text:
\begin{document}

\begin{frontmatter}
\title{The length of the shortest closed geodesic on positively curved 2-spheres}
\author{Ian Adelstein and Franco Vargas Pallete}
\address{Department of Mathematics, Yale University \\ New Haven, CT 06520 United States}
\begin{abstract} We show that the shortest closed geodesic on a 2-sphere with non-negative curvature has length bounded above by three times the diameter. We prove a new isoperimetric inequality for 2-spheres with pinched curvature; this allows us to improve our bound on the length of the shortest closed geodesic in the pinched curvature setting.
\end{abstract}
\begin{keyword} closed geodesics; geodesic nets; spheres
\MSC[2010]  53C22 \sep 53C23
\end{keyword}
\end{frontmatter}

\section{Introduction}

Gromov \cite{Gro83} has asked if there exist constants $c(n)$ such that the length of the shortest closed geodesic $L(M^n)$ on a closed Riemannian manifold $M^n$ is bounded above by $c(n) D(M^n)$, where $D(M^n)$ is the diameter of the manifold. On non-simply connected manifolds the shortest non-contractible closed curve is a geodesic with length bounded above by $2D(M^n)$. On manifolds homeomorphic to the 2-sphere Croke \cite{croke1} provided the first bound of $L(S^2,g) \leq 9D(S^2,g)$, which was improved by Maeda \cite{maeda}, and finally by Nabutovsky and Rotman \cite{NR2002} and independently Sabourau \cite{Sab} to $L(S^2,g) \leq 4D(S^2,g)$. Rotman \cite{rot5} has further proved that $L(S^2,g) \leq 4R(S^2,g)$ where $$R(S^2,g)= \min_{x \in S^2} \max_{y \in S^2} d(x,y) \leq D(S^2,g) =  \max_{x \in S^2} \max_{y \in S^2} d(x,y)$$ is the radius of the Riemannian sphere.
Gromov's question is open for simply connected manifolds in dimensions $n \geq 3$.

An attractive conjecture is that $L(M^n) \leq 2D(M^n)$ for all closed Riemannian manifolds $M^n$. To explore this bound one might consider Zoll spheres:~metrics on the 2-sphere all of whose geodesics are closed and of the same length. The conjecture turns out to be overly optimistic, as Balacheff, Croke, and Katz \cite{bck} have produced Zoll spheres with $L(S^2, Zoll) > 2D(S^2, Zoll)$. These examples are not constructive, and it is unknown how much longer than $2D(S^2, g)$ the shortest closed geodesic could be. In this paper we prove the following:

\begin{theorem}\label{main} Non-negatively curved 2-spheres have $L(S^2, g) \leq 3R(S^2, g) \leq 3D(S^2, g)$.
\end{theorem}

While one does not expect the inequality $L(S^2, g) \leq 3D(S^2, g)$ to be sharp, we note that the inequality $L(S^2, g) \leq 3R(S^2, g)$ is realized by the metric space formed when gluing two equilateral triangles along their common boundaries, the so called Calabi-Croke sphere.
The centers of the triangles realize the radius whereas the vertices realize the diameter. The shortest closed geodesic is a doubled altitude which has length exactly 3 times the radius and $\sqrt{3} $ times the diameter.

We should note that the results cited above are curvature free bounds, whereas our bounds require non-negative curvature. Calabi and Cao \cite{calabi} studied simple closed geodesic on non-negatively curved 2-spheres, and showed that any closed geodesic of shortest length must be simple. We can therefore improve our Theorem~\ref{main} to the following:~on non-negatively curved $(S^2,g)$ the shortest closed geodesic is simple and has length bounded above by three times the radius.

Additional work in the positive curvature setting includes \cite{recent} where pinched metrics on the 2-sphere are studied and an upper bound for the length of the shortest closed geodesic is provided in terms of the area. We combine this result with a new isoperimetric inequality for pinched metrics on the 2-sphere (Theorem~\ref{iso}) to yield the following:

\begin{theorem}\label{pinched}
Let $(S^2,g)$ be a $\delta$-pinched metric with $\delta > \frac{4+\sqrt{7}}{8} \approx 0.83$. Then

\[L(S^2,g) \leq  \frac{2}{\sqrt{\delta}}D(S^2,g)
\]

with equality if and only if the sphere is round.

\end{theorem}

For $\delta = .83$ our theorem yields the bound $L(S^2,g) \leq 2.19 D(S^2,g)$. The theorem is optimal in the sense that the constant $\frac{2}{\sqrt{\delta}}$ converges to $2$ as $\delta$ approaches $1$. Finally, we note that this theorem can be used to further comment on the Zoll spheres due to Balacheff, Croke, and Katz \cite{bck} where we can now say that $$2D(S^2,Zoll) < L(S^2, Zoll) < \frac{2}{\sqrt{\delta}}D(S^2,Zoll).$$

The paper proceeds as follows. In Section~\ref{rot} we present a proof due to Rotman \cite{video} of the fact that $L(S^2,g) \leq 4D(S^2,g)$. A crucial step in the proof uses a weighted length shortening to avoid stationary theta-graphs (critical points of the length functional on nets). This weighted flow increases the bound on the shortest closed geodesic from $3D(S^2,g)$ to $4D(S^2,g)$. In Section~\ref{deets} we show for positive curvature metrics on the 2-sphere that non-trivial stationary theta-graphs are never local minima. The weighted flow in Rotman's proof can therefore be avoided, and Theorem~\ref{main} follows. In Section~\ref{pinch} we prove a new isoperimetric inequality for pinched metrics on the 2-sphere and combine this with the main result of \cite{recent} to prove Theorem~\ref{pinched}. The new isoperimetric inequality is further refined in the Appendix by studying a Sturm-Liouville problem (SLP) related to lower bounds on the first eigenvalue of the Laplace-Beltrami operator.

\textbf{Acknowledgements:}~The authors would like to recognize Christina Sormani and her NSF grant DMS-1612049. The grant made possible a conference at Yale where the authors learned of many of the results cited in this paper and had productive conversations with Alexander Nabutovsky, Regina Rotman, and Frank Morgan. The authors would also like to thank Stéphane Sabourau for helpful suggestions, as well as Ben Andrews for his helpful advice on how to linearly approximate eigenvalues of the related SLP. We also acknowledge Wolfgang Ziller for helpful discussions about short closed geodesics in the pinched curvature setting, the fruits of which will appear in a forthcoming paper. Finally, we want to acknowledge the generous contribution of the referee who indicated how our proof of the inequality $L(S^2,g) \leq 3D(S^2,g)$ could be extended to the sharp bound $L(S^2,g) \leq 3R(S^2,g)$.

\begin{comment}
\section{Background on Zoll spheres}\label{background}

Zoll spheres are metrics on the 2-sphere all of whose geodesics are closed. It was shown by Gromoll and Grove \cite{GG} that these geodesics all have the same length. The first examples were given by Zoll in 1903 who produced surfaces of revolution with the Zoll property. By the triangle inequality one can show that the poles of these surfaces of revolution are diameter realizing, hence $L=2D$ for these examples. See \cite{besse} for a comprehensive treatment of manifolds all of whose geodesics are closed.

Another class of Zoll spheres are due to Funk and Guillemin. In 1913 Funk showed that a smooth conformal variation $g_t = \Phi_t g_0$ of the round metric through smooth Zoll metrics must have $(d \Phi_t / dt) |_{t=0} \in C^{\infty}_{odd}(S^2)$. Guillemin in 1976 \cite{gui} proved the converse, namely that for any odd function $f \in C^{\infty}_{odd}(S^2)$ there exists a smooth conformal variation $g_t = \Phi^f_t g_0$ where $(d \Phi^f_t / dt) |_{t=0} =f $ and $\Phi^f_0=1$ is through smooth Zoll metrics.

The Zoll spheres a la Guillemin can have $L(S^2, Zoll) > 2D(S^2, Zoll)$ as was shown by Balacheff, Croke, and Katz \cite{bck}. They define and show existence of a class of \emph{amply negative} odd functions with the following property:~for sufficiently small $t>0$, every direction from every point on $(S^2, g_t )= (S^2, \Phi^f_t g_0)$ decreases distance to the antipode. The diameter is therefore strictly less than $\pi$, whereas the odd condition guarantees that the length of closed geodesics remains $2\pi$. As their proof is not constructive it is unknown how much longer than twice the diameter the length of a Zoll geodesic can be; our main result demonstrates that $L(S^2, Zoll)  \leq 3D(S^2, Zoll)$.

\end{comment}

\section{Proof that $L(S^2,g) \leq 4D(S^2,g)$}\label{rot}

We start with some preliminaries that will be used both in the proof of the bound $L(S^2,g) \leq 4D(S^2,g)$ and throughout the remainder of the paper.

\begin{definition}
A \emph{geodesic net} is a finite graph immersed in a Riemannian manifold such that each edge is a geodesic segment. A geodesic net is said to be \emph{stationary} if at each vertex the sum of the unit vectors tangent to the incident edges equals zero.
\end{definition}

As such, stationary geodesic nets are critical points of the length functional on the space of nets. Closed geodesics are the first examples of stationary geodesic nets. A figure eight curve is a stationary geodesic net if each loop is geodesic and if the stationarity condition is satisfied at the vertex. In dimension two the stationarity condition implies that a geodesic net based on the figure eight curve will be a self-intersecting closed geodesic.

An example of a stationary geodesic net that is not a closed geodesic is the stationary theta-graph. A theta-graph is a net consisting of exactly two vertices joined by exactly three edges. The stationarity condition ensures that these edges pairwise meet at angle $\frac{2\pi}{3}$ at each vertex. Hass and Morgan \cite{morgan} gave one of the only known existence results for geodesic nets, demonstrating that convex metrics on the 2-sphere nearby the round metric admit stationary theta-graphs.

Nabutovsky and Rotman \cite{NR2002} and independently Sabourau \cite{Sab} gave the original proofs that $L(S^2,g) \leq 4D(S^2,g)$. In working to improve this bound Rotman obtained alternate unpublished proofs, one of which we present here \cite{NR2005, rot2, video, rot3}. This proof uses a pseudo-filling technique, analogous to the technique introduced by Gromov \cite{Gro83} in proving bounds for essential manifolds on the length of the shortest closed geodesic in terms of volume.

\begin{theorem}[\cite{NR2002, Sab}]\label{4d} Riemannian 2-spheres have $L(S^2,g) \leq 4D(S^2,g)$.
\end{theorem}

\begin{proof}[Proof \cite{video}.]
Let $M=(S^2,g)$ be a Riemannian 2-sphere and $f \colon (S^2, std) \to M$ a diffeomorphism. We attempt to extend $f$ to a map $\tilde{f} \colon (D^3, std) \to M$. As $M$ is a 2-sphere such a map should not exist, and as an obstruction to this extension we obtain a periodic geodesic on $M$ with length $\leq 4D(M)$.

First triangulate $(S^2, std)$ such that the diameter of the triangulation on $M$ induced by $f$ is less than $\delta$. Next triangulate $(D^3, std)$ as a cone over the triangulated $(S^2, std)$, i.e.~add a single vertex $p \in D^3$ at the center of the ball and the corresponding 1, 2, and 3-simplexes. We attempt to extend the map $f$ inductive to this skeleton.

{\bf 0-skeleton}:~We need only choose a point $\tilde{p} \in M$ with $\tilde{f}(p) = \tilde{p}$.

{\bf 1-skeleton}:~Let $v_i $ be the vertices of the triangulation of $S^2$ and $f(v_i)=\tilde{v}_i$ the corresponding vertices of the induced triangulation of $M$. We send the 1-simplex between $p$ and $v_i$ on $D^3$ to a minimizing geodesic between $\tilde{p}$ and $\tilde{v}_i$ on $M$ with length less than the diameter of $M$.

{\bf 2-skeleton}:~We attempt to send the 2-simplex on $D^3$ associated to the triple $(p, v_i, v_j)$ to a 2-simplex on $M$ associated to the triple $(\tilde{p}, \tilde{v}_i, \tilde{v}_j)$. The triple on $M$ is already connected by 1-simplexes that form a piecewise smooth closed curve with length less than $2D(M)+\delta$. We use Birkhoff curve shortening process to deform this closed curve without increasing its length, either to a closed geodesic with length less than $2D(M)+\delta$, or to a point in which case we have swept out the desired 2-simplex.

{\bf 3-skeleton}:~We attempt to send the 3-simplex on $D^3$ associated to the tuple $(p, v_i, v_j, v_k)$ to a 3-simplex on $M$ associated to the tuple $(\tilde{p}, \tilde{v}_i, \tilde{v}_j , \tilde{v}_k)$. By the previous steps we know where the boundary of this 3-simplex is sent; call this boundary 2-sphere $S^2_0 \subset M$. If we are able to contract $S^2_0$ to a point, i.e.~construct a homotopy $S^2_t$ with $S^2_1 = \{x\}$, then we will have succeeded in sending 3-complexes to 3-complexes, thus extending the map $f$ to $\tilde{f}$. Such an extension is not possible, and as an obstruction we obtain a short periodic geodesic on $M$.

We first contract the small 2-simplex associated to the triple $(\tilde{v}_i, \tilde{v}_j , \tilde{v}_k)$ to a point which we call $\tilde{v} \in M$ (c.f.~\cite{rot3}, Remark ending Section 1). We then have a theta-graph between the pair of points $(\tilde{p}, \tilde{v})$ consisting of three 1-simplexes, which we call $e_1, e_2, e_3$. As before, the Birkhoff curve shortening process on each pair of 1-simplexes $\{ e_i , e_j \}$ yields the boundary 2-sphere $S^2_0$.

In order to construct the homotopy $S^2_t$ we first use length shortening flow for nets to deform the theta-graph to a point, c.f.~\cite[Section 3]{NR2005}. At each time in this deformation we apply the Birkhoff curve shortening process to each pair of edges, sweeping out the desired $S^2$. The continuity of the Birkhoff curve shortening process (with respect to the initial pair $\{ e_i , e_j \}$ of edges) in the absence of short closed geodesics is what allows us to extend the homotopy which contracts the theta-graph to the desired homotopy $S^2_t$.

We therefore need only study the situation in which the theta-graph gets stuck on a stationary geodesic net before contracting to a point during the length shortening process. There are three cases to consider:

Case 1: The theta-graph degenerates to a periodic geodesic; this geodesic will have length less than $3D(M)$.

Case 2: One of the edges disappears during the length shortening yielding a stationary figure eight with length less than $3D(M)$. The stationarity condition in dimension two implies that this is a (self-intersecting) periodic geodesic.

