text:
\section{\textbf{Introduction}
Let $(M^n,g)$ be a closed manifold. ${\mathrm R}, {\mathrm W}, {\mathrm {Rc}}, {\mathrm \dot{Rc}}, {\mathrm S}$ denote the Riemann curvature, Weyl tensor, Ricci tensor, traceless Ricci, and scalar curvature, respectively.

The study of the Einstein equation and its various curvature generalizations has a vast literature, see \cite{besse} for an overview. In fact, the Riemann curvature decomposes into three orthogonal components, namely the Weyl curvature, the traceless Ricci part, and a scalar multiple of the identity. All could be seen as operators on the space of two-forms. The Einstein condition says that the traceless Ricci part vanishes and the scalar is just a constant. That is, for an Einstein manifold of dimension at least four, the problem of understanding the Riemann curvature reduces to investigate the Weyl tensor.

Our interest here is essentially about a generalization in terms of that tensor, namely harmonic Weyl curvature, and rigidity results.

For an Einstein manifolds, S. Tachibana shows that non-negative curvature operator implies locally symmetric \cite{tachibana}. Furthermore, S. Brendle extends the result by only assuming nonnegative isotropic curvature \cite{brendle10einstein}. In dimension four, non-negative isotropic curvature is equivalent to a point-wise pinch of an eigenvalue of ${\mathrm W}$ by scalar curvature. In addition, rigidity of Weyl tensor, when its $L^{n/2}$-norm is small, has been intensively investigated, such as \cite{singer92, is02, catino15integral}

A generalization of the Einstein condition is to assume harmonic curvature. If the former corresponds to Ricci curvature being a constant multiple of the metric, then the later does to Ricci curvature a Codazzi tensor. This condition is also of interests due to its connection to the theory of Yang-Mills equation. Consequently, the topic has been studied extensively, such as \cite{derd82, kim11, hv96, tv05, fu15pinched}. In particular, rigidity results due to integral pinched conditions are obtained in \cite{hv96, fu15pinched}. Also, A. Gray shows that, if $(M,g)$ additionally has positive sectional curvature then it must be Einstein. In connection with our interests, harmonic curvature is characterized by harmonic Weyl curvature and constant scalar curvature.

Consequently, it is natural to investigate the condition of harmonic Weyl curvature. In this direction, most of the research has focused on the case of K\"{a}hler or dimension four. A K\"{a}hler manifold with harmonic Weyl must have parallel Ricci tensor \cite{gray78, tanno72, mat72}. In dimension four, Weyl decomposes into self-dual and anti-self-dual parts (${\mathrm W}^{\pm}$). A. Polombo \cite{polombo92} derived Bochner-Weitzenb\"ock formula for a Dirac operator and obtained applications. Then, M. Micallef and M. Wang \cite{mw93} showed that a closed four-dimensional manifold with harmonic self-dual and non-negative isotropic curvature (on the self-dual part) must be either conformally flat or quotient of a K\"{a}hler metric with constant scalar curvature. As mentioned earlier, that curvature assumption is equivalent to a point-wise bound, for $\omega$ any eigenvalue of ${\mathrm W}^+$, $$\omega \leq \frac{{\mathrm S}}{6}.$$
Furthermore, Gursky \cite{g00} made use of the point-wise Bochner-Weitzenb\"ock  formula for Weyl tensor to derive optimal $L^2$-gap theorems. \\

Unfortunately, in higher dimension, very little is known. Our purpose here is to fill in some of those gaps, deriving analogous rigidity results when assuming harmonic Weyl tensor. First, we show a generalization of Tachibana's theorem.
\begin{theorem}
\label{nonnegative}
Let $(M^n,g)$, $n\geq 5$, be a closed Riemannian manifold with harmonic Weyl tensor and non-negative curvature operator. Then $(M,g)$ must be either locally conformally flat or locally symmetric.
\end{theorem}
\begin{remark} In dimension four, the same statement holds and is a weaker version of Micallef and Wang's theorem. In higher dimension, we also obtain more general results in terms of pinching eigenvalues and norms, see Theorems \ref{mainpointwise} and \ref{normthm}.
\end{remark}

Next, by using the solution to the Yamabe's problem, we obtain an integral gap.
\begin{theorem}
\label{mainintegral}
Let $(M^n,g)$, $n\geq 6$, be a closed Riemannian manifold with harmonic Weyl tensor and positive Yamabe invariant $\lambda(g)$. Then,
\[a_1(n)||{\mathrm W}||_{L^{n/2}}+a_2(n)||{\mathrm \dot{Rc}}||_{L^{n/2}}\geq \frac{n-2}{4(n-1)}{\lambda(g)},\]
unless $(M,g)$ is locally conformally flat. Here, for $\alpha$ the biggest root of the quadratic,
\[8(n-1)^2\alpha^2-2n(n-1)(n-2)\alpha+n(n-2)(n-3)=0,\] we have
\begin{align*}
a_1(n) &=\frac{10(n-1)\alpha^2}{2(n-1)\alpha-n+3} \text{,} &a_2(n) &=\frac{2(n-1)\alpha^2}{2(n-1)\alpha-n+3}\sqrt{\frac{n-1}{n}}.
\end{align*}
\end{theorem}
\begin{remark} For large $n$, $\alpha \approx n/4$ and $a_1\approx 5n/4$ and $a_2\approx n/4$.
\end{remark}
\begin{remark} The casse $n=5$ and $n=4$ are treated somewhat differently, see Theorem \ref{case5} and Section \ref{fourdim} respectively.
\end{remark}
It should be noted that, if just diffeomorphism type is desired, there are general rigidity results. Several authors, G. Huisken \cite{Huisken} C. Margerin \cite{margerin86} and S. Nishikawa \cite{nishi86}, independently showed that point-wise pinching on the Weyl and traceless Ricci curvature implies diffeomorphism to a space form. E. Hebey and M. Vaugon \cite{hv96} extended it to $L^{n/2}$ pinching. Dimension four is also special; Margerin \cite{margerin98} obtained the optimal/sharp estimate in dimension four. Furthermore, by using a solution to a fourth order fully nonlinear equation in conformal geometry, S. Chang, M. Gursky, and P. Yang \cite{CGY03} obtained an $L^2$-sharp version.

Putting everything together, the picture becomes clearer. If ${\mathrm W}$ and ${\mathrm \dot{Rc}}$ are small, either in point-wise or integral sense, then $(M,g)$ must be diffeomorphic to a space form. If, furthermore, it has harmonic Weyl curvature then, isometrically, $(M,g)$ must be conformally flat or locally symmetric.\\

The key ingredients in our proof are new Bochner-Weitzenb\"ock-Lichnerowicz (BWL) type identities connecting the gradient and divergence of ${\mathrm W}$. Here is a prototype.
\begin{theorem}
\label{main1}
For a closed manifold $(M^n,g)$, $n\geq 4$, we have:
\begin{equation}
\label{BWintegral}
\int_M |\nabla {\mathrm W}|^2-\frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2=2\int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle -\int_M\left\langle{{\mathrm {Rc}}\circ g, {\mathrm W}^2}\right\rangle.
\end{equation}
\end{theorem}
\begin{remark}
For relevant notation, see Section \ref{notation}.
\end{remark}
\begin{remark}
In dimension four, the result could be deduced from the work of Chang, Gursky, and Yang \cite[Equation 3.23]{CGY03}, for details see Section \ref{fourdim}. Also A. Lichnerowicz \cite{lich58} derived an identity relating the gradient and divergence of Riemann curvature, which is a vital step in the proof of Tachibana's theorem.
\end{remark}
\begin{remark} For more identities, see Section \ref{gradvsdiv}.
\end{remark}

A BWL type formula basically implies that zero-order terms could control derivative terms, see \cite[Chapter 7]{pe06book} or \cite{caotran1}. As a consequence, it is fundamental to obtain rigidity and gap theorems: assumptions of smallness on curvature quantity leads to vanishing results. Thus, our argument relies on the new identities to conclude that $\nabla {\mathrm W}\equiv 0$. Then, by a result of A. Derdzinski and W. Roter \cite[Theorem 2]{dr77}, for a positive definite metric, parallel Weyl implies either locally conformally flat or locally symmetric.

The organization of the paper is as follows. In Section \ref{notation}, we explain our convention and recalls several preliminaries. Section \ref{gradvsdiv} discusses the connection between gradient and divergence of the Weyl tensor and proves several identities including Theorem \ref{main1}.  In Section \ref{gap}, we obtain rigidity theorems, point-wise and integral.
Finally, the Appendix collects some useful computation and explain why dimension four is special.

{\bf Acknowledgments:} We thank Professors Xiaodong Cao, Richard Schoen, and Bennett Chow for various discussion and suggestions.

\section{\textbf{Preliminaries}}
\label{notation}
Throughout the paper, we adopt the following notation.
\begin{itemize}
\item $\{e_i\}_1^n$ is a normal orthonormal frame.
\item $\wedge^k$ is the space of $k$-forms, or fully anti-symmetric $(k,0)$ tensor.
\item $S^2(.)$ is the space of self-adjoint/symmetric operators.
\item $S_0^2(.)$ is the space of traceless symmetric operators.
\end{itemize}
The Riemann curvature $(3,1)$ tensor and $(4,0)$ tensor are defined as follows,
\begin{align*}
{\mathrm R}(X,Y,Z)&=-\nabla^2_{X,Y}Z +\nabla^2_{Y,X}Z\\
{\mathrm R}(X,Y,Z,W)&=-\left\langle{\nabla^2_{X,Y}Z -\nabla^2_{Y,X}Z, W}\right\rangle_g.
\end{align*}
\begin{remark}
Our (3,1) curvature sign agrees with \cite{besse, brendlebook10} and opposite to \cite{chowluni, H3, pe06book}. Our (4,0) curvature convention is the same as \cite{besse, H3, brendlebook10} and opposite to \cite{chowluni, pe06book}
\end{remark}
Consequently, we have the Ricci identity, for a $p$-tensor S,
\begin{align}
\label{ricciidentity}
&(\nabla^2_{X,Y} S)(Z_1,...,Z_p)-(\nabla^2_{Y,X} S)(Z_1,...,Z_p),\\
&=\sum_{i,k}{\mathrm R}(X,Y,Z_i, e_k)S(...Z_{i-1},e_k, Z_{i+1}..).\nonumber
\end{align}

For any $(m,0)$-tensor $T$, its divergence is defined as
\[(\delta T)_{p_{2}...p_{m}}\doteqdot \sum_{i} \nabla_{i}T_{p_{2}...p_{m}i}. \]

 For $T\in S^2(\wedge^2)$, $|T|$ denotes the operator norm (which is $1/2$ of the tensor norm). Accordingly, the convention of norms is as follows.
 \begin{itemize}
 \item If $T\in T^{*}\otimes S^2(\wedge^2)$ , $|T|^{2}=\sum_{i}\sum_{a<b; c<d}(T_{iabcd})^2. $
\item If $T \in \wedge^2 \otimes T^{*} $, $|T|^2=\sum_{i}\sum_{a<b}(T_{abi})^2. $
\item If $T\in \Lambda^3\otimes \Lambda^2$, $|T|^{2}=\sum_{i<j<k}\sum_{a<b}(T_{ijkab})^2.$
\end{itemize}

\subsection{Algebraic curvature} The exposition here follows from \cite[Chapter I.G]{besse}.

Let V be a real vector space equipped with a non-degenerate quadratic form $g$ and an orthonormal basis $\{e_i\}_{i=1}^n$.
Associated is an orthogonal group $O(g)$.
The irreducible orthogonal decomposition of the $O(g)$-module $\otimes ^2 V$ is the following,
\[ \otimes^2 V= \wedge^2 V\oplus S_0^2(V) \oplus \mathbb{R}g.\]

We recall the Bianchi map on $\otimes^4 (V)$,
\[b(R)(X,Y,Z,W)\doteqdot \frac{1}{3}(R(X,Y,Z,W)+R(Y,Z,X,W)+R(Z,X,Y,W)).\]

So we have the $GL(V)$-equivariant decomposition,
\[S^2 (\wedge^2 V)=\text{Ker}b\otimes \text{Im}b.\]

\begin{definition} $\mathfrak{C}V\doteqdot \text{Ker}(b)\cap S^2(\wedge^2 V)$ is the space of algebraic curvature tensor.
\end{definition}

We recall the Ricci contraction, an $O(g)$-equivariant map $S^2 (\wedge^2 V)\mapsto S^2(V)$,
\[ rc(R)(X,Y)=\text{tr}_g R(X,.,Y,.).\]

Conversely, the Kulkarni-Nomizu product gives a canonical way to build an element of $S^2(\wedge^2 V)$ from two elements of $S^2(V)$.
\begin{align*}
(h\circ k) (X,Y,Z,W)=& h(X,Z)k(Y,W)+ k(X,Z)h(Y,W)\\
&-h(X,W)k(Y,Z)-k(X,W)h(Y,Z).\end{align*}

The map $g\circ .: S^2(V)\mapsto \mathfrak{C}V$ is injective and its adjoint is precisely $rc$.
\begin{equation}
\label{selfadjoint}
 \left\langle{g\circ k, R}\right\rangle=\left\langle{k, rc(R)}\right\rangle .\end{equation}

If $n\geq 4$, the $O(g)$-module $\mathfrak{C}V$ has the following decomposition into unique irreducible subspaces,
\begin{align*}
\mathfrak{C}V &= \mathfrak{S}V\oplus \mathfrak{Rc}V\oplus \mathfrak{W}V;\\
 \mathfrak{S}V &=\mathbb{R}g\circ g,\\
 \mathfrak{Rc}V &=q\circ (S_0^2 V),\\
 \mathfrak{W}V &=\text{Ker}(rc)\cap \text{Ker}b.
\end{align*}
Conveniently, $\mathfrak{S}V$ is just a multiple of the identity; $\mathfrak{Rc}V$ is the traceless Ricci part; $\mathfrak{W}V$ is the Weyl curvature. By symmetry, $\mathfrak{W}V$ is trivial in dimension less than four.

