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olivier dot begassat dot cours at gmail dot com


1d
comment Distributive lattices (interpretation of distributivity)
@Eran you are right! I just started looking into lattice theory a little more seriously, and the modular law never made any sense to me when worded as $$\forall a,b,x\in L,\:\big(x\leq b\longrightarrow x\vee(a\wedge b)=(x\vee a)\wedge b\big)$$ because it seems like you should think of $x$ as a variable, but it means just what I was looking for when you think of $a$ as a variable instead: the two natural projections $L\to [x,b]$ are equal.
1d
comment Distributive lattices (interpretation of distributivity)
@Eran Mine states that for any $a,b\in L$, $a\wedge\bullet:[b,a\vee b]\to[a\wedge b,a]$ and $b \vee\bullet:[a\wedge b,a]\to[b,a\vee b]$ are inverse bijections. But your equation is familiar to me...
1d
comment Distributive lattices (interpretation of distributivity)
@Eran I'm aware of modular lattices, but there one asks that intervals be isomorphic through specific maps, which, a priori, is different from what I ask. Are they the same?
2d
comment The automorphism group of the real line with standard topology
@IttayWeiss I was just throwing stuff out there, I'm not convinced this actually works in a compatible way with the group structure.
2d
comment The automorphism group of the real line with standard topology
@IttayWeiss I'm not talking about deformation retractions that respect the group structure, only about deformation retractions of the underlying topological space, I don't think the dynamic properties of the automorphisms plays any role in that. As for something that is compatible with the group structure, how about scaling in the domain? What about the maps $H:A\times[0,1]\to A, (f,t)\mapsto \lbrace x\mapsto\frac{f(tx)-f(0)}{t}+f(0)\rbrace$? At least when one considers the subgroup of $C^1$ diffeomorphisms, this should give a strong deformation retraction onto the affine functions.
2d
comment The automorphism group of the real line with standard topology
@IttayWeiss I was only talking about the topology of this space, not the group structure. I remember proving that it is a topological group with the compact open topology years ago. don't you think that the space should strongly deformation retract onto the subset of all affine homeomorphisms with slope $\pm1$? From there it will deformation retract onto $\lbrace \pm1\rbrace$ I think...
2d
comment The automorphism group of the real line with standard topology
It seems to me, at first sight, reasonable, that there should be two path components, and a strong deformation retraction onto $\lbrace 1,-1\rbrace$, where $1$ is shorthand for the identity of $\Bbb R$. After all, the homeomorphisms of $\Bbb R$ are precisely the surjective stricly monotone functions, and there are obvious "convex homotopies".
Apr
20
comment Bounded linear functionals on $L^\infty$.
Are you sure $C^0[0,1]$ isn't the space of continuous functions on the unit interval? The notation is standard, and they are all measurable and bounded.
Apr
20
comment Can a cube always be fitted into the projection of a cube?
Are you allowing for all projections, or only for the orthogonal ones (i.e. those projections with orthogonal image and kernel)?
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
Ok, if you say so ^^ I awarded you the bounty, I guess I need to read the articles more in depth to believe their claims and my own explanation ^^
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
Also, are the spaces $X$ and $Z$ the actual spaces he uses to find homology equivalent (in $\pi_1$ and $H_*$), non homotopy equivalent spaces?
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
@studiosus But as I understand it, it is only required that the Euler characteristics be equal and $m=1$, and it seems to me he is saying that $m=1$ whenever $\chi>\chi_{\min}(G,2)$. Since we know by example (the complex associated to the presentation $\langle a,b\mid a^3=b^2\rangle$) that there is a $[G,2]$-complex with $\chi=0$ we have $0\geq\chi_{\min}(G,2)$, and so, for any $[G,2]$-complex with $\chi\geq 1$, $m=1$. My point is that it apparantly doesn't matter that $\chi_{\min}(G,2)=0$ or not, it only matters that it be less than our $\chi$. Am I correctly interpreting the article?
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
I left an answer to tell you what I understood from your examples, could you take a look at it and correct my understanding of it?
Apr
20
comment Euler characteristic and free action
What action of $G$ on $K^2$ are you considering?
Apr
19
comment Let $A$ a matrix with real or complex entries. Proof that $\displaystyle\lim_{n\rightarrow\infty}(E+\frac{A}{n})^n=e^A, E=$indentity.
You can probably show that this sequence of functions converges uniformly on every compact set (equipping $M_n(\Bbb C)$ with some submultiplicative norm for greater convenience), so that the function $A\mapsto\lim_n\left(I+\frac{A}{n}\right)^n$ is continuous. It is then easy to show that it agrees with the usual exponential on the dense set (in $M_n(\Bbb C)$) of diagonalizable matrices, and hence the standard exponential and this agree on all of $M_n(\Bbb C)$.
Apr
19
comment Help trying to identify a set and determine whether it is a subspace of $\Bbb{R}^n (n>2)$
Recall the definition of a subspace, and verify the axioms. You only have to prove $S$ is non empty, stable under sum and scalar multiplication.
Apr
18
comment Proving an inequality for convex and increasing functions
How differentiable do you assume $f$ to be? Do you assume that it is convex aswell? Please state clearly all your hypotheses.
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
It's not a vector bundle over $T^*E$! It's a vector bundle over $E$!!!
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
Usually you'd include the 0-th exterior power $\Lambda^0(T^*E)$ (which is a line bundle) in the definition of $\Lambda(T^*E)$; in any case, $\Lambda(T^*E)$ (with the correct definition) is a vector bundle of dimension $2^n$ over $E$ (not $T^*E$) where, I take it, $E$ is a manifold of (real) dimension $n$.
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
There seems to be a lot of confusion in your question. Are you sure that is what you mean to ask? As I understand it, the answer is trivially no. A vector bundle that decomposes as a direct sum of vector bundles has dimension equal to the sum of the dimensions of the vector bundles that appear in the sum...