Case 3: The theta-graph gets stuck on a stationary theta-graph. In this situation we apply a weighted length shortening process. Let $(w_1, w_2, w_3)$ be the triple of unit direction vectors at a vertex of a theta-graph. We consider a weighted length shortening flow where we double the weight of the third vector. The stationarity condition is then $| w_1 + w_2 + 2w_3 | = 0$ which implies that stationary theta-graphs are not critical points of the the weighted flow. Critical points occur when $w_1$ and $w_2$ collapse to a single edge or one of these edges disappears, which means we are in one of the two previous cases. Because we doubled the weight of one of the edges we now produce a (potentially self-intersecting) periodic geodesic with length bounded above by $4D(M)$.
\end{proof}

\section{Positive metrics on the 2-sphere} \label{deets}

In this section we indicate how the proof of Theorem~\ref{4d} adapts in the positive curvature setting to yield our Theorem~\ref{main}. Under the positive curvature assumption we show that stationary theta-graphs are never local minima of the length functional on nets, allowing us to avoid weighted length shortening. Once we prove the theorem in the positive setting, we show how it extends to the non-negative setting by considering conformally close positive metrics.
We begin by recalling the first and second variations of length, which can be found for instance in \cite[Section 5.1]{Jost}, see also \cite{morgan2} and \cite{morgan3} for formula that apply more directly in the setting of stationary nets.

\begin{proposition}[First variation of length, Lemma 5.1.1 \cite{Jost}]\label{1stvar} Given a smooth curve $\gamma:[a,b]\rightarrow M$ parametrized by arc-length and a vector field $V$ on $\gamma$, let $H$ be a variation of $\gamma$ in the direction of $V$ so that $H:[a,b]_t\times[-\epsilon,\epsilon]_s \rightarrow M$ is smooth, $H(\cdot,0)=\gamma,\, \frac{dH}{ds}|_{s=0}(\cdot,0)=V$. If we denote by $L(s):= \ell(H(\cdot,s))$ then
\[L'(0) = \langle V, \gamma'\rangle|_a^b - \int_a^b \langle V(t),\nabla_{\gamma'}\gamma'(t) \rangle dt
\]

\end{proposition}

\begin{proposition}[Second variation of length, Theorem 5.1.1 \cite{Jost}]\label{2ndvar} Given a smooth geodesic $\gamma:[a,b]\rightarrow M$ parametrized by arc-length and a vector field $V$ on $\gamma$, let $H$ be a variation of $\gamma$ in the direction of $V$ so that $H:[a,b]_t\times[-\epsilon,\epsilon]_s \rightarrow M$ is smooth, $H(\cdot,0)=\gamma,\, \frac{d}{ds}|_{s=0}H(\cdot,0)=V$. If we denote by $V^\perp$ the perpendicular projection of $V$ with respect to $\gamma'$ and by $L(s):= \ell(H(\cdot,s))$ then
\[L''(0) = \biggl\langle\frac{D}{ds}\frac{dH}{ds}, \gamma'\biggr\rangle\bigg|_{(a,0)}^{(b,0)} + \int_a^b \Vert \nabla_{\gamma'}V^\perp(t)\Vert^2 - \langle R(V^\perp(t),\gamma'(t))\gamma'(t), V^\perp(t) \rangle dt
\]

\end{proposition}

\begin{lemma}\label{lem} Any stationary theta-graph on a positively curved 2-sphere admits directions of decrease (within the space of nets) for the length shortening flow.
\end{lemma}

\begin{proof}
We simply demonstrate a variation with negative second variation of length. Give each edge a unit speed parametrization $\gamma_i \colon [a_i,b_i] \to (S^2,g)$ and define  vector fields $V_i$ so that $V_1^\perp$ and $V_2^\perp$ are of constant size $1$ (hence parallel), $V_3^\perp\equiv0$, and the $V_i$ all agree at the vertices of the theta-graph. For example: \\

$V_1(t) = \frac{1}{\sqrt{3}}\cos{(\frac{t-a_1}{b_1-a_1}\pi)}\dot{\gamma}_1+1\dot{\gamma}_1^{\perp}$ \\

$V_2(t) = \frac{1}{\sqrt{3}}\cos{(\frac{t-a_2}{b_2-a_2}\pi)}\dot{\gamma}_2-1\dot{\gamma}_2^{\perp}$ \\

$V_3(t) = \frac{-2}{\sqrt{3}}\cos{(\frac{t-a_3}{b_3-a_3}\pi)}\dot{\gamma}_3+0\dot{\gamma}_3^{\perp}$ \\

For the given variational fields $V_i$, we choose variations $H_i(\cdot,s)$ which agree at the vertices; for example, one could set $H_i(t,s)=\exp_{\gamma_i(t)}sV_i(t)$. The fact that the $H_i(\cdot,s)$ agree at the vertices ensures that we are deforming through theta-graphs. Moreover, as the variations keep each edge embedded, and maintain angles close to the initial $\frac{2\pi}{3}$ angles, we are guaranteed the theta-graphs remain embedded during the deformation.

If we denote by $L(s)$ the sum of lengths of $H_1(\cdot,s), H_2(\cdot,s), H_3(\cdot,s)$, then by the first variation formula (Proposition \ref{1stvar}) we have that

\begin{equation}\label{eq:length} L'(0)= \sum_{i=1}^3\langle V_i, \gamma_i'\rangle|_{a_i}^{b_i},
\end{equation}\noindent
since $\lbrace\gamma_i\rbrace_{i=1}^3$ are geodesics. And because the vector fields $\lbrace V_i\rbrace_{i=1}^3$ agree at the endpoints and the geodesics meet at angles $\frac{2\pi}{3}$, the summands in Equation~\ref{eq:length} cancel out for each vertex. Hence $L'(0)=0$.

For the second variation, note that because $\lbrace V^\perp_i\rbrace_{i=1}^3$ are parallel and $M$ has positive curvature we have that for $1\leq i\leq3$

\[\int_{a_i}^{b_i} \Vert \nabla_{\gamma_i'}V_i^\perp(t)\Vert^2 - \langle R(V_i^\perp(t),\gamma_i'(t))\gamma_i'(t), V^\perp_i(t) \rangle dt < 0
\]

Applying the second variation formula (Proposition \ref{2ndvar}) we have then

\begin{equation}\label{eq:2ndvar}
L''(0)< \sum_{i=1}^3\biggl\langle\frac{D}{ds}\frac{dH_i}{ds}, \gamma_i'\biggr\rangle\bigg|_{(a_i,0)}^{(b_i,0)}
\end{equation}

Given our choice of $H_i(t,s)=\exp_{\gamma_i(t)}sV_i(t)$ we see that $\frac{D}{ds}\frac{dH_i}{ds}=0$ for each $i \in \{1,2,3\}$ and therefore that the right hand side of Equation~\ref{eq:2ndvar} vanishes. Thus under the described variation we have $L''(0)<0$, which together with $L'(0)=0$, implies that length decreases for small values of $s$ in $\lbrace H_i(\cdot,s)\rbrace_{i=1}^3$.
\end{proof}

We show by example that Lemma~\ref{lem} is sharp in the sense that there exist stationary theta-graphs on non-negatively curved 2-spheres which do not admit directions of decrease. Consider the metric space formed by gluing two equilateral triangles along their common boundaries, so that a geodesic on the top face billiards around an edge to the bottom face. This \emph{doubled triangle} is a 2-sphere with flat metric and three conical singularities; it is sometimes called the Calabi-Croke sphere \cite{Sab2}.

The doubled triangle admits a degenerate stationary theta-graph. Connect the center of the top face with the center of the bottom face via three geodesic segments which pass perpendicularly through each edge of the triangular boundary (see Figure~\ref{graph}). Such an arrangement ensures that the edges of the graph meet at the vertices at angle $2\pi /3$. By moving the vertices of the graph towards one of the vertices of the triangle, and keeping the edges of the graph perpendicular to the edges of the triangle, one produces a degenerate family of stationary theta-graphs, all having the same total length, and failing to admit directions of decrease.

\begin{figure}[h]
  \includegraphics[scale=.25]{theta2}
  \centering
  \caption{A degenerate theta-graph on the doubled triangle, where edges on the top face are solid and edges on the bottom face are dashed.}
  \label{graph}
\end{figure}

Because these theta-graphs on the doubled triangle avoid the vertices of the triangle, this degenerate family also exists on smooth non-negative 2-sphere metrics close to the doubled triangle metric. This example illustrates that the proof of Theorem~\ref{main} in the non-negative setting can not rely directly on Lemma~\ref{lem}, and instead will follow by first proving the theorem in the positive setting, and then extending by considering conformally close positive metrics.

\begin{proof}[Proof of Theorem~\ref{main}]

Let us assume first $M = (S^2, g)$ to be a metric of positive curvature. We follow the proof of Theorem~\ref{4d}, trying to extend the map $f \colon (S^2, std) \to M$ to a map $\tilde{f} \colon (D^3, std) \to M$. When extending the 0-skeleton we now choose a point $\tilde{p} \in M$ that realizes the radius, i.e.~such that $B(\tilde{p},R)$ covers $M$. This choice ensure that $\tilde{p}$ is at distance at most $R(M)$ from the points in the triangulation; it is the same choice that Rotman \cite{rot5} makes when improving her bound from $4D(S^2, g)$ to $4R(S^2, g)$.

Next we follow the proof of Theorem~\ref{4d} until Case 3 where the theta-graph gets stuck on a stationary theta-graph, a critical point of the length shortening flow for nets. By Lemma~\ref{lem} this critical point admits a direction of decrease, allowing us to continue the contraction past the stationary theta-graph, and eliminating the necessity of the weighted length shortening.

While we are required to make a choice about the direction we deform from a stationary theta-graph, this choice can be made independently in each $3$-cell; we do not expect this extended (restarted) flow to depend continuously on the initial theta-graph. Indeed, we only need the fact \cite[Lemma 4]{NR2005} that the space of theta-graphs with length less than $3R(M)+\delta$ is connected in order to contract the initial theta-graph to a point. The continuity of the Birkoff curve shortening process (with respect to an initial pair of edges) in the absence of short closed geodesics allows us to extend this contraction to the entire 3-cell.

.

Upon resuming the flow, it is possible that we encounter another stationary theta-graph. Note that this stationary theta-graph will be distinct from the first, as the length shortening flow is strictly decreasing (preventing us from visiting the same sequence of theta-graphs repeatedly). It is possible running the flow in this extended (restarted) manner that we encounter a sequence of stationary theta-graphs accumulating to some limit object. In this case an application of transfinite induction ensures that this limit object is again a stationary theta-graph from which we can continue the flow.

As it is impossible to contract all theta-graphs to a point (this would extend the map $f \colon (S^2, std) \to M$ to a map $\tilde{f} \colon (D^3, std) \to M$) we must end in Case 1 or 2. By avoiding the weighted length shortening  we produce a (potentially self-intersecting) closed geodesic with length bounded above by $3R(M) \leq 3D(M)$.

The theorem is thus proved in the positive case. In the case where the sphere is non-negatively curved, we proceed as follows. Choose a smooth function $\varphi \colon S^2 \to \mathbb{R}$ such that $\Delta\varphi < 0$ on the set $K_{0} = \lbrace x\in S^2 \,|\, K_x=0 \rbrace$. The existence of such a $\varphi$ is possible because $K_0$ is a proper subset of the sphere (via Gauss-Bonnet). We consider the metrics $g_t = e^{2t\varphi}g$ which have strictly positive curvature $K_t = e^{-2t\varphi}(K-t\Delta\varphi)$ for $t>0$ small. Applying the result for positively curved metrics to $(S^2,g_t)$, and letting $t\rightarrow 0^+$, we use that the convergence $(S^2,g_t)\rightarrow(S^2,g)$ is smooth  and that $L(S^2,.)$ is lower-semicontinuous (this last property follows from the fact that in smooth convergence closed geodesics converge to closed geodesics) to conclude that the inequality $L(S^2,g) \leq 3R(S^2,g)$ holds in the non-negative setting.
\end{proof}

Finally, note that in \cite{calabi} it is proved that the shortest closed geodesic on a non-negatively curved $(S^2,g)$ is simple. Therefore, if the obstruction in the above proof yields a self-intersecting periodic geodesic (Case 2), then there also exists a simple closed geodesic with length bounded above by three times the radius. In short, we can use the main result of \cite{calabi} to improve our Theorem~\ref{main} to the following: on non-negatively curved $(S^2,g)$ the shortest closed geodesic is simple and has length bounded above by three times the radius.

\section{Pinched metrics on the 2-sphere}\label{pinch}

The main goal of this section is to prove the following new isoperimetric inequality for pinched metrics on the 2-sphere. A positive metric $(S^2,g)$ is said to be $\delta$-pinched if $K_{\min}/K_{\max}\geq \delta$.

\begin{theorem}\label{iso}
Let $(S^2,g)$ be a $\delta$-pinched metric. Then $$\pi A \leq \frac{4D^2}{\delta}$$ with equality if and only if the sphere is round.
\end{theorem}

We combine this inequality with the main result from \cite{recent} in order to prove Theorem~\ref{pinched}. The main result of \cite{recent} is achieved via a combination of techniques from Riemannian and symplectic geometry.

\begin{theorem}[\cite{recent}]\label{thm:L2Aineq}
Let $(S^2,g)$ be a $\delta$-pinched metric with $\delta > \frac{4+\sqrt{7}}{8} \approx 0.83$. Then $L^2(S^2, g) \leq \pi A(S^2, g)$ where $A(S^2,g)$ is the area.
\end{theorem}

By combining Theorem~\ref{iso} with Theorem~\ref{thm:L2Aineq}, we have for $(S^2,g)$ a $\delta$-pinched metric with $\delta > \frac{4+\sqrt{7}}{8} \approx 0.83$, that  $$L^2(S^2, g) \leq \pi A(S^2, g) \leq   \frac{4D^2(S^2,g)}{\delta} $$ and therefore that $$L(S^2,g) \leq  \frac{2}{\sqrt{\delta}}D(S^2,g).$$

Equality here implies equality in the isoperimetric inequality, and therefore that the sphere is round by Theorem~\ref{iso}. We have thus proved Theorem~\ref{pinched} and all that remains in this section is the proof of Theorem~\ref{iso}.