Next we collect some definitions of quadratic product on algebraic curvature tensors (they are also related to Ricci flow, see, for example, \cite{H3, bohmwilking, brendlebook10}).
\begin{definition} For algebraic curvature tensors $R,S$, we define,
\begin{align}
\label{Rsquared}
(R.S)_{ijkl}&=\frac{1}{2}\sum_{p,q}R_{ijpq}S_{klpq},\\
\label{Rsharp}
(R\sharp S)_{ijkl}&=\frac{1}{2}\sum_{p,q}\Big[R_{ipkq}S_{jplq}+S_{ipkq}R_{jplq}\\
&-R_{iplq}S_{jpkq}-S_{iplq}R_{jpkq}];\nonumber\\
R^2(X,Y,Z,W)&=(R.R)(X,Y,Z,W)=\frac{1}{2}\sum_{p,q}R(X,Y,e_{p},e_{q})R(Z,W,e_{p},e_{q})\nonumber \\
R^{\sharp}(X,Y,Z,W) &=(R\sharp R)(X,Y, Z,W)\nonumber\\
&=\sum_{p,q}(R(X,e_{p},Z,e_{q})R(Y,e_{p},W,e_{q})-R(X,e_{p},W,e_{q})R(Y,e_{p},Z,e_{q}))\nonumber.
\end{align}
\end{definition}
\begin{remark} It is clear that both products are distributive with respect to addition. Also $\sharp$ is commutative.
\end{remark}
Even though $R^2, R^\sharp$ are not necessarily in $\mathfrak{C}V$, $R^2+R^\sharp$ is. Furthermore,
\begin{equation}
\label{rcsimplify}
rc(R^2+R^\sharp)(X,Y)=\sum_{p,q}R(X,e_p,Y,e_q)rc(R)(e_p, e_q).\end{equation}
Finally, it is observed that, the 3-linear map,
\begin{equation}
\text{tri}(R_1, R_2, R_3)\doteqdot \left\langle{R_1.R_2+R_2.R_1+2R_1\sharp R_2,R_3}\right\rangle,
\end{equation}
is symmetric in all three components $R_1, R_2, R_3\in S^2(\wedge^2 V)$.
\subsection{Curvature Decomposition}
Using the algebraic disucssion above, the Riemann curvature has the following decomposition:
\begin{equation*}
{\mathrm R}={\mathrm W}+\frac{{\mathrm S} g\circ g }{2n(n-1)} + \frac{{\mathrm \dot{Rc}} \circ g}{n-2}={\mathrm W}-\frac{{\mathrm S} g\circ g}{2(n-2)(n-1)} + \frac{{\mathrm {Rc}} \circ g}{n-2}.
\end{equation*}
We note that, as $(4,0)$ tensors at a point $x$, ${\mathrm W}\in \mathfrak{W}(T_xM)$, ${\mathrm \dot{Rc}}\circ g\in \mathfrak{Rc}(T_xM)$, ${\mathrm S} g\circ g\in \mathfrak{S}(T_xM)$ are orthogonal components. Consequently,
\begin{align*}
{\mathrm R} &= {\mathrm W}+\frac{{\mathrm S} g\circ g }{2n(n-1)} + \frac{{\mathrm \dot{Rc}} \circ g}{n-2}={\mathrm W}+U+V, \\
|{\mathrm R}|^2 &=|{\mathrm W}|^2+|U|^2+|V|^2, \\
|U|^2 &=\frac{1}{2n(n-1)}{\mathrm S}^2,\\
|V|^2 &=\frac{1}{n-2}|{\mathrm \dot{Rc}}|^2.
\end{align*}
The Weyl curvatutre is an algebraic curvature which is traceless so it follows that,
\[W_{1212} =\sum_{2<i<j}W_{ijij}.\]
More generally, if the tangent space is decomposed into orthogonal subspaces $N_1, N_2$ then,
\begin{equation}
\label{weylsect}
{\mathrm W}_{N_1}\doteqdot \sum_{i<j, i,j\in N_{1}}W_{ijij} =\sum_{k<l, k,l\in N_{2}}W_{klkl}\doteqdot {\mathrm W}_{N_2}.\end{equation}
That is, the Weyl ``generalized sectional curvature''s of complementing subspaces are equal and then well-defined.

In dimension four the decomposition becomes,
\[{\mathrm R} ={\mathrm W}+\frac{{\mathrm S}}{24} g\circ g+ \frac{1}{2}{\mathrm \dot{Rc}} \circ g.\]

A special feature of dimension four is that the Hodge $\ast$ operator decomposes the space of two-forms orthogonally according to eigenvalues $\pm1$: $\wedge^{2}=\wedge^2_{+}\otimes \wedge^2_{-}$. Also, we observe that ${\mathrm W}(\Lambda^{2}_{\pm})\in \Lambda^{2}_{\pm}$; so it is unambiguous to define ${\mathrm W}^{\pm}\doteqdot {\mathrm W}^{|\Lambda_{\pm}}$

Accordingly, the curvature as an operator on $\wedge^2$ is,
\begin{equation}
\label{curdec}
{\mathrm R}=
 \left( \begin{array}{cc}
A^{+} & C \\
C^{T} & A^{-} \end{array} \right).
\end{equation}
Here C is essentially the traceless Ricci. In addition,
\begin{align*}
A^{\pm} &={\mathrm W}^{\pm}+\frac{S}{12}\text{Id}^{\pm},\\
|A^{\pm}|^{2} &=|{\mathrm W}^{\pm}|^{2}+\frac{{\mathrm S}^{2}}{48}, \\
|{\mathrm {Rc}}|^2-\frac{{\mathrm S}^2}{4}&=|{\mathrm \dot{Rc}}|^2=4|C|^2=4\text{tr}(CC^T).
\end{align*}

\section{\textbf{Gradient Versus Divergent}}
\label{gradvsdiv}
This section investigates the connection between the gradient and divergence of the Weyl curvature.
Let's recall the following operators, for $T\in \mathfrak{C}V$, $A\in \Lambda^2\otimes T^{*}M$
\begin{align}
B(T)_{ijkmn}&=\nabla_i T_{jkmn}+\nabla_j T_{kimn}+\nabla_k T_{ijmn};\\
(A\circ' g)_{ijkmn}&=g_{kn}A_{ijm}+g_{in}A_{jkm}+g_{jn}A_{kim}\nonumber \\
&+g_{km}A_{jin}+g_{im}A_{kjn}+g_{jm}A_{ikn}.
\end{align}
\begin{remark} If $b$ corresponds to the first Bianchi's identity, $B$ does the second. Also, \[B(T), A\circ' g\in \Lambda^3\otimes \Lambda^2.\]
\end{remark}
\begin{lemma}
We have, for $V$ of dimension $n$ at least four,
\[|A\circ' g|^2= (n-3)|A|^2\]
\end{lemma}
\begin{proof} The statement follows from a tedious computation. Below are the key steps.
\begin{align*}
|A\circ' g|^2&=\frac{1}{12}\sum_{i,j,k,m,n}(A\circ' g)_{ijkmn}^2;\\
\sum_{i,j,k, m,n} (g_{kn}A_{ijm})^2&=d\sum_{i,j,m}A_{ijm}^2=2n|A|^2;\\
\sum_{i,j,k, m,n} g_{kn}g_{in}A_{ijm}A_{jkm}&=\sum_{j,m,n}A_{njm}A_{jnm}=-2|A|^2;\\
\sum_{i,j,k, m,n} g_{kn}g_{km}A_{ijm}A_{jin}&=\sum_{i,j,k}A_{ijk}A_{jik}=-2|A|^2;\\
\sum_{i,j,k, m,n} g_{kn}g_{im}A_{ijm}A_{kjn}&=\sum_{i,k,j}A_{iji}A_{kjk}=0.
\end{align*}
\end{proof}

Also, for a Riemann curvature, $B({\mathrm R})=0$. That implies algebraic relations when applying map $B$ on components of the curvature. To see that, first we recall:
\begin{definition}
\label{framedef}
The tensors $P,Q: \wedge^2 TM \otimes T^*M$ are defined as:
\begin{align}
\label{equaP}
P_{ijk}&=\nabla_{i}{\mathrm {Rc}}_{jk}-\nabla_{j}{\mathrm {Rc}}_{ik}.\\
P(X\wedge Y,Z)&=(d_{\nabla}{\mathrm {Rc}})(X,Y,Z)=\delta{{\mathrm R}}(Z,X,Y), \nonumber \\
\label{equaQ}
Q_{ijk}&=g_{ki}\nabla_{j}{\mathrm S}-g_{kj}\nabla_{i}{\mathrm S}.
\end{align}
\end{definition}

\begin{remark} Thus, harmonic Riemann curvature is equivalent to that ${\mathrm {Rc}}$ is a Codazzi tensor.
It is also well-known that, for example see \cite{chowluni},
\[ \delta {\mathrm W}=-\frac{n-3}{n-2}(P+\frac{1}{2(n-1)}Q).
\]
\end{remark}

\begin{lemma}
\label{bianchimap}
We have the followings
\begin{enumerate}
\item $B(Rc\circ g)= P\circ' g.$
\item $B(S g\circ g)=-Q\circ' g$
\item $B({\mathrm W})=\frac{1}{n-3}(\delta {\mathrm W}) \circ' g$
\end{enumerate}
\end{lemma}
\begin{proof} Since we consider equality of tensors, it suffices to show it for a normal orthonormal frame.
\begin{enumerate}
\item We compute:
\begin{align*}
({\mathrm {Rc}}\circ g)_{jkmn}&={\mathrm {Rc}}_{jm}g_{kn}+{\mathrm {Rc}}_{kn}g_{jm}-{\mathrm {Rc}}_{jn}g_{km}-{\mathrm {Rc}}_{km}g_{jn};\\
\nabla_i ({\mathrm {Rc}}\circ g)_{jkmn}&=\nabla_i{\mathrm {Rc}}_{jm}g_{kn}+\nabla_i{\mathrm {Rc}}_{kn}g_{jm}-\nabla_i{\mathrm {Rc}}_{jn}g_{km}-\nabla_i{\mathrm {Rc}}_{km}g_{jn};\\
B({\mathrm {Rc}}\circ g)_{ijkmn}&=\nabla_i{\mathrm {Rc}}_{jm}g_{kn}+\nabla_i{\mathrm {Rc}}_{kn}g_{jm}-\nabla_i{\mathrm {Rc}}_{jn}g_{km}-\nabla_i{\mathrm {Rc}}_{km}g_{jn};\\
&+\nabla_j{\mathrm {Rc}}_{km}g_{in}+\nabla_j{\mathrm {Rc}}_{in}g_{km}-\nabla_j{\mathrm {Rc}}_{kn}g_{im}-\nabla_j{\mathrm {Rc}}_{im}g_{kn}\\
&+\nabla_k{\mathrm {Rc}}_{im}g_{jn}+\nabla_k{\mathrm {Rc}}_{jn}g_{im}-\nabla_k{\mathrm {Rc}}_{in}g_{jm}-\nabla_k{\mathrm {Rc}}_{jm}g_{in}\\
&=g_{kn}P_{ijm}+g_{in}P_{jkm}+g_{jn}P_{kim}\\
&+g_{km}P_{jin}+g_{im}P_{kjn}+g_{jm}P_{ikn}.
\end{align*}
\item We compute:
\begin{align*}
({\mathrm S} g\circ g)_{jkmn}&={\mathrm S} 2g_{jm}g_{kn}-2{\mathrm S} g_{jn}g_{km};\\
\nabla_i ({\mathrm S} g\circ g)_{jkmn}&=\nabla_i {\mathrm S} 2(g_{jm}g_{kn}-2{\mathrm S} g_{jn}g_{km});\\
B({\mathrm S} g\circ g)_{ijkmn}&=\nabla_i{\mathrm S} 2(g_{jm}g_{kn}-g_{jn}g_{km});\\
&+\nabla_j{\mathrm S} 2(g_{km}g_{in}-g_{kn}g_{im})\\
&+\nabla_k{\mathrm S} 2(g_{im}g_{jn}-g_{in}g_{jm})\\
&=g_{kn}Q_{jim}+g_{in}Q_{kjm}+g_{jn}Q_{ikm}\\
&+g_{km}P_{ijn}+g_{im}Q_{jkn}+g_{jm}Q_{kin}.
\end{align*}
\item For the last statement, recall
\begin{align*}
B({\mathrm R})&=0;\\
{\mathrm R} &= {\mathrm W}-\frac{S g\circ g}{2(n-2)(n-1)}+\frac{{\mathrm {Rc}}\circ g}{n-2};\\
(\delta {\mathrm W}) &=-\frac{n-3}{n-2}(P+\frac{Q}{2(n-1)}).
\end{align*}
\end{enumerate}
\end{proof}
\begin{remark} The key observation here is that harmonic Weyl curvature is equivalent to Weyl satisfying the second Bianchi's identity.
\end{remark}
\begin{remark}
Using the decomposition above, we can show that
\[\frac{1}{n-3}|\delta {\mathrm W}|^2=|B(W)|^2\leq 3|\nabla {\mathrm W}|^2.\]
\end{remark}
The inequality above is, however, generally not sharp so we'll expose the connection between gradient and divergence by part integration as follows.
\subsection{Integral formula}
First, observe that by (\ref{rcsimplify}),
\[rc({\mathrm W}^2+{\mathrm W}^\sharp)(.,.)=\sum_{p,q}{\mathrm W}(.,e_p,.,e_q)rc({\mathrm W})(e_p,e_q)\equiv 0.\]
Therefore, ${\mathrm W}^2+{\mathrm W}^\sharp\in \mathfrak{W}$.

