TITLE: How can I find the partitions of an equivalence relation?
QUESTION [3 upvotes]: I have the following equivalence relation:
$$\{(1,1),(1,4), (2,2), (3,3), (4,1), (4,4)\}$$
On the set:
$ A = \{1,2,3,4\}$
How can I find it's partitions? This example will help me understand the more general process of finding partitions for equivalence relations. Thanks.

REPLY [0 votes]: We can also approach this as follows. Define a function $f:A\to S$ where $S$ is any set with at least four elements, which follows the rule $f(x)=f(y)$ if and only if $x$ and $y$ are equivalent. Then, the partition is just the set of nonempty preimages of $f$.
Here, say we have the set $S=\{w,x,y,z\}$, and define $f$ by $f(1)=w$ arbitrarily. Since $2$ and $1$ are not equivalent, define $f(2)=x$, some other element of $S$. $3$ is not equivalent to either $1$ or $2$, so $f(3)=y$ some unused element of $S$. Finally, $4$ is equivalent to $1$, so $f(4)=f(1)=z$.
Now, we list the preimages: $f^{-1}[w]=\{1,4\}$, $f^{-1}[x]=\{2\}$, $f^{-1}[y]=\{3\}$, and $f^{-1}[z]=\emptyset$. Thus our partition is given by the first three sets.


TITLE: lower central series and basic generators
QUESTION [2 upvotes]: After reading the following question:
Third quotient of the lower central series
I have a similar question.
Let us denote $F_n$ the free group with $n$ generators, $x_1,...,x_n$ and let $[a,b] = a^{-1}b^{-1}ab$, and $G_i$ be the $i^{th}$ group in the lower central series. It is known that $G_3/G_4$ is generated by (the basic commutators) $[[x_i,x_j],x_k]$ where $i>j$ and $j \leq k$. Now, let $i,j,l,k$ be indices. How do we express $[[x_i,x_jx_l],x_k]$ in $G_3/G_4$
using only elements of the form $[[a,b],c]$ when $a,b,c \in \{x_i,x_j,x_l,x_k\}$?
Thank you, David

REPLY [2 votes]: $[[x_i,x_jx_l],x_k] = [[x_i,x_l][x_i,x_j]^{x_l},x_k] =$ $[[x_i,x_l][x_l,[x_i,x_j]^{-1}][x_i,x_j],x_k] \equiv [[x_i,x_l],x_k][[x_i,x_j],x_k] \mod G_4$.


TITLE: Differentiability of $h(x)=x^2\sin(1/x)$ for $x\in\mathbb{R} \backslash \{0\}$ and $h(0)=0$ and continuity of derivative
QUESTION [1 upvotes]: Let $h(x)=x^2\sin(1/x)$ for $x\in\mathbb{R} \backslash \{0\}$ and $h(0)=0$.

$h$ is differentiable in $0$ if $\displaystyle\lim_{k \to 0} \frac{h(0+k)-h(0)}{k}$ exists.
$$\lim_{k \to 0} \frac{h(0+k)-h(0)}{k}=\lim_{k \to 0} \frac{k^2\sin\left(\frac{1}{k}\right)}{k} = \lim_{k\to0} k\sin\left(\frac{1}{k}\right).$$
Using the Squeeze Theorem: $\left|\sin\left(\frac{1}{k}\right)\right|\leq1$, so $$\lim_{k\to0} -k = 0 \leq \lim_{k\to0} k\sin\left(\frac{1}{k}\right) \leq 0 = \lim_{k\to 0} k.$$ so
$$\lim_{k\to0} k\sin\left(\frac{1}{k}\right)=0,$$ so $h$ is differentiable at $0$ and $h'(0)=0$.

Is this proof of differentiability correct? I'm especially wary about the conclusion $h'(0)=0$, furthermore, how do I prove that $h'$ is not continuous at $0?$ (Can this be proven without proving what the derivative of $h$ is for all $x?$) I do not know where to start this continuity proof.

REPLY [1 votes]: The proof of differentiability is correct. In fact you do not need to use Squeeze Theorem as $\lim_{x\rightarrow 0} f(x)g(x) = 0$ whenever $\lim_{x\rightarrow 0} f(x) =0$ and $g$ is bounded in a neighborhood of $0$. However, an easier approach to show that $h'$ is not continuous at $x =0$ is to consider the null sequence $\{\frac{1}{n\pi}\}$ and use the sequential criteria of continuity.


text:
\section{Gromov--Witten invariants}\mylabel{sec:defgwi}

Gromov--Witten invariants of a symplectic manifold $X$ are defined
using intersection theory on the moduli space of stable (holomorphic or
pseudo--holomorphic) maps to $X$. They are invariants of the deformation
class of the symplectic structure $\omega$ of $X$, so in particular they ought to be
independent of the (pseudo--)complex structure $J$ compatible with $\omega$.

Unfortunately, even the dimension of the moduli spaces of stable maps
can vary with the (pseudo--)complex structure. However, these moduli spaces
are the pre--image of zero under the $\bar\partial_J$ operator
and we would have $\dim \gdmsp = \dim_\virt \gdmsp$
if $\bar\partial_J$ were transversal to the zero section of
$\Omega_C^{0,1}(f^*TX)$ at each stable $J$--holomorphic map $f:C\rightarrow X$ in $\gmodsp$.

Two different approaches have been developed to solve this problem:
one is to try to make $\bar\partial_J$ transversal to the zero section, the other
is to use principles of excess intersection theory to obtain a cycle in
$H_*(\gmodsp)$ of degree equal to the virtual dimension $\dim_{\virt}\gmodsp$ of the moduli space.
The former has been pursued  by Ruan and Tian (\cite{rt95}) for weakly monotone
symplectic manifolds.

The latter has been developed by Behrend and Fantechi as well as Li and Tian
(\cite{bf97,beh97,lt98a}) for all smooth projective complex varieties, and by Fukaya
and Ono, Li and Tian, Ruan, and Siebert (\cite{fo96,lt96,rua96,sie96}) for all smooth
symplectic manifolds\footnote{Of course, this class of manifolds includes the smooth projective
complex varieties, though the constructions by Behrend and Fantechi, and Lian and Tian
are entirely in the algebro--geometric category. In particular, Behrend and Fantechi construct
a cycle in the Chow ring $A_*(\gmodsp)$ of the moduli space.}. The  basic idea of the
construction is as follows: Consider a smooth variety $W$, two smooth subvarieties
$X,Y$ of $W$, and their intersection $Z$:
\[
\xymatrix{
Z \ar[r] \ar[d] &
X \ar[d]^f \\
Y \ar[r]^g &
W.
}
\]

Now, if $X$ and $Y$ intersect properly, \ie. if $\dim Z=
\dim X + \dim Y - \dim W$ then the fundamental cycle of Z is the
intersection of the fundamental cycles of $X$ and $Y$: $[Z]=[X]\cdot[Y]$.
Otherwise, using excess intersection theory we can find a cycle in the Chow ring $A_*(Z)$
representing $[X]\cdot[Y]$, the {\em virtual cycle} of $Z$: $[Z]^{\virt}$.
Let $s:Z\rightarrow C_{Y/W}\times_Y Z$ be the zero section of the normal cone to
$Y$ in $W$ pulled back to $Z$. Then $[Z]^{\virt}$ is the intersection
of the zero section $s$ with the normal cone $C_{Z/X}$ to $X$ in $Z$:
\[ [Z]^{\virt}= s^*(C_{Z/X}),\]
where $s^*:A_*(C_{Y/W}\times_Y Z)\rightarrow A_*Z$ is the Gysin morphism
induced by $s$.