We first note that Theorem~\ref{iso} is a curvature pinched version of the following result due to Calabi and Cao. Moreover, the proof techniques we use are adaptations of theirs to the pinched curvature setting.

\begin{theorem}[\cite{calabi}, Theorem C]
Let $(S^2,g)$ have non-negative curvature. Then $A \leq \frac{8}{\pi} D^2$.
\end{theorem}

Calabi and Cao proved the above inequality by combining an upper bound on the first eigenvalue $\lambda_1(S^2,g) \leq \frac{8\pi}{A(S^2,g)}$ due to Hirsch  \cite{hersch, YangYau} with a lower bound $ \frac{\pi^2}{D^2(M^n,g)} \leq \lambda_1 (M^n,g)$ due to Zhong and Yang \cite{zhongyang} which holds in the non-negative Ricci setting. This lower bound on $\lambda_1(M^n,g)$ has been improved many times in the setting of positive lower bound on Ricci (see \cite{he} for a survey or \cite{AndrewsClutterbuck} for the optimal bound). We use the version from \cite{AndrewsClutterbuck}, which we state for $\lambda_1(S^2,g)$.

\begin{theorem}[\cite{AndrewsClutterbuck}]\label{thm:evlowerbound}
The quantity
\[\lambda_1(d,k) = \inf\lbrace \lambda_1(S^2,g)\,|\, Diam(S^2,g)\leq d,\,K(S^2,g)\geq k\rbrace
\]
is equal to the first eigenvalue $\mu$ of the following Sturm-Liouville problem (SLP) with Neumann initial conditions
\[ (y'\cos(\sqrt{k}x))' + \mu\cos(\sqrt{k}x)y = 0,\quad y'(\pm d/2)=0
\]
\end{theorem}

We can apply this directly to the case $d=\pi,\, k=1$ to obtain the case ${\rm dim}=2$ of the classical result of Lichnerowicz \cite{Lichnerowicz}.

\begin{lemma}\label{two} For $(S^2,g)$ with $Diam(S^2,g)\leq \pi$ and $\,K(S^2,g)\geq 1$ we have $\lambda_1 \geq 2.$
\end{lemma}
\begin{proof}
The first eigenfunction of $(y'\cos(x))' + \mu\cos(x)y = 0,~ y'(\pm \pi/2)=0$ is $y(x)=\sin(x)$ which has eigenvalue $2$.
\end{proof}

We can now provide a proof of Theorem~\ref{iso}. The main insight is that when considering diameter bounds, we can use Klingenberg's injectivity radius estimate to translate between the positive and the pinched curvature settings.

\begin{proof}[Proof of Theorem~\ref{iso}]

For simplicity we will drop $(S^2,g)$ from our notation. Since the inequality is scale invariant, let us rescale the metric so $K_{min}=1$ and by Myers theorem $D \leq \pi$. Lemma~\ref{two} then gives the bound $ \lambda_1\geq2$. Following the ideas of \cite{calabi} we combine this lower bound on $\lambda_1$ with the upper bound due to Hirsch $\lambda_1 \leq \frac{8\pi}{A}$ (see \cite{hersch}, \cite{YangYau}) to yield $A \leq 4\pi$.
Klingenberg's injectivity radius estimate for positive metrics on 2-spheres says that $$D \geq \frac{\pi}{\sqrt{K_{max}}} = \pi \sqrt{\delta}.$$ We have therefore related both area and diameter to $\pi$, and conclude that $$\pi A \leq 4\pi^2 \leq \frac{4D^2}{\delta}.$$

In the case of equality we have $4\pi^2 = \frac{4D^2}{\delta}$ and therefore $D=\frac{\pi}{\sqrt{K_{max}}}$. This is the limiting case of Klingenberg's injectivity radius estimate, and therefore diameter equals injectivity radius. This is the so called Blaschke condition, and for Blaschke metrics on the 2-sphere Green \cite{green} gives an elementary proof using classical surface geometry that the sphere must be round.
\end{proof}

We finish by observing that the isoperimetric inequality in Theorem~\ref{iso} is sharp in the sense that the estimate $A/D^2$ approaches $4/\pi$ as we move towards the round metric, i.e.~as $\delta$ approaches $1$. We do not recover the Calabi-Cao inequality as $\delta$ approaches 0, i.e.~for $(S^2,g)$ with non-negative curvature. Note that the Calabi-Cao inequality is not sharp for convex metrics on $(S^2,g)$; Alexandrov has conjectured that the sharp inequality is realized by the doubled disk metric where $A/D^2 = \pi/2$.

\section{Appendix}

Using more advanced techniques from Sturm-Liouville theory we can improve the isoperimetric inequality in Theorem~\ref{iso} when the sphere is not round. This improved version of the inequality is not needed for the proof of Theorem~\ref{pinched} that was presented in Section~\ref{pinch}.

\begin{theorem}\label{ap_thm}
Let $(S^2,g)$ be a $\delta$-pinched metric and denote by $\eta=D\sqrt{K_{min}}$.  Then
\begin{equation*}\label{eq:CCpinched}

\pi A \leq \frac{4D^2}{\delta(2-\sin(\eta/2))}.
\end{equation*}
\end{theorem}

First note that $2- \sin(\eta/2) \geq 1$ so that this isoperimetric inequality is indeed an improvement on that of Theorem~\ref{iso}. Moreover, because $2- \sin(\eta/2) > 1$ when $\eta \neq \pi$ we note that Theorem~\ref{ap_thm} together with Cheng's \cite{cheng} rigidity result when $D=\pi/\sqrt{K_{min}}$ provide an alternate proof of the equality case of Theorem~\ref{pinched} that does not depend on the Blaschke ideas from Section~\ref{pinch}.

This new inequality follows as before by combining the inequality due to Hirsch with an improved lower bound on the first eigenvalue. We therefore set out to calculate the linear approximation of $\lambda_1(d,1)$ at $d=\pi$. The main idea is that because the coefficients of the SLP are constant while the domain varies, we can restrict and extend eigenfunctions to compare the values $\lambda_1(d,1)$ as $d\rightarrow\pi$. Let us then define by $\mu(d)$ the SLP eigenvalue $\lambda_1(d,1)$. We then have

\begin{lemma}\label{lem:1sttaylorlambda} For $\pi\geq d >0$
\[2 + 2(1-\sin(d/2)) \leq \mu(d).
\]
\end{lemma}
\begin{proof}
Define the change of variable $\tanh(u)=\sin(x)$ on the SLP, helpfully suggested to us by Ben Andrews. This becomes

\begin{equation}\label{eq:modSLP}
    y'' + \mu.\sech^2(u)y = 0,\quad y'(\pm T)=0,\, \quad \tanh(T)=\sin(d/2)
\end{equation}
In particular, for $d=1, T=+\infty$, the eigenvalue $\mu=2$ is realized by $y_{+\infty}=\tanh(u)$.

Recall that the first eigenvalue for a SLP can be written as a Rayleigh quotient

\begin{equation*}
    \mu(d) = \inf_{y\neq 0, y'(\pm T)=0} R[y,T]
\end{equation*}
where

\[R[y,t]= \frac{\int_{-T}^T -y.y''du}{\int_{-T}^T y^2\sech^2(u)du}
\]

Denote by $\hat{y}_T$ the eigenfunction of the SLP (\ref{eq:modSLP}) normalized so $\int_{-T}^T \hat{y}_T^2\sech^2(u)du=1$. Extended $\hat{y}_T$ as a constant to the entire real line so (while keeping the same notation):

\[  \hat{y}_T(u) =
  \begin{cases}
     \hat{y}_T(u) & \text{if $|u|\leq T$} \\
     \hat{y}_T(T) & \text{if $u>T$}\\
     \hat{y}_T(-T) & \text{if $u<-T$}

  \end{cases}
\]

In order to use $\mu(1)\leq R[\hat{y}_T,+\infty]$ we will need to bound $\hat{y}_T(\pm T)$. Observe first that given the symmetries of the SLP (\ref{eq:modSLP}) the function $\hat{y}_T$ is an odd function. Since $\mu(d)$ is the first eigenvalue, we know that $\hat{y}'_T$ does not vanish in the open interval $(-T,T)$. Hence $|\hat{y}_T(\pm T)| = \max_{-T\leq u \leq T}  |\hat{y}_T(u)|$, so then

\begin{equation*}
    1=\int_{-T}^T \hat{y}^2_T\sech^2(u)du \leq |\hat{y}_T(\pm T)|^2 \int_{-T}^T \sech^2(u)du \leq 2|\hat{y}_T(\pm T)|^2
\end{equation*}

Replacing now $R[\hat{y}_T,+\infty]$

\begin{equation*}
    2=\mu(\pi)\leq R[\hat{y}_T,+\infty] = \frac{\mu(d)}{1 + 2|\hat{y}_T(T)|^2\int_T^{\infty} \sech^2(u)du}
\end{equation*}

Hence we obtain

\begin{equation*}
    2 + 2(1-\tanh(T)) \leq \mu(d)
\end{equation*}

Replacing now $\tanh(T)=\sin(d/2)$ and using the Taylor series of sine at $\pi/2$ we finally get

\begin{equation*}
    2 + 2(1-\sin(d/2)) \leq \mu(d).
\end{equation*}\end{proof}

We are now ready to prove Theorem~\ref{ap_thm}.

\begin{proof}
The proof proceeds exactly as the proof for Theorem~\ref{iso} and we begin by rescaling the metric so that $K_{min}=1$ and by Myers theorem $\eta = D \leq \pi$. Lemma~\ref{lem:1sttaylorlambda} then gives the bound $  2 + 2(1-\sin(d/2)) \leq \mu(d) = \lambda_1(d,1)$. Following the ideas of \cite{calabi} we combine this lower bound on $\lambda_1$ with the upper bound due to Hirsch $\lambda_1 \leq \frac{8\pi}{A}$ (see \cite{hersch}, \cite{YangYau}) to yield

\begin{equation*}
    2 + 2(1-\sin(\eta/2)) \leq \frac{8\pi}{A}
\end{equation*}
which we can manipulate to
\begin{equation*}
    \pi A \leq \frac{8\pi^2}{2 + 2(1-\sin(\eta/2))}.
\end{equation*}

Klingenberg's injectivity radius estimate for positive metrics on 2-spheres says that $$D \geq \frac{\pi}{\sqrt{K_{max}}} = \pi \sqrt{\delta}.$$

We have therefore related both area and diameter to $\pi$, and conclude that

\begin{equation*}\label{eq:sdependance}
    \pi A \leq \frac{8\pi^2}{2 + 2(1-\sin(\eta/2))} = \frac{4\pi^2}{K_{min}(2-\sin(\eta/2))} \leq \frac{4D^2}{\delta(2-\sin(\eta/2))}.
\end{equation*} \end{proof}

Finally, while not necessary for the proof of Theorem \ref{ap_thm}, the following Lemma of independent interest expands the approach of Lemma \ref{lem:1sttaylorlambda} and gives an idea of the sharpness of its inequality, both in general and at the limiting case of $d=\pi$.

\begin{lemma}
For $\pi\geq d \geq 2\arcsin(\tanh(3/2)) \approx2.263$
\[\mu(d) \leq 2 + \cos^2(d/2) A(d),
\]
where $$A(d) = \frac{6[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}{\sin^3(d/2) - 3\cos^2(d/2)[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}.$$
Moreover, at the limiting case $d=\pi$ we have $\mu'(\pi)=0$ and $\frac14\leq\mu''(\pi)\leq\frac32$.
\end{lemma}

\begin{proof}
Define the test function $y_T = \tanh(u) - u\sech^2(T)$, which by design satisfies the Neumann initial conditions of (\ref{eq:modSLP}). By direct calculation
\begin{eqnarray*}
\begin{split}
    \int_{-T}^T -y_T.y_T''du &= \int_{-T}^T 2\tanh^2(u)\sech^2(u)du - \sech^2(T)\int_{-T}^T 2u\tanh(u)\sech^2(u)du\\
    &= \frac23[2\tanh^3(T)] - \sech^2(T)[2\tanh(T) - 2T\sech^2(T)]
\end{split}
\end{eqnarray*}

\begin{eqnarray*}
\begin{split}
    \int_{-T}^T y_T^2\sech(u)du &= \int_{-T}^T \left(\tanh^2(u) -2u\tanh(u))\sech^2(T) + u^2\sech^4(T)\right)\sech^2(u)du\\
    &= \int_{-T}^T \tanh^2(u)\sech^2(u)du -\sech^2(T)\int_{-T}^T 2u\tanh(u)\sech^2(u)du \\& +  \sech^4(T)\int_{-T}^T u^2\sech^2(u)du\\
    &= \frac13[2\tanh^3(T)] - \sech^2(T)[2\tanh(T) - 2T\sech^2(T)] \\&+ \sech^4(T) \int_{-T}^T u^2\sech^2(u)du.
\end{split}
\end{eqnarray*}
Hence our test function $y_T$ gives us the inequality
\begin{eqnarray*}
\begin{split}
    \mu(d) &\leq \frac{\int_{-T}^T 2\tanh^2(u)\sech^2(u)du - \sech^2(T)\int_{-T}^T 2u\tanh(u)\sech^2(u)du}{\int_{-T}^T \tanh^2(u)\sech^2(u)du -\sech^2(T)\int_{-T}^T 2u\frac{\tanh(u)}{\cosh^2(u)}du + \sech^4(T)\int_{-T}^T u^2\sech^2(u)du}\\
    &\leq \frac{\frac23[2\tanh^3(T)] - \sech^2(T)[2\tanh(T) - 2T\sech^2(T)]}{\frac13[2\tanh^3(T)] - \sech^2(T)[2\tanh(T) - 2T\sech^2(T)]}
\end{split}
\end{eqnarray*}
where the last inequality is valid if $T\geq\frac32$, so for $d\geq2\arcsin(\tanh(3/2))\approx 2.263$.

Using that $\sech^2(T)=1-\tanh^2(T) = 1-\sin^2(d/2)=\cos^2(d/2)$ we have the desired inequality:
\begin{eqnarray*}
\begin{split}
    \mu(d) &\leq \frac{2\sin^3(d/2) - 3\cos^2(d/2)[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}{\sin^3(d/2) - 3\cos^2(d/2)[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}\\
    &= 2 + \cos^2(d/2) \frac{6[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}{\sin^3(d/2) - 3\cos^2(d/2)[\sin(d/2) - \arctanh(\sin(d/2))\cos^2(d/2)]}
\end{split}
\end{eqnarray*}

Turning our attention to the limiting case of $d=\pi$, we Taylor expand cosine at $\pi/2$ to yield
\begin{equation*}\label{eq:upper2deriv}
    \mu(d) \leq 2 + \frac32(\pi-d)^2 + O((\pi-d)^3).
\end{equation*}

Similarly, we combine Lemma \ref{lem:1sttaylorlambda} and the Taylor expansion of sine at $\pi/2$ to yield

\begin{equation*}\label{eq:lower2deriv}
    2 + \frac14(\pi-d)^2 + O((\pi-d)^3) \leq \mu(d).
\end{equation*}

From this pair of Taylor expansions we calculate that

\begin{equation*}
    \mu'(\pi)=0,\quad\frac14 \leq \mu''(\pi) \leq \frac32 .
\end{equation*}
\end{proof}

config: arxiv


TITLE: Intuition behind parallel lines intersecting in projective space
QUESTION [1 upvotes]: There are answers to this elsewhere, but my question is in the context of a particular problem, that being 1.4.21 from Garrity's Algebraic Geometry: A Problem Solving Approach.

Exercise 1.4.21. Once we have homogenized an equation, the original variables $x$ and $y$ are no more important than the variable $z.$ Suppose we regard $x$ and $z$ as the original variables in our homogenized equation. Then the image of the $xz$-plane in $\mathbb{P}^2$ would be $\{(x:y:z) \in \mathbb{P}^2 : y = 1\}.$
(1) Homogenize the equations for the parallel lines $y=x$ and $y = x + 2.$

The homogenizations are $y = x$ and $y = x+ 2z.$

(2) Now regard $x$ and $z$ as the original variables and set $y = 1$ to sketch the image of the lines in the $xz$-plane.

