal and McDuff prove is that if the action is by weak homology equivalences, then $\phi$ is a weak homology equivalence. This is what is typically used to established the above results. To prove that 1,2,3 are strong homology equivalences, one can invoke an extra argument which is specific to each case.</p> <p>Now, in McDuff-Segal, I find the claim (Remark 2) that their methods give that $M_{\infty} \to \Omega BM$ is a strong homology equivalence and I want to understand this.</p> <p>I convinced myself that the whole argument goes through with strong homology equivalences (and the corresponding notion of "strong homology fibration"). Proposition 2 loc.cit. then has the assumption that $M$ acts on $M_{\infty}$ by strong homology equivalences (one needs the notion of homology equivalences one wants to prove in the end - which I find plausible). </p> <p>This amounts, say in example 4, to prove that the stable stabilization map $B Out(F_{\infty}) \to B Out(F_{\infty})$ is a homology equivalence in the strong sense. For "weak homology equivalence", one invokes the usual homology stability theorem (Hatcher-Vogtmann-Wahl). But it seems that for the map being a strong homology equivalence, one needs a stronger homological stability result. I can imagine how the homological stability arguments can be modified to include abelian coefficient system, but that is not a satisfying solution.</p> <p>Here are, finally, some questions: </p> <ol> <li><p>McDuff and Segal refer to ''argument by Wagoner'' in his paper ''Delooping classifying spaces in algebraic K-Theory''. I am unable to find an argument in Wagoners paper that proves under general assumptions quasiperfectness. What argument do McDuff and Segal refer to?</p></li> <li><p>If $M$ is a topological monoid and if $M_{\infty} \to \Omega BM$ is a weak homology equivalence, is it always a strong homology equivalence?</p></li> <li><p>If not, do you know a counterexample?</p></li> <li><p>If 2 is not true, is there a useful general criterion to prove that the group completion map is acyclic (besides the trivial case $H_1 (M_{\infty})=0$ and besides proving quasiperfectness of $\pi_1 (M_{\infty})$ by hands).</p></li> </ol> <p>A related, but not central question:</p> <ol> <li>What are good counterexamples to the ''group-completion'' theorem in general that explain why the hypothesis is essential?</li> </ol> http://mathoverflow.net/questions/123624/nearby-homomorphisms-from-compact-lie-groups-are-conjugate/123823#123823 Answer by Johannes Ebert for Nearby homomorphisms from compact Lie groups are conjugate Johannes Ebert 2013-03-06T23:34:53Z 2013-03-06T23:43:16Z <p>Here is a proof sketch using cohomological ideas. The argument is in four main steps: </p> <p>I. General theory of families of Lie algebra homomorphisms.</p> <p>II. The case of a semisimple $G$.</p> <p>III. The case of a torus (here is a major gap in my argument)</p> <p>IV. combining both cases.</p> <p>A preliminary observation: if $H$ is the full linear group of a complex vector space, then the result is well-known, because up to conjugacy, a homomorphism $G \to H$ is given by its character; and the set of characters is a discrete subspace of the space of all smooth maps $G \to \mathbb{C}$.</p> <p>I. For arbitrary Lie algebras, $Hom_{Lie -alg} (\mathfrak{g},\mathfrak{h})$ is a real algebraic variety and thus it is locally path- connected. Therefore, nearby homomorphisms can be connected by smooth families of homomorphisms (I am not entirely sure whether this is true, but it seems so).</p> <p>Now consider a smooth family $f_t$, $t \in \mathbb{R}$, of Lie algebra homomorphisms. We study the problem of finding $h: \mathbb{R} \to H$ such that $f_t (X) = Ad (h(t)) f_0 (X)$ holds for all $t$ and $X \in \mathfrak{g}$. If $f_t$ is the derivative of a smooth family of group homomorphisms $\phi_t$, then $h(t)$ conjugates $\phi_0$ to $\phi_t$ and thus solves the original problem.</p> <p>Let $F_t$ be the derivative of $f_t$ with respect to $t$. Differentiating the equation $[f_t X,f_t Y]=f_t [X,Y]$ shows that $F_t\in Hom (\mathfrak{g},\mathfrak{h})$ satisfies $F_t ([X,Y])= [F_t (X);Y]-[F_t (Y);X]$. This means that $F_t$ is a $1$-cocycle in the Chevalley-Eilenberg complex for $H^{\ast}(\mathfrak{g};f_t)$. By the cohomology I mean cohomology of $\mathfrak{g}$ with coefficients in $\mathfrak{h}$, viewed as a $\mathfrak{g}$-module via $f_t$.</p> <p>We can consider the collection of all Chevalley-Eilenberg complexes $C^{\ast} (\mathfrak{g},f_t)$ as a complex of vector bundles on the real line; denote the vector bundles by $C^{\ast}(\mathfrak{g},f)$. The derivatives $F_t$ are a smooth family of $1$-cocycles and $[F_t]$ is a family of cohomology classes, smooth in a certain sense. I say that $[F_t]$ is uniformly trivial if there is a smooth family $H_t$ of $0$-cochains such that $[f_t (X);H_t]=F_t (X)$ for all $t$ and all $X \in \mathfrak{g}$ (this means that $d H_t =F_t$, but in a ''uniform way'').</p> <p>Suppose that the cohomology class $[F_t]$ is uniformly trivial. Then</p> <p>$$f_t (X) = \int_{0}^{t} F_s (X) ds = - \int_{0}^{1}[H_s;f_s (X)] ds;$$</p> <p>in other words $f_t (X)$ solves the ODE $\frac{d}{dt} f_t (X) = - [H_t;f_t(X)]$ with initial value $f_0$. Another solution of the same ODE is $Ad (h(t)) f_0(X)$, where $h(t) \in H$ solves $\frac{d}{dt} h(t)= H_t$. So $f_t$ is conjugate to $f_0$. Vice versa, if $Ad (h(t)) f_0(X)$, then $[F_t]$ is uniformly trivial. </p> <p>If $f_t$ is the derivative of a group homomorphism $G \to H$ and $G$ is compact, then pointwise triviality ($[F_t]=0$ for each $t$) implies uniform triviality. This is by the preliminary observation, which implies that $d_0:C^0 (\mathfrak{g},f) \to C^1 (\mathfrak{g},f)$ has constant rank and so its image is a vector bundle (pass to the complexification of $\mathfrak{h}$, which is unproblematic as we are only interested in the dimension of the invariant subspace). Thus we can pick a smooth $r: im (d_0) \to C^0 (\mathfrak{g},f)$ with $d_0 r = id$. Choosing $H_t:= r (F_t)$ solves the problem. Thus we arrive at</p> <p>THEOREM: ''If $f_t: \mathfrak{g} \to \mathfrak{h}$ is a family of homomorphisms of Lie algebras and $H$ a Lie group with Lie algebra $\mathfrak{h}$, then there is a smooth map $h: \mathbb{R} \to H$ with $f_t = Ad (h(t))f_0$ if and only if the obstruction cocycle $[F_t]$ is uniformly trivial.''</p> <p>ADDENDUM: ''If $G$ is a compact Lie group with Lie algebra $\mathfrak{g}$ and if $f_t$ is the derivative of a smooth family of homomorphisms $G \to H$, then pointwise triviality of $[F_t]$ implies uiform triviality.''</p> <p>II.</p> <p>Assume $G$ is semisimple. For each representation $V$ of $\mathfrak{g}$, we have an isomorphism $H^{\ast} (\mathfrak{g};V) \cong H^{\ast} (\mathfrak{g};V^{\mathfrak{g}})$, because of the compactness of $G$. But $H^1 (\mathfrak{g})=0$ since $G$ is semisimple, and so the cohomology class $[F_t]$ is zero, and by the addendum, it is uniformly trivial. Thus by the theorem, nearby homomorphisms are conjugate if $G$ is semisimple.