Unfortunately, for our moduli problem such an ambient space $W$ and maps
$X,Y\rightarrow W$ do not exist naturally such that $X\times_W Y$ is the
moduli space and $[X]\cdot[Y]$ a virtual moduli cycle with the properties
we want. Instead, the construction will use an {\em obstruction theory} for
$\gmodsp$, a two--term complex $E^\bullet$ on $\gmodsp$ with $\rk E^\bullet=
\dim_{\virt}\gmodsp$.

We will sketch the definition in some generality following
\cite{bf97,beh97}, and then apply it
to the moduli space of stable maps and Gromov--Witten invariants.

\subsection{Perfect obstruction theory and virtual fundamental class}

Let $Y$ be a Deligne--Mumford stack, that is an algebraic stack with unramified
diagonal. Let $L^\bullet_Y$ be the cotangent complex of
$Y$ (see for example \cite{buc81,ill71} for its
definition and properties on schemes, and \cite{lm92} for its
generalization to algebraic stacks). The {\em intrinsic
normal sheaf ${\mathfrak N}_{Y}$} is defined to be the quotient stack
\[ {\mathfrak N}_{Y} := h^1/h^0((L^\bullet_{Y})^\vee)=
[\ker (L_1\rightarrow L_2)/L_0].\]
The {\em intrinsic normal cone ${\mathfrak C}_Y$ of $Y$} is the unique
closed subcone stack ${\mathfrak C}_{Y} \hookrightarrow
{\mathfrak N}_Y$ such that for a local embedding
\[
\xymatrix{
U \ar[r]^f \ar[d]_i&
M\\
Y
}
\]
of $Y$, we have ${\mathfrak C}_{Y}|_U=[C_{U/M}/f^*TM]$
(\cite[Corollary 3.9]{bf97}).  The intrinsic normal cone
${\mathfrak C}_Y$ is of pure dimension zero
(\cite[Theorem 3.11]{bf97}).

\begin{defi}
Let  $Y$ be a Deligne--Mumford stack, that is,
an algebraic stack with unramified diagonal.

Let $E^\bullet=[E^{-1}\rightarrow E^0]$ be a two--term complex of
vector bundles on $Y$. Then a morphism in the derived category from
$E^\bullet$ to the cotangent complex $L^\bullet_Y$
\[ \phi: E^\bullet \rightarrow L^\bullet_Y\]
is called a {\em perfect obstruction theory} for $Y$
if $h^0(\phi)$ is an isomorphism and $h^{-1}(\phi)$ is surjective.
\end{defi}

\begin{rema}
The definition of a perfect obstruction theory in
\cite{bf97} is more general than the one given here, that is they consider
two--term complexes of locally free sheaves ${\mathcal E}^\bullet$.
A two--term complex of vector bundles $E^\bullet$ as above that is
isomorphic to ${\mathcal E}^\bullet$ in the derived category is then called
a global resolution.
\end{rema}

The morphism $\phi$ induces a closed immersion
$\phi^\vee:{\mathfrak N}_Y \rightarrow h^1/h^0((E^\bullet)^\vee)$
(Proposition 2.6 in \cite{bf97}), so $E_1={E^{-1}}^\vee$ is a
global presentation of the quotient stack $h^1/h^0((L^\bullet)^\vee)$ and ${\mathfrak
C}_Y\rightarrow {\mathfrak N}_Y$ embeds into $E_1$. Consider the
fibered product
\[
\xymatrix{
C(E^\bullet) \ar[r] \ar[d] &
E_1 \ar[d]\\
{\mathfrak C}_Y \ar[r] &
[E_1/E_0].
}
\]
Hence, $C(E^\bullet)$ is a closed subcone of the vector bundle $E_1$.
Locally, for a local embedding
$U\rightarrow M$ as above, $i^*C(E^\bullet)$ is those just given by
\[  i^*C(E^\bullet) = \left(C_{U/M}\times_U (i^*E_0)\right)/f^*TM.\]

By this construction, $C(E^\bullet)\rightarrow {\mathfrak C}_Y$ is
smooth of relative dimension $\rk E_0$. Since the intrinsic normal
cone ${\mathfrak C}_Y$ is of pure dimension zero, $C(E^\bullet)$
is thus of pure dimension $\rk E_0$.

\begin{defi}
Let $Y$, $C(E^\bullet)$ and $E_1$ be as above. Let $n=\rk E^\bullet=\rk E^0-
\rk E^{-1}=\rk E_0-\rk E_1$ be the {\em virtual dimension} of $Y$ with respect to
the obstruction theory $E^\bullet$.
The {\em virtual fundamental class}
$[Y,E^\bullet]\in H_n(Y,\Q)$ of $Y$ is the intersection of $C(E^\bullet)$ with the
zero section of $E_1$.
\end{defi}

\begin{rema}
The virtual fundamental class is independent of the choice of the perfect obstruction
theory within a quasi--isomorphism class. That is, if $F^\bullet$ is another perfect
obstruction theory and $\psi:F^\bullet\rightarrow E^\bullet$ a quasi--isomorphism, $\psi$
naturally induces the identity map for the virtual fundamental classes associated to
$F^\bullet$ and $E^\bullet$ (\cite[Proposition 5.3]{bf97}).

By abuse of notation, we will often write $[Y]^{\virt}$ for the virtual
fundamental class $[Y,E^\bullet]$ when it is understood which obstruction theory
is used.
\end{rema}

\subsection{The obstruction complex for the definition of GW invariants}
\mylabel{subsec:obstrcplxGW}

We will now describe the obstruction theory used for the definition
of the Gromov--Witten invariants of a smooth projective complex variety $X$. Moreover, if there
is an action by a torus $T_N$ on the variety $X$, this obstruction theory will be $T_N$--equivariant.

Let $X$ be a smooth projective complex variety, $A\in H_2(X;\Z)$ an integral degree--2
homology class of $X$, and $\gmodsp$ the corresponding moduli stack of stable $m$--marked
genus--$g$ maps to $X$. Let $\pi: \ucgmodsp \longrightarrow \gmodsp$ be the universal
curve, and let $x_i: \gmodsp \longrightarrow \ucgmodsp$ ($i=1,\ldots, m$) be the marked
point sections. We will denote by $D$ the divisor defined by the images of the marked point
sections $x_i$. If no confusion can arise, we will also use the notation $\cM:=\gmodsp$ and
$\cC:=\ucgmodsp$.  We will consider the following complex:
\[ E^\bullet:=R\pi_*\left(\left(f^*\Omega^1_X[1] \oplus \Omega^1_{\cC/\cM}(D)\right)\stackrel{L}{\otimes}
\omega_{\cC/\cM}\right), \]
where $\omega_{\cC/\cM}$ is the relative dualizing sheaf.
We will first show that there is a canonical morphism $\phi: E^\bullet
\longrightarrow L^\bullet_\cM$ and then prove that this morphism is an obstruction
theory.