This gives us $x = 1$ and $z = -\frac{1}{2}x + \frac{1}{2}.$ The graph looks like the following: https://www.desmos.com/calculator/4ncyzir9kg

(3) Explain why the lines in part (2) meet at the $x$-axis

We can see in the picture that the lines intersect at the x-axis.
I know that the original lines intersect at the line at infinity. I can even show this algebraically. If we set the equations $x = 1$ and $z = -\frac{1}{2}x + \frac{1}{2}$ equal to each other, then we get that $z = 0.$ We already know that $x = 1$ and $y = 1,$ so the intersection point is $(1:1:0),$ which is the line at infinity. My question is how parts (2) and (3) are supposed to show this. I'm looking for some geometric intuition as to why intersecting at the $x$-axis means that the original lines intersect at the line at infinity.

REPLY [3 votes]: Taking a “top-down” approach, there is no distinguished line at infinity in the pure projective plane, nor is there a notion of parallel lines. Any line can be singled out as the line at infinity, often implicitly by the coordinate system that you impose on $\mathbb P^2$, which then in turn defines which lines are parallel. (In doing so you have imposed an affine geometry on the projective plane.) That’s more or less what the text is trying to communicate by switching the line at infinity from $0:0:1$ to $0:1:0$.
In the “bottom-up” model of $\mathbb P^2$ being used by your text, points in $\mathbb P^2$ correspond to lines through the origin in $\mathbb R^3$ (strictly speaking, in $\mathbb R^3\setminus\{0\}$), and lines in $\mathbb P^2$ correspond to planes through the origin in $\mathbb R^3$. Obviously, any two such distinct planes intersect in a line through the origin—a point in $\mathbb P^2$. The correspondence between lines/planes in $\mathbb R^3$ and point/lines, respectively, in $\mathbb P^2$ is established by choosing a plane in $\mathbb R^3$ that doesn’t go through the origin—conceptually, we embed $\mathbb P^2$ into $\mathbb R^3$ as that plane. The intersections of the lines/planes through the $\mathbb R^3$ origin are the corresponding projective points/lines; the parallel plane through the origin maps to the projective line at infinity and the lines that lie on this plane are the projective points at infinity.
The standard coordinate system used in the text corresponds to choosing the plane $z=1$ as the embedded projective plane. If two projective lines intersect at infinity, then the intersection of their corresponding planes in $\mathbb R^3$ lies on the $x$-$y$ plane. In this exercise, you are then asked to switch to $y=1$ as the embedded projective plane. If two lines on this plane intersect on the $x$-axis, then the line in $\mathbb R^3$ that represents this point lies entirely within the $x$-$y$ plane, but per the preceding that means that the projective lines intersected at infinity in the original coordinate system that has $0:0:1$ as the line at infinity. On the other hand, they clearly don’t intersect at infinity in the second embedding. This change of embedding is equivalent to applying a projective transformation to $\mathbb P^2$ and illustrates that parallelism of lines is not a projective invariant—it’s not a property that’s preserved by projective transformations.
Here are the planes (in red) that correspond to the two lines in the problem and their line of intersection (orange). This line lies entirely within the $x$-$y$ plane (gray).

In the original embedding, the two lines represented by the red planes are their intersection with the plane $z=1$:

Here, the plane $z=0$ corresponds to the line at infinity. The red planes’ intersection lies on this plane, so they are parallel and intersect at a point at infinity.
In the second embedding, we instead intersect the two red planes with $y=1$:

In this embedding, the plane $y=0$ (not shown) corresponds to the line at infinity, and the lines represented by the red planes are no longer parallel. The orange line intersects $y=1$ at a single point—the point of intersection of the two lines on the embedded plane. The black lines are not parallel in this embedding, but their intersection does lie on the embedded plane’s $x$-axis, which is its intersection with the $x$-$y$ plane in $\mathbb R^3$, the original line at infinity.


TITLE: The number of odd-degree vertices is even in a finite graph?
QUESTION [1 upvotes]: if you add up the degrees of all the vertices of a finite graph then the result will be twice the number of edges, provided that the vertices are odd-degree. What is the name of this theorem? why does it require that the graph be undirected?

REPLY [2 votes]: It is known as the handshaking lemma because if a bunch of people shake hands and you add up the number of hands each person shook, you get an even number.  Each handshake adds two to the total.  It does not require that each vertex be of odd degree, but it shows there are an even number of vertices of odd degree.  For a directed graph, this is still true if you add all the indegrees and outdegrees for the same reason, but the sum of indegrees need not be even.  Think of a directed graph with two vertices and one directed edge connecting them.


TITLE: Number rings as free module over base ring
QUESTION [2 upvotes]: Let $K \subset L$ be number fields and $\mathcal{O_K}, \mathcal{O_L}$ the corresponding rings of algebraic integers. Further let dimension$(L/K)$ = n as a vector space.
If $K$ is a PID, then $\mathcal{O_L}$ is a free module of dimension n over $\mathcal{O_K}$ (I believe this is true, please correct me if I am wrong). Is this also true in the general case and if not, can we describe the structure of $\mathcal{O_L}$ over $\mathcal{O_K}$ as a module?

REPLY [3 votes]: If $\mathcal{O_K}$ is a PID, then $\mathcal{O_L}$ is a free module over $\mathcal{O_K}$ (just use the fundamental theorem of f.g. modules over PID together with the fact that $\mathcal{O_L}$ is torsion free).
In general, it is not true that $\mathcal{O_L}$ is free over $\mathcal{O_K}$. There is a similar theorem for f.g. modules over Dedekind domains, but there the torsion free part is a direct sum of fractional ideals (which are all free if the domain is actually a PID). See for example here.
For rings of integers, there are examples where the extension is not free. Moreover, in "On Integral Bases" by Henry B. Mann, it is shown that if $\mathcal{O}_K$ isn't a PID, then $K$ has a quadratic extension $L$ such that
$\mathcal{O}_L$ is not free over $\mathcal{O}_K$, or in other words, $\mathcal{O}_K$ is a PID iff all the extensions $\mathcal{O}_L$ of integer rings are free over it.


TITLE: Computing $\liminf A_{n}$ and $\limsup A_{n}$ for a given sequences of sets
QUESTION [1 upvotes]: For each $n \in \mathbb{N}$, let
$$A_{2n-1}=\left\lbrace (x,y) \in \mathbb{R}^{2} \: | \: y\geq 0, x\geq 1 +\frac{1}{n} \right\rbrace$$
and
$$A_{2n}=\left\lbrace (x,y) \in \mathbb{R}^{2} \: | \: x \geq 0, 0 \leq y \leq
 \frac{1}{n} \right\rbrace$$
Calculate  $\liminf A_{n}$ and $\limsup A_{n}$. I don't know how to calculate the mentioned limits manually but my intuition says this sequence converges and it converges to
$$A= \lbrace (x,y) \in \mathbb{R}^{2} \: | \: x \geq 1, y=0\rbrace.$$ So if this is true then
$$\liminf A_{n}=A=\limsup A_{n}?$$
I need help with the technicalities of the proof as I've never have worked with limits of sequences involving sets. Thanks!

REPLY [1 votes]: As @Marios Gretsas indicated in a comment, $\liminf$ is the set $\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k$,
and $\limsup$ is the set $\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$.
In other words, $x$ belongs to $\limsup$ if $x$ belongs to infinitely
many $A_k$, and $x$ belongs to $\liminf$ if $x$ belongs to all but finitely
many $A_k$. See also https://en.wikipedia.org/wiki/Set-theoretic_limit
For the sets in the question we do not have that $\limsup=\liminf$.
It is not clear how you have used your intuition to come up with the result you state in the question. Perhaps you could have provided more details.
$\limsup$ might be easier to consider first. From the definition of $A_{2n-1}$
it is easily seen that $\limsup A_n\supseteq O:=\{(x,y)|y\ge0,x>1\}$. In fact,
$\limsup A_{2n-1}=O$ since every point $x$ in $O$ belongs to all but finitely many
$A_{2n-1}$ (and only points $x$ in $O$ have this property). Since $A_{2n-1}$ is
an increasing sequence (in particular monotone), we have moreover that
$\limsup A_{2n-1}=\liminf A_{2n-1}=\lim A_{2n-1}=O$. For an increasing sequence of sets we always have $\limsup =\liminf=\lim=$ the union of the sets, so
$\lim A_{2n-1}=\cup_n A_{2n-1}=O$ (explaining why $O$ looks the way it does, as defined earlier).
Similarly, $A_{2n}$ is a decreasing sequence and
$\limsup A_{2n}=\liminf A_{2n}=\lim A_{2n}=\cap_n A_{2n}=E:=\{(x,y)|x\ge0,y=0\}$.
This is the positive part of the $x$-axis, including the origin.
It follows that $\limsup A_n=O\cup E$.
On the other hand, $\liminf A_n=O\cap E=\{(x,y)|x>1,y=0\}$.


TITLE: Let $V$ an inner product space and $a,b\in V$ prove that $((a,d)=(b,d))\forall d\in V\implies a=b$
QUESTION [1 upvotes]: I've been working on this problem that required a bi-implication, and one side is pretty simple, but then when I get to the implication in the title, I keep running into problems with circular reasoning. Does anyone here have an idea on how to prove this?
The original question: Let $V$ be an inner product space and let $a, b$ be vectors in $V$. Show that $a=b$ if and only if $(a,d)=(b,d)$ for every $d$ in $V$

REPLY [8 votes]: Hint:
$$(a,d) = (b,d) \Longleftrightarrow (a-b,d) = 0. $$
There is a special choice of $d$ which makes it clear that $a-b=0$, i.e. $a=b$. (Think about the properties that define an inner product..)


TITLE: Two Hausdorff Compactifications are equivalent iff the sets of bounded continuous real valued functions in X which extend to Y and Z coincide.
QUESTION [2 upvotes]: Let $(g, Y)$ and $(h, Z)$ be two Hausdorff compactifications of $X$.
We have that $(g, Y)$ is equivalent to $(h, Z)$ if $\exists \text{ homeomorphism } \pi : Z \to Y$ such that $\pi \circ h = g$ by definition.
If this is the case, then if $f$ extends to $Y$ so there is a continuous $F:Y \to \Bbb{R}$ such that $F \circ g = f$ we see that $F \circ \pi$ extends $f$ to $Z$. Similarly the symmetrical case, so we have that if $(h, Z)$ and $(g, Y)$ are equivalent then the sets of bounded continuous real valued functions on $X$ which extend $Y$ and $Z$ coincide.
I am really struggling to show the converse. Any help would be appreciated.

REPLY [3 votes]: Embedding is the key to answer your question. For any space $A$ let $C(A)$ denote the $\mathbb{R}$-algebra of continuous bounded real-valued functions on $A$. We can endow it with the $\sup$-norm, but that is irrelevant here.
The embedding $g : X \to Y$ induces the "restriction" $g^\ast : C(Y) \to C(X), g^\ast(\varphi) = \varphi \circ g$ which is a homomorphism of algebras. Define
$$C(X,g) = g^\ast(C(Y)). $$
This is the set of continuous bounded real-valued functions on $X$ which "extend" to $Y$. Since $g(X)$ is dense in $Y$, extensions are unique so that $g^\ast : C(Y) \to C(X,g)$ is an isomorphism. Let $P = \mathbb{R}^{C(X,g)}$ denote the product of copies $\mathbb{R}_\psi$ of the real line indexed by the set of all $\psi \in C(X,g)$. Define
$$u_g : Y \to P, u_g(y) = (\psi'(y))_{\psi \in C(X,g)} ,$$
where $g^\ast(\psi') =\psi$. Then $u_g$ is continuous since $p_\psi \circ u_g  = \psi'$ is continuous for all projections $p_\psi : P \to \mathbb{R}_\psi = \mathbb{R}$. Moreover $u_g$ is an injection because for any two distinct $y_1, y_2 \in Y$ there is a continuous $\varphi : Y \to \mathbb{R}$ such that $\varphi(y_1 ) \ne \varphi(y_2)$ (now take $\psi = g^\ast(\varphi)$). Therefore $u_g$ is an embedding because $Y$ is compact. By construction
$$g' = u_g \circ g : X \to P$$
is an embedding such that $\overline{g'(X)} = u_g(Y)$. For $x \in X$ we obtain
$$g'(x) = u(g(x)) = (\psi'(g(x)))_{\psi \in C(X,g)} = (\psi(x))_{\psi \in C(X,g)} .$$
Now consider $h : X \to Z$. By assumption we have $C(X,g) = C(X,h)$. We therefore obtain an embedding $u_h : Z \to P$ such that $\overline{h'(X)} = u_h(Z)$, where $h' = u_h \circ h$. But $g'= h'$ so that $u_g(Y) = u_h(Z)$ and $\pi = u_h^{-1} \circ u_g : Y \to Z$ is the desired homeomorphism.


TITLE: Find $\int{\arctan x}\,\mathrm dx$ without substitution
QUESTION [4 upvotes]: 2018-08-15: I'm still looking for an answer that does not rely on
$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$

I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to find $\int{\arctan x}\,\mathrm dx$.
I start with integration by parts:
$$\int{\arctan x}\,\mathrm dx
= \int{1\cdot\arctan x}\,\mathrm dx
= x\cdot\arctan x - \int{x\cdot \frac{1}{x^2+1}}\,\mathrm dx$$
Next, it would be natural to use substitution, and I can do that to get the answer:
$$x\cdot\arctan x - \frac{\ln(x^2+1)}{2}+c$$
Which is correct, but it's clear from context that the book wants me to do it without using substituation. I feel like I've tried everything, looking at the book's own examples, but I must be missing some essential trick.
Here are some of the approaches I've tried:
1. Integration by parts, again
1.1. $u = x$ , $v' = \frac{1}{x^2+1}$
$$x\cdot\arctan x - \left(x\cdot\arctan x - \int{1\cdot\arctan x}\,\mathrm dx\right)
= \int{\arctan x}\,\mathrm dx$$
Back where I started.
1.2. $u = \frac{1}{x^2+1}$ , $v' = x$
$$x\cdot\arctan x - \left(\frac{x^2}{2} \cdot \frac{1}{x^2+1} - \int{\frac{x^2}{2} \cdot \frac{-2x}{(x^2+1)^2}}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2(x^2+1)} - \int{\frac{x^3}{(x^2+1)^2}}\,\mathrm dx$$
I've tried hacking away at this, but it doesn't look like it's getting any easier.
2. Adding and subtracting $x^2$ in the numerator
$$x\cdot\arctan x - \int{x\cdot \frac{(x^2+1)-x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x\cdot \left(1 - \frac{x^2}{x^2+1}\right)}\,\mathrm dx$$
$$= x\cdot\arctan x - \int{x - x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \int{x\cdot\frac{x^2}{x^2+1}}\,\mathrm dx$$
The book uses a similar trick (adding and subtracting 1 in the numerator) to solve $\int{x\cdot\arctan x}\,\mathrm dx$. In the process it finds that $\int{\frac{x^2}{x^2+1}}\,\mathrm dx = x - \arctan x$. Perhaps I need to use this result:
Partial integration with $u = x$ , $v´ = \frac{x^2}{x^2+1}$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x\cdot(x-\arctan x) - \int{1\cdot (x-\arctan x)}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + \left(x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx\right)$$
$$= x\cdot\arctan x - \frac{x^2}{2} + x^2 - x\cdot\arctan x - \frac{x^2}{2} + \int{\arctan x}\,\mathrm dx$$
$$= \int{\arctan x}\,\mathrm dx$$
Back where I started, again. I've been trying for days, different tricks and manipulations. The book doesn't even list it as a particularly tricky question, so I'm feeling a bit dumb.
3. Searching for the answer
I've looked at some other questions, including

Integration of arctan(x) is itself?
Indefinite integral of $\arctan(x)$, why consider $1\cdot dx$?