Let $A\in S^2(TM)$ and $B=A\circ g\in \mathfrak{C}(TM)$. We have the following.
\begin{lemma}\label{productW}
We have, for a basis diagonalizing both $A$ and $g$ at a point then,
\begin{align}
\left\langle{B, {\mathrm W}^2+{\mathrm W}^{\sharp}}\right\rangle &=0;\\
\left\langle{{\mathrm W}^2, B}\right\rangle=-\left\langle{{\mathrm W}, B\sharp {\mathrm W}}\right\rangle &=\frac{1}{2}\sum_{i,j,p,q} A_{ii}{\mathrm W}_{ijpq}^2;\\
\left\langle{{\mathrm W}, {\mathrm R}\sharp {\mathrm W}}\right\rangle &= \frac{1}{2}{\mathrm W}_{ijkl}{\mathrm W}_{jplq}{\mathrm R}_{ipkq}.
\end{align}
\end{lemma}
\begin{proof}
By (\ref{selfadjoint}), we have,
\begin{align*}
\left\langle{B, {\mathrm W}^2+{\mathrm W}^{\sharp}}\right\rangle &=\left\langle{A, rc({\mathrm W}^2+{\mathrm W}^{\sharp})}\right\rangle.
\end{align*}
The first statement follows since ${\mathrm W}^2+{\mathrm W}^\sharp\in \mathfrak{W}$.

Next, if we choose a basis diagonalizing both $A$ and $g$ then,
\begin{align*}
8\left\langle{B,{\mathrm W}^2}\right\rangle &=\sum_{i,j,k,l} B_{ijkl}\sum_{p,q}{\mathrm W}_{ijpq}{\mathrm W}_{pqkl}\\
&=2\sum_{i,j,p,q} B_{ijij}{\mathrm W}_{ijpq}{\mathrm W}_{ijpq}=2\sum_{i,j,p,q}(A_{ii}+A_{jj}){\mathrm W}_{ijpq}{\mathrm W}_{ijpq}\\
&=4\sum A_{ii}{\mathrm W}_{ijpq}{\mathrm W}_{ijpq}.
\end{align*}

Finally, by re-indexing,
\begin{align*}
8\left\langle{{\mathrm W}, {\mathrm R}\sharp {\mathrm W}}\right\rangle &= {\mathrm W}_{ijkl}\Big[{\mathrm R}_{ipkq}{\mathrm W}_{jplq}+{\mathrm W}_{ipkq}{\mathrm R}_{jplq}\\
&-{\mathrm R}_{iplq}{\mathrm W}_{jpkq}-{\mathrm R}_{jpkq}{\mathrm W}_{iplq}\Big]\\
&=2{\mathrm W}_{ijkl}({\mathrm R}_{ipkq}{\mathrm W}_{jplq}+{\mathrm W}_{ipkq}{\mathrm R}_{jplq});\\
&=4{\mathrm W}_{ijkl}{\mathrm W}_{jplq}{\mathrm R}_{ipkq}.
\end{align*}

\end{proof}
Now, we are ready to prove Theorem \ref{main1}. The proof is inspired by some arguments of \cite[Proposition 3.3]{CGY03}.
\begin{proof}
We have,
\begin{align*}
4\int_M |\nabla {\mathrm W}|^2 &= \int_M \sum_{m, i,j,k,l} \nabla_m {\mathrm W}_{ijkl}(B(W)_{mijkl}-\nabla_i {\mathrm W}_{jmkl}-\nabla_j {\mathrm W}_{mikl}),\\
\int_M \sum_{m, i,j, k,l} \nabla_m {\mathrm W}_{ijkl}B(W)_{mijkl} &=\int_M \sum_{m, i,j, k,l} \nabla_m {\mathrm W}_{ijkl}\frac{1}{n-3}(\delta {\mathrm W}\circ' g)_{mijkl}\\
&=\frac{1}{n-3}\int_M \sum_{i,j, k,l}\nabla_m{\mathrm W}_{ijkm}(\delta{{\mathrm W}})_{ijk}+\nabla_m{\mathrm W}_{ijml}(\delta{\mathrm W})_{jil}\\
&=\frac{4}{n-3}\int_M |\delta{\mathrm W}|^2;\\
-2\int_M\sum_{m,i,j, k,l} \nabla_m {\mathrm W}_{ijkl}\nabla_i {\mathrm W}_{jmkl}&=-2\int_M\sum_{m,i,j, k,l} \nabla_m {\mathrm W}_{ijkl}\nabla_j {\mathrm W}_{mikl},\\
&=2\int_M\sum_{m,i,j, k,l}{\mathrm W}_{ijkl}\nabla_m\nabla_i{\mathrm W}_{jmkl}.
\end{align*}

Using (\ref{ricciidentity}),
\begin{align*}
\nabla_{m}\nabla_i{\mathrm W}_{jmkl}-\nabla_{i}\nabla_m{\mathrm W}_{jmkl} &= \sum_s {\mathrm R}_{mijs}{\mathrm W}_{smkl}+{\mathrm R}_{mims}{\mathrm W}_{jskl}+{\mathrm R}_{miks}{\mathrm W}_{jmsl}+{\mathrm R}_{mils}{\mathrm W}_{jmks};\\
2\int_M\sum_{m,i,j, k,l}{\mathrm W}_{ijkl}\nabla_{i}\nabla_m{\mathrm W}_{jmkl} &=-\int_M\sum_{m,i,j, k,l}\nabla_{i}{\mathrm W}_{ijkl}\nabla_m{\mathrm W}_{jmkl}=4\int_M (\delta{\mathrm W})^2.
\end{align*}
For the zero order terms we compute, using the Bianchi's first identity and re-indexing,
\begin{align*}
2\sum_{m,i,j,k,l,s}{\mathrm W}_{ijkl}{\mathrm R}_{mijs}{\mathrm W}_{smkl}&=\sum_{m,i,j,k,l,s}{\mathrm R}_{mijs}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}+\sum_{m,i,j,k,l,s}{\mathrm R}_{mijs}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}\\
&=\sum_{m,i,j,k,l,s}{\mathrm R}_{mijs}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}+\sum_{m,i,j,k,l,s}{\mathrm R}_{mjis}{\mathrm W}_{jikl}{\mathrm W}_{smkl}\\
&=\sum_{m,i,j,k,l,s}{\mathrm R}_{mijs}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}+\sum_{m,i,j,k,l,s}{\mathrm R}_{jmis}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}\\
&=\sum_{m,i,j,k,l,s}({\mathrm R}_{mijs}+{\mathrm R}_{jmis}){\mathrm W}_{ijkl}{\mathrm W}_{smkl}\\
&=-\sum_{m,i,j,k,l,s}{\mathrm R}_{ijms}{\mathrm W}_{ijkl}{\mathrm W}_{smkl}\\
&=8\left\langle{{\mathrm R},{\mathrm W}^2}\right\rangle.
\end{align*}
Next, if we choose a basis diagonalizing both ${\mathrm {Rc}}$ and $g$ then, by Lemma \ref{productW}, for $T=Rc\circ g$,
\begin{align*}
2\sum_{m,i,j,k,l,s}{\mathrm W}_{ijkl}{\mathrm R}_{mims}{\mathrm W}_{jskl}&=2\sum_{i,j,k,l}{\mathrm R}_{ii}{\mathrm W}_{ijkl}{\mathrm W}_{jikl}\\
&=-4\left\langle{T,{\mathrm W}^2}\right\rangle.
\end{align*}
Next, also by Lemma \ref{productW},
\begin{align*}
2\sum_{m,i,j,k,l.s}{\mathrm W}_{ijkl}({\mathrm R}_{miks}{\mathrm W}_{jmsl}+{\mathrm R}_{mils}{\mathrm W}_{jmks})&=4{\mathrm W}_{ijkl}{\mathrm R}_{miks}{\mathrm W}_{jmsl}\\
&=8\left\langle{{\mathrm W},{\mathrm R}\sharp{\mathrm W}}\right\rangle.
\end{align*}
Summing equations above yields,
\begin{align}
\label{supmain1}
4\int_M |\nabla {\mathrm W}|^2 &=\frac{4(n-2)}{n-3}\int_M (\delta{\mathrm W})^2\\
&+8\int_M \left\langle{{\mathrm R},{\mathrm W}^2}\right\rangle-4\int_M \left\langle{T,{\mathrm W}^2}\right\rangle+8\int_M \left\langle{{\mathrm W},{\mathrm R}\sharp {\mathrm W}}\right\rangle\nonumber
\end{align}
Recalling that
\[ {\mathrm R}={\mathrm W}+\frac{{\mathrm S} g\circ g }{2n(n-1)} + \frac{{\mathrm \dot{Rc}} \circ g}{n-2}={\mathrm W}+B.\]

Since the curvature products are distributive, we obtain,
\begin{align}
&\int_M \left\langle{{\mathrm R},{\mathrm W}^2}\right\rangle+\int_M \left\langle{{\mathrm W},{\mathrm R}\sharp {\mathrm W}}\right\rangle \nonumber\\
=&\int_M \left\langle{{\mathrm W},{\mathrm W}^2}\right\rangle+\int_M \left\langle{B,{\mathrm W}^2}\right\rangle+\int_M \left\langle{{\mathrm W},{\mathrm W}\sharp {\mathrm W}}\right\rangle+\int_M \left\langle{B,{\mathrm W}\sharp}\right\rangle \nonumber\\
\label{supmain2}
=& \int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle.
\end{align}
The last equality is due to Lemma \ref{productW} again. Finally, the theorem follows from (\ref{supmain1}) and (\ref{supmain2}).
\end{proof}

Inspired by Tachibana \cite{tachibana}, the theorem could be restated as follows.
\begin{theorem}
\label{tachi}
For a closed manifold $(M^n,g)$, $n\geq 4$, we have:
\begin{equation*}
\int_M |\nabla {\mathrm W}|^2-\frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2=-2\int_M \left\langle{{\mathrm W}^{\flat},\delta(\nabla {\mathrm W})-\nabla (\delta{\mathrm W})}\right\rangle.
\end{equation*}
Here, ${\mathrm W}^\flat_{iklj}={\mathrm W}_{jikl}$.
\end{theorem}
\begin{proof}
The zero order terms in the proof above are essentially, for a normal coordinate,
\begin{align*}
\sum_{i,j,k,l,m}{\mathrm W}_{ijkl}(\nabla_m\nabla_i{\mathrm W}_{jmkl}-\nabla_i\nabla_m{\mathrm W}_{jmkl})&= \sum_{i,j,k,l}{\mathrm W}_{ijkl}[(\delta(\nabla {\mathrm W}))_{iklj}-(\nabla (\delta{\mathrm W}))_{iklj}],\\
&=-4\left\langle{{\mathrm W}^{\flat},\delta(\nabla {\mathrm W})-\nabla (\delta{\mathrm W})}\right\rangle.
\end{align*}
So the result follows.
\end{proof}
\subsection{Harmonic Weyl}
In this section, using the method similar to above, we obtain identities for a manifold with harmonic Weyl curvature. First, we show the following identity for the Ricci tensor.
\begin{proposition}
\label{ricci}
For a closed manifold $(M^n,g)$ with harmonic Weyl tensor, $n\geq 4$,
\begin{equation}
\int_M |\nabla {\mathrm {Rc}}|^2-\frac{n}{4(n-1)}\int_M (\nabla {\mathrm S})^2=\int_M {\mathrm W}({\mathrm \dot{Rc}},{\mathrm \dot{Rc}})-\frac{n}{n-2}{\mathrm \dot{Rc}}^3-\frac{1}{n-1}{\mathrm S}|{\mathrm \dot{Rc}}|^2.
\end{equation}
Here, for an orthonormal frame, ${\mathrm W}({\mathrm \dot{Rc}},{\mathrm \dot{Rc}})=\sum_{i,j,k,l}{\mathrm W}_{ijkl}{\mathrm \dot{Rc}}_{ik}{\mathrm \dot{Rc}}_{jl}$, ${\mathrm \dot{Rc}}^3=\sum_{i,j,k}{\mathrm \dot{Rc}}_{ij}{\mathrm \dot{Rc}}_{jk}{\mathrm \dot{Rc}}_{ki}$.
\end{proposition}
\begin{proof}
First, we recall, by the second Bianchi's identity,
\[ \delta{\mathrm {Rc}}=\frac{1}{2}\nabla {\mathrm S}.\]
Also, since $\delta {\mathrm W}=0$,
\[-P=\frac{1}{2(n-1)}Q.
\]
We have,
\begin{align*}
\int_M |\nabla {\mathrm {Rc}}|^2 &= \int_M \sum_{i,j,k} \nabla_i {\mathrm R}_{jk}(P_{ijk}+\nabla_j {\mathrm R}_{ik}),\\
\int_M \sum_{i,j, k} \nabla_i {\mathrm R}_{jk}P_{ijk} &=-\frac{1}{2(n-1)}\int_M \sum_{i,j, k} \nabla_i {\mathrm R}_{jk}Q_{ijk}\\
&=-\frac{1}{2(n-1)}\int_M \sum_{i,j, k}\nabla_i{\mathrm R}_{jk}(g_{ki}\nabla_j {\mathrm S}-g_{kj}\nabla_i {\mathrm S})\\
&=\frac{1}{4(n-1)}\int_M |\nabla{\mathrm S}|^2;\\
\int_M\sum_{i,j, k} \nabla_i {\mathrm R}_{jk}\nabla_j {\mathrm R}_{ik}&=-\int_M\sum_{i,j, k} {\mathrm R}_{jk}\nabla_i\nabla_j {\mathrm R}_{ik}.
\end{align*}

