User olivier bégassat - MathOverflow

Polynomial roots and convexity

2011-03-18T06:46:59Z

A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but noone had ever heard of the question before, and noone knew the answer.

I hope this is an original question, but seeing how natural it is, I doubt this is the first time someone has asked it.

First, some motivation. Take $P$ any non zero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P=a\cdot\prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots.

Now fix $P$ a certain non zero complex polynomial, and consider $\Pi$ it's primitive (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex $\omega$, write $\Pi_{\omega}=\Pi-\omega$ so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, to be the convex hull of $Q$'s roots.

$\mathrm{MAIN~QUESTION}$: describe $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathbb{C}}\mathrm{Conv}(\Pi_{\omega})$.

By the above quoted property, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

Here are some easy observations:

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ be unitary. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence.

$2)$ the intersection defining $\mathrm{Hull}(P)$ can be taken for $\omega$ ranging in a compact subset of $\mathbb{C}$: as $|\omega|\rightarrow\infty$, the roots of $\Pi_{\omega}$ will tend to become close to the $(\deg (P)+1)$-th roots of $\omega$, so for large enough $\omega$, their convex hull will always contain , say, $\mathrm{Conv}(\Pi)$.

$3)$ $\mathrm{Hull}(P)$ can be explicitely calculated in the following cases: $P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: 2 simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$.

Also if $\Pi$ is a real polynomial of odd degree $n+1$, that has all its roots real and simple, say $\lambda_1<\mu_1<\lambda_2<\dots<\mu_n<\lambda_{n+1}$, where I have also placed $P$'s roots $\mu_1,\dots,\mu_n$, and if you further assume that $\Pi(\mu_{2j})\leq\Pi(\mu_n)\leq\Pi(\mu_1)\leq\Pi(\mu_{2j+1})$ for all suitable $j$ (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n),\Pi(\mu_1)]$, the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get

$[\mu_1,\mu_n]=\mathrm{Conv}(P)\subset\mathrm{Hull}(P)\subset \mathrm{Conv}(\Pi_{\Pi(\mu_1)})\cap \mathrm{Conv}(\Pi_{\Pi(\mu_n)})=$ $[\mu_1,\dots]\cap [\dots,\mu_n]=[\mu_1,\mu_n]$

$4)$ The equation $\Pi_{\omega}(z)=\Pi(z)-\omega=0$ defines a Riemann surface, but I don't see how that could be of any use.

Computing $\mathrm{Hull}(P)$ for the next most simple polynomial $P=X^3-1$ has proven a challenge, and I can only conjecture what it might be.

Computing $\mathrm{Hull}(X^3-1)$ requires factorizing degree 4 polynomials, so one naturally tries to look for good values of $\omega$, the $\omega$ that allow for easy factorisation of $\Pi_{\omega}=X^4-4X-\omega$. For instance the $\omega$ that produce a double root. All that remains to be done afterwards is to factor a polynomial of degree $2$. The problem is symmetric, and you can focus on the case where 1 is the double root (i.e. $\omega=-3$). Plugging the result in the intersection, and rotating twice, you obtain the following superset of $\mathrm{Hull}(X^3-1)$: a hexagon that is the intersection of 3 similar isocele triangles with their main vertex located on the three third roots of unity $1,j,j^2$

$\mathrm{QUESTION}$: is this hexagon equal to $\mathrm{Hull}(X^3-1)$?

Here's why I think this might be.

Consider the question of how the convex hulls of the roots of $\Pi_{\omega}$ vary as $\omega$ varies. When $\omega_0$ is such that all roots of $\Pi_{\omega_0}$ are simple, then the inverse function theorem shows that the roots of $\Pi_{\omega}$ with $\omega$ in a small neighborhood of $\omega_0$ vary holomorphically $\sim$ linearly in $\omega-\omega_0$: $z(\omega)-z(\omega_0)\sim \omega-\omega_0$. If however $\omega_0$ is such that $\Pi_{\omega_0}$ has a multiple root $z_0$ of multiplicity $m>1$, then a small variation of $\omega$ about $\omega_0$ will split the multiple root $z_0$ into $m$ distinct roots of $\Pi_{\omega}$ that will spread out roughly as $z_0+c(\omega-\omega_0)^{\frac{1}{m}}$, where $c$ is some non zero coefficient. This means that for small variations, these roots will move at much higher velocities than the simple roots, and they will do the major contribution to the variation of $\mathrm{Conv}(\Pi_{\omega})$, also, they spread evenly out, and (at least if the multiplicity is greater or equal to $3$) they will tend to increase the convex hull around $z_0$. Thus it seems not too unreasonable to conjecture that the convex hull $\mathrm{Conv}(\Pi_{\omega})$ has what one can only describe as critical points at the $\omega_0$ that produce roots with multiplicities. I'm fairly certain there is a sort of calculus on convex sets that would allow one to make the above statement precise, but I don't know it.