Remember that Behrend and Fantechi have given an obstruction theory for the problem
relative to the stack of prestable curves:

\begin{theo}[\cite{beh97, bf97, bm96}]
Let $p: \gmodsp \rightarrow {\mathfrak M}_{g,m}$ be the ca\-noni\-cal
mor\-phism from the stack of stable maps to the stack of prestable curves
given by forgetting the map and retaining the curve without
stabilizing. Then $\gmodsp\rightarrow {\mathfrak M}_{g,m}$ is an
open substack of a relative space of morphisms, hence it has a
relative obstruction theory which is given by
\[ \psi: \left(R\pi_*f^*TX\right)^{\vee} \longrightarrow L^\bullet_{\gmodsp/{\mathfrak M}_{g,m}}.
\]
Here $\pi: \ucgmodsp \longrightarrow \gmodsp$ is the universal curve and $f:{\mathcal C}
\rightarrow X$ is the universal stable map. \proofend
\end{theo}

\begin{coro}
There exists a canonical morphism $\phi: E^\bullet \longrightarrow L^\bullet_\cM$
in the derived category induced by the morphism $\psi$.
\end{coro}

\begin{proof}
Consider the following cartesian diagram where $\pi: {\mathcal C} \longrightarrow {\mathcal M}$
is a stable map to $X$, and $p:{\mathcal M} \longrightarrow S$ is the forgetting
map, \ie. $Z\longrightarrow S$ is a prestable curve:
\[
\xymatrix{
Z \ar[d]^\sigma &
{\cC} \ar[l]_\tau \ar[r]^f \ar[d]^\pi &
X\\
S &
{\cM.} \ar[l]_p&
}
\]
Remember that if we have two morphisms of schemes (or stacks) $U\stackrel{h}{\longrightarrow}V\longrightarrow W$
we get a distinguished triangle of cotangent complexes:
\[ h^*L^\bullet_{V/W} \mapright L^\bullet_{U/W} \mapright L^\bullet_{U/V} \mapright h^*L^\bullet_{V/W}[1].\]
Moreover $f^*\Omega^1_X=f^*L^\bullet_X$ naturally maps to $L^\bullet_{\mathcal C}$, so we get the following
diagram:
\[
\xymatrix{
f^*\Omega^1_X \ar[r] &
L^\bullet_{\cC} \ar[r] \ar[d] &
L^\bullet_{\cC/\cM} \ar[r]^{\sim} &
\tau^* L^\bullet_{Z/S} \ar[d]\\
&
L^\bullet_{\cC/Z} \ar[d]^{\sim}&
&
\tau^*\sigma^*L^\bullet_S[1] \ar[d]^{\sim}\\
&
\pi^*L^\bullet_{\cM/S} \ar[rr] &
&
\pi^*p^*L^\bullet_S[1].
}
\]
This diagram is in fact commutative since $\sigma$ is flat, and so by \cite[(9.2.5)]{lm92} we have
\[ \tau^* L^\bullet_{Z/S} \oplus \pi^*L^\bullet_{\cM/S} \MRE{\sim} L^\bullet_{\cC/S}, \]
and the morphisms in the diagram above are just the morphism induced by the distinguished triangle
\[ \pi^* p^* L^\bullet_S \mapright L^\bullet_\cC \mapright L^\bullet_{\cC/S} \mapright \pi^*p^*L^\bullet_S[1].\]
Applying the cut--off functor $\tau_{\ge 0}$ to $f^*L^\bullet_X \longrightarrow L^\bullet_{\cC/\cM}$
and taking the mapping cone yields the following diagram in the derived category:
\[
\xymatrix{
f^*\Omega^1_X \ar[r] \ar[d] &
\Omega^1_{\cC/\cM}(D) \ar[r] \ar[d] &
f^*\Omega^1_X[1] \oplus \Omega^1_{\cC/\cM}(D) \ar[d]\\
\pi^*L^\bullet_{\cM/S} \ar[r] &
\pi^*p^*L^*_S[1] \ar[r] &
\pi^* L^\bullet_\cM [1].
}
\]
The projection formula yields the desired morphism $\phi$.
\end{proof}

\begin{prop}\mylabel{claim:obstrthy}
Let $\omega_{\cC/\cM}$ be the relative dualizing sheaf of $\pi:\cC\rightarrow \cM$.
The morphism
\[ \phi: E^\bullet \mapright L^\bullet_\cM\]
is a perfect obstruction theory for the moduli stack of stable maps $\gmodsp$. If there is a torus $T_N$ acting
on $X$, this obstruction theory is $T_N$--equivariant.
\end{prop}

\begin{proof}
First we will construct
a two--term resolution of $R\pi_*E^\bullet$ that is $T_N$--equiva\-riant if such an
action exists on $X$.  We will use similar arguments as Behrend does for $R\pi_*f^*TX$
(\cf. \cite[Proof of Proposition 5]{beh97}). Let $M$ be an ample invertible sheaf on
$X$ and let $L=\omega_{\cC/\cM}(D)\otimes f^*M^{\otimes3}$. Then by \cite[Proposition 3.9]{bm96},
for $N$ sufficiently large and $V$ a vector bundle on $\cC$ we have that
\begin{enumerate}
\item $\pi^*\pi_*(V\otimes L^{\otimes N})\longrightarrow V\otimes L^{\otimes N}$ is surjective,
\item $R^1\pi_*(V\otimes L^{\otimes N})=0$,
\item for all $s\in S$ we have that $H^0(C_s, L_s^{\otimes -N})=0$.
\end{enumerate}
Let us set
$ F:= \pi^*\pi_*(f^*\Omega^1_X \otimes L^{\otimes N})\otimes L^{\otimes -N}$ and
$H:=\ker (F\longrightarrow f^*\Omega^1_X)$, and
consider the complexes (\cf. \cite[section 4]{lt98a}) indexed at $-1$ and $0$
\[ \cA^\bullet = [H\otimes\omega_{\cC/\cM} \longrightarrow 0] \quad \text{and} \quad
\cB^\bullet = [F\otimes\omega_{\cC/\cM} \longrightarrow \Omega^1_{\cC/\cM}(D)\otimes\omega_{\cC/\cM}],\]
where the morphism within the complex $\cB^\bullet$ is induced from the composition map $F \longrightarrow f^*\Omega^1_X \longrightarrow
\Omega_{\cC/\cM}(D)$. Hence there are morphisms
\[ R^1\pi_* (F\otimes\omega_{\cC/\cM}) \mapright R^1\pi_* (f^*\Omega^1_X\otimes\omega_{\cC/\cM}) \MRE{\alpha}
R^1\pi_*\cB^\bullet \]
where $\alpha$ is surjective, by lemma \ref{lem:stableinj} and duality. As before we also have
\begin{eqnarray*}
H^0(C_t, F\otimes\omega_{\cC/\cM})&=&H^0(C_t, \pi_*(f^*\Omega^1_X\otimes L^{\otimes N})_t\otimes
L_t^{\otimes-N}\otimes \O_{\cM,t})\\
&=&H^0(C_t, L_t^{\otimes -N})\otimes \pi_*(\Omega^1_X\otimes L^{\otimes N})\otimes \O_{\cM,t}=0,
\end{eqnarray*}
so $R^0\pi_* (H\otimes\omega_{\cC/\cM})=R^0\pi_*(F\otimes\omega_{\cC/\cM})=0$.
Observe that the complex $\cB^\bullet$ fits into the short exact sequence
\[ 0 \mapright \Omega_{\cC/\cM}(D)\otimes\omega_{\cC/\cM} \mapright \cB^\bullet
\mapright F\otimes\omega_{\cC/\cM}[1] \mapright 0,\]
therefore we get a corresponding long exact sequence of higher direct image sheaves:
\begin{multline*}
0 \longrightarrow R^{-1}\pi_* \cB^\bullet \longrightarrow
\underbrace{R^0\pi_*(F\otimes\omega_{\cC/\cM})}_{=0}
\longrightarrow R^0\pi_*(\Omega^1_{\cC/\cM}(D)\otimes\omega_{\cC/\cM}) \longrightarrow\\
\longrightarrow R^0\pi_*\cB^\bullet \longrightarrow R^1\pi_*(F\otimes\omega_{\cC/\cM})
\MRE{\text{surj.}} R^1\pi_*(\Omega^1_{\cC/\cM}(D)\otimes\omega_{\cC/\cM})\longrightarrow\\
\longrightarrow R^1\pi_* \cB^\bullet \longrightarrow 0.
\end{multline*}
Hence $R^i\pi_* \cB^\bullet = 0$ for $i\neq 0$. Moreover, since $R^i\pi_* (H\otimes\omega_{\cC/\cM}) = 0$
for $i\neq 1$, we also get $R^i\pi_* \cA^\bullet = R^{i+1}\pi_* (H\otimes\omega_{\cC/\cM})= 0$ for $i\neq 0$.
Now note that these two complexes fit into the following short exact sequence:
\[ 0 \mapright \cA^\bullet \mapright \cB^\bullet \mapright
(f^*\Omega^1_X[1]\oplus\Omega^1_{\cC/\cM}(D))\otimes\omega_{\cC/\cM} \mapright 0,\]
yielding the long exact sequence
\[ 0 \mapright h^{-1}(E^\bullet) \mapright R^0\pi_*\cA^\bullet \mapright R^0\pi_*\cB^\bullet
\mapright h^0(E^\bullet) \mapright 0.\]

Thus we have found a two--term resolution of $E^\bullet$ by locally free sheaves:
\[ E^\bullet \cong [R^0\pi_* \cA^\bullet \longrightarrow R^0\pi_* \cB^\bullet].\]
Moreover, the entire construction is $T_N$--equivariant, so we actually have found a $T_N$--equivariant
resolution of $E^\bullet$, if such an action exists on $X$.