But they don't seem to answer my particular question.

REPLY [1 votes]: I don't know if this is what you are looking for or not, but functions $f$ and their inverses $f^{-1}$ are pretty cool. If you know the derivative of $f$, you know the derivative of $f^{-1}$. And if you know the antiderivative of $f$, you know the antiderivative of $f^{-1}$ as well via this interesting formula on integration of inverse functions . I write it out here
$$ \int f^{-1}(y) \;dy = yf^{-1}(y) - F \;\circ \; f^{-1}(y) + C$$
So if you know that the antiderivative of $\tan{x}$ is $-\ln(\cos{x})$, then the result follows. This is a trick, so I'm not sure it's what you are looking for. Also, in order to find the antiderivative of $\tan{x}$, you use u-substitution which goes against our goal of integrating without u-substitution. Anyways, hopes this helps


TITLE: What's the upper bound for sofa problem?
QUESTION [9 upvotes]: I have seen a claim that for the sofa problem, an upper bound for the area of a sofa is $2 \sqrt 2$, and that this can be proved by a "simple" argument. But I can't find a proof. What that argument?
(The articles and papers I found all seem to point back to On the enfeeblement of mathematical skills by modern mathematics and by similar soft intellectual trash in schools and universities, but that paper gives this bound as a problem, and I can't find the solution.)

REPLY [6 votes]: It does not seem useful to me to just repeat the solution in e.g. Ian Stewart's book Another Fine Math You've Got Me Into... as linked from the Wikipedia Moving sofa problem page, so I'll just describe how I understood it.
(Note: This is take 2. Edited on 2017-12-12 to fix the +2 to -2 in the formula, and to expand a bit on how the areas are calculated.)

Consider the situation when the sofa is rotated 45 degrees with respect to the corner. In order to fit into the straight part of the corridor, the sofa must lie within the shaded area (corresponding to a straight corridor of width 1, denoted by dashed lines) in the illustrations below.
We can parametrize the different possibilities by the distance $h$ between the outer corner and the part that fits into the straight corridor.
There are two basic cases:

The sofa fits within the corner triangle, $0 \le h \lt \sqrt{2} - 1$:

Note that the area of both triangles is $(2 h + 2)(h + 1)/2 = h^2 + 2 h + 1$, the area of the upper, white triangle is $2hh/2 = h^2$.
In this case, the area is $$A_1(h) = h^2 + 2 h + 1 - h^2 = 2 h + 1$$
which reaches its maximum $2 \sqrt{2} - 1 \approx 1.828427$ at the upper limit, $h = \sqrt{2} - 1$.
The sofa does not fit into the corner triangle; $\sqrt{2} - 1 \le h \le \sqrt{2}$:

Note that the area of the triangle above the blue line is $2\sqrt{2}\sqrt{2}/2 = 2$, the area of the white triangle is $2 h h / 2 = h^2$, and the area of each of the parallelograms below the blue line is $\sqrt{2}(1 - (\sqrt{2} - h)) = \sqrt{2} h + \sqrt{2} - 2$.
In this case, the shaded area, the area the sofa can possibly occupy, is
$$A_2(h) = (2 - h^2) + 2 (\sqrt{2} h + \sqrt{2} - 2) = 2\sqrt{2} - 2 + 2\sqrt{2} h - h^2$$
which reaches its maximum, $2\sqrt{2} \approx 2.828427$, at the upper end of the range, $h = \sqrt{2}$.

Note that $h \lt 0$ makes no physical sense (we'd just be rejecting part of the available width for the sofa for zero gain), and that $h \gt \sqrt{2}$ means the sofa would split into two in the cornering.
In short, the maximum area the sofa can occupy, as a function of $h$, is
$$A(h) = \begin{cases}
0, & h \lt -1 \quad \text{(no sofa at all)} \\
(h + 1)^2, & -1 \le h \lt 0 \\
2 h + 1, & 0 \le h \lt \sqrt{2} - 1 \\
2\sqrt{2} - 2 + 2\sqrt{2} h - h^2, & \sqrt{2} - 1 \le h \le \sqrt{2} \\
2 \sqrt{2}, & h \gt \sqrt{2} \quad \text{ but split in two } \end{cases}$$
which reaches a maximum at $h = \sqrt{2}$ (still connected by an infinitesimally thin part at the middle), with area $2 \sqrt{2}$:

Indeed, all sofa variants, convex and concave, and even cars (that turn both ways), must reside within the shaded area -- just with different values of $0 \le h \le \sqrt{2}$. If they do not, they either do not fit into the straight corridor, or they do not fit into the corner at a 45 degree angle.
This means the maximum area for the moving sofa (and cars) cannot exceed $2\sqrt{2}$.
(Note that because we only consider the non-rotated and rotated-by-45-degrees cases, this does not prove that there is a sofa (or car) with area $2\sqrt{2}$ that can turn the corner. We only proved that no sofa (or car) cannot be larger than $2\sqrt{2}$ in area, because it would be impossible for a sofa (or car) to fit into the corridor when rotated by 45 degrees.)


TITLE: How is ${n \choose n/2}$ approximated in this manner?
QUESTION [2 upvotes]: In this question, the author asked how to get an asymptotic growth of ${n \choose n/2}$, to which the answer is as follows. First:
$$ {n \choose n/2} = \frac{n!}{(n/2)!(n- n/2)!} = \frac{n!}{(\tfrac{n}{2}!)^2} $$
Now, to find an upper bound on this, take the upper bound of Stirling's approximation for $n!$ and the lower bound for $\tfrac{n}{2}!$ which if I'm not mistaking gives the following:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} $$
However, the answer states this should give:
$$ {n \choose n/2} \leq \frac{e \cdot n^{n + \tfrac{1}{2}} e^{-n}}{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}}\Big )^2} $$
I'm confused as to how one would get from the former to the latter. Up to now I get stuck around here:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi n} \cdot n^n e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} \cdot \tfrac{n}{2}^\tfrac{n}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi} \cdot n^{n + \tfrac{1}{2}} e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2}
$$
Now, do the latter terms just disappear due to $\lim_{n \rightarrow \infty} \frac{e^{\tfrac{1}{12n}}}{\Big ( e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} = 0$? But in this case, why should there still be an $e$ term in the numerator? Also what happened to the $\sqrt{2 \pi}$ in the numerator? It cannot have canceled out with the one in the denominator since the latter is still there...

REPLY [1 votes]: Not an answer but a general hint, too long for comment: it's good to practice, but a math software like Maxima is free and it helps a lot when things get messy - at least to check results. In this case, to get the upper bound
   stU(n) := sqrt(2*%pi*n) * (n/%e)^n * exp(1/(12*n)); /*  stirling approx, upper bound */

   stL(n) := sqrt(2*%pi*n) * (n/%e)^n * exp(1/(12*n+1)); /*  stirling approx, lower bound */

   cfU(n) := stU(n) /( stL(n/2)^2);

   factor(cfU(n))

outputs
$$\frac{{{2}^{n+\frac{1}{2}}} {{e}^{\frac{1}{12 n}-\frac{2}{6 n+1}}}}{\sqrt{{n\pi} }} $$
Of course, the exponential tends to one. IF you want to refine that:
taylor( ev( %e^(1/(12*n)-2/(6*n+1)) , n = 1/x) , x , 0 , 4);

outputs
$$1-\frac{x}{4}+\frac{25 {{x}^{2}}}{288}-\frac{89 {{x}^{3}}}{3456}+\frac{1211 {{x}^{4}}}{165888}$$


text:
\begin{document}
\maketitle

\begin{abstract}
Let $M$ be an $n$-dimensional $d$-{\em bounded Stein manifold $M$}, i.e., a complex
$n$-dimensional manifold $M$ admitting a smooth strictly plurisubharmonic exhaustion $\rho$ and endowed with the K\"ahler metric whose fundamental
form is $\omega=i\del\delbar\rho$, such that $i\delbar\rho$ has bounded $L^\infty$ norm. We prove a vanishing result for $W^{1,2}$ harmonic forms
with respect to the Bott-Chern Laplacian on $M$.
\end{abstract}

\section{Introduction}
Let $M$ be an $n$-dimensional complex manifold endowed with a Hermitian metric $g$. Then, on the space of $(p,q)$-forms on $M$ there are defined
several self-adjoint elliptic differential operators of order two, as the {\em Dolbeault Laplacian} $\Delta^g_{\delbar}$ and of order four as
{\em Bott-Chern} and
{\em Aeppli Laplacians}, respectively denoted by $\tilde{\Delta}^g_{BC}$ and $\tilde{\Delta}^g_{A}$, involving $\del,\delbar,\del^*,\delbar^*$ and their
suitable combinations. If $M$ is compact, then, in particular, according to Schweitzer \cite{S}, it  turns out that a
$(p,q)$-form $\varphi$ on $M$ satisfies $\tilde{\Delta}^g_{BC}\varphi=0$
if and only if
$$\del\varphi=0,\quad\delbar\varphi=0,\quad \delbar^*\del^*\varphi=0.
$$
Furthermore, if the Hermitian metric $g$ is also K\"ahler on the compact complex manifold $M$, then according to the ellipticity, the kernels
of such differential
operators are finite dimensional and, as a consequence of K\"ahler identities, they coincide.

For a complex non compact manifold $M$, one can consider smooth $L^2$ forms and study the space of $L^2$ harmonic $(p,q)$-forms for the above Laplacians.
In \cite{DF}, the space of $L^2$ harmonic forms with respect to the Dolbeault Laplacian $\Delta^g_{\delbar}$ on a bounded strictly
pseudoconvex domain $\Omega$ in $\C^n$ with
smooth boundary, endowed with the Bergman metric, is studied. Precisely, denoting by $\mathcal{H}^{p,q}_{\delbar,2}$ the space
of $L^2$ harmonic forms with respect to $\Delta^g_{\delbar}$, Donnelly and Fefferman proved that under the assumptions as above, it
holds that
$$\left\{
\begin{array}{lll}
\dim\mathcal{H}^{p,q}_{\delbar,2}=0 &\hbox{if}&\,\,\, p+q\neq n\\[10pt]
\dim\mathcal{H}^{p,q}_{\delbar,2}=\infty&\hbox{if}&\,\,\, p+q= n.
\end{array}
\right.
$$
In \cite{O} Ohsawa proved that the dimension of the middle $L^2$ $\delbar$-cohomology of a domain in a complex manifold, admitting
a non-degenerate regular boundary point and whose defining function satisfies some suitable assumptions, is infinite dimensional.
Later, Gromov in \cite{G} introduced the notion of K\"ahler hyperbolicity showing that, if $X$ is a K\"ahler
complete simply-connected manifold whose K\"ahler
form $\omega$ is $d$-bounded, admitting a uniform discrete subgroup of isometries, then
$$\left\{
\begin{array}{lll}
\mathcal{H}^{p,q}_{\delbar,2}=\{0\} &\hbox{if}&\,\,\, p+q\neq n\\[10pt]
\mathcal{H}^{p,q}_{\delbar,2}\neq\{0\}&\hbox{if}&\,\,\, p+q= n.
\end{array}
\right.
$$

In the present paper we are interested in studying the space of $L^2$ $(p,q)$-harmonic forms with respect to the
Bott-Chern Laplacian on Hermitian complete manifolds. More precisely, we consider a $d$-{\em bounded Stein manifold $M$}, i.e., a complex
$n$-dimensional manifold $M$ admitting a smooth strictly plurisubharmonic exhaustion $\rho$ and endowed with the K\"ahler metric whose fundamental
form is $\omega=i\del\delbar\rho$, such that $i\delbar\rho$ has bounded $L^\infty$ norm; examples of such manifolds
are bounded strictly pseudoconvex domains in $\C^n$ with
smooth boundary, endowed with the Bergman metric (see \cite{DO}). Denoting by $\mathcal{H}^{p,q}_{BC,2}$ the space of $W^{1,2}$
Bott-Chern harmonic $(p,q)$-forms, where $W^{1,2}$ denotes the Sobolev Space, we prove the following vanishing result\vskip.2truecm\noindent
{\bf Theorem} (see Theorem \ref{GromovBC}) {\em
Let $M$ be a $d$-bounded Stein manifold of complex dimension $n$. Then
$$\mathcal{H}^{p,q}_{BC,2}=\{0\},\quad \hbox{for}\,\, p+q\neq n.
$$
}
\vskip.2truecm\noindent
The paper is organized as follows: in Section \ref{preliminaries} we fix some notation and recall some well known results in K\"ahler geometry.
In Section \ref{cut-off}, adapting an argument by Demailly \cite{DE}, we construct cut-off functions on a $d$-bounded Stein manifold, proving estimates
of second order derivatives. Section \ref{pezzi} is mainly devoted to prove that a smooth $W^{1,2}$ $(p,q)$-form $\varphi$ satisfies
$\tilde{\Delta}^g_{BC}\varphi=0$ if and only if
$\del\varphi=0,\quad\delbar\varphi=0,\quad \delbar^*\del^*\varphi=0$ (see Theorem \ref{teo:pezzi}). Basic tools in the proof of
such a theorem are the estimates of second order
derivatives of cut-off functions ensured by Stein $d$-bounded assumption (see Lemma \ref{estimates-derivatives}).
As a corollary, we also derive that (see Theorem \ref{teo:harmonic}),
\begin{equation*}
\mathcal{H}^{p,q}_{BC,2}\subset\mathcal{H}^{p,q}_{\delbar,2}=\mathcal{H}^{p,q}_{\del,2}=\mathcal{H}^{p,q}_{d,2},
\end{equation*}
where the last two sets are the spaces of $L^2$-harmonic forms with respect to the $\del$-Laplacian $\Delta^g_{\del}$ and
to the Hodge-de Rham Laplacian $\Delta^g_{d}$. Then, combining Theorem \ref{teo:harmonic} with the results by Gromov, we obtain the
proof of the vanishing Theorem \ref{GromovBC}.