Back to $X^3-1$: explicit calculations suggest that up to second order the double root $1$ of $X^4-4X+3-h$ for $|h|<<1$ splits in half nicely (here $\omega=-3+h$) and the convex hull will continue to contain the aforementioned hexagon.

$\mathrm{QUESTION\Conjecture}$: is it true that $\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where $\mathrm{MR}$ is the set of all $\omega_0$ such that $\Pi_{\omega_0}$ has a multiple root, i.e. the set of all $\Pi(\alpha_i)$ where the $\alpha_i$ are the roots of $P$?

All previous examples of calculations agree with this, and I have tried as best I can to justify this guess heuristically.

Are you aware of a solution? Is this a classical problem? Is anybody brave enough to make a computer program that would compute some intersections of convex hulls obtained from the roots to see if my conjecture is any good?

Shrinking Group Actions

2011-06-11T05:27:01Z

This is a repost from stackexchange. I didn't get an answer, so I figured I'd ask it here.

Suppose $H\subset G$ is a subgroup of a topological group $G$, and $Y\subset X$ is a subspace of a topological space $X$. Suppose we are given a continuous group action $\rho : G\times X\rightarrow X$ on $X$, and suppose that $Y$ is $H$ stable, that is $h.y \in Y$ for all $h\in H$ and $y\in Y$. You can form the quotient spaces $H\setminus Y$ and $G \setminus X$, and there is a natural, continuous, in general neither injective nor surjective map $\theta : H\setminus Y\rightarrow G\setminus X$. I am looking for conditions that assure this is a homeomorphism.

You can show easily that $\theta$ is onto $\mathrm{iff}~Y$ intersects all orbits, and one to one $\mathrm{iff} ~ \forall y\in Y, H.y=G.y\cap Y$. So I'll suppose these two conditions.

$\mathrm{QUESTION:}$ When is $\theta$ a homeomorphism?

All spaces $X$ I have in mind are Hausdorff, but not necessarily locally compact. Also, the groups $G$ I consider are Lie groups, but I am interested in weaker conditions too, and don't want to restrict myself to that case. I am looking for practical $sufficient$ conditions on $X,Y,H,G$ and $\rho$.

One way to make $\theta$ into a homeomorphism is to have compact (Hausdorff) $Y$ and $H$, and Hausdorff $G \setminus X$.

References would be perfect!

Answer by Olivier Bégassat for two complex polynomials with equal modulus on a parabola

2011-06-06T16:42:02Z

Yes, take $P$ and $Q=e^{it}P$.

Answer by Olivier Bégassat for Is G-bundle over 1-skeleton trivial

2011-05-18T23:00:53Z

Hi,

I don't know your background knowledge, so I'll try to post an elementary explanation to your question, which is also the way I understand it. I am not 100% reliable, and there might be more or less serious mistakes in what follows, but I think the main ideas are there.

You know that a principal $G$ bundle is trivial $\mathrm{iff}$ it admits a section. So suppose you have a (locally finite) cell complex $X$, put $X_n$ the $n$-squeleton of $X$ and let $P$ be a principal $G$-bundle over $X$. We'll also call $P_n$ the restriction of $P$ to $X_n$, which is still a princial $G$ bundle. You want to know why $P_1$ is trivial, and how (non) triviality of the $P_n$ is related to the fundamental groups of $G$.

Well, what is $P_0$? It is nothing more than a discrete collection of (homeomorphic) copies of $G$, one above each vertex of $X_0$. Now, for each vertex $x\in X_0$, chose a point $g_x\in G$. This gives you a trivialisation of $P_0$ (the (continuous) map $X_0\rightarrow P_0,~x\mapsto (x,g_x)$), and $P_0$ is trivial.