Finally, we observe that $\delta:R\pi_*(\Omega^1_{\cC/\cM}(D)\otimes\omega_{\cC/\cM})\cong p^*L^\bullet_{S}$ in the
derived category. Then by using the fact that $\psi:(R\pi_* f^*TX)^\vee\longrightarrow L^\bullet_{\cM/S}$ is an
obstruction theory for the relative problem, and by applying the five lemma we get that $h^0(\phi)$ is an
isomorphism and that $h^{-1}(\phi)$ is surjective:
\[
\xymatrix@C-0.5ex{
0 \ar[r] \ar[d]_{=} &
h^{-1}(E^\bullet) \ar[r] \ar[d]_{\phi^{-1}} &
\left(R^1\pi_*f^*TX\right)^\vee \ar[r] \ar[d]_{\psi^{-1}} &
R^0\pi_*(\Omega_{\cC/\cM}(D)\otimes\omega_{\cC/\cM}) \ar[r] \ar[d]_{\delta^0}^{\sim} & \\
0 \ar[r] &
h^{-1}(L^\bullet_{\cC}) \ar[r] &
h^{-1}(L^\bullet_{\cC/{\mathfrak M}}) \ar[r] &
h^0(p^*L^\bullet_{{\mathfrak M}}) \ar[r] &\\
\ar[r] &
h^0(E^\bullet) \ar[r] \ar[d]_{\phi^0} &
\left(R^0\pi_*f^*TX\right)^\vee \ar[r] \ar[d]_{\psi^0}^{\sim} &
R^1\pi_*(\Omega_{\cC/\cM}(D)\otimes \omega_{\cC/\cM}) \ar[d]_{\delta^1}^{\sim}\\
\ar[r] &
h^0(L^\bullet_{\cC}) \ar[r] &
h^0(L^\bullet_{\cC/{\mathfrak M}}) \ar[r] &
h^1(p^*L^\bullet_{{\mathfrak M}}).
}
\]
Hence $\phi:E^\bullet \longrightarrow L_{\gmodsp}^\bullet$ is indeed a ($T_N$--equivariant) perfect
obstruction theory for the moduli stack of stable maps $\gmodsp$.
\end{proof}

That is exactly the obstruction theory we will use for the definition of the
Gromov--Witten invariants.

\begin{lemm}
The virtual fundamental class $[E^\bullet, \phi]$ of the obstruction theory
$\phi: E^\bullet \longrightarrow L^\bullet_\cM$ coincides with the virtual
fundamental class coming from Behrend's relative obstruction theory \cite{beh97}
$\psi: (R\pi_*f^*TX)^\vee \longrightarrow L^\bullet_{\cM/{\mathfrak M}}$.
\end{lemm}

\begin{defi}
The obstruction theory $\phi: E^\bullet \longrightarrow L^\bullet_\cM$
is the {\em obstruction theory for the Gromov--Witten invariants}.
\end{defi}

\begin{rema}
Behrend has proven in \cite{beh97} that his relative obstruction
theory defines Gromov--Witten invariants, so the definition is good. Moreover,
there has been considerable progress in showing that the different versions
of Gromov--Witten invariants all coincide, \eg. \cite{lt98b,sie98}.
\end{rema}

\begin{proof}[Proof of the Lemma]
The equality of the two virtual fundamental classes can easily be seen by
looking at the complex $E_\bullet$ dual to $E^\bullet$,
\begin{align*}
E_\bullet&= R \underline\Hom (E^\bullet, \O_{\gmodsp})\\
&\cong [(R^0\pi_*\cB^\bullet)^\vee \mapright (R^0\pi_*\cA^\bullet)^\vee]\\
&\cong \left[R^1\pi_*[F\longrightarrow\Omega^1_{\cC/\cM}(D)]^\vee \mapright R^1\pi_*[H\longrightarrow 0]^\vee \right]
\quad \text{(by duality)}\\
&=\left[ {\underline\Ext}_\pi^1([F\longrightarrow \Omega^1_{\cC/\cM}(D)],\O_\cC)\mapright
{\underline\Ext}_\pi^1([H\longrightarrow 0],\O_\cC)\right]\\
&\cong R\pi_*{\underline\Hom} ([f^*\Omega^1_X \longrightarrow \Omega^1_{\cC/\cM}(D)], \O_\cC).
\end{align*}

Here we have used that by \cite[lemma II.3.1, proposition I.5.4]{har66} there exists
a morphism of functors
\[ \zeta: R(\pi_*\circ\underline\Hom(\phantom{M}, \O_\cC)) \mapright R\pi_* \circ R\underline\Hom(\phantom{M}, \O_\cC),\]
and that this morphism $\zeta$ is an isomorphism. For convenience we also use the notation
\[ \underline\Ext^i_\pi(\phantom{M}, \O_\cC) := R^i(\pi_*\circ\underline\Hom(\phantom{M}, \O_\cC)).\]

\noindent Therefore, the $E_i$'s fit into an exact sequence
\[ 0 \longrightarrow {\mathcal T}^0 \longrightarrow E_0 \longrightarrow E_1
\longrightarrow {\mathcal T}^1 \longrightarrow 0,\]

where the sheaves ${\mathcal T}^i$ are given by taking cohomology of $E_\bullet$:
\mycomment{eq:defnti}
\begin{equation}\label{eq:defnti}
{\mathcal T}^i = \underline\Ext^i_\pi([f^*\Omega_X^1 \rightarrow
\Omega_{{\mathcal C}_{g,m}^A(X)/\gmodsp}(D)], \O_{{\mathcal C}_{g,m}^A(X)}).
\end{equation}
\end{proof}

\begin{rema} Contrary to \cite{lt98b}, the complex $[f^*\Omega_X^1
\rightarrow \Omega_{\cC/\cM}(D)]$ in (\ref{eq:defnti})
is indexed at $0$ and $1$, instead of $-1$ and $0$, moving the ${\mathcal T}^i$ complex
to the left.
\end{rema}

We will end this subsection with a lemma about how this obstruction theory
behaves under base change.
This lemma will be used when we pass to the fixed point components
of the torus action on the moduli space in section \ref{sec:virtnormbdl}.