Using (\ref{ricciidentity}),
\begin{align*}
\nabla_{i}\nabla_j{\mathrm R}_{ik}-\nabla_{j}\nabla_i{\mathrm R}_{ik} &= \sum_s {\mathrm R}_{ijis}{\mathrm R}_{sk}+{\mathrm R}_{ijks}{\mathrm R}_{is};\\
-\int_M\sum_{i,j,k}{\mathrm R}_{jk}\nabla_{j}\nabla_i{\mathrm R}_{ik} &=\int_M\sum_{i,j, k}\nabla_{j}{\mathrm R}_{jk}\nabla_i{\mathrm R}_{ik}=\frac{1}{4}\int_M (\nabla{\mathrm S})^2.
\end{align*}
For the zero order terms we compute, using the Bianchi's first identity and re-indexing,
\begin{align*}
-\sum_{i,j,k,s}{\mathrm R}_{jk}{\mathrm R}_{ijis}{\mathrm R}_{sk}&=-\sum_{s,j,k}{\mathrm R}_{jk}{\mathrm R}_{js}{\mathrm R}_{sk};\\
-\sum_{i,j,k,s}{\mathrm R}_{jk}{\mathrm R}_{ijks}{\mathrm R}_{is}&=\sum_{i,j,k,s}{\mathrm R}_{ijks}{\mathrm R}_{ik}{\mathrm R}_{js}.\end{align*}
By a tedious computation,
\begin{align*}
{\mathrm {Rc}}^3\doteqdot \sum_{s,j,k}{\mathrm R}_{jk}{\mathrm R}_{js}{\mathrm R}_{sk} &={\mathrm \dot{Rc}}^3+\frac{3}{n}{\mathrm S}|{\mathrm {Rc}}|^2-\frac{2{\mathrm S}^3}{n^2}\\
&={\mathrm \dot{Rc}}^3+\frac{3}{n}{\mathrm S}|{\mathrm \dot{Rc}}|^2+\frac{{\mathrm S}^3}{n^2};\\
{\mathrm R}({\mathrm {Rc}},{\mathrm {Rc}})\doteqdot \sum_{i,j,k,s}{\mathrm R}_{ijks}{\mathrm R}_{ik}{\mathrm R}_{js}&={\mathrm W}({\mathrm {Rc}},{\mathrm {Rc}})-\frac{2}{n-2}{\mathrm \dot{Rc}}^3+\frac{{\mathrm S}^3}{n^2}+\frac{2n-3}{n(n-1)}{\mathrm S}|{\mathrm \dot{Rc}}|^2.
\end{align*}
\end{proof}
\begin{remark}
Huisken \cite[Lemma 4.3]{Huisken} shows that for any Riemannian manifold, $|\nabla{\mathrm {Rc}}|^2\geq \frac{3n-2}{2(n-1)(n+2)}|\nabla {\mathrm S}|^2$.
\end{remark}
Furthermore, we have a point-wise identity for Weyl tensor. This is a generalization of one in dimension four, see \cite[16.73]{besse}.
\begin{theorem}
\label{pointwiseBW}
For a closed manifold $(M^n,g)$, $n\geq 4$, with harmonic Weyl curvature, we have:
\begin{equation}
\label{pointwiseW}
\Delta |{\mathrm W}|^2=2|\nabla {\mathrm W}|^2-4\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle +2\left\langle{{\mathrm {Rc}}\circ g, {\mathrm W}^2}\right\rangle.
\end{equation}
\end{theorem}
\begin{proof}
We have, by Lemma \ref{bianchimap} and harmonic Weyl curvature, $B({\mathrm W})\equiv 0$.
\begin{align*}
2\Delta |{\mathrm W}|^2&= 4|\nabla {\mathrm W}|^2+\sum_{m,i,j,k,l}{\mathrm W}_{abcd}\nabla_m\nabla_m {\mathrm W}_{ijkl};\\
\sum_{m,i,j,k,l}{\mathrm W}_{ijkl}\nabla_m \nabla_m{\mathrm W}_{ijkl} &=\sum_{m, i,j,k,l}  {\mathrm W}_{ijkl}\nabla_m(-\nabla_i {\mathrm W}_{jmkl}-\nabla_j {\mathrm W}_{mikl}),\\
&=-2\sum_{m,i,j, k,l} {\mathrm W}_{ijkl}\nabla_m\nabla_i {\mathrm W}_{jmkl}.
\end{align*}

Using (\ref{ricciidentity}),
\begin{align*}
\nabla_{m}\nabla_i{\mathrm W}_{jmkl}-\nabla_{i}\nabla_m{\mathrm W}_{jmkl} &= \sum_s {\mathrm R}_{mijs}{\mathrm W}_{smkl}+{\mathrm R}_{mims}{\mathrm W}_{jskl}+{\mathrm R}_{miks}{\mathrm W}_{jmsl}+{\mathrm R}_{mils}{\mathrm W}_{jmks};\\
2\sum_{m,i,j,k,l}{\mathrm W}_{ijkl}\nabla_{i}\nabla_m{\mathrm W}_{jmkl}
&=\sum_{m,i,j,k,l}{\mathrm W}_{ijkl}\nabla_i(\delta {\mathrm W})_{klj}\equiv 0.
\end{align*}
For the zero order terms we compute as in the proof of Theorem \ref{main1},
\begin{align*}
-2\sum_{m,i,j,k,l,s}{\mathrm W}_{ijkl}{\mathrm R}_{mijs}{\mathrm W}_{smkl}&=
=-8\left\langle{{\mathrm R},{\mathrm W}^2}\right\rangle;\\
-2\sum_{m,i,j,k,l,s}{\mathrm W}_{ijkl}{\mathrm R}_{mims}{\mathrm W}_{jskl}&=4\left\langle{T,{\mathrm W}^2}\right\rangle;\\
-2\sum_{m,i,j,k,l}{\mathrm W}_{ijkl}({\mathrm R}_{miks}{\mathrm W}_{jmsl}+{\mathrm R}_{mils}{\mathrm W}_{jmks})&=8\left\langle{{\mathrm W},{\mathrm R}\sharp{\mathrm W}}\right\rangle.
\end{align*}
Summing equations above yields,
\begin{align}
\sum_{m,i,j,k,l}{\mathrm W}_{ijkl}\nabla_m \nabla_m{\mathrm W}_{ijkl} &=-8\int_M \left\langle{{\mathrm R},{\mathrm W}^2}\right\rangle+4\int_M \left\langle{T,{\mathrm W}^2}\right\rangle-8\int_M \left\langle{{\mathrm W},{\mathrm R}\sharp {\mathrm W}}\right\rangle\nonumber\\
&=-8\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle+4\int_M \left\langle{T,{\mathrm W}^2}\right\rangle
\end{align}
The result then follows.

\end{proof}
\section{\textbf{Rigidity Theorems}}
\label{gap}
In this section, we derive applications of new Bochner-Weitzenb\"ock-Lichnerowicz type formulas. Precisely, results include point-wise and integral rigidity theorems.

First, we'll investigate the quantity, $\left\langle{{\mathrm W}, {\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle$. Recall that, by re-indexing,
\begin{align*}
8\left\langle{{\mathrm W}, {\mathrm W}^\sharp}\right\rangle &= 2\sum_{i,j,k,l,p,q}{\mathrm W}_{ijkl}({\mathrm W}_{ipkq}{\mathrm W}_{jplq}-{\mathrm W}_{iplq}{\mathrm W}_{jpkq}),\\
&= 4\sum_{i,j,k,l,p,q}{\mathrm W}_{ijkl}{\mathrm W}_{ipkq}{\mathrm W}_{jplq};\\
8\left\langle{{\mathrm W}, {\mathrm W}^2}\right\rangle &=\sum_{i,j,k,l,p,q}{\mathrm W}_{ijkl}{\mathrm W}_{ijpq}{\mathrm W}_{klpq}.
\end{align*}

When $4\leq n\leq 5$, due to \cite[A1]{jp87}, (beware there is a typo in (A1))
\begin{equation}\label{sharpcubic}
\left\langle{{\mathrm W}, {\mathrm W}^\sharp}\right\rangle=2\left\langle{{\mathrm W},{\mathrm W}^2}\right\rangle\doteqdot 2{\mathrm W}^3.
\end{equation}
Thus,
\begin{equation}
\label{five}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle= 3{\mathrm W}^3. \end{equation}
\begin{remark} Subsection \ref{pure} gives a computation for pure curvature.
\end{remark}
For a general dimension, we follow the argument in \cite{Huisken,tachibana} and define, for fixed m, n, p, q, a local skew symmetric tensor,
\begin{align*}
u_{ij}^{(mnpq)} &= {\mathrm W}_{inpq}g_{jm}+{\mathrm W}_{mipq}g_{jn}+{\mathrm W}_{mniq}g_{jp}+{\mathrm W}_{mnpi}g_{jq}\\
&-{\mathrm W}_{jnpq}g_{im}-{\mathrm W}_{mjpq}g_{in}-{\mathrm W}_{mnjq}g_{ip}-{\mathrm W}_{mnpj}g_{iq}.
\end{align*}
It follows that,
\begin{align*}
8|u|^2=\sum_{m,n,p,q}\left\langle{u_{ij}^{(mnpq)},u_{ij}^{(mnpq)}}\right\rangle &= 32(n-1)|{\mathrm W}|^2,\\
8\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle &=\frac{1}{8}\left\langle{W_{ijkl} u_{ij}^{(mnpq)},u_{kl}^{(mnpq)}}\right\rangle.
\end{align*}
If $\omega$ be the largest magnitude of any eigenvalue of ${\mathrm W}$, then by Berger's estimate \cite{berger60b},
\[{\mathrm W}_{ijkl}\leq \frac{4}{3}\omega.\]
It follows that,
\begin{equation}
\label{six}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle =\frac{1}{64}\left\langle{W_{ijkl} u_{ij}^{(mnpq)},u_{kl}^{(mnpq)}}\right\rangle \leq \frac{4}{3}\frac{1}{64}\omega 8|u|^2= \frac{2}{3}(n-1)|{\mathrm W}|^2 \omega.
\end{equation}
Thus, the computation above proves the following result.
\begin{lemma}
\label{atleast5}
Let $\omega$ be the largest magnitude of an eigenvalue of ${\mathrm W}$ then, for $n\geq 5$,
\begin{equation}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle\leq \frac{2(n-1)}{3} \omega|{\mathrm W}|^2.
\end{equation}
\end{lemma}
\begin{remark} \label{fivecase} For $n=5$, the result holds with the same constant for largest eigenvalue of ${\mathrm W}$ due to (\ref{five}) and Lemma \ref{Wcubic}.
\end{remark}

We also have an estimate just in terms of norm. First we have the following result, for a proof see \cite[Lemma 2.4]{Huisken}.
\begin{lemma} \label{eigenestimate}
Let $T$ be a symmetric trace-free operator on an $m$-dimensional vector space $V$. If $\lambda$ is an eigenvalue then,
\[\lambda^2 \leq \frac{m-1}{m} |T|^2.\]
\end{lemma}
Then the following holds.
\begin{lemma}
\label{atleast5norm}
For $n\geq 5$,
\begin{equation}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle\leq c(n)|{\mathrm W}|^3.
\end{equation}
Here, we find $c(5)=\frac{8}{\sqrt{10}}$ and $c(n)=5$ for $n\geq 6$.
\end{lemma}
\begin{proof}
For any dimension, by Cauchy-Schwartz,
\begin{align*}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle &=\frac{1}{8}(\sum_{i,j,k,l,p,q}4{\mathrm W}_{ijkl}{\mathrm W}_{ipkq}{\mathrm W}_{jplq}+{\mathrm W}_{ijkl}{\mathrm W}_{ijpq}{\mathrm W}_{klpq}),\\
&\leq 5|{\mathrm W}|^3.
\end{align*}
In dimension five, ${\mathrm W}$ is an operator on a $10$-dimensional vector space. By Lemma \ref{Wcubic} and \ref{eigenestimate},
\[\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle=3{\mathrm W}^3\leq 3\sqrt{\frac{9}{10}}\frac{8}{9}|{\mathrm W}|^3=\frac{8}{\sqrt{10}}|{\mathrm W}|^3.\]
\end{proof}
\begin{remark} For $n>5$, $c(n)$ might be improved slightly, see \cite[Lemma 2.3]{catino15integral}.
\end{remark}

\subsection{Point-wise}
In this subsection, we'll show results under point-wise assumptions.
First, we prove Theorem \ref{nonnegative}.

\begin{proof}
By the work of Tachibana \cite{tachibana} (see also \cite[Lemma 7.33]{chowluni}), non-negative curvature operator implies that,
\begin{align*}
\left\langle{{\mathrm W}^{\flat},\delta(\nabla {\mathrm W})-\nabla (\delta{\mathrm W})}\right\rangle &\geq 0.
\end{align*}
Therefore, by Theorem \ref{tachi} and harmonicity of Weyl, $\nabla{\mathrm W}\equiv 0$. Since a Riemannian metric is always positive definite, due to Derdzinski and Roter \cite[Theorem 2]{dr77}, the metric must be either locally conformally flat or locally symmetric.
\end{proof}
The next result is concerned with bounding eigenvalues by the scalar curvature.