In \cite{G} Gromov gave an $L^2$ Hodge decomposition Theorem for complete Riemannian manifolds, (see also \cite[Chap.VIII, Thm.3.2]{DE}).
It seems to be harder to prove a link between $W^{1,2}$
Bott-Chern harmonic forms and $L^2$ reduced cohomology. For other results on $L^2 $ cohomological decomposition in the complete almost Hermitian setting
see \cite{HT}. Thanks to elliptic regularity, we consider only $L^2$ forms that are also smooth.
It is sufficient for our results.

\medskip

\noindent{\sl Acknowledgments.} We are grateful to Professor Boyong Chen for useful suggestions and remarks for a
better presentation of the results.
\section{Preliminaries}\label{preliminaries}
Let $M$ be an $n$-dimensional complex manifold. Denote by $\Omega^r(M;\C)$, respectively $\Omega^{p,q}(M)$ the space of smooth complex $r$-forms,
respectively
smooth $(p,q)$-forms on $M$. Let $g$ be a Hermitian metric on $M$; denote by $\omega$ the fundamental form of $g$ and by
$\hbox{Vol}_g:=\frac{\omega^n}{n!}$ the
standard volume form. Let $\langle,\rangle$ be the pointwise Hermitian inner product induced by $g$ on the space of $(p,q)$-forms.
Given any $\varphi\in\Omega^r(M;\C)$, set
$$\vert\varphi(x)\vert^2_g=\langle\varphi,\varphi\rangle(x)$$
and
\begin{gather*}
\lVert \varphi\rVert_{L^2}:=\int_M\vert\varphi\vert^2_g\hbox{Vol}_g,\\
\lVert \varphi\rVert_{W^{1,2}}:=\int_M(\vert\varphi\vert^2_g+\vert\nabla\varphi\vert^2_g)\hbox{Vol}_g,
\end{gather*}
where $\nabla$ is the Levi-Civita connection of $(M,g)$ and $\vert\nabla\varphi\vert_g$ is the pointwise Hermitian norm of the
covariant derivative of the $(p,q)$-form $\varphi$, induced by $g$ on the space of complex covariant tensors on $M$.
Define
$$
L^2(M):=\left\{\varphi\in\Omega^r(M;\C)\,\,\,\vert\,\,\, 0\le r\le 2n,\ \lVert \varphi\rVert_{L^2}<\infty \right\}
$$
and
$$
W^{1,2}(M):=\left\{\varphi\in\Omega^r(M;\C)\,\,\,\vert\,\,\, 0\le r\le 2n,\ \lVert \varphi\rVert_{W^{1,2}}<\infty \right\}
$$
For any given $\varphi\in\Omega^r(M;\C)$, we also set
$$
\lVert \varphi\rVert_{L^\infty}:=\sup_{x\in M}\vert\varphi(x)\vert_g
$$
and we call $\varphi$ {\em bounded} if $
\lVert \varphi\rVert_{L^\infty}<\infty$. Furthermore, if $\varphi=d\eta$, then $\varphi$ is said to be {\em d-bounded}, if $\eta$ is bounded.
For any $\varphi,\psi\in\Omega_c^{p,q}(M)$ denote by $\la,\ra$ the $L^2$-Hermitian product defined as
$$
\la\varphi,\psi\ra =\int_M\langle\varphi,\psi\rangle \hbox{Vol}_g
$$\newline
Denoting by $*:\Omega^{p,q}(M)\to \Omega^{n-p,n-q}(M)$ the complex anti-linear Hodge operator
associated with $g$, the {\em Bott-Chern Laplacian} and {\em Aeppli Laplacian} $\tilde\Delta_{BC}^g$ and $ \tilde\Delta_{A}^g$
are the $4$-th order elliptic
self-adjoint differential operators defined respectively as (see \cite[p.8]{S})
$$
\tilde\Delta_{BC}^g \;:=\;
\del\delbar\delbar^*\del^*+
\delbar^*\del^*\del\delbar+\delbar^*\del\del^*\delbar+
\del^*\delbar\delbar^*\del+\delbar^*\delbar+\del^*\del $$
and
$$ \tilde\Delta_{A}^g \;:=\; \del\del^*+\delbar\delbar^*+
\delbar^*\del^*\del\delbar+\del\delbar\delbar^*\del^*+
\del\delbar^*\delbar\del^*+\delbar\del^*\del\delbar^*\,,
$$
where, as usual
$$
\del^*=-*\del\, *\,,\qquad \delbar^*=-*\delbar\, *.
$$
\begin{rem}\label{bc-kh-id}{If $g$ is a K\"ahler metric on $M$, then as a consequence of the K\"ahler identities, (see \cite[p.120]{H}
and \cite[Prop.2.4]{S}), it is
$$
\tilde\Delta_{BC}^g=\Delta^g_{\bar\partial}\Delta^g_{\bar\partial}+\delbar^*\delbar+\del^*\del,\quad
\tilde\Delta_{A}^g=\Delta^g_{\bar\partial}\Delta^g_{\bar\partial}+\del\del^*+\delbar\delbar^*,
$$
$\Delta^g_{\bar\partial}$ being the Dolbeault Laplacian on $(M,g)$. }
\end{rem}
Finally, we define the space of $W^{1,2}$ Bott-Chern harmonic forms by setting
$$
\mathcal{H}^{p,q}_{BC,2}:=\left\{\varphi\in\Omega^{p,q}(M)\,\,\,\vert\,\,\,\tilde\Delta_{BC}^g\varphi=0,\ \varphi\in W^{1,2}(M)\right\}.
$$

In the sequel we will need a local expression for the operators $\delbar^*$. To this purpose, let $(z^1,\ldots, z^n)$
be local holomorphic coordinates on $M$,
$$
g=\sum_{\alpha,\beta}g_{\alpha\bar\beta}dz^\alpha\otimes d\bar{z}^\beta
$$
be the local expression of the Hermitian metric $g$ and $(g^{\bar\alpha\beta})=(g_{\bar\alpha\beta})^{-1}$.
Given $\psi\in\Omega^{p,q}(M)$, for
$$
A_p=(\alpha_1,\ldots,\alpha_p),\qquad
B_q=(\beta_1,\ldots,\beta_q)$$
multiindices of length $p$, $q$ respectively, with
$\alpha_1<\cdots <\alpha_p$ and $\beta_1<\cdots <\beta_q$, denote by
$$
\psi=\sum_{A_p, B_q}\psi_{A_p\bar{B}_q}dz^{A_p}\wedge d\bar{z}^{B_q}
$$
the local expression of $\psi$ and by
$$
\psi^{\bar{A}_pB_q}=\sum_{\Gamma_p,\Lambda_q}g^{\bar{\alpha}_1\gamma_1}\cdots g^{\bar{\alpha}_p\gamma_p}
g^{\bar{\lambda}_1\beta_1}\cdots g^{\bar{\lambda}_q\beta_q}\psi_{\gamma_1\ldots \gamma_p\bar{\lambda}_1\ldots \bar{\lambda}_q}.
$$
Then, if $\varphi\in\Omega^{p,q}(M)$, locally the pointwise Hermitian inner product $\langle,\rangle$ on $\Omega^{p,q}(M)$ is given by
$$
\langle \varphi,\psi\rangle=\sum_{A_p, B_q}\varphi_{A_p\bar{B}_q}\overline{\psi^{\bar{A}_pB_q}}.
$$
The local expression of the pointwise Hermitian inner product induced by $g$ on the space of complex covariant tensors on $M$ is similar.

According to \cite[Prop.2.3]{MK}, we recall the local formula for $\delbar^*$, that is for any given $\psi\in\Omega^{p,q+1}(M)$, it is
$$
(\delbar^*\psi)^{\bar{A}_pB_q}=-\sum_{\gamma=1}^n
\left(\frac{\partial}{\partial z^\gamma}+\frac{\partial\log \det(g_{\alpha\bar\beta})}{\partial z^\gamma}\right)\psi^{\gamma\bar{A}_pB_q}.
$$
In the special case that $g$ is a K\"ahler metric on $M$, then as a consequence of the last formula, for every fixed $x_0\in M$, denoting by
$(z^1,\ldots, z^n)$ local normal holomorphic coordinates at $x_0$, we obtain
\begin{equation}\label{delbar-normal}
(\delbar^*\psi)_{\alpha_1\ldots\alpha_p\bar{\beta}_1\ldots\bar{\beta}_q}(x_0)=-\sum_{\gamma=1}^n
\frac{\partial \psi_{\bar{\gamma}\alpha_1\ldots\alpha_p\bar{\beta}_1\ldots\bar{\beta}_q}}{\partial z^\gamma}(x_0).
\end{equation}
Finally, for any given $\varphi\in\Omega^{p,q}(M)$, still using local normal holomorphic coordinates at $x_0$, we have
\begin{equation}\label{nabla-normal}
|\nabla\varphi|^2_g(x_0)=2\sum_{A_p, B_q}\sum_{\gamma=1}^n
\biggl(\biggl|\de{\varphi_{A_p\bar{B}_q}}{z^\gamma}\biggr|^2+\biggl|\de{\varphi_{A_p\bar{B}_q}}{\bar{z}^\gamma}\biggr|^2\biggl)(x_0).
\end{equation}

\section{Construction of cut-off functions in $d$-bounded Stein manifolds}
\label{cut-off}
Let $M$ a be Stein manifold and let $\rho$ be a strictly plurisubharmonic exhausting smooth function. Denote $\omega=i\del\delbar\rho$ the
fundamental form with the K\"ahler metric $g$ associated. We say that $M$ is {\em d-bounded} if $\omega=d\eta$ and $\eta=i\delbar\rho$ is
bounded. In particular, $\omega$ is $d$-bounded.  In the following, $\rho$, $\omega$, $g$, $\eta$ are considered fixed.

We remark that any $d$-bounded Stein manifold is complete, see \cite[Chap.VIII, Lemma 2.4]{DE}.

Examples of $d$-bounded Stein manifolds are bounded strictly pseudoconvex domains in $\C^n$ with smooth boundary, endowed with the Bergman metric,
see \cite[Prop.3.4]{DO}.

Now we prove the existence of cut-off functions with specific bounds on the second order derivatives on a d-bounded Stein manifold.
We need the following known lemma.

\begin{lemma}
Let $a,b\in\R$, $a<b$. Then there exists a \Cinf function $\psi:\R\rightarrow [0,1]\subset\R$ such that the following properties hold:
\begin{itemize}
\item $\psi(t)=1 \ \iff\  t\le a$;
\item $\psi(t)=0 \ \iff\  t\ge b$;
\item $\exists C\in\R$ such that $|\psi'(t)|,|\psi''(t)|\le C\psi(t)^{\frac{1}{2}}\ \forall t\in\R$.
\end{itemize}
\end{lemma}
\begin{proof}
Let us define $\phi:\R\rightarrow\R$, a \Cinf function such that
\begin{equation*}
\phi(t)=
\begin{cases}
\exp(-\frac{1}{t^2}) & \text{ if } t>0\\
0 & \text{ if } t\le 0.
\end{cases}
\end{equation*}
Then we define
\begin{equation*}
\psi(t)=\frac{\phi(b-t)}{\phi(b-t)+\phi(t-a)}.
\end{equation*}
Note that $\psi(t)=1$ iff $t\le a$ and $\psi(t)=0$ iff $t\ge b$. After some calculations we obtain
\begin{equation*}
\begin{split}
\psi'(t) & = \frac{-2\frac{\phi(b-t)}{(b-t)^3}(\phi(b-t)+\phi(t-a))-
\phi(b-t)\biggl(-2\frac{\phi(b-t)}{(b-t)^3}+2\frac{\phi(t-a)}{(t-a)^3}\biggr)}{(\phi(b-t)+\phi(t-a))^2}\\
& = -2\frac{\phi(b-t)\phi(t-a)}{(\phi(b-t)+\phi(t-a))^2}\biggl(\frac{1}{(b-t)^3}+\frac{1}{(t-a)^3}\biggr)\\
& = -2\psi(t)^\frac{1}{2}\frac{\phi(b-t)^\frac{1}{2}\phi(t-a)}{(\phi(b-t)+\phi(t-a))^\frac{3}{2}}
\biggl(\frac{1}{(b-t)^3}+\frac{1}{(t-a)^3}\biggr).
\end{split}
\end{equation*}
This implies that $\exists C\in\R$ such that $|\psi'(t)|\le C\psi(t)^{\frac{1}{2}}\ \forall t\in\R$. The
calculations of the estimate on $\psi''$ are analogous.
\end{proof}

The following lemma is inspired by \cite[Chap.VIII, Lemma 2.4]{DE}.

\begin{lemma}\label{estimates-derivatives}
Let $M$ be a d-bounded Stein manifold of complex dimension $n$. Then there exists a sequence $\{K_\nu\}_{\nu\in\N}$ of
compact subsets of $M$ and a sequence $a_\nu : M \rightarrow [0,1]\subset\R$, $\nu \in \N$, of \Cinf functions with compact support,
called cut-off functions, such that the following properties hold:
\begin{itemize}
\item $\bigcup_{\nu\in\N} K_\nu=M$ and $K_\nu\subset \mathop K\limits^ \circ$$_{\nu+1}$;
\item $\forall\nu \in \N$ $a_\nu=1$ in a neighbourhood of $K_\nu$
and $\supp{a_\nu}\subset \mathop K\limits^ \circ$$_{\nu+1}$;
\item $\exists C\in\R$ such that $|\del a_\nu(x)|_g,|\delbar a_\nu(x)|_g,|\del\delbar a_\nu(x)|_g\le 2^{-\nu}C a_\nu(x)^\frac{1}{2}$ $\forall x\in M$.
\end{itemize}
\end{lemma}
\begin{proof}
We define
\begin{equation*}
a_\nu(x)=\psi(2^{-\nu}\rho(x)) \ \forall x\in M \text{ and } K_\nu=\overline{\{x\in M\ |\ \rho(x)< 2^{\nu}\}},
\end{equation*}
where $\psi$ is the function of the previous lemma, with $a=1.1$ and $b=1.9$.

Let us check that the claimed properties hold.
The subsets $K_\nu$ are compact because of the definition of exhausting function. If $x\in M$,
then $\exists\nu\in\N$ such that $\rho(x)<2^\nu$, so $x\in K_\nu$, thus $\bigcup_{\nu\in\N} K_\nu=M$.
The inclusions $K_\nu\subset \mathop K\limits^ \circ$$_{\nu+1}$ hold by the construction of $K_\nu$, in fact if $x\in K_\nu$, then $\rho(x)\le 2^\nu$
by continuity and $x\in\{y\in M\ |\ \rho(y)<2^\nu\cdot 1.5\}\subset\mathop K\limits^ \circ$$_{\nu+1}$.

