Search Results

What can be said about the tensor product $T\otimes S$ of two Fredholm operators $T:X_1\to Y_1$ and $S:X_2 \to Y_2$ where $X_1,X_2,Y_1, Y_2$ are Banach spaces and tensor product of operat …

I just realize that the answer is negative: Radu Pantilie, Some remarks on harmonic Riemannian submersion, Bulletin mathématique de la Société des Sciences Mathématiques de Roumanie, Nouvelle Série T …

Is every Riemannian submersion necessarily a Harmonic map? If not under what condition that is true? The motivation: the linear part of a Riemannian submersion is the direct sum og an isometry a …

Is there a Riemannian structure on $\mathbb{R}^n $with two non homeomorphic compact hypersurfaces $M,N$ such that both satisfy the isoperimetric inequality. I precisely meanthe following: $$\el …

Motivation: First I present my motivation for this question but this motivational part is not my main question. I participated in a talk on knot theory. Then I presented the following …

Let $M$ be a compact Riemannian manifold. The norm of vector fields are computed with respect to the metric. Moreover for every distribution $D$, the orthogonal projection on $D$ is den …

My apology in advance if this question is obvious: I know that an Einstein manifold need not have a constant sectional curvature example $\mathbb{C}P^n$. But this space has a constan …

Inspired by this question about conjugation of reql analytic maps to a holomorphic function and with a group action view point we ask the following question. The complex Lie group $H=\math …

Inspired by comment discussions in this MO post smooth version of splitting principle we ask: Are there two compact real analytic manifolds $M,N$ of dimension $m>n$ such that there is not any analyti …

Inspried by this MO question A manifold whose tangent space is a sum of line bundles and higher rank vector bundles we pose the following question as a possible smooth version of the splitting p …

To have a wide familly of counter example lets consider the Yamabe problem which says every compact manifold admite a metric of constant scalar curvature. So every manifold which does not …

Let $(M,g)$ be a Riemannian manifold. So $TM$ has a natural structure of a symplectic manifold. The zero section is denoted by $Z$. We define a Hamiltonian on $T^0 M=TM\setminus Z$ via $$ …

$\newcommand{\Ric}{\operatorname{Ric}}\newcommand{\Iso}{\operatorname{Iso}}$Let $(M,g)$ be a Riemannian manifold with corresponding LC connection and Ricci tensor. Is there an obvious description of …

Some Particular cases are explained in Papaghiuc - On an Einstein structure on the tanent bundle of a space form.

Motivation: The aim of this post is to extend the Lie algebra of a foliation to a bigger Lie algebra. We assume that a manifold $M$ is foliated by compat leaves. The Lie algebra of the foliation is th …