</p> <p>III. </p> <p>Assume $G=T$ is a torus (sketch). Let $V$ be the universal cover (equal to $\mathfrak{t}$) and $\Gamma \subset V$ be the kernel; this is a lattice. Smooth families $f_t:\mathfrak{t} \to \mathfrak{h}$ are in bijection with smooth families $\psi_t: V \to H$ and induce families of group homomorphisms $g_t: \Gamma \to H$. As Misha indicates, there is a parallel obstruction theory for such families; with an obstruction in $H^{1}_{group}(\Gamma;\mathfrak{h})$. Consult Weil's paper quoted in Mishas answer.</p> <p>There is the Van Est isomorphism $H_{Lie}^{\ast} (\mathfrak{t},\mathfrak{h}) \cong H_{smooth} (V,\mathfrak{h})$ to smooth group cohomology and furthermore a restriction $H_{smooth} (V,\mathfrak{h}) \to H^{\ast}_{group}(\Gamma; \mathfrak{h})$; this latter map is an isomorphism. This isomorphism should map the corresponding obstructions onto each other (this is the part of the argument where I do not know the details).</p> <p>So a family of group homomorphisms $V \to H$ is constant up to conjugacy iff the restriction to the lattice $\Gamma$ is constant up to conjugacy. If the family $V \to H$ is the universal cover of a family $T \to H$, then the restriction to $\Gamma$ is constant; thus $T \to H$ is constant up to conjugacy.</p> <p>IV. </p> <p>Consider an arbitrary compact $G$. Without loss of generality, we can pass to a finite cover and thus assume $G=T \times K$, $T$ a torus and $K$ semisimple. Consider a family of group homomorphisms $\phi_t:G \to H$, with Lie algebra maps $f_t$ and obstruction cocycle $F_t$ as above. By the solution of the problem for $T$, the restriction $F_t|_{\mathfrak{t}}$ is uniformly trivial. But by the Künneth formula, the restriction $H^1 (\mathfrak{g} ) \cong H^1 (\mathfrak{k})\oplus H^1 (\mathfrak{t})\to H^1 (\mathfrak{t})$ is an isomorphism. Therefore, $[F_t]$ is trivial and thus uniformly trivial, again by the addendum.</p> <p>Afterthought: It is probably better to study the whole question in the context of smooth cohomology. A family $\phi_t:G \to H$ should give an obstruction class in $H^{1}_{smooth} (G; \mathfrak{h})$. If $G$ is compact, this space is trivial by invariant integration.</p> http://mathoverflow.net/questions/58131/parallelizability-of-the-milnors-exotic-spheres-in-dimension-7/58140#58140 Answer by Johannes Ebert for Parallelizability of the Milnor's exotic spheres in dimension 7 Johannes Ebert 2011-03-11T10:48:05Z 2013-03-04T21:14:14Z <p>A much more general result is true.</p> <p>Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma$ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$.</p> <p>It says that exotic spheres cannot be distinguished by looking at the tangent bundle. This result is one of the hidden gems of the golden age of topology and the proof invokes the whole plethora of topology of the 1950s.</p> <p>The argument can be recollected from the old literature, but I do not know a coherent reference.</p> <p>To start with, there are several invariants of the tangent bundles that do not depend on the smooth structure. Let $\Sigma$ be a homotopy sphere. Then:</p> <ol> <li>the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (Gauss-Bonnet, relatively easy).</li> <li>$T \Sigma^n \oplus \mathbb{R}$ is trivial. This is a deep result by Kervaire and Milnor (not in the Annals paper, but a small note published before). $T \Sigma \oplus \mathbb{R}$ is given by an element of $\pi_{n-1} (O)$, which is known, by Bott periodicity, to be either $Z$, $0$ or $Z/2$. The $Z$ groups are detected by the Pontrjagin class, which has to vanish by Hirzebruchs signature formula because the sphere evidently has signature $0$. In the $Z/2$ case, the argument is more delicate. Essentially, the normal spherical fibration of a manifold does not depend on the smooth structure. Since the normal fibration of the standard sphere is trivial, so is that of $T \Sigma$. The process of associating to a vector bundle its spherical fibration is the J-homomorphism which is injective in these dimensions by Adams' $J(X)$ paper.</li> </ol> <p>Now look at the homotopy sequence of the fibration $O(n)\to O(n+1) \to S^n$, i.e. the piece</p> <p>$$\mathbb{Z}=\pi_n (S^n) \to \pi_{n-1} (O(n)) \to \pi_{n-1} (O(n+1)) = \pi_{n-1} (O).$$</p> <p>It is known that $TS^n$ (for the standard smooth structure) is the image of the generator of $\pi_n (S^n)$ (not hard, see Steenrods book). By the above deep result, $T \Sigma$ lies in the kernel of $\pi_{n-1}(O(n)) \to \pi_{n-1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n)$. The image of $Z \to \pi_{n-1} (O(n)$ can be computed. It is $Z$ if $k$ is even (using the Euler class), it is $0$ if $n=1,3,7$ (follows directly from Adams' result on the parallelizability of the standard spheres) and it is $Z/2$ in the remaining cases. You can find the (not so hard, but clever) argument for the last assertion in Levine's lectures on homotopy spheres (which can be viewed as the sequel to the Kervaire-Milnor paper).</p> <p>How to proceed? If $n=1,3,7$, it follows that $T \Sigma$ is trivial, as $TS^n$. If $n$ is even, then the kernel of $\pi_{n-1}(O(n))\to \pi_{n-1}(S^{n+1})$ is detected by the Euler class, and by Gauss-Bonnet, this characterizes $T \Sigma$.</p> <p>It remains the case of odd $n$ apart from the "Adams dimensions". One has to argue that in these dimensions, $T \Sigma$ is nontrivial. In the introduction to his Hopf invariant paper, Adams attributes to Dold the result ''$T \Sigma$ parallel implies that $\Sigma$ (and hence $S^n$) is an H-space''. But he (Adams) proved that is not the case $n=1,3,7$. Adams does not give a reference for Dolds result, but in his answer to this question (and the subsequent comments), John Klein sketches a proof that looks like a 1950s argument.</p> http://mathoverflow.net/questions/21024/what-is-the-exterior-derivative-intuitively/56005#56005 Answer by Johannes Ebert for What is the exterior derivative intuitively? Johannes Ebert 2011-02-19T18:52:06Z 2013-02-21T08:53:42Z <p>The exterior derivative is the unique (sequence of) linear map $d: \mathcal{A}^p (M) \to \mathcal{A}^{p+1}$, such that the following axioms hold:</p> <ol> <li>for a function $f$, $df$ is the total differential.</li> <li>For any function $f$ and any differential form $a$, the Leibniz rule $d(fa)= df \wedge a + f da$ holds.</li> <li>For any diffeomorphism $\phi: M \to N$, you have $\phi^{\ast} \circ d = d \circ \phi^{\ast}$.</li> </ol> <p>I think that 3 is more natural or at least easier to motivate than the usual $dd=0$. But both properties are really equivalent.</p> <p>Proof (of uniqueness): 2. implies locality, i.e. the value of $d a$ at a point $x \in M$ only depends on the value of $a$ in a neighborhood of $x$. This, together with the axiom 3, shows that it is enough to consider $M =\mathbb{R}^n$.