The functions $a_\nu$ are \Cinf because $\psi$ and $\rho$ are \Cinf. The function $\psi$ takes values in
the interval $[0,1]$, and so is true for $a_\nu$. Choosing a sufficiently small neighbourhood of $K_\nu$, we can assume that $2^{-\nu}\rho<1.1$,
so $a_\nu=1$ in that neighbourhood.

In order to prove $\supp{a_\nu}\subset \mathop K\limits^ \circ$$_{\nu+1}$, let us take $\tilde{x}\in\supp{a_\nu}$
and a sequence $\{x_k\}_{k\in\N}$ of points in $M$ such that $a_\nu(x_k)>0\ \forall k\in \N$
and $x_k\rightarrow\tilde{x}$ as $k\rightarrow\infty$. By the construction of $\psi$,
we have that, $\forall x\in M$, $a_\nu(x)>0$ if and only if $2^{-\nu}\rho(x)< 1.9$.
Therefore, by the continuity of $\rho$, $2^{-\nu}\rho(\tilde{x})\le 1.9$, so that $\rho(\tilde{x})< 2^{\nu}\cdot 1.95$
and $\tilde{x}\in \mathop K\limits^ \circ$$_{\nu+1}$. Because $\supp{a_\nu}$ is a close set contained in a compact set, then it is compact.

Finally we have to prove the estimates on the differentials of $a_\nu$. Let $x\in M$, then
\begin{equation*}
\del a_\nu(x)= 2^{-\nu}\psi'(2^{-\nu}\rho(x)) \sum\limits_{i=1}^n \de{\rho}{z^i}(x)dz^i,
\end{equation*}
and
\begin{equation*}
\begin{split}
|\del a_\nu(x)|_g & = 2^{-\nu}|\psi'(2^{-\nu}\rho(x))| |\del\rho(x)|_g \\
& \le 2^{-\nu}C(\psi(2^{-\nu}\rho(x)))^{\frac{1}{2}} ||\del\rho||_{L^\infty} \\
& \le 2^{-\nu}Ca_\nu(x)^{\frac{1}{2}},
\end{split}
\end{equation*}
where the constant $C$ is taken big enough
and may not be the same at every passage of the calculations. In the last passage we used the
hypothesis that $\omega$ is $d$-bounded and the fact that $\rho$ is real,
so $\de{\rho}{\bar{z}^i}(x)=\overline{\de{\rho}{z^i}}(x)$ and $|\del\rho(x)|_g=|\delbar\rho(x)|_g$.
By the same calculations, we also obtain the estimate of $|\delbar a_\nu(x)|_g$. Moreover,
\begin{equation*}
\begin{split}
\delbar\del a_\nu(x) & = 2^{-\nu} \sum\limits_{i,j=1}^n \biggl(2^{-\nu}\psi''(2^{-\nu}\rho(x)) \de{\rho}{\bar{z}^j}(x) \de{\rho}{z^i}(x)+\\
& + \psi'(2^{-\nu}\rho(x)) \de{^2\rho}{\bar{z}^j \del z^i}(x)\biggr) d\bar{z}^j\land dz^i,
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
|\delbar\del a_\nu(x)|_g & \le 2^{-\nu} \biggl(2^{-\nu}|\psi''(2^{-\nu}\rho(x))||\delbar\rho(x)|_g|\del\rho(x)|_g+\\
& + |\psi'(2^{-\nu}\rho(x))| |\sum\limits_{i,j=1}^n \de{^2\rho}{\bar{z}^j \del z^i}(x) d\bar{z}^j\land dz^i|_g \biggr)\\
& \le 2^{-\nu}C(\psi(2^{-\nu}\rho(x)))^{\frac{1}{2}} \biggl(2^{-\nu}||\delbar\rho||_{L_\infty}||\del\rho||_{L_\infty}+|\omega(x)|_g\biggr)\\
& \le 2^{-\nu}Ca_\nu(x)^{\frac{1}{2}}.
\end{split}
\end{equation*}
In fact $\omega$ is $d$-bounded as before and $|\omega(x)|_g$ is constant, due to the definition of the metric. The proof is complete.
\end{proof}

\section{Vanishing of $L^2$ Bott-Chern harmonic forms}\label{pezzi}
\label{vanishing}
Our main theorem states that the following property, true in the compact case, also holds in our non-compact case.
\begin{theorem}
\label{teo:pezzi}
Let $M$ be a d-bounded Stein manifold of complex dimension $n$. Let $\varphi\in \Omega^{p,q}(M)\cap W^{1,2}(M)$.
If $\tilde{\Delta}_{BC}^g\varphi=0$, then
\begin{equation*}
\del\varphi=0,\quad \delbar\varphi=0,\quad \delbar^*\del^*\varphi=0.
\end{equation*}
\end{theorem}
The following Lemma will be useful for the proof of Theorem \ref{teo:pezzi}.
\begin{lemma}\label{estimates}
Let $M$ be a K\"ahler manifold of complex dimension $n$ and denote its metric with $g$.
If $\varphi\in\Omega^{p,q}(M)$ and $\{a_\nu\}_\nu$ are \Cinf functions on $M$, then $\exists C>0$ such that
$\forall\nu\in\N$
\begin{equation}
\begin{split}
\label{lemma:delbar}
&|\delbar^*(\delbar a_\nu\land*\varphi)|_g\le C\bigl(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g\bigr),\\
&|\delbar^*(\delbar a_\nu\land\varphi)|_g \le C\bigl(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g\bigr).
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
The pointwise Hermitian norm on forms is invariant under change of coordinates,
so we can prove the inequalities locally with a uniform constant $C$. For every fixed $x_0\in M$, denoting by $(z^1,\ldots, z^n)$
local normal holomorphic coordinates at $x_0$, we have
\begin{gather*}
\delbar a_\nu=\sum_\beta \de{a_\nu}{\bar{z}^\beta}d\bar{z}^\beta, \\
\varphi=\sum_{A_p,B_q} \varphi_{A_p\bar{B}_q} dz^{A_p}\land d\bar{z}^{B_q}, \\
\delbar a_\nu\land\varphi =\sum_{A_p,B_q,\beta} \de{a_\nu}{\bar{z}^\beta} \varphi_{A_p\bar{B}_q}
d\bar{z}^\beta\land dz^{A_p}\land d\bar{z}^{B_q}
\end{gather*}
We can write
\begin{equation*}
\delbar a_\nu\land\varphi= \frac{1}{p!(q+1)!}\sum_{\substack{\alpha_1,\dots,\alpha_p,\\ \beta_0,\beta_1,\dots,\beta_q}}
(\delbar a_\nu\land\varphi)_{\alpha_1\dots\alpha_p\bar{\beta}_0\bar{\beta}_1\dots\bar{\beta}_q}
dz^{\alpha_1\dots\alpha_p}\land d\bar{z}^{\beta_0\beta_1\dots\beta_q}
\end{equation*}
where the coefficients $(\delbar a_\nu\land\varphi)_{\alpha_1\dots\alpha_p\bar{\beta}_0\bar{\beta}_1\dots\bar{\beta}_q}$ are antisymmetric
in the indices $\alpha_1,\dots,\alpha_p,\bar{\beta}_0,\bar{\beta}_1,\dots,\bar{\beta}_q$, so
\begin{equation*}
\begin{split}
(\delbar a_\nu\land\varphi)_{\bar{\beta}_0\alpha_1\dots\alpha_p\bar{\beta}_1\dots\bar{\beta}_q} & =
(-1)^p (\delbar a_\nu\land\varphi)_{\alpha_1\dots\alpha_p\bar{\beta}_0\bar{\beta}_1\dots\bar{\beta}_q} \\
& = \sum_{j=0}^q (-1)^{qj} \de{a_\nu}{\bar{z}^{\beta_j}}  \varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}},
\end{split}
\end{equation*}
where we set $\beta_{j+q}:=\beta_{j-1}$ for $j=1,\dots,q$.
Now we apply (\ref{delbar-normal}) and obtain
\begin{equation*}
\begin{split}
(\delbar^*(\delbar a_\nu\land\varphi))&_{\alpha_1\ldots\alpha_p\bar{\beta}_1\ldots\bar{\beta}_q}(x_0) = \\
& = -\sum_{\beta_0=1}^n
\de{}{z^{\beta_0}} \bigl( (\delbar a_\nu\land\varphi)_{\bar{\beta}_0\alpha_1\ldots\alpha_p\bar{\beta}_1\ldots\bar{\beta}_q}\bigr)(x_0)\\
& = -\sum_{\beta_0=1}^n \sum_{j=0}^q \de{}{z^{\beta_0}} \bigl(
(-1)^{qj} \de{a_\nu}{\bar{z}^{\beta_j}}  \varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}}
\bigr) (x_0) \\
& = -\sum_{\beta_0=1}^n \sum_{j=0}^q (-1)^{qj} \bigl(
\de{^2a_\nu}{z^{\beta_0}\partial\bar{z}^{\beta_j}}\varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}}(x_0)+\\
& \quad\quad\quad\quad\quad\quad\quad\quad
+ \de{a_\nu}{\bar{z}^{\beta_j}}\de{\varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}}}{z^{\beta_0}}(x_0)
\bigr) .
\end{split}
\end{equation*}
This yields
\begin{equation*}
|\delbar^*(\delbar a_\nu\land\varphi)|_g(x_0) \le C\bigl(|\del\delbar a_\nu|_g(z_0)|\varphi|_g(x_0)+|\delbar a_\nu|_g(x_0)|\nabla\varphi|_g(x_0)\bigr),
\end{equation*}
where $C$ depends only on $n,p,q$. To prove it, we have to do some calculations. We set
\begin{equation*}
\gamma_{\beta_0j}:=\de{^2a_\nu}{z^{\beta_0}\partial\bar{z}^{\beta_j}}\varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}}
\text{ and }
\lambda_{\beta_0j}:=\de{a_\nu}{\bar{z}^{\beta_j}}\de{\varphi_{\alpha_1\dots\alpha_p\bar{\beta}_{j+1}\dots\bar{\beta}_{j+q}}}{z^{\beta_0}},
\end{equation*}
with $\alpha_1,\dots,\alpha_p,\beta_1,\dots,\beta_q=1,\dots,n$, $\beta_0=1,\dots,n$ and $j=0,\dots,q$.
So we have, using (\ref{nabla-normal}),

\begin{equation*}
\begin{split}
|\delbar^*(\delbar a_\nu\land\varphi)|_g^2(x_0) & =
\frac{1}{p!q!}\sum_{\substack{\alpha_1,\dots,\alpha_p,\\ \beta_1,\dots,\beta_q}} \sum_{\beta_0,\beta'_0=1}^n \sum_{j,j'=0}^q
(\gamma_{\beta_0j}+\lambda_{\beta_0j})(\bar{\gamma}_{\beta'_0j'}+\bar{\lambda}_{\beta'_0j'})(x_0) \\
& = \frac{1}{p!q!}\sum (\gamma_{\beta_0j}\bar{\gamma}_{\beta'_0j'}+\gamma_{\beta_0j}\bar{\lambda}_{\beta'_0j'}+\lambda_{\beta_0j}
\bar{\gamma}_{\beta'_0j'}+\lambda_{\beta_0j}\bar{\lambda}_{\beta'_0j'})(x_0) \\
& = \frac{1}{p!q!}\sum
(\real(\gamma_{\beta_0j}\bar{\gamma}_{\beta'_0j'})+2\real(\gamma_{\beta_0j}\bar{\lambda}_{\beta'_0j'})+
\real(\lambda_{\beta_0j}\bar{\lambda}_{\beta'_0j'}))(x_0) \\
& \le \frac{2}{p!q!}\sum (|\gamma_{\beta_0j}|^2+|\lambda_{\beta_0j}|^2)(x_0) \\
& \le C \bigl(|\del\delbar a_\nu|_g(x_0)|\varphi|_g(x_0)+|\delbar a_\nu|_g(x_0)|\nabla\varphi|_g(x_0)\bigr)^2.
\end{split}
\end{equation*}

To prove the other inequality in (\ref{lemma:delbar}), we want to estimate $|\delbar^*(\delbar a_\nu\land*\varphi)|_g(x_0)$,
so the calculations are analogous except for the fact that we use the following formula (see \cite[p.94]{MK}) for the Hodge star operator:
\begin{equation*}
*\varphi=(i)^n(-1)^{\frac{1}{2}n(n-1)+qn}\sum_{A_p,B_q} \sgn A\sgn B \det(g_{h\bar{k}}) \bar{\varphi}^{\bar{B}_qA_p}
dz^{A_{n-p}}\land d\bar{z}^{B_{n-q}},
\end{equation*}
where $A$ and $B$ are respectively the permutations that send $(1,\dots,n)$ in $(A_p,A_{n-p})$ and in $(B_q,B_{n-q})$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{teo:pezzi}]
First of all, by remark \ref{bc-kh-id}, we have
$$\tilde\Delta_{BC}^g=\Delta^g_{\bar\partial}\Delta^g_{\bar\partial}+\delbar^*\delbar+\del^*\del.
$$
Thanks to the cut-off functions of the previous lemma, now we can integrate by part, using the Stokes Theorem.
\begin{equation*}