\begin{lemm}\mylabel{lem:restrobth}
Let $\pi: \cC \longrightarrow \cM$ be a stable map to $X$ that is
an atlas for $\gmodsp$. Furthermore, let $\iota:\cM_\mygraph
\longrightarrow \cM$ be a subscheme, and look at the cartesian
diagram
\[
\xymatrix{
{\cC_\mygraph} \ar[r]^{\tilde\iota} \ar[d]^{\pi_\mygraph} &
{\cC} \ar[r]^f \ar[d]^{\pi} &
X\\
{\cM_\mygraph} \ar[r]^{\iota} &
{\cM}. &
}
\]
Let $f_\mygraph:= f\circ \tilde\iota$. Then the restrictions of
the obstruction theory $E^\bullet$ and its dual $E_\bullet$ are
given by
\begin{align*}
E^\bullet|_{\cM_\mygraph}&=R\pi_*\left( \left( f_\mygraph^*
\Omega^1_X[1] \oplus
\Omega^1_{\cC_\mygraph/\cM_\mygraph}(D_\mygraph) \right) \lotimes
\omega_{\cC_\mygraph/\cM_\mygraph}\right)\\
E_\bullet|_{\cM_\mygraph}&=R\underline\Hom_\pi([f_\mygraph^*
\Omega_X^1 \longrightarrow
\Omega^1_{\cC_\mygraph/\cM_\mygraph}(D_\mygraph) ],
\O_{\cC_\mygraph}).
\end{align*}
\end{lemm}

\begin{proof}
We will prove the lemma for the obstruction complex $E^\bullet$, the arguments for the dual
complex $E_\bullet$ are similar. There is a natural morphism
\[ R\pi_*\left( \left( f_\mygraph^* \Omega^1_X[1] \oplus \Omega^1_{\cC_\mygraph/\cM_\mygraph}(D_\mygraph)
\right) \lotimes \omega_{\cC_\mygraph/\cM_\mygraph}\right)
\mapright E^\bullet|_{\cM_\mygraph}, \] and we have to show that
this morphism is a isomorphism in the derived category, \ie. a
quasi--isomorphism between complexes. Let $\cK^\bullet :=
[f^*\Omega_X^1 \otimes \omega_{\cC/\cM} \longrightarrow
\Omega^1_{\cC/\cM}(D)\otimes\omega_{\cC/\cM}]$, indexed at $-1$
and $0$. We then have to show that
\[ \left.\left(R^i \pi_* \cK^\bullet\right)\right|_{\cM_\mygraph} = R^i{\pi_\mygraph}_* (\cK^\bullet|_{\cC_\mygraph}).\]
Now $\cK^\bullet$ fits into a short exact sequence of complexes
\[ 0 \mapright \cA^\bullet \mapright \cB^\bullet \mapright \cK^\bullet \mapright 0 \]
such that $R^i\pi_* \cA^\bullet$ and $R^i\pi_* \cB^\bullet$ are locally free and
\[ R\pi_* \cK^\bullet \cong [R^0\pi_* \cA^\bullet \longrightarrow R^0\pi_* \cB^\bullet]\]
(see above). Since $\pi$ is a proper flat morphism, we have by Grauert's continuity theorem (see for example
\cite[Th\'eor\`eme 4.12(ii)]{bs77})  that
\[ \left.\left(R^i\pi_* \cA^\bullet\right)\right|_{\cM_\mygraph} = R^i{\pi_\mygraph}_* (\cA^\bullet|_{\cC_\mygraph})
\quad \text{and} \quad \left.\left(R^i\pi_*
\cB^\bullet\right)\right|_{\cM_\mygraph} = R^i{\pi_\mygraph}_*
(\cB^\bullet|_{\cC_\mygraph}).
\]
This yields the same property for the complex $\cK^\bullet$.
\end{proof}

\subsection{Definition of the Gromov-Witten invariants}

In the previous section we have constructed a ($T_N$--equivariant) perfect obstruction theory
for the moduli stack $\gmodsp$.
Hence we get a virtual fundamental class $[\gmodsp]^{\virt}:=
[\gmodsp, E^\bullet]\in H_n(\gmodsp,\Q)$, where
$n=\rk E^\bullet$ is equal to the virtual dimension of $\gmodsp$: $n=(1-g) (\dim X -3)
+ \langle c_1(X), A\rangle +m$. So for cohomology classes $\alpha_1, \ldots, \alpha_m
\in H^*(X, \Z)$ and $\beta \in H^*(\gdmsp)$ we define the Gromov--Witten invariant
$\Psi^A_{m,g}(\beta;\alpha_1, \ldots, \alpha_k)$ by:
\[ \Psi^A_{m,g}(\beta;\alpha_1, \ldots, \alpha_m) :=
\int_{[\gmodsp]^{\text{vir}}}
\ev^*(\alpha_1\otimes\ldots\otimes\alpha_m)\wedge \pi^*\beta,\]
where $\ev: \gmodsp \rightarrow X^{\otimes m}$ is the $m$--point
evaluation map, and $\pi$  the natural
forgetting (and stabilization) morphism $\pi: \gmodsp \rightarrow \gdmsp$.

In the remaining part of the paper we will restrict ourselves to the genus--zero
case, \ie. when $g=0$, and to the invariants $\Phi^A_m$ where moreover
the class $\beta=1$ is trivial:
\[ \Phi^A_m(\alpha_1,\ldots,\alpha_m) := \Psi^A_{m,0}(1;\alpha_1,\ldots,\alpha_m).\]
Note that for $m=3$ and $g=0$,
the Deligne--Mumford space of stable curves is a point, hence $\beta=1$ is
the only class that exists.

config: arxiv


text:
\section{Lattice homomorphisms} \label{Sdrum2}

In this section we consider lattice homomorphisms between
$L_p$-spaces on two Riemannian manifolds without the assumption
that the manifolds are regular in capacity.
The aim is to prove that the associated $H^1_0$-spaces are
equivalent, under the conditions of Theorem~\ref{tdrum101}.

The first step is to use elliptic regularity of the Laplace--Beltrami operator
to reduce to smooth functions.

\begin{lemma} \label{ldrum201}
Let $(M_1,g_1)$ and $(M_2,g_2)$ be two Riemannian manifolds.
Let $p \in [1,\infty)$.
For all $j \in \{ 1,2 \} $ let $\Delta_j$ be the Dirichlet Laplace--Beltrami operator
on $M_j$ and let $S^{(j)}$ be the associated semigroup on $L_p(M_j)$.
Let  $U \colon L_p(M_1) \to L_p(M_2)$ be a lattice homomorphism such that
\begin{equation}
U S^{(1)}_t = S^{(2)}_t U
\label{eldrum201;1}
\end{equation}
for all $t > 0$.
Then
\begin{subtabel}
\item \label{ldrum201-1}
$U C_c^\infty(M_1) \subset C^\infty(M_2)$.
\item \label{ldrum201-2}
$U \varphi \geq 0$ for all $\varphi \in C_c^\infty(M_1)$ with $\varphi \geq 0$.
\item \label{ldrum201-3}
$(U \varphi) (U \psi) = 0$ for all $\varphi,\psi \in C_c^\infty(M_1)$ with $\varphi \, \psi = 0$.
\item \label{ldrum201-4}
$\Delta_2 U \varphi = U \Delta_1 \varphi$ for all $\varphi \in C_c^\infty(M_1)$.
\end{subtabel}
\end{lemma}
\proof\
It follows from (\ref{eldrum201;1}) that
$U D(\Delta_1) \subset D(\Delta_2)$ and $\Delta_2 U \varphi = U \Delta_1 \varphi$
for all $\varphi \in D(\Delta_1)$.
Hence by iteration
$U \bigcap_{n=1}^\infty D(\Delta_1^n) \subset \bigcap_{n=1}^\infty D(\Delta_2^n)$.
But $C_c^\infty(M_j) \subset \bigcap_{n=1}^\infty D(\Delta_j^n) \subset C^\infty(M_j)$
for all $j \in \{ 1,2 \} $ by elliptic regularity.
This shows \ref{ldrum201-1} and \ref{ldrum201-4}.
Property~\ref{ldrum201-2} follows since $U$ is a lattice homomorphism.
Moreover,
$|U \varphi| = U |\varphi|$ for all $\varphi \in L_p(M_1)$.
Hence if $\varphi,\psi \in C_c(M_1)$ and $\varphi \, \psi = 0$ then
$|\varphi| \wedge |\psi| = 0$ and
$|U \varphi| \wedge |U \psi| = U |\varphi| \wedge U |\psi|
= U(|\varphi| \wedge |\psi|) = 0$.
Therefore $|(U \varphi) (U \psi)| = |U \varphi| \, |U \psi| = 0$
and $(U \varphi) (U \psi) = 0$.
This implies Property~\ref{ldrum201-3}.\hfill$\Box$

