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olivier dot begassat dot cours at gmail dot com


Jun
21
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
edited title
Jun
21
comment Defining principal elements of every poset. Is this a new idea?
You don't use $b$ in the definition of "intersecting elements".
Jun
21
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
edited tags
Jun
21
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
added 14 characters in body
Jun
21
comment showing $\int _a^b\left(f'\left(x\right)\right)dx\:=\:f\left(b\right)-f\left(a\right)$
$f'$ need not be continuous in general.
Jun
21
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
Clarification of the question
Jun
21
asked $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
Jun
21
answered If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that
Jun
19
comment Characteristic classes not defined on vector bundles
It seems that if $G$ admits a faithful linear representation then there should be an equivalence between principal $G$ bundles and $G$-vector bundles of dimension the dimension of (the chosen faithful linear representation of $G$). So I guess you can most of the time frame your problem (!) in terms of frame bundles. But I'll admit this is silly. In any case, there are such characteristic classes when one studies principal $G$ bundles with finite $G$, there is a paper on the topic by Freed and Quinn (Chern Simons theory with finite gauge group). They recover some character formulas.
Jun
19
comment How to embbed $S^1\times [0, 1]$ in $\mathbb R^2$?
What is $S^1\times[0,1]$ geometrically? How could you embed this geometric figure in the plane?
Jun
17
answered The property of an integral manifold of differential form
Jun
17
answered Measurability of a function in $\mathcal{B}(\mathcal{C}([0,1],\mathbb{R}))$
Jun
17
comment Measurability of a function in $\mathcal{B}(\mathcal{C}([0,1],\mathbb{R}))$
You are considering the Borel $\sigma$-algebra associated to the topology of uniform convergence on the continuous functions? It seems $\Lambda_a$ may even upper or lower semicontinuous. Indeed, if $\epsilon>0$ and $f$ is in a sufficiently small neighborhood of $f_0$, then $\lambda_a(f)$ will stay $\geq\lambda_a(f_0)-\epsilon$, so the map will be lower semicontinuous. I think this implies measurability.
Jun
17
comment Calculate Geodesic Path of $N\times N$ matrix on Riemannian manifold of fixed rank
What studiosus is getting at is that you need to define what riemannian metric (RM) you want to consider: a manifold, such as the manifold of matrices of given rank, may admit many different RMs, and usually different metrics will define different geodesics. About the induced metric, if $S$ is a submanifold of a manifold $M$ with a RM, then the RM of $M$ induces a RM on $S$ called the induced metric.
Jun
17
comment Accepted symbol (or way of writing) “A is a subset of B or B is a subset of A”
But $A$ cannot contain $B$ (or vice versa) unless they are equal, because finite dimension etc.
Jun
16
revised $f: SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$ show that $Im(f)=SL_4(\mathbb{R})$
added 341 characters in body
Jun
16
answered $f: SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$ show that $Im(f)=SL_4(\mathbb{R})$
Jun
15
answered Induced bijections of combinatorial species
Jun
15
comment Is there a diffeomorphism $f$ such that $V(B) < V(f(B)) <V(B)+V(B)^2$
Because for any $x\in\Bbb R^n$, you should have $$\lim_{r\to 0}\frac{V(f(B(x,r)))}{V(B(x,r))}=|d_xf|$$ and your hypothesis forces this limit to be one everywhere.
Jun
15
comment Is there a diffeomorphism $f$ such that $V(B) < V(f(B)) <V(B)+V(B)^2$
It would seem that such a diffeomorphism has its differential of constant determinant equal to one. But then the change of variable formula tells you that for any measurable set $A$, in particular for any ball $B$, the measure of $f(A)$ is equal to that of $A$, which contradicts the assumption $V(B)<V(f(B))$ for open balls.