\begin{split}
0 & = \la\tilde{\Delta}^g_{BC}\varphi,a_\nu\varphi\ra \\
  & =\la\delbar\delbar^*\delbar\delbar^*\varphi,a_\nu\varphi\ra+\la\delbar^*\delbar\delbar^*\delbar\varphi,a_\nu\varphi\ra+
	\la\delbar^*\delbar\varphi,a_\nu\varphi\ra+\la\del^*\del\varphi,a_\nu\varphi\ra \\
  & =\la\delbar\delbar^*\varphi,\delbar\delbar^*(a_\nu\varphi)\ra+\la\delbar^*\delbar\varphi,\delbar^*\delbar(a_\nu\varphi)\ra+
	\la\delbar\varphi,\delbar(a_\nu\varphi)\ra+\la\del\varphi,\del(a_\nu\varphi)\ra
\end{split}
\end{equation*}
Now we calculate every differential in the right sides of the inner products.
\begin{equation*}
\begin{split}
\delbar\delbar^*(a_\nu\varphi) & = -\delbar*(\delbar a_\nu\land*\varphi+a_\nu\delbar*\varphi) \\
  & =(-1)^{p+q}*\delbar^*(\delbar a_\nu\land*\varphi) +\delbar a_\nu\land \delbar^*\varphi+a_\nu\delbar\delbar^*\varphi \\
\delbar^*\delbar(a_\nu\varphi) & = \delbar^*(\delbar a_\nu\land\varphi+a_\nu\delbar\varphi) \\
  & =\delbar^*(\delbar a_\nu\land\varphi) -*(\delbar a_\nu\land *\delbar\varphi)+a_\nu\delbar^*\delbar\varphi \\
\delbar(a_\nu\varphi) & =\delbar a_\nu\land\varphi+a_\nu\delbar\varphi \\
\del(a_\nu\varphi) & =\del a_\nu\land\varphi+a_\nu\del\varphi
\end{split}
\end{equation*}
Therefore,
\begin{equation*}
0 = \la\tilde{\Delta}^g_{BC}\varphi,a_\nu\varphi\ra= I_1(\nu)+I_2(\nu),
\end{equation*}
where
\begin{equation*}
I_1(\nu)= \int_M a_\nu\bigl(|\delbar\delbar^*\varphi|_g^2+|\delbar^*\delbar\varphi|_g^2+
|\delbar\varphi|_g^2+|\del\varphi|_g^2\bigr)\,\vol,
\end{equation*}
and
\begin{equation*}
\begin{split}
I_2(\nu) & = \int_M \bigl(
\langle\delbar\delbar^*\varphi,(-1)^{p+q}*\delbar^*(\delbar a_\nu\land*\varphi) +\delbar a_\nu\land \delbar^*\varphi\rangle+\\
& + \langle\delbar^*\delbar\varphi,\delbar^*(\delbar a_\nu\land\varphi) -*(\delbar a_\nu\land *\delbar\varphi)\rangle+\\
& + \langle\delbar\varphi,\delbar a_\nu\land\varphi\rangle+\langle\del\varphi,\del a_\nu\land\varphi\rangle\bigr)\,\vol.
\end{split}
\end{equation*}
We have $I_1(\nu)=|I_2(\nu)|$ and, by the monotone convergence theorem, as $\nu\rightarrow\infty$,
\begin{equation*}
I_1(\nu)\rightarrow \int_M \bigl(|\delbar\delbar^*\varphi|_g^2+|\delbar^*\delbar\varphi|_g^2+
|\delbar\varphi|_g^2+|\del\varphi|_g^2\bigr)\,\vol.
\end{equation*}
Thus, if we show that $I_1(\nu)\rightarrow 0$ as $\nu\rightarrow\infty$, we have
\begin{equation}\label{I1}
\del\varphi=0,\qquad \delbar\varphi=0,\qquad\delbar\delbar^*\varphi=0,\qquad\delbar^*\delbar\varphi=0.
\end{equation}
Estimating $|I_2(\nu)|$, we obtain
\begin{equation*}
\begin{split}
|I_2(\nu)| & \le \int_M \bigl(
   |\delbar\delbar^*\varphi|_g(|\delbar^*(\delbar a_\nu\land*\varphi)|_g+|\delbar a_\nu|_g|\delbar^*\varphi|_g)+\\
& +|\delbar^*\delbar\varphi|_g(|\delbar^*(\delbar a_\nu\land\varphi)|_g+ |\delbar a_\nu|_g|\delbar\varphi|_g)+\\
& +|\delbar\varphi|_g|\delbar a_\nu|_g|\varphi|_g+|\del\varphi|_g|\del a_\nu|_g|\varphi|_g\bigr)\,\vol.
\end{split}
\end{equation*}
By Lemma \ref{estimates} there exists a constant $C>0$ such that
\begin{equation*}
\begin{split}
&|\delbar^*(\delbar a_\nu\land*\varphi)|_g\le C\bigl(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g\bigr),\\
&|\delbar^*(\delbar a_\nu\land\varphi)|_g \le C\bigl(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g\bigr).
\end{split}
\end{equation*}
Therefore we have
\begin{equation*}
\begin{split}
|I_2(\nu)| & \le C \int_M \bigl(
   |\delbar\delbar^*\varphi|_g(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g)+\\
& +|\delbar^*\delbar\varphi|_g(|\del\delbar a_\nu|_g|\varphi|_g+|\delbar a_\nu|_g|\nabla\varphi|_g)+\\
& +|\delbar\varphi|_g|\delbar a_\nu|_g|\varphi|_g+|\del\varphi|_g|\del a_\nu|_g|\varphi|_g\bigr)\,\vol.
\end{split}
\end{equation*}
The estimates on the cut-off functions, i.e.,
$$|\del a_\nu(x)|_g,|\delbar a_\nu(x)|_g,|\del\delbar a_\nu(x)|_g\le 2^{-\nu}C a_\nu(x)^\frac{1}{2},$$
yield
\begin{equation*}
\begin{split}
I_1(\nu)=|I_2(\nu)| & \le 2^{-\nu}C \int_M a_\nu(x)^\frac{1}{2} \bigl(
   |\delbar\delbar^*\varphi|_g(|\varphi|_g+|\nabla\varphi|_g)+\\
& +|\delbar^*\delbar\varphi|_g(|\varphi|_g+|\nabla\varphi|_g)+\\
& +|\delbar\varphi|_g|\varphi|_g+|\del\varphi|_g|\varphi|_g\bigr)\,\vol \\
& \le 2^{-\nu}C \int_M a_\nu(x)^\frac{1}{2}
        \bigl(|\delbar\delbar^*\varphi|_g+|\delbar^*\delbar\varphi|_g+|\delbar\varphi|_g+|\del\varphi|_g\bigr)\cdot \\
& \cdot \bigl(|\varphi|_g+|\nabla\varphi|_g\bigr)\,\vol \\
& \le 2^{-\nu}C \biggl( \int_M a_\nu \bigl(
|\delbar\delbar^*\varphi|_g+|\delbar^*\delbar\varphi|_g+|\delbar\varphi|_g+|\del\varphi|_g\bigr)^2\,\vol\biggr)^{\frac{1}{2}}\cdot\\
& \cdot \biggl( \int_M \bigl(|\varphi|_g+|\nabla\varphi|_g\bigr)^2\,\vol\biggr)^{\frac{1}{2}} \\
& \le 2^{-\nu}C\bigl(I_1(\nu)\bigr)^{\frac{1}{2}}\cdot \bigl(||\varphi||_{L_2}+||\nabla\varphi||_{L_2}\bigr),
\end{split}
\end{equation*}
and consequently
$$
\bigl(I_1(\nu)\bigr)^{\frac{1}{2}}\leq C\,2^{-\nu}.
$$
Thus $\varphi$ is $\del$-closed and $\delbar$-closed. This implies
$$\tilde{\Delta}_{BC}^g\varphi=\del\delbar\delbar^*\del^*\varphi.$$
Now we substantially reapply the argument as above to this form of the Bott-Chern Laplacian. We have
\begin{equation*}
0 = \la\tilde{\triangle}_{BC}\varphi,a_\nu\varphi\ra = \la\del\delbar\delbar^*\del^*\varphi,a_\nu\varphi\ra =
\la\delbar^*\del^*\varphi,\delbar^*\del^*(a_\nu\varphi)\ra,
\end{equation*}
and
\begin{equation*}
\begin{split}
\delbar^*\del^*(a_\nu\varphi) & = \delbar^*(-*(\del a_\nu\land*\varphi)+a_\nu\del^*\varphi) \\
  & =(-1)^{p+q-1}*(\delbar\del a_\nu\land*\varphi) -(-1)^{p+q-1}*(\del a_\nu\land\delbar*\varphi)+\\
	& -*(\delbar a_\nu\land*\del^*\varphi)+a_\nu\delbar^*\del^*\varphi,
\end{split}
\end{equation*}
thus
\begin{equation*}
0 = \la\tilde{\Delta}^g_{BC}\varphi,a_\nu\varphi\ra= I'_1(\nu)+I'_2(\nu),
\end{equation*}
where
\begin{equation*}
I'_1(\nu)= \int_M a_\nu|\delbar^*\del^*\varphi|_g^2\,\vol,
\end{equation*}
and
\begin{equation*}
\begin{split}
|I'_2(\nu)| & \le 2^{-\nu}C \int_M a_\nu(x)^\frac{1}{2}
|\delbar^*\del^*\varphi|_g(|\varphi|_g+|\delbar^*\varphi|_g+|\del^*\varphi|_g)\,\vol \\
& \le 2^{-\nu}C \bigl(I'_1(\nu)\bigr)^{\frac{1}{2}}\cdot \bigl(||\varphi||_{L_2}+||\nabla\varphi||_{L_2}\bigr).
\end{split}
\end{equation*}
Thus we also have $\delbar^*\del^*\varphi=0$. This ends the proof.
\end{proof}
\begin{rem}
From Theorem \ref{teo:pezzi} we immediately obtain that
$$
\varphi\in \mathcal{H}^{p,q}_{BC,2}\quad\hbox{\rm if and only if}\qquad \varphi\in W^{1,2}(M),\qquad \del\varphi=0,\qquad \delbar\varphi=0,
\qquad \delbar^*\del^*\varphi=0,
$$
which ``extends'' the characterization of the space of Bott-Chern harmonic forms on a compact Hermitian manifold to any $d$-bounded Stein manifold.
\end{rem}

As a straightforward consequence of Theorem \ref{teo:pezzi}, we obtain the following

\begin{theorem}
\label{teo:harmonic}
Let $M$ be a d-bounded Stein manifold of complex dimension $n$. Then
\begin{equation*}
\mathcal{H}^{p,q}_{BC,2}\subset\mathcal{H}^{p,q}_{\delbar,2}=\mathcal{H}^{p,q}_{\del,2}=\mathcal{H}^{p,q}_{d,2},
\end{equation*}
where the last three sets are the spaces of $L^2$-harmonic $(p,q)$-forms with respect to $\Delta^g_{\delbar}$, $\Delta^g_{\del}$ and $\Delta^g_{d}$.
\end{theorem}
\begin{proof}

We note that the last three equalities of the thesis hold because $M$ is K\"ahler, in fact $\Delta^g_d=2\Delta^g_{\del}=2\Delta^g_{\delbar}$ by K\"ahler identities. Therefore it is enough to prove $\mathcal{H}^{p,q}_{BC,2}\subset\mathcal{H}^{p,q}_{\delbar,2}$.

Let $\varphi\in\mathcal{H}^{p,q}_{BC,2}$; from (\ref{I1}) in the proof of Theorem
\ref{teo:pezzi} we have
$$\delbar\delbar^*\varphi=0,\quad \delbar^*\delbar\varphi=0.
$$
Thus $\Delta^g_{\bar\partial}\varphi=0$ and $\varphi\in\mathcal{H}^{p,q}_{\delbar,2}$.
\end{proof}
We are ready to prove the following
\begin{theorem}\label{GromovBC}
Let $M$ be a $d$-bounded Stein manifold of complex dimension $n$. Then
$$\mathcal{H}^{p,q}_{BC,2}=\{0\},\quad \hbox{for}\,\, p+q\neq n.
$$
\end{theorem}
\begin{proof} By Theorem \ref{teo:harmonic}, it is enough to prove that
$\mathcal{H}^{p,q}_{d,2}=\{0\}$.
This last fact is a consequence of \cite[Thm.1.2.B.]{G}. For the sake of completeness we remind the argument by Gromov.

Let us consider the Lefschetz operator
$$
L:\Omega^{(p,q)}(M)\lr\Omega^{(p+1,q+1)}(M)
$$ defined by
$$L\varphi=\omega\land\varphi
$$
for every $\varphi\in\Omega^{(p,q)}(M)$. By \cite[rem.3.2.7.iii)]{H}, the map
\begin{equation*}
L^{n-p-q}:\mathcal{H}^{p,q}_{d}\lr\mathcal{H}^{n-q,n-p}_{d}
\end{equation*}
is an isomorphism for $p+q\le n$, where $\mathcal{H}^{p,q}_{d}$ denotes the space of
$\Delta^g_{d}$-harmonic $(p,q)$-forms. Now set $k=n-p-q$ and consider the form $L^k\varphi$, where $\varphi\in\Omega^{(p,q)}(M)\cap L^2(M)$
is a $d$-closed form. Since $\omega=d\eta$, if $k>0$, then we get
\begin{equation*}
L^k\varphi=\omega^k\land\varphi=(d\eta)^k\land\varphi=d(\eta\land(d\eta)^{k-1}\land\varphi).
\end{equation*}
Furthermore,
\begin{itemize}
\item $\eta\land(d\eta)^{k-1}$ is bounded, since $\eta$ is bounded and $|d\eta|_g$ is constant;
\item $\eta\land(d\eta)^{k-1}\land\varphi\in L^2(M)$, since $\varphi\in L^2(M)$.
\end{itemize}

Moreover, if $\varphi\in\Omega^{(p,q)}(M)\cap L^2(M)$ is a $\Delta^g_d$-harmonic form,
then $L^k\varphi$ is also $\Delta^g_d$-harmonic. Thus, in view of the $L^2$
Hodge decomposition theorem (see \cite[Chap.VIII, Thm.3.2]{DE}), we obtain that $L^k\varphi=0$.

Now let $\varphi\in\mathcal{H}^{p,q}_{d,2}$. If $p+q<n$, then $k>0$ and $L^k\varphi=0$; therefore $\varphi=0$ since $L^k$ is injective.
Conversely, if $p+q>n$, then $*\varphi\in\mathcal{H}^{n-p,n-q}_{d,2}$ and $(n-p)+(n-q)<n$; by the previous argument $*\varphi=0$
and consequently $\varphi=0$.

Summing up, we showed that $\mathcal{H}^{p,q}_{d,2}=\{0\}$ for $p+q\ne n$. This ends the proof.
\end{proof}
\begin{rem}
The Lefschetz argument, in the proof of Theorem \ref{GromovBC}, uses only the K\"ahler and the $d$-bounded assumptions.
\end{rem}
\begin{rem} By  similar computations, Theorem \ref{GromovBC} can be stated and proved also for the Aeppli Laplacian.
Indeed, it is sufficient to repeat the proof of Theorem \ref{teo:pezzi}
with $\tilde\Delta_{A}^g=\Delta^g_{\bar\partial}\Delta^g_{\bar\partial}+\del\del^*+\delbar\delbar^*$.
\end{rem}


TITLE: Why is F[x] a UFD?
QUESTION [0 upvotes]: When reading the proof for if $R$ is a UFD, then $R[x]$ is a UFD, the author uses a fact that $F[x]$ is a UFD. I don't quite understand this.

Why $F[x]$ is a UFD? ($F$ is the fraction field of $R$).

Can someone explain to me why? Thanks.

REPLY [3 votes]: In the book Ideals, Varieties, and Algorithms by Cox D., Little J., O'Shea D. you can see :

Note that every PID is UFD.

REPLY [0 votes]: We want to define a degree function for $F[X]$ and deduce that $F[X]$ is an euclidean domain and so an UFD.
Hou to construct our function?
Look at $\mathbb{Q}[x]$:
The constant terms have degree zero and the polynomials with leading monomial $x^n$ have degree $n$.
Let's set now the same values of degree function for $\mathbb{F}[x]$.
We have to prove:

If $a$ and $b$ are in $\mathbb{F}[x]$ and $b$ is nonzero, then there are $q$ and $r$ in $\mathbb{F}[x]$ such that $a = bq + r$ and either $r = 0$ or $deg(r) < deg(b)$.
For all nonzero $a$ and $b$ in $\mathbb{F}[x]$, we have $deg(a) \leq deg(ab)$

but now the proofs of these facts are the same as the case $\mathbb{Q}[x]$.