\begin{proposition}
\label{p1} Let $(M^n,g)$, $n\geq 5$, be a closed Riemannian manifold. Let $\omega$ be the largest magnitude of an eigenvalue of ${\mathrm W}$ and $-\ell$ the smallest eigenvalue of ${\mathrm \dot{Rc}}$. Suppose at each point,
$$\frac{2(n-1)}{3}\omega+\ell\leq \frac{{\mathrm S}}{n},$$ then the following holds,
\[\int_M |\nabla {\mathrm W}|^2\leq \frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2.\]
In dimension five, the same result holds if $\omega$ is the largest eigenvalue of ${\mathrm W}$.
\end{proposition}

\begin{proof}
By Theorem \ref{main1},
\begin{align*}
\int_M |\nabla {\mathrm W}|^2 -\frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2 &=2\int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle -\int_M\left\langle{{{\mathrm {Rc}}\circ g},{\mathrm W}^2}\right\rangle,\\
\int_M \left\langle{{\mathrm W}^2,{\mathrm {Rc}}\circ g}\right\rangle &=\left\langle{({\mathrm \dot{Rc}}+\frac{{\mathrm S}}{n}g)\circ g,{\mathrm W}^2}\right\rangle,\\
&=\left\langle{{\mathrm \dot{Rc}}\circ g,{\mathrm W}^2}\right\rangle+\frac{2{\mathrm S}}{n}|{\mathrm W}|^2.
\end{align*}
By Lemma \ref{atleast5},
\[\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle\leq \frac{2(n-1)}{3}\omega|{\mathrm W}|^2.\]
By standard inequality, an eigenvalue of $E\circ g$ must be greater than or equal $-2\ell$. Thus,
\[\left\langle{{\mathrm \dot{Rc}}\circ g,{\mathrm W}^2}\right\rangle\geq -2\ell |{\mathrm W}|^2.\]
Therefore, if $$\frac{4(n-1)}{3}\omega+2\ell\leq \frac{2{\mathrm S}}{n},$$
then,
\[\int_M |\nabla {\mathrm W}|^2 -\frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2\leq 0.\]

In dimension five, we could use Remark (\ref{fivecase}) instead of Lemma \ref{atleast5}.
\end{proof}

Restricted to the case of harmonic Weyl, we obtain the following result.
\begin{theorem}
\label{mainpointwise} Let $(M^n,g)$, $n\geq 5$, be a closed Riemannian manifold with harmonic Weyl tensor. Let $\omega$ be the largest magnitude of an eigenvalue of ${\mathrm W}$ and $-\ell$ the smallest eigenvalue of ${\mathrm \dot{Rc}}$. Suppose at each point,
$$\frac{2(n-1)}{3}\omega+\ell\leq \frac{{\mathrm S}}{n},$$ then $(M,g)$ must be either conformally flat or locally symmetric. In dimension five, the same result holds if $\omega$ is the largest eigenvalue of ${\mathrm W}$.
\end{theorem}
\begin{proof}
For harmonic Weyl tensor, $\delta {\mathrm W}\equiv 0$. Then Prop. \ref{p1} implies that $\nabla {\mathrm W}\equiv 0$. Thus, $(M,g)$ has parallel Weyl curvature. Since a Riemannian metric is always positive definite, the work of Derdzinski and Roter \cite[Theorem 2]{dr77} says that the metric must be either locally conformally flat or locally symmetric.
\end{proof}
The next theorem yields a similar conclusion under an assumption on the norm.
\begin{theorem}\label{normthm}
Let $(M^n,g)$, $n\geq 5$, be a closed Riemannian manifold. Suppose at each point,
$$c(n)|{\mathrm W}|+\sqrt{\frac{n-1}{n}}|{\mathrm \dot{Rc}}|\leq \frac{{\mathrm S}}{n},$$ then then the following holds,
\[\int_M |\nabla {\mathrm W}|^2\leq \frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2.\]
Here $c(n)$ is the same as in Lemma \ref{atleast5norm}.
\end{theorem}
\begin{proof}
The proof is similar to one above. The only difference is to obtain following inequalities in terms of norm.
\begin{align*}
\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle &\leq c(n)|{\mathrm W}|^3;\\
-\left\langle{{\mathrm \dot{Rc}}\circ g,{\mathrm W}^2}\right\rangle &\leq 2\ell |{\mathrm W}|^2\leq 2\sqrt{\frac{n-1}{n}}|{\mathrm \dot{Rc}}||{\mathrm W}|^2.
\end{align*}
The first line is just Lemma \ref{atleast5norm} while the second line is justified by Lemma \ref{eigenestimate}.
\end{proof}

\subsection{Integral}
This subsection will focus on integral gap results using the solution to the Yamabe problem.

\textbf{Yamabe problem}: Given a compact Riemannian manifold $(M,g)$ of dimension $n\geq 3$, find a conformal metric with constant scalar curvature.

For a change of metric, the volume form and scalar curvature transform as follows.
\begin{align*}
\overline{g}&=u^{4/(n-2)}g,\\
d\overline{\mu}&=u^{2n/(n-2)}d\mu,\\
\overline{{\mathrm S}}&=u^{-(n+2)/(n-2)}\Big(-\frac{4(n-1)}{n-2}\Delta +{\mathrm S}\Big)u=u^{-(n+2)/(n-2)}L_gu,\\
L_g&\doteqdot-\frac{4(n-1)}{n-2}\Delta +{\mathrm S}.
\end{align*}
Yamabe \cite{yam60} observed that the problem is equivalent to find the minimizer of the functional,
\[Q[\overline{g}]=\frac{\int_M \overline{{\mathrm S}} d\overline{\mu}}{(\int_M d\overline{\mu})^{(n-2)/n}}=Y[u]=\frac{\int_{M} u Lu d\mu}{(\int_M u^{2n/(n-2)}d\mu)^{(n-2)/n}}.\]
The Yamabe invariant is defined accordingly,
\[\lambda(g)=\inf\{Y[u]:~ u>0,~ u \in C^2(M)\}.\]
The problem has an extensive history and was finally settled by R. Schoen using the positive mass theorem\cite{schoen84}, see \cite{yam60, tru68, aub76} for partial results and \cite{lp87} for an expository account.
When the Yamabe invariant is positive, a consequence of the solution is a conformal Sobolev inequality, for $s_n = \frac{n-2}{4(n-1)}$,
\begin{equation}
\label{sobolev}
s_n\lambda (g)(\int_M |u|^{\frac{2n}{n-2}}d\mu_g)^{\frac{n-2}{n}}\leq \int_M|\nabla_g u|^2 d\mu_g+s_n\int_M{\mathrm S} |u|^2 d\mu_g.  \end{equation}
That inequality will play an crucial role in the proof of Theorem \ref{mainintegral}.

\begin{proposition}
\label{integralrigid}
Let $(M^n,g)$, $n\geq 5$, be a closed Riemannian manifold with positive Yamabe invariant $\lambda(g)$. Assume that,
\[c_1(n)||{\mathrm W}||_{L^{n/2}}+c_2(n)||{\mathrm \dot{Rc}}||_{L^{n/2}}\leq d(n){\lambda(M)}.\]
Then the following holds,
\[(1-\frac{d(n)}{s_n})\int_M |\nabla {\mathrm W}|^2 \leq \frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2+(d(n)-\frac{2}{n})\int_M {\mathrm S}|{\mathrm W}|^2.\]
Here for $c(n)$ the same as in Lemma \ref{atleast5norm},
\begin{align*}
c_1(n) &=2c(n), &c_2(n) &=2\sqrt{\frac{n-1}{n}}.
\end{align*}
\end{proposition}
\begin{proof}
From Theorem \ref{main1},
\begin{align*}
\int_M |\nabla {\mathrm W}|^2-\frac{(n-2)}{n-3}\int_M (\delta{\mathrm W})^2 &=2\int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle -\int_M\left\langle{{{\mathrm {Rc}}\circ g},{\mathrm W}^2}\right\rangle.
\end{align*}
By Lemma \ref{atleast5norm},
\begin{align*}
2\int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle &\leq \int_M 2c(n)|{\mathrm W}|^3= c_1(n)|{\mathrm W}|^3.\\
\int_M \left\langle{{\mathrm W}^2, E \circ g}\right\rangle &\leq \int_M 2\ell|{\mathrm W}|^2\leq \int_M 2\sqrt{\frac{n-1}{n}}|{\mathrm \dot{Rc}}||{\mathrm W}|^2= c_2(n)|{\mathrm \dot{Rc}}||{\mathrm W}|^2.
\end{align*}

Applying the H\"{o}lder's inequality and the conformally invariant Sobolev's inequality,
\begin{align*}
\int_M |{\mathrm W}|^3 =||{\mathrm W}||_{L^3}^3 &\leq ||{\mathrm W}||_{L^{n/2}}||{\mathrm W}||^2_{L^{2n/(n-2)}}\\
 &\leq \frac{||{\mathrm W}||_{L^{n/2}}}{\lambda(g)}(\frac{1}{s_n}\int_M|\nabla |{\mathrm W}||^2+\int_M {\mathrm S}|{\mathrm W}|^2);\\
\int_M |{\mathrm \dot{Rc}}||{\mathrm W}|^2  &\leq ||{\mathrm \dot{Rc}}|_{L^{n/2}}||{\mathrm W}||^2_{L^{2n/(n-2)}}\\
 & \leq \frac{||{\mathrm \dot{Rc}}||_{L^{n/2}}}{\lambda(g)}(\frac{1}{s_n}\int_M|\nabla |{\mathrm W}||^2+\int_M {\mathrm S}|{\mathrm W}|^2).
\end{align*}
Also we recall the classical Kato inequality,
$$|\nabla {\mathrm W}|^2\geq |\nabla |{\mathrm W}||^2.$$
Combining inequalities above yield,
\begin{align*}
\int_M|\nabla {\mathrm W}|^2+ \frac{2{\mathrm S}}{n}|{\mathrm W}|^2 &\leq c_1(n)\int_M |{\mathrm W}|^3+c_2(n)\int_M |{\mathrm \dot{Rc}}||{\mathrm W}|^2,\\
&\leq \frac{c_1(n)||{\mathrm W}||_{L^{n/2}}+c_2(n)||{\mathrm \dot{Rc}}||_{L^{n/2}}}{\lambda(g)}\Big(\frac{1}{s_n}|\nabla |{\mathrm W}||^2+\int_M {\mathrm S}|{\mathrm W}|^2\Big),\\
&\leq d(n)(\frac{1}{s_n}|\nabla {\mathrm W}|^2+\int_M {\mathrm S}|{\mathrm W}|^2).
\end{align*}

\end{proof}
The proof of Theorem \ref{mainintegral} is a refinement of the above method.
\begin{proof}
For harmonic Weyl tensor, $\delta {\mathrm W}\equiv 0$. Also we recall the improved Kato inequality for harmonic Weyl tensor \cite{branson00},
$$|\nabla {\mathrm W}|^2\geq \frac{n+1}{n-1}|\nabla |{\mathrm W}||^2 .$$
By Theorem \ref{pointwiseBW},
\begin{align*}
\Delta |{\mathrm W}|^2 &=2|\nabla {\mathrm W}|^2-4\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle +2\left\langle{{\mathrm {Rc}}\circ g, {\mathrm W}^2}\right\rangle;\\
|{\mathrm W}|\Delta|{\mathrm W}| &=\frac{1}{2}\Delta |{\mathrm W}|^2-|\nabla |{\mathrm W}||^2\\
&\geq \frac{2}{n-1}|\nabla |{\mathrm W}||^2-2\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle +\left\langle{{\mathrm {Rc}}\circ g, {\mathrm W}^2}\right\rangle,\\
&\geq \frac{2}{n-1}|\nabla |{\mathrm W}||^2+\frac{2{\mathrm S}}{n}|{\mathrm W}|^2-c_1(n)|{\mathrm W}|^3-c_2(n)|{\mathrm \dot{Rc}}||{\mathrm W}|^2.
\end{align*}
Here $c_1(n), c_2(n)$ are as in Prop. \ref{integralrigid}. Now we suppose $|{\mathrm W}|>0$ everywhere and let $u=|{\mathrm W}|$ (if ${\mathrm W}$ assumes value zero at some point, we could replace $u=|{\mathrm W}|+\epsilon$, $\epsilon>0$ and use a standard limit argument for $\epsilon\rightarrow 0$).
\begin{align*}
u^\alpha\Delta u^\alpha &=\frac{\alpha-1}{\alpha}|\nabla u^\alpha|^2+\alpha u^{2\alpha-2} u\Delta u,\\
&\geq (1-\frac{n-3}{(n-1)\alpha})|\nabla u^\alpha|^2+\frac{2\alpha}{n}{\mathrm S} u^{2\alpha}-\alpha c_1(n)u^{2\alpha+1}-c_2(n)\alpha |{\mathrm \dot{Rc}}|u^{2\alpha}
\end{align*}
Integration by parts and applying H\"{o}lder's inequality yield,
\begin{align*}
0 &\geq (2-\frac{n-3}{(n-1)\alpha})\int_M|\nabla u^\alpha|^2+\frac{2\alpha}{n}\int_M{\mathrm S} u^{2\alpha}-\alpha c_1(n)\int_M u^{2\alpha+1}-\alpha c_2(n) \int_M|{\mathrm \dot{Rc}}|u^{2\alpha},\\
&\geq (2-\frac{n-3}{(n-1)\alpha})||\nabla u^\alpha||_{L^2}^2+ \frac{2\alpha}{n}||{\mathrm S} u^{2\alpha}||_{L^1}\\
&-\alpha c_1(n)||{\mathrm W}||_{L^{n/2}} ||u^{\alpha}||_{L^{2n/(n-2)}}^2-\alpha c_2(n)||{\mathrm \dot{Rc}}||_{L^{n/2}} ||u^{\alpha}||_{L^{2n/(n-2)}}^2
\end{align*}
Applying the Sobolev inequality (\ref{sobolev}), we have,
\begin{align*}
 & A(n,\alpha) \doteqdot (2-\frac{n-3}{(n-1)\alpha}); \\
& \alpha c_1(n)||{\mathrm W}||_{L^{n/2}} ||u^{\alpha}||_{L^{2n/(n-2)}}^2+\alpha c_2(n)||{\mathrm \dot{Rc}}||_{L^{n/2}} ||u^{\alpha}||_{L^{2n/(n-2)}}^2 \\
&\geq A(n,\alpha) \int_M|\nabla u^\alpha|^2+\frac{2\alpha}{n}\int_M{\mathrm S} u^{2\alpha}\\
&\geq s_n\lambda(g)A(n,\alpha)||u^{\alpha}||_{L^{2n/(n-2)}}^2+(\frac{2\alpha}{n}-A(n,\alpha)s_n)\int_M{\mathrm S} u^{2\alpha}.
\end{align*}
We consider,
\begin{align*}
\frac{2\alpha}{n}-A(n,\alpha)s_n &=\frac{8(n-1)^2\alpha^2-2n(n-1)(n-2)\alpha+n(n-2)(n-3)}{4n(n-1)^2\alpha}.
\end{align*}
It follows that if $n\neq 5$, we could choose $\frac{n-3}{2(n-1)}<\alpha$ such that $\frac{2\alpha}{n}-A(n,\alpha)s_n=0$. In that case, if the assumption in the theorem does not hold, then ${\mathrm W}\equiv 0$.
\end{proof}