\ruimte

We frequently need the following sufficient condition for point evaluations.

\begin{lemma} \label{ldrum201.3}
Let $M$ be a manifold and $F \colon C_c^\infty(M) \to \Ri$ a positive linear functional
such that
\begin{equation}
F(\varphi) \, F(\psi) = 0
\mbox{ for all } \varphi,\psi \in C_c^\infty(M) \mbox{ with } \varphi \, \psi = 0
\;\;\; .
\label{eldrum201.3;1}
\end{equation}
Then there exist $c \in [0,\infty)$ and $p \in M$ such that
$F(\varphi) = c \, \varphi(p)$ for all $\varphi \in C_c^\infty(M)$.
\end{lemma}
\proof\
Arguing as in \cite{EvG} Corollary~1.8.1 it follows that
there exists a unique Radon measure $\mu$ on $M$ such that
$F(\varphi) = \int \varphi \, d \mu$ for all $\varphi \in C_c^\infty(M)$.
Then it follows from (\ref{eldrum201.3;1}) that $\mu$ is a point measure.
Hence there exist $p \in M$ and $c \in [0,\infty)$ such that
$F(\varphi) = c \, \varphi(p)$ for all $\varphi \in C_c^\infty(M)$.\hfill$\Box$

\ruimte

The next proposition is a manifold version of \cite{AB} Theorem~7.22.

\begin{prop} \label{pdrum202}
Let $(M_1,g_1)$ and $(M_2,g_2)$ be two  Riemannian manifolds.
Suppose there exists a
linear map $U \colon C_c^\infty(M_1) \to C^\infty(M_2)$
such that
\begin{subtabel}
\item \label{pdrum202-1}
$U \varphi \geq 0$ for all $\varphi \in C_c^\infty(M_1)$ with  $\varphi \geq 0$, and,
\item \label{pdrum202-2}
$(U \varphi) (U \psi) = 0$ for all $\varphi,\psi \in C_c^\infty(M_1)$ with  $\varphi \, \psi = 0$.
\end{subtabel}
Then there exist an open set $M_2' \subset M_2$, a function
$\tau \colon M_2 \to M_1$ and a function
$h \colon M_2 \to [0,\infty)$ such that
$M_2' = \{ q \in M_2 : h(q) > 0 \} $ and
$U \varphi = h \cdot (\varphi \circ \tau)$ pointwise for all
$\varphi \in C_c^\infty(M_1)$.
Moreover, the restrictions $\tau|_{M_2'}$ and $h|_{M_2'}$ are both $C^\infty$.
\end{prop}
\proof\
Set $M_2' = \{ q \in M_2 : (U \varphi)(q) \neq 0 \mbox{ for some } \varphi \in C_c^\infty(M_1) \} $.
Then $M_2'$ is open.
Let $q \in M_2$.
Then the map $\varphi \mapsto (U \varphi)(q)$ from $C_c^\infty(M_1)$ into
$\Ri$ is linear, positive and satisfies (\ref{eldrum201.3;1}).
Hence it follows from Lemma~\ref{ldrum201.3} that
there exist $\tau(q) \in M_1$ and $h(q) \in [0,\infty)$ such that
\[
(U \varphi)(q)
= h(q) \, \varphi(\tau(q))
\]
for all $\varphi \in C_c^\infty(M_1)$.
So one obtains functions $\tau \colon M_2 \to M_1$ and
$h \colon M_2 \to [0,\infty)$.
Moreover, $M_2' = \{ q \in M_2 : h(q) > 0 \} $.
It remains to show the smoothness of the restrictions of $\tau$ and $h$
to the set $M_2'$.

First we show that the function $\tau|_{M_2'}$ is continuous.
Otherwise there are $q,q_1,q_2,\ldots \in M_2'$
and $\varepsilon > 0$
such that $\lim q_n = q$ and $d(\tau(q_n),\tau(q)) \geq \varepsilon$ for all $n \in \Ni$,
where $d$ is a distance on $M_1$.
There exists a $\varphi \in C_c^\infty(M_1)$ such that  $\varphi(\tau(q)) = 1$ and
$\varphi(p) = 0$ for all $p \in M_1$ with $d(p,\tau(q)) > \varepsilon$.
Then $U \varphi$ is continuous, so
$h(q) = (U \varphi)(q) = \lim (U \varphi)(q_n) = 0$, which is a contradiction.

Secondly, let $\Omega$ be a relatively compact open subset of $M_2'$ with $\overline{\Omega} \subset M_2'$.
Then $\tau(\overline \Omega)$ is compact, so there is a
$\psi \in C_c^\infty(M_1)$ such that $\psi|_{\tau(\Omega)} = 1$.
Then $h|_\Omega = (U \psi)|_\Omega$ is a $C^\infty$-function.
Hence $h|_{M_2'}$ is a $C^\infty$-function from $M_2'$ into $(0,\infty)$.
Then $(\varphi \circ \tau)|_\Omega = (h^{-1} \cdot U \varphi)|_\Omega \in C^\infty(\Omega)$ for all
$\varphi \in C_c^\infty(M_1)$ and $\tau|_{M_2'}$ is a $C^\infty$-function.\hfill$\Box$

\ruimte

Using the fact that $U$ intertwines with the Laplace--Beltrami operators
implies that $\tau$ is almost an isometry.

\begin{prop} \label{pdrum202.3}
Let $(M_1,g_1)$ and $(M_2',g_2)$ be two  Riemannian manifolds with
Dirichlet Laplace--Beltrami operators $\Delta_1$ and $\Delta_2$.
Let $\tau \colon M_2' \to M_1$ be a $C^\infty$-map and
$h \colon M_2' \to (0,\infty)$ a $C^\infty$-function.
Define the map $U \colon C_c^\infty(M_1) \to C^\infty(M_2')$ by
$U \varphi = h \cdot (\varphi \circ \tau)$.
Suppose that
$\Delta_2 U = U \Delta_1$.
Then
\begin{equation}
g_1|_{\tau(q)}(\alpha,\beta) = g_2|_q(\tau^*(\alpha),\tau^*(\beta))
\label{epdrum202.3;4}
\end{equation}
for all $q \in M_2'$
and $\alpha,\beta \in T^*_{\tau(q)} M_1$.
In particular, $\dim M_2' \geq \dim M_1$.