Similarly what is $P_1$? It is a graph with, along each line, tightly packed (homeomorphic) copies of $G$. To show it is trivial we need only construct a section of $P_1$. Let's start with the section of $P_0$ we already constructed. This was nothing more than a collection of pretty random points in the fibres over the vertives of $X$. What would it mean to extend this (very partial) section of $P_1$ (only defined on the zero squeleton $X_0$) to the whole of $X_1$? It means constructing, for each edge $e$ with endpoints $x$ and $y$ a continuous map from $e$ (which is basically $[-1,+1]$) to $P_1$ that takes on the values $g_x$ at $x$ and $g_y$ at $y$ (that is with values imposed on $\lbrace -1,+1\rbrace=\partial [-1,+1]$. This is, a priori, possible $\mathrm{iff}$ $g_x$ and $g_y$ lie in the same path component of $G$, but such will always be the case [EDIT: actually, it's not, consider the $\lbrace -1,+1\rbrace$ principal fibre bundle over the circle which is trivial on the circle minus $-1$ and on the circle minus $+1$, and has a cocycle that is $-1$ on the left and $+1$ on the right (this is also $\mathbb{S}^1\rightarrow\mathbb{S}^1,~z\mapsto z^2$): it is not trivial].

You can, for instance cut the edge $e$ into many little closed intervals that each trivialize the (induced) bundle [Edit: and only overlap at their extremities], and glue sections together, imposing the first and final value of the section, and working your way from $-1$ to $+1$. [EDIT: What may happen now is that the final two values you have to link within $G$ may not lie in the same path component, and so there is a dependence on $\pi_0 (G)$] Now we have a map from the $1$-squeleton $X_1$ to $P_1$ that is a section of the bundle, and is continuous along each edge, and prolongs the first section we constructed earlier on. But because $X$ is locally finite, this map is continuous. (Actually I'm pretty sure that condition is unnecessary, because you can impose your sections to be 'locally constant around the vertices')

So $P_1$ is trivial,always, regardless of any restrction. This technique can be carried on. Suppose you have shown $P_n$ to be trivial, let $s_n$ be a global section. Can you extend $s_n$ to a global section $s_{n+1}$ of $P_{n+1}$? Consider a $(n+1)$-face $F$ and its border $\partial F$. The desired section is already defined along the border $\partial F$ (it's the restriction of $s_n$ to $\partial F$), so you have what is essentially a (continuous) map $c_{\partial F}$ from $\mathbb{S}^n$ (the border $\partial F$) to $G$, and you want to extend it to a map from the euclidean $(n+1)$-unit ball $\mathbb{B}^{n+1}$ (which would be $F$) to $G$. This problem is equivalent to nulhomotopy of the map $c_{\partial F}$, and you thus find that $P_{n+1}$ is trivial $\mathrm{iff}$ all such maps are nullhomotopic. In particular, if $\pi_n (G)$ is trivial, the above problem is always solvable.

You might want to have a look at http://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups if you are unfamiliar with the fact that the second fundamental group of a Lie group is trivial. This fact implies, as Thomas Nikolaus has already pointed out, that a principal $G$ bundle over a surface or $3$ manifold with simply connected $G$ is always trivial. Another good reference is Milnor and Stasheff's excellent book "Characteristic Classes", where everything I list is explained in better and shorter form.

why study Lie algebras?

2011-03-17T00:24:39Z

I don't mean to be rude asking this question, I know that the theory of Lie groups and Lie algebras is a very deep one, very aesthetic and that has broad applications in various areas of mathematics and physics. I visited a course on Lie groups, and an elementary one on Lie algebras. But I don't fully understand how those theories are being applied. I actually don't even understand the importance of Lie groups in differential geometry.

I know, among others, of the following facts:

$1)$ If $G$ and $H$ are two Lie groups, with $G$ simply connected, and $\mathfrak{g,h}$ are their respective Lie algebras, then there is a one to one correspondance between Lie algebra homomorphisms $\mathfrak{g}\rightarrow\mathfrak{h}$ and group homomorphisms $G\rightarrow H$.

$2)$ The same remains true if we replace $H$ with any manifold $M$: any Lie algebra homomorphism from $\mathfrak{g}$ to the Lie algebra $\Gamma(TM)$ of smooth vector fields on $M$ gives rise to a local action of $G$ on $M$.