</p> <p>The group $\mathbb{R}^n$ acts by translations on $\mathbb{R}^n$. By axiom 3, for any translation-invariant form $a$ on $\mathbb{R}^n$, the form $da$ is again translation-invariant. </p> <p>On the other hand, each nonzero $\lambda \in \mathbb{R}$ gives rise to the diffeomorphism $h_{\lambda}:x \mapsto \lambda x$ of $\mathbb{R}^n$. It is easy to check that it acts on translation-invariant $p$-forms by multiplication with $\lambda^p$. Thus for any translation-invariant $p$-form $a$, you get</p> <p>$$\lambda^p d a = d (\lambda^p a) = d (h_{\lambda}^{\ast} a ) = h_{\lambda}^{\ast} d a = \lambda^{p+1} da,$$</p> <p>which implies that any translation-invariant form is closed. Finally, note that any $p$-form on $\mathbb{R}^n$ can be written as a linear combination of translation-invariant form, with coefficients in $C^{\infty}(\mathbb{R}^n)$ (a basis for the translation-invariant forms is formed by the usual elements $dx_{i_1} \wedge \ldots \wedge x_{i_p}$). </p> <p>From axioms 1 and 2, you now conclude that $d$ must be the exterior derivative that you knew before. This, of course, implies all the other properties of $d$.</p> http://mathoverflow.net/questions/120184/how-to-write-down-explictly-the-isomorphism-of-two-finite-dimensional-representat/120192#120192 Answer by Johannes Ebert for How to write down explictly the isomorphism of two finite dimensional representation of compact groups? Johannes Ebert 2013-01-29T10:03:52Z 2013-02-21T08:51:12Z <p>I know this fact from a homological algebra background, so forgive me if I am using a language that is foreign to you.</p> <p>Easy Lemma: Let If $(X,d)$ be an exact chain complex of vector spaces over a field of characteristic $0$ $K$ and $e:X \to X$ a contraction, i.e. $de +ed =c$, with $0 \neq c \in K$. Then $d+e :\bigoplus X_{2j} \to \bigoplus X_{2j-1}$ is an isomorphism.</p> <p>If $e$ and $d$ are explicit, this is reasonably explicit. The appropriate chain complex for your problem is constructed as follows. Fix a vector space $V$ of finite dimension, write $E^p := \Lambda^{p} V^{\ast}$ and $S^q := Sym^q V^{\ast}$. Let $R^{\ast} := \bigoplus_{p,q} E^p \otimes S^q$. This is a graded commutative algebra if you give $E^p$ the degree $p$ and $S^q$ the degree $2q$.</p> <p>There are canonical, mutually inverse, isomorphism $d:E^1 \to S^1$ and $e:S^1 \to E^1$. Extend $d$ to all of $R$ by the property $d(xy)= (dx)y + (-1)^{deg(x) } x (dy)$ and the requirement that $d(S^q)=0$. Do the same with $e$ (but here $e(E^p)=0$). The formulas $d^2=e^2=0$ hold.</p> <p>Explicit formulas are</p> <p>$$d(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = \sum_{i=1}^{p} (-1)^{i-1} (v_1 \wedge v_{i-1} \wedge v_{i+1} \ldots v_p \otimes (dv_i w_1 \ldots w_q))$$</p> <p>and</p> <p>$$e(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = (-1)^p \sum_{i=1}^{q} (ew_i \wedge v_1 \wedge \ldots v_p \otimes ( w_1 \ldots w_{i-1} w_{i+1} \ldots w_q)).$$</p> <p>Now on the piece $E^p \otimes S^q$, the formula $de+ed=p+q$ holds, as you prove using the product formulae and induction. The sequence (maps given by $d$)</p> <p>$$0\to E^n \to E^{n-1} \otimes S ^1 \to \ldots E^1 \otimes S^{n-1} \to S^n \to 0.$$</p> <p>is a chain complex and $e$ is a contraction, as in the above easy lemma. The sum $d+e$ is your desired isomorphism.</p> <p>Remark: the construction is completely natural and therefore $GL(V)$-equivariant.</p> http://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers/120361#120361 Answer by Johannes Ebert for Equivariant Cohomology for actions with finite stabilizers Johannes Ebert 2013-01-30T21:56:22Z 2013-01-30T22:02:55Z <p>The argument in the quoted paper is a bit too sketchy. An actual proof will have two parts:</p> <ol> <li><p>Let $f:X \to Y$ be map. Assume that for each $y \in Y$, $\tilde{H}^{\ast} (f^{-1}(y);\mathbb{Q})=0$ (this is what the phrase ''$\mathbb{Q}$-acyclic'' means). Find conditions that guarantee that $f$ induces an isomorphism in cohomology. Essentially, one needs to provide assumptions that the Leray spectral sequence of the map is as nice as claimed in Marks answer.</p></li> <li><p>Provide conditions for group action of a topological group $G$ on a space $X$ such that $EG \times_G X \to X/G$ has the properties found in 1 (and the answer in 1 depends on the example you want to consider).</p></li> </ol> <p>ad 1. Just having finite stabilizers is not enough. Consider the action of $G=\mathbb{Z}$ on $X=S^1$ by an irrational rotation. Then $EG \times_G S^1 \simeq S^1 \times S^1$. The quotient space is an uncountable set with the trivial topology. Because each map into this space is continuous, it is contractible and has trivial (Cech or singular) cohomology. </p> <p>One needs criteria that show that a map $f:X \to Y$ with contractible/acyclic/$\mathbb{Q}$-acyclic fibres is itself a weak equivalence/acyclic/$\mathbb{Q}$-acyclic. Having contractible fibres does not suffice: look at the identity map $[0,1]^{\delta} \to [0,1]$, where $\delta$ means ''discrete topology''.</p> <p>Such problems are solved by what I would call ''fibre theorems''. You wish to know how close the natural map $f^{-1}(y) \to hofib_f (y)$ from the point-preimages to the homotopy fibres is to be a weak homotopy equivalence. What one needs is a suitable local condition and then a local-to-global result. Many technical results are of this type: ''fibre bundles are Serre-fibrations'', ''local Hurewicz fibrations are Hurewicz fibrations'', Dold-Thom's result on quasifibrations, the Vietoris-begle mapping theorem, Quillens Theorems A and B and McDuff-Segals formulation of the ''group-completion'' theorem. A variation of the latter solves your problem:</p> <p>Assume, as above, that for each $y \in Y$, $f^{-1}(y)$ is $\mathbb{Q}$-acyclic. Assume moreover that $Y$ is Hausdorff and for each $y \in U \subset Y$, there is $y \in V \subset U$ ($U$ and $V$ open) such that $V$ is contractible and $f^{-1} (V)$ is $\mathbb{Q}$-acyclic (let us call these sets ''good''). Then, I claim, $f$ induces an isomorphism in rational cohomology.</p> <p>Proof: Let $P:=\coprod_U U$, where $U$ runs through all good sets. This is a topological poset: $(U,x) \leq (V,y)$ iff $x=y$ and $U \subset V$. Take the nerve $|N_{\bullet} P|$ and consider the map $g:|N_{\bullet} P| \to Y$. It has contractible point-preimages: the preimage of $x$ is the poset of all good sets containing $x$, ordered by inclusion and the assumption says that this has contractible realization. In their recent paper arXiv:1201.3527, Galatius and Randal-Williams prove a lemma that under the present assumptions, $g$ is a Serre fibration with contractible fibres, in particular a weak homotopy equivalence.</p> <p>Apply the same argument to get a poset $Q$, formed not from good subsets of $Y$, but their preimages in $X$. Thus up to weak equivalence, we can replace $f:X \to Y$ by $|N_{\bullet} Q| \to |N_{\bullet} P|$. By assumption, on the $p$th simplicial stage, the map $N_p Q \to N_p P$ is a rational homology equivalence (because this is just a disjoint union of the maps $f^{-1}(U) \to U$). The spectral sequence of simplicial spaces gives then that $|N_{\bullet} Q| \to |N_{\bullet} P|$ is a rational homology equivalence.</p> <p>ad 2. There are several instances where the above argument can be applied. Example 1: let $G$ be a discrete group that acts simplicially on a simplicial complex, with finite stabilizers. Example 2: let $G$ be a Lie group that acts properly on a smooth manifold with finite stabilizers. By some standard results, there is a $G$-invariant Riemannian metric and $G$-equivariant tubular neighborhoods of orbits $Gx\cong G/G_x$. The image of such a tubular neighborhood $U \subset M$ is a contractible subset of $M/G$. The preimage of $U/G$ under $EG \times_G M \to M/G$ is $EG \times_G U \simeq EG \times_G G/G_x \simeq BG_x$, which is $\mathbb{Q}$-acyclic.