If, in addition, $\dim M_1 = \dim M_2'$ then $\tau$ is locally an isometry and
$h$ is locally constant.
\end{prop}
\proof\
It follows from the identity $\Delta_2 U = U \Delta_1$ that
\begin{equation}
\Delta_2( h \cdot (\varphi \circ \tau))
= h \cdot ( (\Delta_1 \varphi) \circ \tau)
\label{epdrum202;2}
\end{equation}
on $M_2'$ for all $\varphi \in C_c^\infty(M_1)$.
Let $q \in M_2'$.
There exists a chart $(V,x)$ on $M_1$ such that
$\tau(q) \in V$ and $x(\tau(q)) = 0$.
Let $\Omega \subset M_1$ be a relatively compact subset such that
$\tau(q) \in \Omega \subset \overline\Omega \subset V$.
Let $d_1 = \dim M_1$ and $d_2 = \dim M_2'$.
Let $\lambda_1,\ldots,\lambda_{d_1} \in \Ri$.
For all $t > 0$ there exists a $\varphi_t  \in C_c^\infty(M_1)$
such that
\[
\varphi_t |_\Omega
= e^{t \sum_{k=1}^{d_1} \lambda_k x^k} |_\Omega
\;\;\, .  \]
Since
\[
\Delta_1
= - \sum_{i,j=1}^{d_1} \frac{1}{\sqrt{g_1}} \, \frac{\partial}{\partial x^i} \,
           g_1^{ij} \, \sqrt{g_1} \, \frac{\partial}{\partial x^j}
\]
on $V$ it follows that
\[
\Delta_1 \varphi_t
= - \sum_{i,j=1}^{d_1} t^2 g_1^{ij} \, \lambda_i \, \lambda_j \, \varphi_t
   + t \, \frac{\lambda_j}{\sqrt{g_1}} \, \varphi_t \,
     \frac{\partial}{\partial x^i} ( g_1^{ij} \, \sqrt{g_1} )
\]
on $\Omega$.
Hence
\[
\lim_{t \to \infty} t^{-2} \Big( h \cdot ( (\Delta_1 \varphi_t) \circ \tau) \Big) (q)
= - h(q) \sum_{i,j=1}^{d_1} g_1^{ij}|_{\tau(q)} \, \lambda_i \, \lambda_j
\;\;\, .  \]
Next, let $(W,y)$ be a chart on $M_2'$ such that $q \in W$.
Then it follows similarly that
\begin{eqnarray*}
\lim_{t \to \infty} t^{-2} \Big( \Delta_2( h \cdot (\varphi_t \circ \tau)) \Big) (q)
& = & - \sum_{k,l=1}^{d_2} h(q) \, g_2^{kl}|_q
   \Big( \frac{\partial}{\partial y^k} \sum_{i=1}^{d_1} \lambda_i x^i \circ \tau \Big)(q)
   \Big( \frac{\partial}{\partial y^l} \sum_{j=1}^{d_1} \lambda_j x^j \circ \tau \Big)(q)  \\
& = & - \sum_{i,j=1}^{d_1} h(q) \, g_2|_q(\tau^*(dx^i),\tau^*(dx^j)) \, \lambda_i \, \lambda_j
\;\;\, .
\end{eqnarray*}
But then (\ref{epdrum202;2}) gives
\[
\sum_{i,j=1}^{d_1} g_1^{ij}|_{\tau(q)} \, \lambda_i \, \lambda_j
= \sum_{i,j=1}^{d_1} g_2|_q(\tau^*(dx^i),\tau^*(dx^j)) \, \lambda_i \, \lambda_j
\]
for all $\lambda_1,\ldots,\lambda_{d_1} \in \Ri$ and
$g_1^{ij}|_{\tau(q)} = g_2|_q(\tau^*(dx^i),\tau^*(dx^j))$ for all
$i,j \in \{ 1,\ldots,d_1 \} $.
Hence
$g_1|_{\tau(q)}(\alpha,\beta) = g_2|_q(\tau^*(\alpha),\tau^*(\beta))$ for all
and $\alpha,\beta \in T^*_{\tau(q)} M_1$.
In particular, $\tau^*$ is injective and $d_2 \geq d_1$.

Finally suppose that $d_1 = d_2$.
Since $\tau^*|_{\tau(q)}$ is injective for all $q \in M_2'$ it follows that
$\tau^*|_{\tau(q)}$ is bijective for all $q \in M_2'$.
Hence $\tau$ is locally a $C^\infty$-diffeomorphism.
Moreover, $\tau^*|_{\tau(q)}$ is a orthogonal map and therefore also
$\tau_*|_q$ is an orthogonal map for all $q \in M_2'$.
In particular, $\tau$ is locally an isometry.

It then also follows that
$\Delta_2(\varphi \circ \tau)
= (\Delta_1 \varphi) \circ \tau$
on $M_2'$ for all $\varphi \in C_c^\infty(M_1)$.
If $q \in M_2'$ and $\Omega$ is a relatively compact open subset of $M_2'$
with $q \in \Omega \subset \overline \Omega \subset M_2'$, and if one chooses
$\varphi \in C_c^\infty(M_1)$ such that $\varphi|_{\tau(\Omega)} = 1$,
then it follows from (\ref{epdrum202;2}) that
$(\Delta_2 h)(q) = 0$.
Then for all $\varphi \in C_c^\infty(M_1)$ one deduces from (\ref{epdrum202;2}) that
\begin{eqnarray*}
h \cdot ( (\Delta_1 \varphi) \circ \tau)
& = & \Delta_2( h \cdot (\varphi \circ \tau))  \\
& = & (\Delta_2 h) \cdot (\varphi \circ \tau) + 2 (\nabla_2 h) \cdot (\nabla_2 (\varphi \circ \tau))
          + h \cdot \Delta_2(\varphi \circ \tau)  \\
& = & 2 (\nabla_2 h) \cdot (\nabla_2 (\varphi \circ \tau)) + h \cdot ( (\Delta_1 \varphi) \circ \tau)
\end{eqnarray*}
on $M_2'$.
So $(\nabla_2 h) \cdot (\nabla_2 (\varphi \circ \tau)) = 0$ on $M_2'$.
Since $\tau$ is locally a diffeomorphism it follows that $\nabla_2 h = 0$ and
$h$ is locally constant.
This completes the proof of Proposition~\ref{pdrum202.3}.\hfill$\Box$

\ruimte

In the next lemmas we consider injectivity and the density of the range of $U$.

\begin{lemma} \label{ldrum202.5}
Assume the hypothesis of Lemma~$\ref{ldrum201}$.
Moreover, assume  $U \neq 0$, the manifold $M_1$ is connected
and  the manifolds
have equal dimension.

Then there exist open sets $M_1' \subset M_1$ and $M_2' \subset M_2$, a
map $\tau \colon M_2 \to M_1$ and a bounded function $h \colon M_2 \to [0,\infty)$
such that $M_2' = \{ q \in M_2 : h(q) > 0 \} $, $M_1' = \tau(M_2')$,
$\tau|_{M_2'}$ is a local isometry, $h|_{M_2'}$ is a $C^\infty$-function and
\[
U \varphi = h \cdot (\varphi \circ \tau)
\]
pointwise for all $\varphi \in C_c^\infty(M_1)$ and a.e.\ for all $\varphi \in L_p(M_1)$.
Moreover, $U$ is injective and $|M_1 \setminus M_1'| = 0$.
\end{lemma}
\proof\
Since $C_c^\infty(M_1)$ is dense in $L_p(M_1)$ and
$U \neq 0$ it follows that the restriction of $U$
to $C_c^\infty(M_1)$ does not vanish.
So we can apply Lemma~\ref{ldrum201} and Proposition~\ref{pdrum202}.
We use the notation of Proposition~\ref{pdrum202}.
Set $M_1' = \tau(M_2')$.
It follows from the inverse function theorem that $M_1'$ is
open in $M_1$.
Moreover, since $M_2' \neq \emptyset$ it follows that
$M_1' \neq \emptyset$ and $|M_1'| \neq 0$.