$3)$ Under some conditions like (I think) compactness, the cohomology of $\mathfrak{g}$ is isomorphic to the real cohomology of the group $G$. I know that calculating the cohomology of $\mathfrak{g}$ is tractable in some cases.

$4)$ There is a whole lot to be said of the representation theory of Lie algebras

$5)$ Compact connected centerless Lie groups $\leftrightarrow$ complex semisimple Lie algebras

How do people use Lie groups and Lie algebras? What questions do they ask for which Lie groups or algebras will be of any help? And if a geometer reads this, how (if at all) do you use Lie theory? How is the representation theory of Lie algebras useful in differential geometry?

Thank you for your time

Answer by Olivier Bégassat for Are there any mathematical objects that exist but have no concrete examples?

2011-04-22T23:47:43Z

If I remember correctly, there is a theorem that asserts that all but possibly zero, one or two prime numbers generate infinitely many of the (cyclic) multiplicative groups $\mathbb{Z}/q\mathbb{Z}^{\times}$ where $q$ varies among the primes. Yet not even one such prime is known, not even $2$ or $3$. Thus, among $2,3$ and $5$, at least one of them has the property, but no one knows which do.

Compact open topology on $\mathrm{Homeo}(X)$

2011-03-16T22:09:19Z

Let $X$ and $Y$ be topological spaces. Define the compact open topology on the set $\mathrm{M}(X,Y)$ of contiunous maps from $X$ to $Y$ via the subbase $[K,O]$ of all maps $f:X\rightarrow Y$ s.t. $f(K)\subset O$, where $K$ is any compact subset of $X$, and $O$ is any open subset of $Y$. So a basis of open sets is given by the following subsets: $[K_1,\dots,K_n,O_1,\dots,O_n]=[K_1,O_1 ]\cap\dots\cap [K_n,O_n]$, the collection of continuous maps $f:X\rightarrow Y$ that send each $K_i$ into $O_i$ for some specified collcetion of compact $K_i$'s and open $O_i$'s.

This topology has some nice properties: the exponential law holds under some hypotheses on the spaces $X$ and $Y$, and is certainly true if all spaces involved are locally compact Hausdorff spaces, as will be the case from now on.

My question is as follows: if $X$ is a locally compact Hausdorff space (or even a topological manifold), the compact open topology induces a topology on the set of homeomorphisms of $X$, which is a group. Does this topology turn $\mathrm{Homeo}(X)$ into a topological group? I can show that the product (composition) is continuous, but is the inverse too? $(f\rightarrow f^{-1})$

I was able to prove continuity for compact spaces, where it is very easy to establish. I also managed to prove it for $X=\mathbb{R}$ because all homeomorphisms of $\mathbb{R}$ are monotone, but that's everything so far.

I tried looking it up in several textbooks on topology and algebraic topology where the C.O. topology is usually discussed, but couldn't find a discussion on this topic anywhere.

Answer by Olivier Bégassat for Optimizing a quadratic restricted to the sphere

2011-04-11T02:51:32Z

This is not an answer, but it was getting too long to be a comment. Also, I'm not familiar with such problems, so hopefully someone will be able to give you a complete and more satisfying answer and some references. Here are some general remarks. It's a bit messy and mostly guesswork :S

If $|A^{-1}b|<1$ there might be several solutions, i.e. several $x$ minimizing $\varphi(x):=\frac{1}{2}\langle x|Ax \rangle+\langle b|x \rangle$ (I added $\frac{1}{2}$ for convenience only), precisely, if you call $\lambda_1>\dots>\lambda_p>0$ the distinct eigenvalues of $A$ and you have $e_1,\dots, e_p$ an associated orthonormal frame and $b$ such that $$|A^{-1}b|<1~\mathrm{and}~b\in\mathrm{Span}(e_1,\dots,e_r)\setminus\mathrm{Span}(e_1,\dots,e_{r-1})$$

I would expect the set of minimizing $x$ to be a sphere of dimension $p-r-1$ and a point when $r=p$

