bio  website  normalesup.org/~cornulier 

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visits  member for  3 years, 10 months 
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13h

comment 
Closed geodesics avoiding points in hyperbolic surfaces
Do you know for a single point? 
20h

comment 
A KrullSchmidt Theorem for Lie groups?
Then each direct summand is a connected Lie subgroup, and the chains of connected closed subgroups have bounded length (bounded by the dimension). Is this enough to make the proof of KrullSchmidt work? 
22h

comment 
Geometric automorphism of free group respect to nonorientable suface
Even if the answer to the precise question were true, it would not mean that the study can boil down to the orientable case. For instance if we have a geometric pair of outomorphisms (in the sense that they are geometric for a common surface), this would possibly not imply that the latter can be chosen orientable. Still the question is reasonable (though I regret nonorientable surfaces are often considered as bad guys). 
1d

comment 
Categorical proof subgroups of free groups are free?
It's not free in the categorical sense, but for the free $K$fields $K(x_1,\dots,x_n)$, discussing whether its $K$subfields are free amounts to the question whether unirational implies rational. For instance for $K=\mathbf{C}$, the answer is yes for small $n$ (at least $n\ge 2$) but not for large $n$. 
1d

comment 
Maximal compact subgroup of SL(2,H)
Every compact subgroup of $GL_2(H)$ preserves a hermitian form. (Note that a hermitian form on $V=H^n$ has to be carefully defined, e.g. it means $b:V\times V\to H$ satisfying $b(xt,ys)=\bar{t}b(x,y)s$ for all $x,y\in V$, $s,t\in H$, and $b(y,x)=\overline{b(x,y)}$.) 
1d

comment 
A KrullSchmidt Theorem for Lie groups?
I guess you mean connected? Because otherwise, all discrete groups are real Lie groups... 
2d

revised 
Irreducibility of a polynomial
Slightly rephrased. The question was unfairly closed and should be reopened. 
2d

comment 
Irreducibility of a polynomial
The answer is yes when $f$ has degree 1, and in particular when $k$ is algebraically closed. It also seems correct when $f$ is separable of degree 2, and in particular when $k$ is the real field. 
2d

comment 
Irreducibility of a polynomial
sorry, that's correct. I voted to reopen. 
Jan 25 
revised 
Irreducible/prime/indivisible elements
changed type "domain" into "ring" 2 times 
Jan 25 
answered  Irreducible/prime/indivisible elements 
Jan 23 
comment 
Special linear groups over function fields
@Matthias: you mean "any field, in particular for $k=\mathbb{F}_q$" (for the amalgam decomposition of $SL_2(k[t])$) 
Jan 23 
comment 
morphism from a compact group to Z ?
@LSpice I thought of a surjective homomorphism, which is no restriction... but $x\notin\phi^{1}(\{0\})$ is fine. 
Jan 22 
comment 
morphism from a compact group to Z ?
Very nice. Let me slightly restate your argument: 1) replacing $G$ by the closure of some $x\in\phi^{1}(\{1\})$, we can suppose that $G$ is abelian with a dense cyclic subgroup. 2) if $G$ is any connected compact abelian group then the result holds because $G$ is divisible (as a projective limit of tori) 3) now supposing $G$ has dense cyclic subgroup, $\phi$ vanishes on its unit connected component because of (2), hence we can suppose $G$ totally disconnected, hence a quotient of the profinite completion $\hat{Z}\simeq\prod_p\mathbf{Z}_p$, and your last argument finishes the job. 
Jan 22 
comment 
Fermat's last theorem over larger fields
@Pablo if it's due to Faltings, it's maybe not so clear :) I think it was worth mentioning. (I only knew Faltings' statement for the field $\mathbf{Q}$, not for arbitrary number fields.) 
Jan 22 
comment 
Fermat's last theorem over larger fields
It's a general comment about this question and all partial answers: if $K$ is an infinite extension of $\mathbf{Q}$ such as $\mathbf{Q}^{ab}$ or $(\mathbf{Q}^{ab})^{ab}$ and $X$ a affine variety, the statement "$X(K)$ is infinite" is weaker than "$X(L)$ is infinite for some finite subextension $L\subset K$", and both could be addressed. 
Jan 21 
comment 
Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry?
@TerryTao $Gal(\mathbf{R}/(\bar{\mathbf{Q}}\cap\mathbf{R}))=\{1\}$ 
Jan 21 
comment 
Intersection of two real polynomial surfaces
What do you call the 3sphere? The unit sphere in the 3dimensional real space is usually called 2sphere. 
Jan 21 
comment 
Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry?
It's still a bit vague what you ask. Do you require conditions on $X,Y$ implying that it holds for every $f$? Otherwise a "condition" is to assume that $f$ is an isometry :) 
Jan 20 
comment 
Limits of conjugated subgroups
actually you can say that if $H_\infty$ is f.g. then it's contained in a conjugate of $H$ (same trivial argument). Of course this applies in the noetherian case, but not only. 