TITLE: Fiber bundle and fibration of classifying space
QUESTION [3 upvotes]: Let $BG$ is classifying space of $G$ topological group.
If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the
inclusion map $i:H\rightarrow G$ induces
\begin{equation*}
G/H\rightarrow BH\rightarrow BG
\end{equation*}
a fiber bundle?
If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the
inclusion map $i:H\rightarrow G$ induces
\begin{equation*}
G/H\rightarrow BH\rightarrow BG
\end{equation*}
a fibration?
If $G$ is any compact group and $N$ is a closed normal subgroup of $G$, then
the quotient map $\pi :G\rightarrow G/N$ induces
\begin{equation*}
BN\rightarrow BG\rightarrow B\left( G/N\right)
\end{equation*}
a fiber bundle?
If $G$ is any compact group and $H$ is a closed normal subgroup of $G$, then
the quotient map $\pi :G\rightarrow G/N$ induces
\begin{equation*}
BN\rightarrow BG\rightarrow B\left( G/N\right)
\end{equation*}
a fibration?

REPLY [5 votes]: No, this is not true for Q1 and Q2.
To be specific we probably need to pick a particular model for $BG$, and there are several. For the argument I will give, the following property is sufficient: there exist isomorphisms $\pi_{i+1}(BG) \cong \pi_i(G)$ of homotopy groups.
Let $G = S^1$ and let $H$ be the subgroup of torsion elements (the roots of unity). The group $G$ is abelian so $H$ is normal, and $H$ is totally disconnected so $\pi_0(H) \cong H$ as sets. Any open set in $G$ intersects any coset $gH$ nontrivially, so the quotient space $G/H$ has the indiscrete topology and its homotopy groups vanish.
If the answer to any of the first two questions that you asked was affirmative, there would be an exact sequence in homotopy groups of the form
$$
\cdots \to \pi_1(G/H) \to \pi_1(BH) \to \pi_1(BG) \to \cdots
$$
and this would be an exact sequence
$$
\cdots \to 0 \to H \to 0 \to \cdots
$$
which is not possible.


TITLE: Simple way to find the trace or the bound of the trace of $ ABA^T$?
QUESTION [0 upvotes]: Suppose $A$ and $B$ are square matrices, is there any simple way to find the
trace or the bound of the trace of $ABA^T$?
Thanks.
$B$ has the form of
\begin{align*}
\begin{pmatrix}
0 & a_{1} & a_{2} & a_{3} \\
0 & 0 & a_{1} & a_{2} \\
0 & 0 & 0 & a_{1} \\
0 & 0 & 0 & 0
\end{pmatrix}.
\end{align*}
$A$ is unrestricted.

REPLY [1 votes]: One bound you might like (using the Cauchy Schwartz inequality on the Hilbert Schmidt inner product) is
$$
|tr(ABA')| = |tr(A'AB)| \leq tr(A'A)\sqrt{tr(B'B)}
$$


text:
\newcommand{\ud}{\mathrm{d}}
\newcommand{\cref}[1]{(\ref{#1})}
\newcommand{\arcsinh}{\mathrm{arcsinh}}
\newcommand{\distmax}{D}

\section{The Toponogov point of view}
\label{sec:toponogov}

\subsection{Introduction}

In this appendix we discuss a different approach to finding conditions that guarantee that a Riemannian simplex is non-degenerate, that is diffeomorphic to the standard simplex. It is based on the Toponogov comparison theorem, instead of Rauch's theorem. This comparison theorem says that if the sectional curvatures of a manifold $M$ are bounded from above by $\curvupbnd$ and below by $\curvlowbnd$ and there is a geodesic triangle in $M$ of which we know the lengths of the three geodesics, then the angles of the triangle are bounded by the angles for a geodesic triangle in a space of constant curvature $\curvlowbnd$ or $\curvupbnd$ whose geodesics have the same lengths. Similarly if we are given lengths of two geodesics and the enclosed angle in a geodesic triangle in $M$, the length of the third geodesic is bounded by the lengths of the third geodesic in a geodesic triangle with the same lengths for two geodesics and enclosed angle in a space of constant curvature $\curvlowbnd$ or $\curvupbnd$. The dimension of the manifold shall be denoted by $n$.

\begin{figure}[!htb]
    \centerline{
\mbox{\includegraphics[height=6.2in]{pictures/ToponogovPictureComplexCombination}}
    }
\caption{ Pictorial overview of our approach: Given any point (red) and some vertices (black) we can first compare the angles between the geodesics at the red point on a surface of arbitrary bounded curvature (the ellipsoid in the middle) to the angles in the space of constant curvature. In these spaces of constant curvature, study the simplex by lifting to the tangent space at one of the vertices by the exponential map (bottom figure). }
\label{OverviewFig}
\end{figure}

Our non-degeneracy conditions are established in two steps:
First we note that if for any point $x$ in a neighbourhood of the vertices in the manifold there are $n$ tangents to geodesics connecting this point $x$ to some
subset of the vertices
(the choice of subset does depend on $x$) that are linearly independent, then the Riemannian simplex is non-degenerate. This condition is equivalent with the tangent vectors being affinely independent, because for a point $x$ in a Riemannian simplex with barycentric coordinates $\lambda_i$ we have
$ \sum \lambda_i v_i(x)=0,$ with $\sum \lambda_i= 1$,
remember that we write $v_i(x) = \exp^{-1}_x(v_i)$.
Secondly one has to find conditions on the vertex set in combination with geometric properties of the manifold, to be precise bounds on the sectional curvature, such that we can guarantee linear independence. We do so by looking at the angles between the tangent vectors discussed above. Using the Toponogov comparison theorem we can give estimates on the difference between these angles and the angles one would expect in a space of constant curvature $\curvupbnd$ and $\curvlowbnd$ respectively. These small neighbourhoods in spaces of constant curvature are in turn well approximated by small subsets in Euclidean space. To be precise we compare to the simplex we find by lifting the vertices to the tangent space at one of these vertices via the exponential map. Using these estimates we can prove that if the simplex is small enough compared to the quality of the Euclidean simplex, there is a linearly independent set of tangent vectors so that we have non-degeneracy.
This approach puts the emphasis on the geodesics as apposed to barycentric coordinate functions, which provides us with a very concrete geometric picture, see for example figure \ref{OverviewFig}.

The result to which this method leads to the following result:
\begin{dup}[Theorem~\ref{ConditionSimplices}]
Let $v_0, \ldots ,v_n$ be a set of vertices lying in a Riemannian manifold $M$,
whose sectional curvatures are bounded in absolute value by
$\curvabsbnd$, within a convex geodesic ball of radius $\distmax$ centred at one of the vertices ($v_r$) and such that $\sqrt{\curvabsbnd} \distmax <1/2$. If $\sigma^{\mathbb{E}} (v_r )$, the convex hull of $(\exp^{-1}_{v_r}(v_i))_{i=0}^{n}= (v_i(v_r))_{i=0}^{n}$,
satisfies
\begin{align}
\left(\frac{ (n-1)! \textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{
(2\distmax)^n } \right)^2 > 160 n \sqrt{\curvabsbnd} \distmax, \tag{\ref{QualityCrit}}
\end{align}
then the Riemannian simplex with vertices $v_0, \ldots , v_n$ is non-degenerate, that is diffeomorphic to the standard $n$-simplex.
\end{dup}

\input{prelim_topon}
\input{intrin_topon}

config: arxiv