It is noted from the proof that if $n=5$ then the quadratic above is always positive. So it remains to choose $\frac{1}{4}<\alpha$ to minimize $\frac{2\alpha^2}{4\alpha-1}$. It follows that we could choose $\alpha=\frac{1}{2}$ and use Lemma \ref{atleast5norm} to obtain the following result.
\begin{theorem}\label{case5}
Let $(M^5,g)$ be a closed Riemannian manifold with harmonic Weyl tensor and positive scalar curvature and Yamabe invariant $\lambda(g)$. Then,
\[\frac{8}{\sqrt{10}}||{\mathrm W}||_{L^{n/2}}+\frac{2}{\sqrt{5}}||{\mathrm \dot{Rc}}||_{L^{n/2}}\geq \frac{3}{16}{\lambda(g)},\]
unless $(M,g)$ is locally conformally flat.
\end{theorem}

\section{\textbf{Appendix}}
First, we'll show the following general estimate.
\begin{lemma}
\label{Wcubic}
Let $\sum_{i=1}^n x_i=0$ and $x_i\leq s$ then,
\begin{equation}
x^3=\sum_{i=1}^n x_i^3\leq \frac{s(n-2)}{n-1} \sum_{i=1}^n x_i^2=\frac{s(n-2)}{n-1}x^2.
\end{equation}
\end{lemma}
\begin{proof}
When $x^2=\sum_{i=1}^n x_i^2=0$, the statement follows vacuously.

When $x^2>0$, the problem reduces to maximize $f(x)=\frac{x^3}{x^2}$ given constraints
\begin{itemize}
\item $\sum_{i=1}^n x_i=0$;
\item $x_i\leq s>0$.
\end{itemize}

By compactness of the constraints, a maximum exists.
Also we observe that the maximum must be at least
\begin{equation}\label{atleast}\frac{s^3-(n-1)(\frac{s}{n-1})^3}{s^2+(n-1)(\frac{s}{n-1})^2}=\frac{s(n-2)}{n-1}.
\end{equation}
By Karush-Kuhn-Tucker extension of the Lagrange multiplier method, a maximum of $f(x)$ is obtained at $x$ such that:
\begin{itemize}
\item $\nabla f(x)=\sum_{i=1}^n \mu_i \nabla (x_i-s)+\lambda \nabla (\sum_{i=1}^n x_i)$.
\item $\mu_i\geq 0$ and $\mu_i=0$ if $x_i<s$.
\end{itemize}
The first condition implies that, for any $i$,
\[3x_i^2 x^2-2x_i x^3=x^4 (\mu_i+\lambda).
\]
By the second condition, if $x_i<s$, $x_i$ can only assume one of two values, the roots $y_1< y_2$ of the quadratic on y: $3x^2 y^2-2x^3 y_i-\lambda x^4=0$. Therefore, there are two cases.

\textbf{Case 1:} There are $m$ $y_1$'s, $p$ $y_2$'s, $q$ $s$ and $y_2<s$. Then we have,
\begin{align*}
my_1+py_2+qs &=0;\\
f(x)=\frac{x^3}{x^2} &=\frac{p y_2^3+q s^3+my_1^3}{py_2^2 +qs^2+my_1^2}\\
&=\frac{py_2^3+qs^3-y_1^2(py_2+qs)}{py_2^2+qs^2-y_1(py_2+qs)}=\frac{py_2(y_2^2-y_1^2)+qs(s^2-y_1^2)}{py_2(y_2-y_1)+qs(s-y_1)}.
\end{align*}
We observe that,
\begin{align*}
\frac{py_2(y_2^2-y_1^2)+qs(s^2-y_1^2)}{py_2(y_2-y_1)+qs(s-y_1)} &\leq \frac{p s(s^2-y_1^2)+qs(s^2-y_1^2)}{ps(s-y_1)+qs(s-y_1)}\\
\leftrightarrow p^2 s y_2 (s-y_1)(y_2^2-y_1^2)&+qspy_2(y_2^2-y_1^2)(s-y_1)\\
\leq p^2 sy_2(s^2-y_1^2)(y_2-y_1) &+qspy_2(s^2-y_1^2)(y_2-y_1)\\
\leftrightarrow qspy_2(y_2-y_1)(s-y_1)(y_2-s)&\leq p^2 sy_2(y_2-y_1)(s-y_1)(s-y_2).
\end{align*}
So $f(x)\leq s+y_1$. As $p, q>0$, $-y_1>\frac{s}{n-1}$ and so $f(x)<\frac{s(n-2)}{n-1}$, a contradiction to (\ref{atleast}). So this case is ruled out.

\textbf{Case 2:} There are $m$ $y_1$'s, $q$ $s$. Then we have,
\begin{align*}
my_1+qs &=0;\\
f(x)=\frac{x^3}{x^2} &=\frac{q s^3+my_1^3}{qs^2+my_1^2}\\
&=\frac{qs^3-y_1^2qs}{qs^2-y_1qs}=s+y_1=s(1-\frac{q}{m})\leq s(1-\frac{1}{n-1})=\frac{s(n-2)}{n-1}.
\end{align*}
That concludes the proof.
\end{proof}
\subsection{Four-manifolds}
\label{fourdim}
In this section, we explain how computation simplifies greatly in dimension four.

When $n=4$, ${\mathrm W}={\mathrm W}^{+}+{\mathrm W}^{-}$. First, we observe the following.
\begin{lemma} In dimension four,
\[\left\langle{{\mathrm \dot{Rc}}\circ g, {\mathrm W}^2}\right\rangle= 0.\]
The statement also holds when replacing ${\mathrm W}$ by ${\mathrm W}^{\pm}$.
\end{lemma}
\begin{proof}
It is well-known that, for example see \cite[Lemma 2.1]{caotran1}
\[\sum_{k,p,q}{\mathrm W}_{ikpq}{\mathrm W}_{j}^{~kpq}=|{\mathrm W}|^2 g_{ij}.\]
Therefore, for an orthonormal frame diagonalizing both $g$ and ${\mathrm \dot{Rc}}$,
\[\left\langle{{\mathrm \dot{Rc}}\circ g, {\mathrm W}^2}\right\rangle= \sum_{i,j,p,q}{\mathrm \dot{Rc}}_{ii}{\mathrm W}_{ijpq}^2=|{\mathrm W}|^2\sum_i {\mathrm \dot{Rc}}_{ii}=0.\]
\end{proof}

As a consequence, our BWL type formula (\ref{BWintegral}) becomes,
\begin{equation}
\label{bw4}
\int_M |\nabla {\mathrm W}|^2-2\int_M (\delta{\mathrm W})^2=\int_M \left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle -\frac{1}{2}\int_M {\mathrm S}|{\mathrm W}|^2.
\end{equation}
\begin{remark} Indeed, the same identity holds when replacing ${\mathrm W}$ by ${\mathrm W}^{\pm}$.
\end{remark}
We observe that ${\mathrm W}^{\pm}$ is each a self-adjoint operator on a dimension three vector space. As ${\mathrm W}^{\pm}$ is traceless and satisfies the first Bianchi identity, there is a normal form by M. Berger \cite{berger61, st69}.
That is, there exists an orthonormal basis $\{e_{i}\}_{i=1}^4$ of $T_{p}M$, consequently $\{e_{12}, e_{13}, e_{14}, e_{34}, e_{42}, e_{23}\}$ being a basis of $\Lambda_{2}$, such that
\begin{align}
\label{normalform}
{\mathrm W} &=
 \left( \begin{array}{cc}
A & B \\
B & A \end{array} \right);\\
A &=\text{diag}(a_{1}, a_{2}, a_{3}), \nonumber\\
B &=\text{diag}(b_{1}, b_{2}, b_{3}),\nonumber\\
0 &=a_1+a_2+a_3=b_1+b_2+b_3.\nonumber
\end{align}
Then, the self-dual and anti-self-dual ${\mathrm W}^{\pm}$ could be computed,
\[{\mathrm W}^{\pm}=
 \left( \begin{array}{cc}
\frac{A\pm B}{2} & \frac{B\pm A}{2} \\
\frac{B\pm A}{2} & \frac{A\pm B}{2} \end{array} \right).\]
Using this set-up, it is immediate to check that,
\begin{align*}
\left\langle{{\mathrm W}^+, ({\mathrm W}^+)^\sharp}\right\rangle &= \frac{1}{2}\sum_{i,j,k,l,p,q}{\mathrm W}^+_{ijkl}{\mathrm W}^+_{ipkq}{\mathrm W}^+_{jplq}=6 \text{det}({\mathrm W}^+);\\
\left\langle{{\mathrm W}^+, ({\mathrm W}^+)^2}\right\rangle &=\frac{1}{8}\sum_{i,j,k,l,p,q}{\mathrm W}^+_{ijkl}{\mathrm W}^+_{ijpq}{\mathrm W}^+_{klpq}=3\text{det}({\mathrm W}^+).
\end{align*}

So another way to write (\ref{bw4}) is,
\begin{equation}
\label{bw41}
\int_M |\nabla {\mathrm W}^+|^2-2\int_M (\delta{\mathrm W}^+)^2=\int_M 18\text{det}({\mathrm W}^+) -\frac{1}{2}\int_M {\mathrm S}|{\mathrm W}^+|^2.
\end{equation}
\begin{remark} This is the same as \cite[Equation 3.23]{CGY03} modulo out norm convention.
\end{remark}

Furthermore, as the self-dual and anti-self-dual part are operators on a vector space of dimension three, Lemma \ref{Wcubic} yields, for $\omega$ the largest eigenvalue of ${\mathrm W}^+$,
 \[({\mathrm W}^+)^3 \leq \frac{\omega}{2}|{\mathrm W}|^2.\]
 So we recover the following result which is originally due to \cite[Theorem 4.2]{mw93}.
\begin{theorem}
Let $(M^4,g)$ be a closed Riemannian manifold with harmonic self-dual Weyl tensor. $\omega$ denotes the largest eigenvalue of ${\mathrm W}^+$. Suppose that, at each point,
$$  6\omega \leq {{\mathrm S}},$$
then $(M,g)$ has parallel Weyl tensor.
\end{theorem}
It is interesting that another condition also leads to the desired rigidity.

\begin{lemma} \label{pinched} Let $\lambda_1\geq \lambda_2\geq \lambda_3$ be eigenvalues of ${\mathrm W}^+$. Suppose that,
\begin{equation}
\label{farapart} \lambda_1\geq \frac{{\mathrm S}}{6} \text{ and } \lambda_3\leq-\frac{\lambda_1}{2}-\lambda_1 \sqrt{\frac{3(\lambda_1-\frac{{\mathrm S}}{6})}{4(3\lambda_1+\frac{{\mathrm S}}{6})}},
\end{equation}
then we have,
\begin{equation*}
{\mathrm S}|{\mathrm W}^+|^2\geq 36\text{det}({\mathrm W}^+).
\end{equation*}
\end{lemma}
\begin{proof}
Let $\mu_i=\lambda_i+\frac{{\mathrm S}}{12}$. By direct computation, for ${\mathrm S}_2=\sum_{i<j}\mu_i\mu_j, {\mathrm S}_3=\prod_{i=1}^3 \mu_i$.
\[
{\mathrm S}|{\mathrm W}^+|^2- 36\text{det}({\mathrm W}^+)={\mathrm S}\SS_2-36{\mathrm S}_3.\]
Without loss of generality, we can assume $\sum_{i=1}^3\mu_i=\frac{{\mathrm S}}{4}=3$. Then we have,
\begin{align*}
{\mathrm S}_2-3{\mathrm S}_3 &= \mu_1\mu_3(1-3\mu_2)+(\mu_1+\mu_3)\mu_2,\\
&= \mu_1\mu_3-3\mu_1\mu_3(3-\mu_1-\mu_3)+(\mu_1+\mu_3)(3-\mu_1-\mu_3),\\
&=(3\mu_1-1)\mu_3^2+(3\mu_1-1)(\mu_1-3)\mu_3+(3-\mu_1)\mu_1.
\end{align*}
Next, we compute the determinant of that quadratic on $\mu_3$,
\begin{align*}
\Box &=(3\mu_1-1)^2(\mu_1-3)^2-4(3\mu_1-1)(3-\mu_1)\mu_1,\\
&=(3\mu_1-1)(\mu_1-3)(3\mu_1^2-10\mu_1+3+4\mu_1),\\
&=3(3\mu_1-1)(\mu_1-3)(\mu_1-1)^2
\end{align*}
Since $\mu_1> 3$, the quadratic above is positive if and only if $\mu_3 \notin [x_1,x_2]$ where
\[x_{1,2}=-\frac{\mu_1-3}{2}\pm(\mu_1-1)\sqrt{\frac{3(\mu_1-3)}{4(3\mu_1-1)}} \]
As $\mu_3<0$ the only possibility is that $\mu_3\leq x_1$. The statement now follows from normalization and the fact that $\mu_i=\frac{{\mathrm S}}{12}+\lambda_i$.
\end{proof}
\begin{remark}
As a consequence, we also obtain a rigidity result assuming condition (\ref{farapart}) and harmonic Weyl tensor.
\end{remark}