</p> http://mathoverflow.net/questions/119803/relation-between-groups-and-classifying-spaces/119823#119823 Answer by Johannes Ebert for Relation between groups and classifying spaces Johannes Ebert 2013-01-25T09:42:59Z 2013-01-28T09:52:53Z <p>What you should take as a model is the homotopy quotient $EG \times_G BG$. From the homotopy sequence of the fibration $EG \times_G BG \to BG$ (projection on first factor), you get that $EG \times_G BG$ is aspherical and a short exact sequence </p> <p>$$1 \to \pi_1 (BG) \to \pi_1 (EG \times_G BG) \to \pi_1 (BG) \to 1.$$</p> <p>Since the action of $G$ on $BG$ has a fixed point, this sequence (which is completely natural in $G$) is split. The induced action of the base $G$ on the fibre $G$ is by conjugation. So: $EG \times_G BG \cong BK$; $K = G \ltimes_{ad} G$, with the conjugation action. </p> <p>EDIT: Tom Goodwillie pointed out that $G \ltimes_{ad} G \cong G \times G$.</p> http://mathoverflow.net/questions/44866/third-stable-homotopy-group-of-spheres-via-geometry third stable homotopy group of spheres via geometry? Johannes Ebert 2010-11-04T19:56:50Z 2013-01-25T09:26:47Z <p>It is ''well-known'' that the third stable homotopy group of spheres is cyclic of order $24$. It is also ''well-known'' that the quaternionic Hopf map $\nu:S^7 \to S^4$, an $S^3$-bundle, suspends to a generator of $\pi_8 (S^5)=\pi_{3}^{st}$. It is even better known that the complex Hopf map $\eta:S^3 \to S^2$ suspends to a generator of $\pi_4 (S^3) = \pi_{1}^{st} = Z/2$. For this, there is a reasonably elementary argument, see e.g. Bredon, Topology and Geometry, page 465 f:</p> <ol> <li>By the long exact sequence, $\pi_3 (S^2)=Z$, generated by $\eta$.</li> <li>By Freudenthal, $\pi_3 (S^2) \to \pi_4 (S^3) = \pi_{1}^{st}$ is surjective.</li> <li>Because $Sq^2: H^2(CP^2;F_2) \to H^4(CP^2;F_2)$ is nonzero, the order of $\eta$ in $\pi_{1}^{st}$ is at least $2$ (the relation between these things is that $\eta$ is the attaching map for the $4$-cell of $CP^2$).</li> <li>By a direct construction, $2\eta$ is stably nullhomotopic. Essentially, $\eta g = r \eta$, where $r,g$ are the complex conjugations on $S^2=CP^1$ and $S^3 \subset C^2$. $g$ is homotopic to the identity, $\eta=r\eta$. The degree of $r$ is $-1$, so after suspension (but not before), composition with $r$ becomes taking the additive inverse. Therefore $\eta=-\eta$ in the stable stem.</li> </ol> <p>My question is whether one can mimick substantial parts of this argument for $\nu$. Here is what I already know and what not:</p> <ol> <li>There is a short exact sequence $0 \to Z \to \pi_7 (S^4) \to \pi_6 (S^3) \to 0$ that can be split by the Hopf invariant. Thus $\nu$ generates a free summand.</li> <li>is the same argument as for $\eta$. </li> <li>using the Steenrod operations mod $2$ and mod $3$ on $HP^2$, I can see that the order of $\nu$ in $\pi_{3}^{st}$ is at least $6$.</li> <li>this is a complete mystery to me and certainly to others-:)). How can I bring $24$ in via geometry? How do I relate the quaternions and $24$? What one sees immediately is that one has to be careful when talking about conjugations in the quaternionic setting, in order to avoid proving the false result ''$2 \nu=0 \in \pi_{3}^{st}$''.</li> </ol> <p>I know that this result goes back to Serre, but I cannot find a detailed computation in his papers and it seems that the calculation using the Postnikov-tower and the Serre spectral sequence is a bit lengthy. There are three other approaches I know but they are much less elementary: Adams spectral sequence, J-homomorphism (enough to show that the order of $\nu$ is $24$), framed bordism (supported by things like Rochlin's theorem and Hirzebruch's signature formula).</p> <p>Any idea? P.S.: if there is a similar argument for the octonionic Hopf fibration $S^{15} \to S^8$ (the stable order is 240), that would be really great. </p> http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119502#119502 Answer by Johannes Ebert for On the Universality of the Riemann zeta-function Johannes Ebert 2013-01-21T19:12:02Z 2013-01-21T19:12:02Z <p>No, it cannot be true. Suppose the theorem holds for an appropriate translation of the annulus $K:={1/2 \leq z \leq 1}$ and the function $f(z)=1/z$. Then there is a sequence $f_n$ of holomorphic functions on the closed unit disc, continuous on the boundary, such that $f_n (z) \to 1/z$ uniformly in $K$ (because $\zeta$ does not have a pole away from $1$). By the maximum principle, the convergence extends to the whole unit disc, contradicting the identity theorem, because $f$ is not holomorphic on the disc.</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/116986#116986 Answer by Johannes Ebert for Awfully sophisticated proof for simple facts Johannes Ebert 2012-12-21T19:13:22Z 2012-12-21T19:13:22Z <p>The fundamental theorem of algebra holds because:</p> <ol> <li><p>For each degree $n$ normed polynomial $p$ over the complex numbers, there is an $n \times n$ matrix $A$ with characteristic polynomial $\pm p$.</p></li> <li><p>We show that $A$ has an eigenvector.</p></li> <li><p>We may assume that $0$ is not an eigenvalue of $A$ (otherwise $p(0)=0$), so $A \in GL_n (\mathbb{C})$.</p></li> <li><p>$A$ induces a self-map $f_A$ of $CP^{n-1}$, and the eigenspaces of $A$ correspond to the fixed points of $f_A$; so we need to show that $A$ has a fixed point.</p></li> <li><p>As $GL_n (\mathbb{C})$ is connected, $f_A$ is homotopic to the identity (this does not depend on the fundamental theorem of algebra; if $A \in GL_n (\mathbb{C})$, then $z 1 + (1-z )A$ is invertible except for a finite number of values of $z$; and the complement of a finite set of points of the plane is path-connected (this follows, for example, from the transversality theorem).</p></li> <li><p>The Lefschetz number of the identity on $CP^{n-1}$ equals $n\neq 0$, thus the Lefschetz number of $f_A$ is not zero.</p></li> <li><p>By the Lefschetz fixed point theorem, $f_A$ has a fixed point.</p></li> </ol> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/42523#42523 Answer by Johannes Ebert for Awfully sophisticated proof for simple facts Johannes Ebert 2010-10-17T17:22:40Z 2012-12-20T06:26:27Z <p>The Gauß-Bonnet theorem and the Riemann-Roch theorem for Riemann surfaces have both reasonably elementary proofs. Of course, they follow from the general Atiyah-Singer index theorem.</p> http://mathoverflow.net/questions/116814/torsion-in-cohomology-of-smooth-manifolds/116819#116819 Answer by Johannes Ebert for Torsion in cohomology of smooth manifolds Johannes Ebert 2012-12-19T21:01:31Z 2012-12-19T21:07:48Z <p>1) You can pick the homology below the middle dimension quite arbitrarily. More precisely, given a finite complex $K$ and a number $n$, there exists a closed, parallelizable $2n$-dimensional manifold $M$ and an $n$-connected map $f:M \to K$. You begin with a constant map $S^{2n} \to K$ and make it more and more connected by surgeries.