If $|A^{-1}b|>1$, a picture convinces me there is only one solution to your problem (and the "projection onto a convex closed set"-theorem in inner product spaces proves it). If $x_0$ is a solution, then we have to have that the differential of $\varphi$ vanishes at $x_0$ which translates into $\langle Ax_0|- \rangle+\langle b|- \rangle$ vanishing on the orthogonal of $x_0$ i.e. there exists some $\mu\in\mathbb{R}$ such that $Ax_0+b=\mu x_0$, and again, a picture suggests there are preciseyl two solutions and that we have to look for the solutions with $\mu < 0 $. Then there is indeed only one solution, because if we look for $x_0\in\mathbb{S}^{n-1}$ satisying above equation, we can write down $x_0=-(A-\mu)^{-1}b$ because the matrix is invertible. The norm of above vector is stricly increasing with $\mu$ and equals $|A^{-1}b|>1$ at $\mu=0$ and tends to $0$ as $\mu$ tends to $-\infty$. So there'll be only one $\mu<0$ such that $-(A-\mu)^{-1}b$ has unit length. But finding that given $\mu$ requires solving the equation $\frac{b_1^2}{(\lambda_1-\mu)^2}+\dots+\frac{b_p^2}{(\lambda_p-\mu)^2}=1$

So if $|A^{-1}b|>1$, you can find the solution as long as you can find the negative $\mu$ that satisfies the above equation...

Answer by Olivier Bégassat for Holomorphically Convex Hull a Subset of the convex hull of

2011-03-29T05:18:19Z

The exponential function grows in module as the exponential of the real part. Therefore, the set of all $z$ such that $|exp(az)|\leq \sup_K |exp(a\times\cdot)|$ is a half space containing $K$, and meeting $K$. You get all such half spaces, if you vary $a$ in $\mathbb{C}$ or even on the unit circle. Their intersection is the convex hull of $K$ by some famous theorem on convex sets (Krein-Milman?).

So by restricting yourself to the exponential functions you get the convex hull of $K$. The hull you're interested in is a subset of that set.

I don't know what $K^*$ stands for, but it won't be the convex hull in general. for instance, if you take $\Omega=\mathbb{C}\setminus\lbrace 0\rbrace$ and $K=$ the unit circle, and $f(z)=z, ~g(z)=\frac{1}{z}$, you see that the hull you're interested in is just $K$ itself. The convex hull may not even be a subset of $\Omega$.

What holomorphic functions are limits of polynomials?

2011-03-25T07:06:14Z

Let $\Omega$ be a connected open set in the complex plane. What is the closure of the polynomials in $\mathcal{H}(\Omega)$ the set of holomorphic functions on $\Omega$? The topology is the usual compact convergence topology. Take, for instance, an annulus such as $D(r,R)$, the set of all complex $z$ such that $r<|z|< R$, you cannot recover the function $z\mapsto \frac{1}{z}$ because of the residue at $0$, so what holomorphic functions are limits of polynomials?

Examples of naturally occurring Quadratic forms or quadrics.

2011-03-18T20:10:19Z

I am always fascinated when a quadratic form (or a quadric) arises naturally. I have some elementary examples, but most of all, I want to learn more examples. I hope this question isn't considered too vague for MO. Most forms I list are really elementary, and all are finite dimensional.

I got most of the following examples from M.Berger, Geometry I & II, and from the truly beautiful book "Eléments de géométrie : actions de groupes" by french author Rached Meinmné.

$(0)$ the discriminant on the affine space of unitary degre 2 polynomials

$(i)$ the determinant on endomorphisms of a 2 dimensional vector space, and $\mathrm{Tr}^2-4\mathrm{det}$

$(ii)$ the radical on the space of quadratic forms on a 2 dimensional vector space, and the isotrope cone (not sure about the name, degenerate cone?).

$(iii)$ the family of hermitian forms (built from the Wronskian) on the solution space of the discrete Schroedinger equation that allow one to show the existence of right and left side $L^2$ solutions, and the Weyl m function.

$(iv)$ If $\Delta$ is any $2$ dimensional complex vector space, then $\mathrm{Herm}(\Delta)$, the real vector space of hermitian forms on $\Delta$, carries a natural quadratic form obtained by constructing an essentially unique morphism $\rho$ from $\mathrm{Herm}(\Delta)$ to $\mathrm{Hom}(\Delta\oplus\overline{\Delta})$ such that for all $h\in\mathrm{Herm}(\Delta),~\rho(h)^2$ is proportional to $\mathrm{Id}$, the proportionality defining the quadratic form. Here, $\rho$ only depends on a choice of a nonzero element $\omega\in\Lambda^2\Delta^*$.