Next, we consider the proof of the integral rigidity statement, Theorem \ref{integralrigid}, in dimension four. First, without loss of generality, we can choose to work with ${\mathrm W}^{+}$.
Due to the discussion above, we have the sharp estimate:
\begin{equation}
2\left\langle{{\mathrm W},{\mathrm W}^2+{\mathrm W}^\sharp}\right\rangle =18\text{det}({\mathrm W}^+)\leq \sqrt{6}|{\mathrm W}^{+}|^3.
\end{equation}
The equality happens only if either ${\mathrm W}^+=0$ or it has exactly two distinct eigenvalues.

Thus, $c_1(4)=\sqrt{6}$. Then proceeding as earlier, H\"{o}lder and Sobolev inequalities, we obtain $\alpha=\frac{1}{3}$.
the rigidity if assuming harmonic self-dual and the following inequality.
\[||{\mathrm W}||_{L^{2}}\leq \frac{1}{2\sqrt{6}}\lambda(g).
\]
 \begin{remark} This result, however, is not new, see \cite{g00, fu16four}. And the inequality is shown to be sharp. \end{remark}

\subsection{Pure curvature}
\label{pure}
The pure curvature condition says that there is an orthonormal frame $\{e_i\}_{i=1}^n$at each point such that the curvature is diagonalized with respect to the basis $\{e_i\wedge e_j\}_{i\neq j}$. Then, the only nontrivial component of the Weyl tensor is ${\mathrm W}_{ijij}\doteqdot w_{ij}$. It is observed that $w$ is a symmetric matrix with the following properties:
\begin{itemize}
\item Each diagonal term is zero;
\item The sum of each row is zero.
\end{itemize}

In this case, we can compute $\left\langle{{\mathrm W},{\mathrm W}^{2}}\right\rangle$ and $\left\langle{{\mathrm W},{\mathrm W}^{\sharp}}\right\rangle$ as follows.
\begin{align*}
\left\langle{{\mathrm W}, {\mathrm W}^\sharp}\right\rangle &= \sum_{i,j,k,l,p,q}{\mathrm W}_{ijkl}{\mathrm W}_{ipkq}{\mathrm W}_{jplq}=6 \sum_{i,j,k}w_{ij}w_{ik}w_{kj};\\
\left\langle{{\mathrm W}, {\mathrm W}^2}\right\rangle &=\frac{1}{4}\sum_{i,j,k,l,p,q}{\mathrm W}_{ijkl}{\mathrm W}_{ijpq}{\mathrm W}_{klpq}=2\sum_{ij}w_{ij}^3.
\end{align*}

Recalling the following identity,
\[a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)(3a^2+3b^2+3c^2-(a+b+c)^2).\]
Applying for each term $w_{ij}w_{ik}w_{kj}$,
\begin{align*}
\left\langle{{\mathrm W}, {\mathrm W}^\sharp}\right\rangle &=6 \sum_{i,j,k}w_{ij}w_{ik}w_{kj},\\
 &= \sum_{i\neq j\neq k}2(w_{ij}^3+w_{ik}^3+w_{jk}^3)+(w_{ij}+w_{jk}+w_{ik})^3\\
 &~~~~~-3(w_{ij}+w_{jk}+w_{ik})(w_{ij}^2+w_{ik}^2+w_{jk}^2),\\
&=2(n-2)\sum_{ij}w_{ij}^3+\sum_{i\neq j\neq k}(w_{ij}+w_{jk}+w_{ik})^3-3(n-4)\sum_{ij}w_{ij}^3,\\
&=\frac{8-n}{2}\left\langle{{\mathrm W}, {\mathrm W}^2}\right\rangle+\sum_{i\neq j\neq k}(w_{ij}+w_{jk}+w_{ik})^3
\end{align*}
By (\ref{weylsect}), the last term is just a summation of the Weyl generalized sectional curvature on 3-planes. In dimension four, as $4-3=1$, that term vanishes. In dimension five,
\[\sum_{i\neq j\neq k}(w_{ij}+w_{jk}+w_{ik})^3=\sum_{mn}w_{mn}^3=\frac{1}{2}\left\langle{{\mathrm W}, {\mathrm W}^2}\right\rangle.
\]
Thus, when $n\leq 5$, $\left\langle{{\mathrm W}, {\mathrm W}^\sharp}\right\rangle =2\left\langle{{\mathrm W}, {\mathrm W}^2}\right\rangle$. There is little evidence to suggest the equation still holds true in higher dimension.

\def$'${$'$}
\bibliographystyle{plain}

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In 28 states, including Texas, queer and trans workers can still be fired due to their sexual orientation and gender identity. Strong union contracts are often the only legally binding workplace protections available to queer workers to fight employment discrimination in these states.
The importance of protections like this cannot be understated, especially in states like Texas that rank poorly in LGBTQIA+ friendliness due to a lack of workplace inclusion and legislative protections. Teachers who can help to make students feel accepted for who they are, instead of condemned, are crucial to the movement.
More queer educators and allies mean more inclusive spaces for LGBTQIA+ students navigating the often overwhelming process of discovering who they are. Unions help to protect these educators, who can then go on to protect our most vulnerable young people.
Fifty years after Stonewall, the LGBTQIA+ experience has changed greatly in public schools, but there is still work to do. And fortunately, there are organizations fighting that fight.
Join the NEA in Houston for a Stonewall Day of Action.
The National Education Association is America's largest labor union representing nearly three million employees, principally teachers.
On July 3, the NEA will be hosting a Stonewall Day of Action in Houston to commemorate the 50th anniversary of the Stonewall riots and to continue their support and advocacy for the LGBTQIA+ community.
The Stonewall Uprising was the impetus for the formation of the Gay Liberation Front as well as other gay, lesbian, and bisexual civil rights organizations. It's also regarded by many as history's first major protest on behalf of equal rights for the LGBTQIA+ community.
These riots marked a major turning point in the modern LGBTQIA+ civil rights movement in the United States and its impact was felt across the world.
For more information about the Stonewall Day of Action event and on how to register to attend, click here.
by Wesley Story
@WesleyStory0 wesley@progresstexas.org
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China threatens 'military mission' north of Taiwan till Sep 4
Chinese President Xi Jinping. (File Photo: Xinhua/Huang Jingwen/IANS)
Taipei/New Delhi, Aug 28 (IANS) China on Friday threatened to conduct a "military mission" anytime from now till September 4 in the north of the Taiwan strait, provoking Taipei to warn Beijing against any attempt to annex the democratic island nation.
Global Times, the mouthpiece of the Xi Jinping government, said that China's Maritime Safety Administration had announced that it was going to conduct a "military mission from Friday to September 4 in the Bohai strait and northern Yellow Sea".
The Global Times also claimed that China had "expelled" and destroyed the US spy plane U-2, which flew in the Chinese-designated "no-fly zone" on Tuesday when China was conducting a live-fire naval drill in the Bohai Sea off its north coast.
Ridiculing the US, the Global Times in a sarcastic piece wrote: "China's military museum invited US pilots to visit the U-2 wreckage following their recent trespassing into China's no-fly zone, which is applauded by Chinese netizens. Some said that it is ok if the US wants to provide more exhibits."
Taking strong offence to the Chinese threats, Taiwan's President Tsai Ing-wen on Friday vowed to defend the island with a "solid" air force while unveiling the island's US-backed maintenance centre for a fleet of upgraded F-16 fighters.
"It takes solid defence capability, not bowing and cringing, to defend the sovereignty of the Republic of China and maintain regional peace and stability," Tsai said at an event in the central city of Taichung.
The tensions in the region have escalated after China fired one intermediate-range ballistic missile, DF-26B, from the Qinghai Province and another medium-range ballistic missile, DF-21D, from Zhejiang Province on Wednesday into the South China Sea.
However, US media reports said that China had fired four medium-range ballistic missiles during a series of military exercises this week.
In response to the increased militarisation of the South China Sea, the US on Wednesday imposed an unprecedented visa ban on the Chinese Communist Party (CCP) regime, People's Liberation Army and their businesses.
Incidentally, the US is working with Taiwan to restructure global supply chains, American Institute in Taiwan (AIT) Director Brent Christensen said.
"The Covid-19 pandemic has exposed the risks of over-reliance on one country or supplier for critical materials, such as medical supplies and pharmaceuticals, and for industries that are of strategic importance to shared economies," Christensen said.
Washington, he said, aims to diversify and bring supply chains closer to the end users, and ensure that countries like China do not hold supply chains hostage for political purposes.
On Wednesday, Chinese company Huawei lost a significant lawsuit over technology licensing in a landmark case in the Supreme Court of the UK. Huawei was accused of infringing intellectual property belonging to the US company Unwired Planet, which had acquired patents covering wireless connectivity from Ericsson.
In 2017, a high court in the UK had ordered the Chinese company to pay a global licence fee for infringing two UK patents. Huawei had appealed against the verdict in the Supreme Court, arguing that it should only pay license fees for the technology in the UK and not on its global sales.
The apex court dismissed Huawei's appeal on Wednesday, at a time when Huawei is already under fire for its alleged links to the Chinese government. It has been banned by the UK government from supplying new kit for UK's 5G mobile phone networks from next year, in view of national security concerns.
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Where can I get the Malleo Sensa?
You can get the Malleo Sensa ankle support from a medical supply store with a prescription. If you have ankle pain, we recommend consulting a doctor who will prescribe the right product for you.
Do I have to wear special shoes with the Malleo Sensa?
No, you can wear the Malleo Sensa in normal, standard shoes. Naturally, you should choose healthy and comfortable shoes with a low heel, especially with an ankle injury.
Can I wear the Malleo Sensa even for extended periods of time?
Many users wear the Malleo Sensa over an extended period and also for prophylaxis. We recommend consulting your doctor, who will advise you in detail.
Is the Malleo Sensa latex free?
Yes, the support provides effective protection against latex allergies as it is free of natural latex.
Is the Malleo Sensa washable?
The support is machine washable at 30°C. The antibacterial effect of the integrated fibres is retained even after repeated laundering.


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Tropical / Challenge
These tropical species are a challenge to germinate and are suitable for experienced growers, people with green fingers or anyone that likes a challenge. They are suitable as indoor plants.
Blue Quandong (Elaeocarpus angustifolius)
The very special seeds resemble the holy Rudraksha, but come from a different rare plant from Australia.
Calabash Nutmeg (Monodora myristica)
The peculiar flowers are spotted red and yellow, and resemble the flowers of an orchid. This species originates in Africa and is very rare.
Camouflage Plant (Aglaonema pictum)
The leaves this species look spectacular with a camouflage pattern of white, dark green and light green.
€ 12,00 per seed
Coco de Mer (Lodoiceae maldivica)
Sea coconuts are the largest seeds in the world and also one of the rarest. They originally grow on two islands of the Seychelles.
€ 0,00 per seed
Durian (Durio zibethinus)
The Durian is a well-known fruit from South-East Asia. The fruit tastes delicious, but has an unpleasant smell.
Frankincense Tree (Boswellia sacra)
Frankincense is the resin of the rare frankincense tree that grows in the Arabian Peninsula. This species is threatened in its existence.
Jackal Food (Hydnora africana)
Bizarre, parasitic plant from South Africa which only appears above ground with its flowers. The flowers occur with sharp teeth in the video games of Super Mario, but aren't...
Joey Palm (Johannesteijsmannia altifrons)
This might be one of the most spectacular and beautiful palm trees in the world. In the rainforest the enormous fan-shaped leaves grow from the ground up to heights of even 6...
Kluay Pa Banana (Ensete sp. 'Kluay Pa')
A rare species from north-western Thailand with rather short, broad leaves and a coloured midrib. Overwintering is possible at a temperature of at least 5 degrees Celsius.
Kukui (Aleurites moluccana)
The kukui or candlenut is used in the Malaysian and Indonesian cuisine. The nuts are very oily, so they can burn like a candle.
Lipstick Palm (Cyrtostachys renda)
This species is seen by many as the most beautiful palm species in the world. This is not only because of its unusual bright red trunk and leaf stem, but the leaves are also...
Pelican Flower (Aristolochia grandiflora)
This large tropical climbing plant produces spectacular flowers with a total length of up to 60 cm. However, the smell is very unpleasant.
Rebecca Tree Fern (Cyathea rebeccae)
Quite a tropical tree fern, with a thin trunk, which originates from the cloud forests of Queensland in Australia.
Rudraksha (Elaeocarpus ganitrus)
The seeds of this species are of great religious significance in Hinduism and Buddhism. They are often worn as a prayer chain and would provide protection by Lord Shiva.
Suarez Baobab (Adansonia suarezensis)
This baobab species is from the most northern area of Madagascar and only has a few small populations left.
Tree Passion Flower (Passiflora macrophylla)
A rare passion flower from the jungles of Ecuador that grows as a tree instead of a climber. The tropical tree can grow leaves up to 1 metre long! Can be kept in a pot.
Vanilla Orchid (Vanilla planifolia)
Vanilla is made from the fruits of the vanilla orchid. A tropical climbing vine from Mexico that can be kept in a heated greenhouse.
Welwitschia (Welwitschia mirabilis)
Very rare and bizarre plant from the desert of Namibia with two leaves that can grow for more than a thousand years.
€ 10,00 per 2 seeds