</p> <p>2) as you said, a necessary condition for the homology of a manifold is Poincare duality. If you have a finite complex $X$ that satisfies Poincare duality, the question of whether there is a smooth manifold homotopy equivalent to $X$ is a basic problem in surgery theory. If $X$ is simply connected, this has largely been solved by Browder. The answer is that if $X$ is odd-dimensional, there is such a manifold; and if the dimension is divisible by $4$, there is a manifold precisely if there is a stable vector bundle on $X$ such that the Hirzebruch signature formula holds with this bundle. In dimensions $2,6,10,\ldots$, there is a subtle problem with the "Kervaire invariant". And: I forgot to say that the dimension has to be at least $5$. For nonsimplyconnected complexes, Wall gave at least a theoretical answer.</p> <p>3) Poincare duality for integral coefficients (and closed oriented $M$) says that $H_i (M) \cong H^{n-i}(M)$. The universal coefficient theorem implies that the torsion subgroups (for each space with finitely generated homology) are $T H^{i+1} = T H_i$ (abstract isomorphism). Combined, these two results tie the torsion subgroups of cohomology together.</p> <p>4) I would not say that compactly supported cohomology contains more information than ordinary cohomology - they contain different information. With rational coefficients, you have an isomorphism $H^i(M) \cong (H^{n-i}_{c}(M))^{\ast}$; the other isomorphism $H^{i}_{c}(M) \cong (H^{n-i}(M))^{\ast}$ holds iff the cohomology vector space are finitely generated.</p> http://mathoverflow.net/questions/113952/connection-transformation-formula-degree-3-cech-cohomology/115655#115655 Answer by Johannes Ebert for Connection Transformation Formula; Degree 3 Cech Cohomology Johannes Ebert 2012-12-07T00:07:07Z 2012-12-07T00:07:07Z <p>Let $P \to M$ be a $G$ principal bundle with connection $\theta$ and let $g:M \to Z/G)$ a function with values in the center of $G$. Let $\omega$ be the left-invariant Maurer-Cartan form on $G$. Then $g$ induces, by left-multiplication, as bundle automorphism of $P$ and the formula $g^{\ast}\theta - \theta = g^{\ast} \omega$ holds. The identity $\omega = g^{-1} dg$ holds for linear groups.</p> <p>Proof: Assume $P=M \times G$. You can write the connection as $\theta=pr_{G}^{\ast} \omega + pr_{M}^{\ast} \eta$, where $\omega$ is the left-invariant Maurer-Cartan form on $G$ and $\eta$ is a $\mathfrak{g}$-valued form on $M$. A $G$-valued function $g:M \to G$ induces a map $\mu_g :P \to P$ by right-multiplication. Then $\mu_{g}^{\ast} \theta - \theta= \mu_{g}^{\ast} (pr_{G}^{\ast} \omega )- pr_{G}^{\ast} \omega$, because $pr_M \circ \mu_g = pr_M$.</p> <p>Furthermore, $pr_G \circ \mu_g (m,h) = hg(m)$, so $pr_G \circ \mu_g$ is the product of the two $G$-valued functions $pr_G$ and $g \circ pr_M$ on $P$.</p> <p>Next you have to invoke the fact: if $f_0,f_1: P \to G$ are two functions, then $(f_0 f_1)^{\ast} \omega = Ad (f_1)^{-1}f_{0}^{\ast} \omega + f_{1}^{\ast} \omega$. This is because $(f_0 f_1)^{-1} d(f_0 f_1 )= f_{1}^{-1} f_{0}^{-1} df_0 f_1 + f_{1}^{-1} df_1=Ad (f_1)^{-1} f_{0}^{\ast} \omega + f_{1}^{\ast} \omega$. This is for linear groups, and holds in general as any Lie group is isogenous to a linear one.</p> <p>Therefore $\mu_{g}^{\ast} (pr_{G}^{\ast} \omega )=pr_{G}^{\ast} \omega + g^{\ast} \omega$, as $g$ was assumed to be central. </p> http://mathoverflow.net/questions/114676/what-are-some-applications-of-teichmuller-theory/114682#114682 Answer by Johannes Ebert for What are some Applications of Teichmüller Theory? Johannes Ebert 2012-11-27T17:47:45Z 2012-11-27T20:05:48Z <p>The proof of the Mumford conjecture by Madsen and Weiss made essential use of Teichm\"uller theory. This becomes especially clear if you state the conjecture as being about the cohomology of the space $\mathfrak{M}_g$, Riemanns moduli space of genus $g$ complex curves. Defining this space does not require Teichmueller theory, it can be done in a purely algebro-geometric way.</p> <p>The first step is that $H_{\ast} (\mathfrak{M}_g;\mathbb{Q})$ is the same as the rational homology of the mapping class group $\Gamma_g$. This uses Teichm\"ullers theorem that Teichm\"ullers space $\mathcal{T}_g$ is homeomorphic to a euclidean space (being a contractible manifold would be enough).</p> <p>The second step is that $B \Gamma_g$ is homotopy equivalent to $B Diff (\Sigma_g)$, the classifying space of the diffeomorphism group. This is a result by Earle and Eells, which uses Teichm\"ullers theorem as well, albeit not so essentially, because there is a purely topological proof of this result as well.</p> <p>The Madsen-Weiss theorem then computes the homology of $B Diff (\Sigma_g)$ in a range of degrees; this is differential topology/homotopy theory and not related to Teichm\"uller theory.</p> <p>Older results on the homology of $\mathfrak{M}_g$ (or the Deligne-Mumford compactification) are very often based on step 1 as well. Some relevant names are Harer, Harer-Zagier, Arbarello-Cornalba- and others.</p> http://mathoverflow.net/questions/113993/infinite-suspension-functor-on-the-pontryagin-thom-construction/114105#114105 Answer by Johannes Ebert for (Infinite) Suspension Functor on the Pontryagin-Thom Construction Johannes Ebert 2012-11-21T21:50:34Z 2012-11-22T08:24:59Z <p>Answer to the quick subquestion: yes, this is functorial, in the following sense. The set of bordism classes of framed submanifolds of $M$ of codimension $n$ is functorial. Let $f:N \to M$ be a map, make it transverse to a given framed submanifold and take the preimage. The Pontrjagin-Thom argument proves that this is well-defined on the level of bordism classes and homotopy classes.</p> <p>Answer to the second question: it is not clear in which sense you consider the suspension of a manifold as a manifold. However, the set of pointed homotopy classes $[\Sigma M_+; S^{n+1} ]$ is in bijection with the set of bordism classes of all framed compact codimension $n+1$ submanifolds of $\mathbb{R} \times M$. The suspension $[M;S^n] \to [\Sigma M_+; S^{n+1} ]$ takes a submanifold of $M$ and considers it as a submanifold of $\mathbb{R} \times M$, with the product framing and of one codimension higher. </p> <p>EDIT: in connection with Pontrjagin-Thom constructions, you should always talk about pointed maps. If the basepoint lies in $M$, then $[M;S^n]$ corresponds to submanifolds of $M$ avoiding the basepoint. If you want to talk about all submanifolds or free homotopy classes, you need $[M_+;S^n]$ (extra basepoint added).</p> http://mathoverflow.net/questions/112546/index-of-a-differential-operator-between-trivial-bundles/112609#112609 Answer by Johannes Ebert for Index of a differential operator between trivial bundles. Johannes Ebert 2012-11-16T19:21:02Z 2012-11-16T19:21:02Z <p>The result is wrong; the case of a point as base manifold creates counterexamples. Here is a less trivial construction in dimension $2$:</p> <p>Let $M$ be a manifold and $V \to M$ be any vector bundle. There is an elliptic differential operator $D$ of order $2$ on $V$, which is self-adjoint and has thus index $0$: take a connection $\nabla$ on $V$ and put $D=\nabla^{\ast} \nabla$ (this is a Laplace type operator).