$(v)$ If $V$ is a 4 dimensional vector space, then $\Lambda^2 V$ carries the natural quadric $Q(v)=v\wedge v$ where $\Lambda^4 V$ is identified with the underlying field, which vanishes exactly when $v$ comes from the canonical map $\mathrm{Gr}(2,V)\rightarrow P\Lambda^2V$.

I remember reading about one on the space of circles, but I forgot the details. What other examples of natural quadratic forms are there?

how many semi direct products are there?

2011-03-18T11:27:58Z

This question was initially proposed to me by two friends. Given an integer $n$, how many isomorphism classes are there for semidirect products $\mathbb{Z}/n\mathbb{Z}\rtimes\mathbb{Z}/2\mathbb{Z}$?

Maybe this is a really trivial question. I can tell that a semidirect product is the same as an integer $r\in\mathbb{Z}/n\mathbb{Z}$ with $r^2=1\mod[n]$, but are there isomorphisms between some of them? What happens for instance when n is squarefree, thus the product of fields.

Trivial fiber bundle

2011-03-16T20:36:00Z

Suppose $(E,p,B;F)$ is a fiber bundle such that $E$ is homeomorphic to $B\times F$, is it true that the fiber bundle is trivial? A non connected counter example has been provided, so I'll ask for E,B and F to be connected (hopefully low dimensional) manifolds.

nullity of the second fundamental group of a Lie group

2011-03-17T00:39:02Z

Can anyone tell me why it is that Lie groups seem to have their second fundamental group $\pi_2(G)$ equal to $0$, or provide me with a link to an article or a book reference?

I came across this fact reading an article where the author considers principal $G$ bundles with $G$ a simply connected simple group.

thank you

Comment by Olivier Bégassat

2011-06-29T01:49:06Z

Wasn't it Marcel Berger who introduced them? I might be wrong...

Comment by Olivier Bégassat

2011-06-11T09:12:21Z

It is even sufficient that $\varphi$ be bijective and holomorphic for it to be biholomorphic.

Comment by Olivier Bégassat

2011-06-11T07:27:00Z

I'm nitpicking, but you need to put parantheses : $x^{x^x}\neq (x^x)^x$.

Comment by Olivier Bégassat

2011-06-11T05:28:25Z

By the way, $G\setminus X$ is the orbit space, sometimes written as $X/G$.

Comment by Olivier Bégassat

2011-06-06T16:43:30Z

maybe you should include that they have leading coefficient $1$.

Comment by Olivier Bégassat

2011-06-01T22:26:17Z

very nice proof!

Comment by Olivier Bégassat

2011-06-01T09:37:08Z

nevermind, it will only give you solutions for big values of $\rho$ but probably not for $\rho<<1$

Comment by Olivier Bégassat

2011-06-01T09:28:49Z

I'm pretty sure putting $a_0(x)=\sin(x),~a_1(x)=\sin(\sqrt{2}x)$ will do the job.

Comment by Olivier Bégassat

2011-06-01T09:20:11Z

have you tried looking for finite sequences (with only 3, 4 or 5 terms) with $a_n(\lambda)$ given explicitely as, say, polynomial functions?

Comment by Olivier Bégassat

2011-05-16T00:59:10Z

what is your definition of a fiber bundle? If it's the usual one, it encompasses covering spaces, and they can very well have finite fiber and be non trivial.

Comment by Olivier Bégassat

2011-05-16T00:40:28Z

is this jibber-jabber?

Comment by Olivier Bégassat

2011-05-15T22:10:32Z

By the way, you got your title wrong!

Comment by Olivier Bégassat

2011-05-15T21:55:07Z

There is no such embedding for even $n$, just consider the orders of the respective groups: you don't have $|S{n}|$ dividing $|A{n+1}|$ by comparing the order of exponents of $2$.

Comment by Olivier Bégassat

2011-05-12T06:18:04Z

What is $T^M$??

Comment by Olivier Bégassat

2011-05-10T18:45:33Z

There has to be something eluding me, why did you tag your question with the large cardinals and set theory tags? Does the following argument not work? $S=\Cup_{n\in\mathbb{N}\setminus \lbrace 0\rbrace} S_n$ where $S_n=\lbrace s\in S\mathrm{~s.t.~} f(s)>\frac{1}{n}\rbrace$, thus one of them is non denumerable and taking a denumerable subset of said $S_{n_0}$ will yield an infinite sum.