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187 Republicans vote against bill to close the gender wage gap
March 27th, 2019 | Casey Quinlan
The House on Wednesday voted 242-187 for a bill that would strengthen protections for female workers and help close the gender wage gap. The vote comes as Republicans are trumpeting themselves as the champions of women's economic mobility — though only seven of them voted for the bill.
Iterations of this legislation have been debated by lawmakers for decades but have never actually been able to pass. The bill, sponsored by Rep. Rosa DeLauro (D-CT), seeks to boost women's pay by prohibiting employers from seeking job applicants' salary histories and preventing them from retaliating against workers for disclosing their wages. It also would require the Equal Employment Opportunity Commission (EEOC) to collect wage data based on sex, race, and national origin to better determine whether employers are responsible for discriminatory practices. The House passed the bill on Wednesday despite Republicans' opposition, but it now faces an uncertain future in the GOP-controlled Senate.
The House Education and Labor Committee voted to advance the legislation earlier this week. Every single Republican opposed moving the bill out of committee, with many saying the focus should instead be on providing more job opportunities for women.
Republicans often like to point to data showing that women gained 58 percent of new, private-sector jobs in 2018. Trump touted the figure in his State of the Union address in February, and Republicans in the Education and Labor Committee again brought it up when discussing the Paycheck and Fairness Act.
But many of the jobs gained by women are part time, and nearly 80 percent of them fell into just four categories: education and health services, professional and business services, leisure and hospitality, and manufacturing. In three of those industries, women make less than 80 cents for every dollar a man earns, or worse than the average national wage gap, according to a 2018 analysis by the Center for American Progress analysis. (Editor's Note: ThinkProgress is an editorially independent newsroom housed at the Center for American Progress Action Fund.)
Jocelyn Frye, a senior fellow at the Center for American Progress who focuses on work-family balance, pay equity, and women's leadership, said, "It's not to discount that women have received jobs and obviously want jobs but there is a disconnect. It's not responsive to the question [of pay inequality]. The fact that you gave the jobs doesn't change the fact that the jobs are underpaying women."
Republicans, meanwhile, have been looking for ways to appeal to greater numbers of women voters, particularly since their support among women plummeted in the 2018 midterm elections.
In November, 59 percent of women voted for Democrats in the congressional elections, according to exit poll data. Only 40 percent of women voted for Republicans. There was no measurement for how nonbinary people voted across race or educational attainment. Black and Latina women overwhelmingly voted for Democratic candidates.
Although there was a roughly even split for how white women voted, 59 percent of college-educated white women and 56 percent of white voters ages 18 to 29 voted for Democrats. Experts say these shifts likely represent a long-term trend.
Kelly Dittmar of the Center for American Women and Politics, part of the Eagleton Institute of Politics at Rutgers University, said the shift likely isn't about Trump alone, but about the broader Republican Party.
"My hypothesis at this moment is that it is actually a trend because there were signs of this trend before Donald Trump, it's just that you saw it through an acceleration I think — the departure of these women," Dittmar said. "I think you'll continue to see it because these women who are particularly upset with how the party has dealt with Donald Trump, it certainly leaves a taste in their mouth about the party overall."
She added, "If you put these women on a scale when it comes to immigration or guns or the environment, their positions on these issues are just not aligned with the current agenda and leadership in the Republican Party."
Democratic pollster Celinda Lake said that when looking at women who vote in the general election, college-educated and suburban women are identifying as more independent and Democratic. She said three major waves encapsulate that movement.
The Republican Party's position on social issues — including birth control, Title IX, and sexual harassment and violence — led to some women moving away from the Republican Party in 2016. The second wave emerged as voters reacted to Trump's racist and sexist behavior, as well as how he governs.
"The third wave, which is more recent, is a sense that the country is going in the wrong direction, that the priorities are wrong, that we are not dealing with everything from health care to climate change," Lake said.
Lake said that for female voters, including Republican women, equal pay is high on the list of concerns, along with domestic violence programs. The reauthorization and expansion of the Violence Against Women Act is on the House agenda this session. But Rep. Brian Fitzpatrick (R-PA) is the only Republican in the House who is cosponsoring the bill and the only Republican who has shown support for the bill by attending its introduction.
"There's a very high correlation between concerns about sexual harassment and concerns about domestic violence and concerns about equal pay." Lake said. "And equal pay is still the most salient of the three with women overall. And it's particularly salient with Republican women who are very adamant about equal pay and that it remains a problem."
Dittmar said that across gender, voters are concerned about economic stability and the well-being of their families. But they are divided over who is responsible. She explained that college-educated women who identify as Democrats tend to say the government plays a role but Republicans tend to say it's up to businesses to address equal pay.
"Broadly I think there is pretty high popularity for wanting to address equal pay but it's in the how where you see the disparity both among legislators as well as the public," she said.
Ariane Hegewisch, program director of employment and earnings for the Institute for Women's Policy Research, said these measures are necessary to ensure workplace fairness.
"What the Equal Pay Act recognizes and what the Paycheck Fairness Act is trying to update 50 years on to more current circumstances is that there is discrimination in the labor market and if you just rely on what people are paid now, you are going to pick up discrimination and import it into your organization," she said. "You have to pay people the same if they do the same job and have similar education, experience and performance. You can qualify their personal performance but it has to be fact based."
According to the Institute for Women's Policy Research, it will take until 2059 for women to reach pay parity if change continues at the current pace. Black women would have to wait until 2119 for equal pay, and Latina women until 2224.
"After what I would call a wave election in 2018 where women were elected to historic numbers in Congress, people have very high expectations of what they are going to get from lawmakers and it is not acceptable simply to say I support equal pay but I have nothing to show for it," said Frye.
This article was originally published at ThinkProgress on March 27, 2019. Reprinted with permission.
About the Author: Casey Quinlan is a policy reporter at ThinkProgress covering gender and sexuality. Their work has also been published in The Establishment, Bustle, Glamour, The Guardian, and In These Times.
Tags: EEOC, gender pay gap, HELP Committee, House, pay gap, Paycheck Fairness Act, Rosa DeLauro, wages
This entry was posted on Wednesday, March 27th, 2019 at 6:33 pm. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
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text:
import React from "react";
import { StepParams, HardwareFlags } from "../interfaces";
import { ToolTips, Content } from "../../constants";
import { StepWrapper, conflictsString, StepWarning } from "../step_ui";
import { AxisStepRadio } from "../step_ui/step_radio";
import { t } from "../../i18next_wrapper";
import { Xyz, Calibrate, FindHome } from "farmbot";
import { some } from "lodash";

export const TileCalibrate = (props: StepParams<Calibrate>) => {
  const { dispatch, currentStep, index, currentSequence } = props;
  const conflicts = settingConflicts(currentStep, props.hardwareFlags);
  return <StepWrapper
    className={"calibrate-step"}
    helpText={ToolTips.CALIBRATE}
    currentSequence={currentSequence}
    currentStep={currentStep}
    dispatch={dispatch}
    index={index}
    resources={props.resources}
    warning={some(conflicts) &&
      <StepWarning
        warning={Content.END_DETECTION_DISABLED + conflictsString(conflicts)}
        conflicts={conflicts} />}>
    <AxisStepRadio
      currentSequence={currentSequence}
      currentStep={currentStep}
      dispatch={dispatch}
      index={index}
      label={t("Find length")} />
  </StepWrapper>;
};

export const settingConflicts =
  (step: Calibrate | FindHome,
    hardwareFlags: HardwareFlags | undefined,
  ): Record<Xyz, boolean> => {
    const conflicts = { x: false, y: false, z: false };
    if (hardwareFlags) {
      const { axis } = step.args;
      const { findHomeEnabled } = hardwareFlags;
      switch (axis) {
        case "x":
        case "y":
        case "z":
          conflicts[axis] = !findHomeEnabled[axis];
          break;
        case "all":
          conflicts.x = !findHomeEnabled.x;
          conflicts.y = !findHomeEnabled.y;
          conflicts.z = !findHomeEnabled.z;
          break;
      }
    }
    return conflicts;
  };

redpajama_set_name: RedPajamaGithub


text:
Acoustic Productions
Home/Show List
Past Show Gallery
Acoustic Productions Acoustic Productions Acoustic Productions
Update: Concerts On Hold
The "John T and the You Know Who's" concert
will be rescheduled for a future date,TBA.
Concert Menu
J.T. & The You Know Who's ----------- Postponed until a TBA date. Stay Tuned!
Andrew Finn Magill (w/ Will MacMorran) Postponed until a TBA date.
J.T. & The You Know Who's ------------ Postponed until a TBA date. Stay Tuned!
John Tinker and his band of merry musicians will be presenting a "Tastefully Unplugged" version of their powerhouse show at Joe's as a part of the pot-luck dinner and concert series. The show will be presented on two consecutive nights, Friday, March 20 and Saturday, March 21, 2020 at 7:30pm with doors opening at 7pm. There will be around 17 musicians on stage before the night is over and you will recognize these artists as some of the best that Macon has to offer. You can reserve your seats for either night by clicking on the appropriate link marked FRIDAY or SATURDAY. This will be a show that you will LOVE!!
($25 Donation goes 100% to the band)
The Tab Below is for FRIDAY, MAR 20 Reservations.
Saturday Reservations are linked in the NEXT article.
The Tab Below is for SATURDAY, MAR 21 Reservations.
For twenty years, fiddler and composer Andrew Finn Magill (Finn) has been pursuing parallel careers in traditional Irish music, Brazilian choro, jazz and American fiddle, the result of many years living abroad and a childhood growing up at The Swannanoa Gathering in Asheville, North Carolina where he studied every summer with dozens of Irish, old-time, bluegrass and swing fiddle masters. His six albums constitute thousands of hours dedicated to each genre and the basis of Finn's original music which is founded on the principle of collaboration. Over the last ten years he has performed with John Doyle, Rising Appalachia, Charm City Junction, Open the Door for Three, Greg Ruby, The Paul McKenna Band, and Peter Mawanga in venues that include Milwaukee Irish Fest, Celtic Connections, Olympia Hall in Paris, France, DisneyWorld, and Malawi's Lake of Stars Festival.
($25 Donation goes 100% to the artist)
Click to See Andrew's Video


Sixties is a small, fine watch collection with 60-year-old watches that sparkle the charm of days gone by. Their brand is classic, inspired by earlier days makes for the trendy retro look.
To use their existing brand collateral to create various campaigns for a new seller. The campaigns will be displayed in multiple jewellers across the country and need to represent the brand in a consistent and engaging way.
We are creating a suite of stylish and engaging collateral for the jewellers to use throughout the year, on various campaigns. The final products are creating brand awareness and solidifying the brand values to consumers.


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Tournament Wins!
Hello Optimus BJJ students, parents, and friends!
This past weekend some of our students attended their very first tournament and we are proud of their performance. We would like to take some time to recognize these hard working individuals and express some of the moments in their long road to success!
Dixon Kavanaugh
This past weekend was a sure confidence booster for 17 year-old Dixon. His situation for this tournament was pretty interesting. The division that Dixon was scheduled to fight in was empty so his only choice was to fight in the open weight division. After long waiting for his fights he not only took the gold medal in his division but he also took gold in the Open Division as well. In his first fight he jumped guard, swept his opponent, and landed an armbar submission to win the fight in just under a minute.
His second and final fight went the distance of the 5 minute time period and Dixon came out on top once again with a winning score of 16 to 0.
Congratulations Dixon!
Lauren Okon
This ferocious female competitor fought her heart out in this tournament. This was Lauren's very first tournament in martial arts and also she is the first female to compete from our Optimus BJJ girl's team, so everyone knew the nerves were tingling, but that didn't stop her from taking the silver medal at this competition. In Lauren's first fight, in under a minute, she jumped guard faster than lightning and landed a triangle choke that was tighter than a lid of a pickle jar! Lauren lost her second match by points, but that did not break the smile off of her face because she did her best!
Congratulations Lauren!
Maryan Malkosh
This little 6 year-old firecracker had everyone in attendance in shock and awe. This was Maryan's first Muay Thai kickboxing tournament and she did phenomenal! The match was scheduled for three 1-minute rounds but it was cut to a short 30 second fight when she beat her opponent by Technical Knockout (TKO). When the bell rang, both fighters came out and touched gloves, and then Maryan turned up the heat and backed her opponent into the corner with a flurry of punches and kicks. Shortly after, the referee had to separate the fighters, and then her opponent threw in the towel, claiming Maryan the winner of the bout. She made her family and coaches very proud when she went to give her opponent a hug to make sure she was alright.