</p> <p>Now let $M= T^2$ and let $W \to T^2$ be a holomorphic line bundle of degree $1$. By Riemann-Roch, the operator $\bar{\partial}_W$ has index $1$; and it goes from sections of $W$ to sections of $W$, since the canonical line bundle of a torus is trivial. Therefore one can form the composite $P:=(\bar{\partial}_W)^2$, and $P$ has index $2$.</p> <p>Now let $V$ be a complex vector bundle such that $V \oplus W$ is trivial; with the operator $D$ constructed above. Consider the operator $D \oplus P$; this is an order $2$ elliptic operator on the trivial vector bundle over a parallelizable manifold and has index $2$.</p> <p>I do not see how to produce an order $1$ operator of index $1$, though.</p> <p>The vanishing theorems in Aityah-Singer, IoEO III, are quite optimal. My construction does not work in odd dimensions; and it is clear that the resulting trivial vector bundle has dimension at least $2$. If the dimension of the trivial vector bundle is too small, each (pseudo)differerential operator will have index $0$, as proven by Atiyah-Singer.</p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112077#112077 Answer by Johannes Ebert for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? Johannes Ebert 2012-11-11T14:59:49Z 2012-11-11T15:31:28Z <p>I can think of several versions, besides those that have been mentioned in the earlier answers:</p> <ol> <li>You can in fact construct $B GL_n (\mathbb{C})$ as a manifold, but of course an infinite-dimensional one. Start with a countably dimensional Hilbert space $H$. Look at the Stiefel manifold $V_n (H)$ of linear embeddings $\mathbb{C}^n \to H$. Being an open subset of $H^n$, it is a secound-countable Hilbert manifold. It can be proven directly that $V_n (H)$ is contractible and that the quotient $V_n (H) \to V_n (H)/CL_n (\mathbb{C})$ is a principal bundle. The proof for the second fact is more or less the same as in the finite-dimensional case, the first fact in proven in an Eilenberg-swindly way.</li> </ol> <p>Now second-countable Hilbert manifolds are a particularly simple type of infinite dimensional manifolds. They have smooth partitions of unity, and as a consequence the proof of the de Rham theorem (for example the one given in Bredon' book) can be carried out without any substantial change. </p> <p>The theory of connections on principal bundle works in the same way for Hilbert manifolds as base space (if the fibre is a finite-dimensional Lie group). So you get a Chern-Weil homomorphism in the universal case.</p> <p>If you replace $GL_n (\mathbb{C})$ by any closed subgroup $G$, then $V_n (H) \to V_n (H)/G$ is a Hilbert manifold model for $BG$; and the same arguments as before work.</p> <ol> <li>There exist a simplicial set model for $BG$, classifying $G$-bundles with connection. The set of $p$-simplices is the set of all triples $(P,\pi,\omega)$, where $\pi:P \to \Delta^p$ is a smooth $G$ principal bundle and $\omega$ a connection $1$-form on $P$. To turn it into a set (and to make the simplicial structure precise), you take those $P$ with $P \subset \Delta^p \times \mathbb{R}^{\infty}$ (as a manifold).</li> </ol> <p>By the ordinary Chern-Weil construction, you get a simplicial differential form on this simplicial set. What do I mean by this? Observe that forms on the standard simplices assemble to a simplicial d.g.a: $q \mapsto \mathcal{A}^{\ast} (\Delta^q)$. For a simplicial set $X_{\bullet}$, you look at the set of simplicial set maps $X_{\bullet} \to \mathcal{A}^q (\Delta^{\bullet})$; which is a vector space, and for varying $q$ gives a d.g.a.; which by definition is the simplicial de Rham complex.</p> <p>There are two things to be proven here: that the simplicial set I described is indeed $BG$ and that the simplicial de Rham complex computes the real cohomology. The second one you find in the book ''Rational homotopy theory'' by Felix, Halperin, Thomas. For the first part, I do not have a reference; this is folklore.</p> http://mathoverflow.net/questions/111232/triviality-of-associated-bundles/111264#111264 Answer by Johannes Ebert for Triviality of Associated Bundles Johannes Ebert 2012-11-02T09:49:12Z 2012-11-02T09:49:12Z <p>Theorem: ''Let $G$ be a compact, connected Lie group and $f: G \to U(n)$ a group homomorphism such that for each principal bundle $P \to M$ on a manifold, the induced vector bundle $P \times_{G,f} \mathbb{C}^n$ is a trivial vector bundle. Then $f$ is the constant homomorphism.''</p> <p>Proof: ''For a given $k$, there exists a compact manifold $M$ and a map $M \to BG$ that is $k$-connected, $dim (M) \geq 2k+1$. This is manufactured using surgery below middle dimensions. Applying this to the assumption, you get that $f$ induces the trivial map on cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$ of any degree.</p> <p>Now assume $f$ is zero on real cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$. By Chern-Weil theory, $H^{\ast}(BG) \cong Sym^{\ast}(\mathfrak{g})^G$, the algebra of Ad-invariant symmetric polynomials on the Lie algebra. There is a symmetric polynomial of degree $2$ on $\mathfrak{u}(n)$ that is nowhere zero: take an invariant scalar product. Therefore, the assumption implies that $f$ has to be zero on the Lie algebra level; hence $f$ is constant on the unit component of $G$.''</p> <p>I think this is true for nonconnected $G$ and believe the argument is similar to the one by Chris Gerig and myself to this question:</p> <p><a href="http://mathoverflow.net/questions/64688" rel="nofollow">http://mathoverflow.net/questions/64688</a></p> <p>But I do not have time to think this through right now.</p> http://mathoverflow.net/questions/111059/relative-de-rham-cohomologies/111063#111063 Answer by Johannes Ebert for Relative De Rham cohomologies Johannes Ebert 2012-10-30T09:46:02Z 2012-10-30T09:46:02Z <p>A chain map $\Theta$ from the Godbillon theory to the Bott-Tu version is given by $\omega \mapsto (\omega,0)$ (note that is a chain map only on $\Omega^{p} (M;N)_{G}$). I claim that this induces an isomorphism on cohomology. A couple of special cases is obvious: if $N=\emptyset$, then both theories agree with absolute de Rham theory. If $N \to M$ is a homotopy equivalence, both theories are trivial by long exact sequences and homotopy invariance of the absolute theory.</p> <p>For the general case, pick a tubular neighborhood $U$ of $N$. You get short exact sequences of chain complexes (in both cases)</p> <p>$$0\to \Omega (M;N) \to \Omega(U;N) \oplus \Omega (M-N) \to \Omega (U-N) \to 0$$</p> <p>(exactness is checked by means of a partition of unity), and $\Theta$ compares the both short exact sequences. The associated (Mayer-Vietoris) exact sequence and the $5$-lemma concludes the proof.</p> http://mathoverflow.net/questions/110987/interesting-results-for-open-riemann-surfaces/111044#111044 Answer by Johannes Ebert for Interesting results for open Riemann surfaces Johannes Ebert 2012-10-30T00:32:30Z 2012-10-30T00:32:30Z <p>The Riemann uniformization theorem.</p> http://mathoverflow.net/questions/110812/what-kind-of-spectral-sequences-come-from-double-complexes/110820#110820 Answer by Johannes Ebert for What kind of spectral sequences come from double complexes? Johannes Ebert 2012-10-27T11:25:17Z 2012-10-29T18:27:44Z <p>There are two different ways to understand the question: </p> <ol> <li><p>If I see an abstract spectral seqeunce, is there a double complex such that its spectral sequence is isomorphic to the given spectral sequence? I do not have an answer to that question and, to be honest, do not believe it is an interesting question.</p></li> <li><p>For wich set of names ''$XY$''; the $XY$-spectral sequence can be derived from a double complex?</p></li> </ol> <p>The answer is that, as a general rule (it might have exceptions), all $XY$-spectral sequences whose $E_2$-terms and $E_{\infty}$ terms are purely homological can be derived from filtered complexes; and most of them in fact from double complexes.</p> <p>Examples:</p> <ol> <li><p>The spectral sequence of a simplicial space (Segal; ''Classfying spaces and spectral sequences'') can be reformulated using a double complex (a simplicial space $X_{\bullet}$ gives rise to a simplicial chain complex $C_{\ast} X_{\bullet}$ and thus a double complex. </p></li> <li><p>The Serre spectral sequence is a special case of the above; a direct construction using a double complex was given by A. Dress, ''Zur Spectralsequenz von Faserungen''.</p></li> <li><p>Special cases of 2. include the Lyndon-Hochschild-Serre spectral sequ. for group extensions; special cases of 1. include the Bousfield-Kan spectral sequ. of a homotopy colimit and some others.</p></li> <li><p>The Eilenberg-Moore spectral sequence comes from a double complex.</p></li> <li><p>Purely algebraic versions: Grothendieck-spectral sequence. Probably the spectral sequence of a Lie algebra extension fits into here. The Van Est spectral sequence for Lie algebra cohomology also comes from a double complex.</p></li> </ol> <p>The Bockstein spectral sequence is a purely homological construction, it can be derived from a filtered complex; but it does not seem to stem from a double complex. Other counterexamples are the typical spectral sequence of stable homotopy theory (Atiyah-Hirzebruch, Adams spectral sequence): they cannot be derived from filtered complexes. In fact, if $E$ is a generalized homology theory, you cannot write $E_{\ast} (X)$ of a space in a sensible way as the homology of a chain complex functorially associated with $X$.</p> http://mathoverflow.net/questions/110552/is-this-a-folk-theorem-about-analytic-functions-of-a-complex-variable/110575#110575 Answer by Johannes Ebert for Is this a "folk theorem" about analytic functions of a complex variable? Johannes Ebert 2012-10-24T19:07:29Z 2012-10-24T19:07:29Z <p>You are probably misunderstanding the following folk theorem: If $D$ is the convergence disc of a power series converging to $f$, then there must be some singularity of $f$ on $\partial D$. In other words, you cannot continue $f$ analytically onto a larger disc. A counterexample that is more explicit than quids example is the power series expansion of $\sqrt{1+z}$ around $z=0$, which has convergence radius $1$. The singularity at $z=-1$ is not a pole.</p> <p>A hint to the proof: if $D \subset U$ is a disc in the domain of definition of a holomorphic function $f$, then the Taylor expansion around the midpoint of $D$ converges in $D$.</p> http://mathoverflow.net/questions/97449/computing-the-euler-characteristic-of-the-complex-projective-plane-using-differen/109623#109623 Answer by Johannes Ebert for Computing the Euler characteristic of the complex projective plane using differential topology Johannes Ebert 2012-10-14T15:59:15Z 2012-10-14T15:59:15Z <p>Take a $3 \times 3$ complex diagonal matrix $A$ with distinct nonzero diagonal entries. The 1-parameter subgroup $exp(At)$ acts on $CP^2$; the fixed points are the lines in $C^3$ containing eigenvectors of $A$. There are $3$ of them and the derivative of the action is a vector field with $3$ zeroes. As the vector field is holomorphic, the index at each zero is $+1$.</p> http://mathoverflow.net/questions/109495/manifolds-with-prescribed-fundamental-group-and-finitely-many-trivial-homotopy-gr/109530#109530 Answer by Johannes Ebert for Manifolds with prescribed fundamental group and finitely many trivial homotopy groups Johannes Ebert 2012-10-13T13:14:46Z 2012-10-13T13:14:46Z <p>No, the answer is negative in general (if you require $M$ to be compact). $M$ comes with a map $M \to BG$ that is, by definition, $n+1$-connected (iso on $\pi_i$ for $i=0,...,n$, epi on $\pi_{n+1}$). You can turn it into a weak equivalence by attaching cells of dimension $\geq n+1$. From that you see, that there is a model for $BG$ having finite $n$-skeleton. This is a special property of a group that is called $F_n$ (for more information, see <a href="http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/" rel="nofollow">http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/</a>). Finitely presented groups are $F_2$ and you find that a necessary condition on your $G$ is that it is of type $F_n$. The are concrete examples of groups that are $F_i$ but not $F_{i+1}$ for each $i$, which are discussed in same blog post (on page 423 in Hatcher's AT, you find the same examples in a slightly different context).</p> <p>On the other hand, let $G$ be $F_n$ and let $K$ be the $n$-skeleton of $BG$; a finite complex. Then I claim there is a closed manifold $M$ with the desired properties. $M$ can be chosen of arbitrary dimension $d \geq 4,2n+1$ and to be stably parallelizable. Start with a sphere $S^d \to K$ and do surgery on $S^d$ to get rid of the homotopy groups in low dimensions. The precise formulation is for example Proposition 4 in Kreck's paper "Surgery and duality".</p> <p>So we can say that a necessary and sufficient condition is that $G$ is of type $F_n$. Caveat: I might have confused $n$ and $n+1$ at various places. </p> <p>If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult. </p> http://mathoverflow.net/questions/108923/euler-characteristics-and-the-difference-bundle-construction/109188#109188 Answer by Johannes Ebert for Euler characteristics and the difference bundle construction Johannes Ebert 2012-10-08T22:10:06Z 2012-10-08T22:10:06Z <p>You missed that the sequence is not exact at $K(X,Y)$ (neither is it exact at $K(Y)$, but that does not matter here). There is an ambiguity coming from $K^{-1} (Y)$, i.e. automorphisms of bundles. If $Y$ is a point, your construction works, and the purpose of the arguments in Atiyah-Bott-Shapiro is to extend it to $Y \neq \ast$.</p> http://mathoverflow.net/questions/108319/number-of-spin-structures/108328#108328 Answer by Johannes Ebert for Number of spin structures Johannes Ebert 2012-09-28T10:20:07Z 2012-09-28T10:20:07Z <p>Your first answer is correct, and the second one is almost correct, but the problem is that they count different things:</p> <p>In the first case, you count all twofold covers of $P = P_{SO} E$ that restrict to a nontrivial cover of each fibre. In other words, you count $Spin (n)$-bundles $Q$, together with an isomorphism $\phi:Q \times_{Spin(n)} SO(n) \cong P$ (up to isomorphism of $(P,\phi)$). Here $Q \times_{Spin(n)} SO(n)$ is just a different way to write $Q/\mathbb{Z}_2$. This is the usual notion of a spin structure on $E$. </p> <p>Your second answer (correctly) counts the number of $Spin (n)$-principal bundles $Q \to X$ such that $Q /\mathbb{Z}_2$ \emph{admits} an isomorphism with $P$.</p> <p>The point is that an abstract $Spin (n)$-bundle $Q \to X$ can yield many different spin structures on $P = Q/\mathbb{Z}_2$. Say if $P$ is trivial, then the set of homotopy classes of isomorphisms $Q /\mathbb{Z}_2 \to P$ is in bijection with $[X;SO(n)]$. This accounts for the division by the image $\delta$ in your second answer. </p> <p>For nontrivial $P$, your second answer is not quite correct, as your exact sequence is only a sequence of sets.</p> <p>To see this is an example, let $X=S^1$ and $n \geq 3$. As $SO(n)$ $Spin(n)$ are connected, all principal bundles on $S^1$ are trivial. In this case $H^0 (X;SO(n))=[S^1;SO(n)]=Z/2$; $\delta$ is an isomorpism and the two terms to the left are null. The generator of $[S^1;SO(n)]$ gives an isomorphism of the trivial $SO(n)$-bundle that transforms the two spin structures into each other.</p> http://mathoverflow.net/questions/101745/relative-index-theorem-for-clifford-linear-dirac-operators Comment by Johannes Ebert Johannes Ebert 2013-04-25T21:03:15Z 2013-04-25T21:03:15Z Ulrich Bunke has written a paper &quot;A K-theoretic relative index theorem...&quot;, available on his webpage <a href="http://www.mathematik.uni-regensburg.de/Bunke/" rel="nofollow">mathematik.uni-regensburg.de/Bunke</a>. He treats the case of an arbitrary (complex or real) C^*-algebra as coeffient algebra. http://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theorem/87414#87414 Comment by Johannes Ebert Johannes Ebert 2013-02-05T23:40:12Z 2013-02-05T23:40:12Z Lie's Third Theorem is hard, but there is a wonderful differential geometric proof whose Lie algebraic part is really simple. I found it in a paper by Van Est, &quot;Une demonstration de E. Cartan du troisieme theoreme de Lie&quot;. Another proof of Lie III without even less Lie algebra theory is in the book by Duistermaat-Kolk. Both proofs use that $H^2(G;\mathbb{R})=0$ for simply-connected Lie group. http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/116986#116986 Comment by Johannes Ebert Johannes Ebert 2013-01-30T22:54:33Z 2013-01-30T22:54:33Z @Ryan: yes, in a sense it is a nice proof. Lefschetz fixed point theorem is a hard result, which depends either on Poincare duality or on simplicial approximation. Most topological proofs I know are considerably more elementary (and use the topology of the complex plane, which is more obviously related to the problem than self-maps of $CP^n$). http://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers/120357#120357 Comment by Johannes Ebert Johannes Ebert 2013-01-30T22:04:26Z 2013-01-30T22:04:26Z @Demin: it does not collapse, unless some condition holds. http://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers/120357#120357 Comment by Johannes Ebert Johannes Ebert 2013-01-30T21:02:20Z 2013-01-30T21:02:20Z You cannot apply Vietoris-Begle here, properness of the map is a crucial assumption. In the present case, the fibres are spaces $BG_x$, and if $1 \neq G_x$, there is no compact model for $BG_x$. http://mathoverflow.net/questions/120184/how-to-write-down-explictly-the-isomorphism-of-two-finite-dimensional-representat/120192#120192 Comment by Johannes Ebert Johannes Ebert 2013-01-29T17:06:46Z 2013-01-29T17:06:46Z But the differentials $d$ and $e$ preserve the number $p+q$ (make a picture of the bigraded algebra to see this). http://mathoverflow.net/questions/120184/how-to-write-down-explictly-the-isomorphism-of-two-finite-dimensional-representat Comment by Johannes Ebert Johannes Ebert 2013-01-29T09:33:38Z 2013-01-29T09:33:38Z &quot;&quot;isomorphic as repn of GL(V)&quot; is equivalent to isomorphic as repn of SU(V) (when chosen a metric) since GL(V) and SU(V) generate the same subalgebra in End(V), by density theorem.&quot; This is not true: you can twist any rep of $GL(V)$ by a power of the determinant; without changing the restriction of the representation to $SL(V)$. http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119502#119502 Comment by Johannes Ebert Johannes Ebert 2013-01-21T21:09:11Z 2013-01-21T21:09:11Z Yes, now the question is clear, I was misunderstanding it. http://mathoverflow.net/questions/119470/naturality-of-the-transfer-in-group-cohomology/119473#119473 Comment by Johannes Ebert Johannes Ebert 2013-01-21T19:14:37Z 2013-01-21T19:14:37Z The problem is that the pullback of $EG'\to BG'$ via $Bf$ is only connected if $f$ is surjective. If $f$ is surjective, the naturality holds. http://mathoverflow.net/questions/116894/cohomology-ring-of-bg Comment by Johannes Ebert Johannes Ebert 2012-12-20T20:40:21Z 2012-12-20T20:40:21Z See <a href="http://mathoverflow.net/questions/61784/cohomology-of-bg-g-compact-lie-group/61796#61796" rel="nofollow" title="cohomology of bg g compact lie group">mathoverflow.net/questions/61784/&hellip;</a> http://mathoverflow.net/questions/116814/torsion-in-cohomology-of-smooth-manifolds/116819#116819 Comment by Johannes Ebert Johannes Ebert 2012-12-19T23:26:17Z 2012-12-19T23:26:17Z Odd-dimensional manifolds have two middle dimensions to control, so in that case, the situation is more complicated. http://mathoverflow.net/questions/116017/delignes-letter-to-looijenga-from-1974 Comment by Johannes Ebert Johannes Ebert 2012-12-10T21:35:37Z 2012-12-10T21:35:37Z Why not asking Deligne, Looijenga or Bessis? http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it Comment by Johannes Ebert Johannes Ebert 2012-12-09T18:52:51Z 2012-12-09T18:52:51Z +1 for the question and for the answers. Here is an example for an application that I will give in a class, which shows that the theorem in question is not purely a no-go-result: the Poincare duality in de Rham theory states that $H^k (M) \cong (H^{n-k}_{cpt} )^{\ast}$ for an oriented manifold. If $M$ is compact, then $H^k (M) \cong (H^k (M))^{\ast \ast}$, so $H^k (M)$ is finite-dimensional.. http://mathoverflow.net/questions/114075/de-rham-isomorphism-without-integration Comment by Johannes Ebert Johannes Ebert 2012-11-21T20:02:53Z 2012-11-21T20:02:53Z For a purist with integral-phobia, there are some more obstructions to overcome, even if you take the Poincare lemma as god-given. It seems unclear to me how to prove that manifolds are homotopy equivalent to CW complexes or have good covers without the construction techniques of differential topology, which rely on two pillars, both deeply connected to calculus: Sards Theorem and the existence of solutions of ODEs. http://mathoverflow.net/questions/114075/de-rham-isomorphism-without-integration Comment by Johannes Ebert Johannes Ebert 2012-11-21T19:55:48Z 2012-11-21T19:55:48Z @Yemon: I had a heated discussion with Dmitri here: <a href="http://mathoverflow.net/questions/43681/motivating-the-de-rham-theorem/43730#43730" rel="nofollow" title="motivating the de rham theorem">mathoverflow.net/questions/43681/&hellip;</a>. In the end, he could cook up a quite involved argument for the Poincare lemma that avoided the explicit use of the integral. His claim, that the Lebesgue integral can be constructed by this is, however, totally unclear to me.