Now let $\varphi \in L_p(M_1)$.
Since $C_c^\infty(M_1)$ is dense in $L_p(M_1)$
there exists a sequence $\varphi_1,\varphi_2,\ldots \in C_c^\infty(M_1)$
such that $\lim \varphi_n = \varphi$ in $L_p(M_1)$.
Since $U$ is continuous it follows that $\lim U \varphi_n = U \varphi$
in $L_p(M_2)$.
Passing to subsequences, if necessary, we may assume that
$\lim \varphi_n = \varphi$ a.e.\ and
$\lim U \varphi_n = U \varphi$ a.e.
But since $M_1'$ is $\sigma$-compact and $\tau|_{M_2'}$ is locally an
isometry it follows that $\tau^{-1}(N)$ is a null-set in $M_2'$
for every null-set $N$ in $M_1'$.
Therefore $\lim \varphi_n \circ \tau = \varphi \circ \tau$ a.e.\ on $M_2'$.
Then
\[
U \varphi
= \lim U \varphi_n
= \lim h \cdot (\varphi_n \circ \tau)
= h \cdot (\varphi \circ \tau)
\]
a.e.\ on $M_2'$.
In addition one has $U \varphi = \lim U \varphi_n = 0$ a.e.\ on $M_2 \setminus M_2'$.
So $U\varphi=h\cdot (\varphi \circ\tau)$ a.e.\ on $M_2$.

Next we show that $U$ is injective.
Let $\varphi \in L_2(M_1)$ and suppose that $U \varphi = 0$.
Then $U |\varphi| = |U \varphi| = 0$, so we may assume that $\varphi \geq 0$.
Fix $t > 0$.
Then
\[
0
= S^{(2)}_t U \varphi
= U S^{(1)}_t \varphi
= h \cdot (S^{(1)}_t \varphi) \circ \tau
\]
a.e.\ on $M_2'$.
Since also $\tau$ maps $M_2'$-null-sets into $M_1'$-null-sets
and $h > 0$ pointwise, one deduces
that $S^{(1)}_t \varphi = 0$ a.e.\ on $M_1'$.
But $M_1$ is connected and if $\varphi \neq 0$ then $(S^{(1)}_t \varphi)(p) > 0$
for all $p \in M_1$ and in particular for all $p \in M_1'$.
So $\varphi = 0$ and $U$ is injective.
It is obvious that this implies that $|M_1 \setminus M_1'| = 0$.

Finally we show that $h$ is bounded by $\|U\|$.
Let $q \in M_2'$ and $\varepsilon > 0$.
Since $\tau|_{M_2'}$ is locally an isometry and $h|_{M_2'}$ is continuous there exists an open
neighbourhood $\Omega$ of $q$ in $M_2'$ such that $\tau|_\Omega \colon \Omega \to \tau(\Omega)$
is an isometry and $h|_\Omega \geq (1 - \varepsilon) h(q)$.
Fix $\varphi \in C_c^\infty(\tau(\Omega))$ with $\varphi \neq 0$.
Then
\begin{eqnarray*}
(1 - \varepsilon) \, h(q) \|\varphi \circ \tau\|_{L_p(\Omega)}
& \leq & \|h \cdot (\varphi \circ \tau)\|_{L_p(\Omega)}
\leq \|U \varphi\|_{L_p(M_2)}  \\
& \leq & \|U\| \, \|\varphi\|_{L_p(M_1)}
\leq \|U\| \, \|\varphi\|_{L_p(\tau(\Omega))}
= \|U\| \, \|\varphi \circ \tau\|_{L_p(\Omega)}
\end{eqnarray*}
where we used (\ref{eSdrum1;1}) in the last step.
So $h(q) \leq \|U\|$.\hfill$\Box$

\begin{lemma} \label{ldrum202.6}
Assume the hypothesis of Lemma~$\ref{ldrum201}$.
Let $M_2'$ be an open subset of $M_2$,
let $\tau \colon M_2 \to M_1$ and
$h \colon M_2 \to [0,\infty)$ be two functions such that
$M_2' = \{ q \in M_2 : h(q)> 0\}$ and
the restrictions $\tau|_{M_2'}$ and $h|_{M_2'}$ are both $C^\infty$-maps.
Suppose that $U \varphi = h \cdot (\varphi \circ \tau)$ for all
$\varphi \in C_c^\infty(M_1)$.

Then the following are equivalent.
\begin{tabel}
\item \label{ldrum202.6-1}
$U L_p(M_1)$ is dense in $L_p(M_2)$.
\item \label{ldrum202.6-2}
$U C_c^\infty(M_1)$ is dense in $L_p(M_2)$.
\item \label{ldrum202.6-3}
For every pair of disjoint measurable
subsets $A_1$ and $A_2$ in $M_2$  with $0 < |A_1|, |A_2| < \infty$
the functionals
\[
\varphi \mapsto \int_{A_1} U \varphi
\;\;\;\; \mbox{and} \;\;\;\;
\varphi \mapsto \int_{A_2} U \varphi
\]
from $C_c^\infty(M_1)$ into $\Ri$ are linearly independent.
\item \label{ldrum202.6-4}
The map $\tau|_{M_2'}$ is injective  and $|M_2 \setminus M_2'| = 0$.
\end{tabel}
Moreover, these conditions imply that the dimensions of $M_1$ and $M_2$ are equal.
\end{lemma}
\proof\
Clearly \ref{ldrum202.6-1}$\Leftrightarrow$\ref{ldrum202.6-2}$\Rightarrow$\ref{ldrum202.6-3}.

Next we show that \ref{ldrum202.6-3} or \ref{ldrum202.6-4}
implies that $\dim M_1 = \dim M_2$.
Obviously $U \neq 0$, so $M_2' \neq 0$ and $d_2 \geq d_1$ by Proposition~\ref{pdrum202.3},
where $d_1 = \dim M_1$ and $d_2 = \dim M_2$.
Fix $q \in M_2'$ and set $p = \tau(q)$.
Let $(V,x)$ be a chart on $M_1$ with $p \in V$ and $(W,y)$ a chart on $M_2'$
with $q \in W$.
Let $V' = x(V)$, $W' = y(W)$, $p' = x(p)$ and $q' = y(q)$.
Define the $C^\infty$-map $f \colon W' \to V'$ by
$f = x \circ \tau \circ y^{-1}$.
Then it follows from (\ref{epdrum202.3;4}) that
$(Df)(q')$ has maximal rank.
Suppose that $k = d_2 - d_1 > 0$.
Then it follows from the inverse function theorem that
there exist open $W'' \subset \Ri^{d_2}$, $\delta > 0$ and
a $C^\infty$-diffeomorphism $F \colon W'' \to B(p'\,;3 \delta) \times (-3\delta,3\delta)^k$
such that $q \in W'' \subset W'$,
$B(p'\,;3\delta) \subset V'$, $F(q') = (p',0)$ and $f(G(u,v)) = u$
for all $(u,v) \in B(p'\,;3 \delta) \times (-3\delta,3\delta)^k$, where $G = F^{-1}$.

In particular, $f$ is not injective.
This contradicts the injectivity of $\tau$ in \ref{ldrum202.6-4}.

In order to obtain a contradiction with Condition~\ref{ldrum202.6-3}
we proceed as follows.
Define the $C^\infty$-function $\Phi \colon B(p'\,;3 \delta) \times (-3\delta,3\delta)^k \to (0,\infty)$
by $\Phi(u,v) = ((h \, \sqrt{g_2}) \circ y^{-1} \circ G)(u,v) \, |(JG)(u,v)|$,
where $JG$ denotes the Jacobian determinant of $G$.
If $A \subset y^{-1}(W'')$ is measurable then
\begin{equation}
\int_A U \varphi
= \int_{(F \circ y)(A)} \Phi(u,v) \, (\varphi \circ x^{-1})(u) \, d(u,v)
\label{eldrum202.6;4}
\end{equation}
for all $\varphi \in C_c^(M_1)$.
By compactness, there are $m,M > 0$ such that
$m \leq \Phi(u,v) \leq M$ for all $(u,v) \in \overline{B(p'\,;2\delta)} \times [-2\delta,2\delta]^k$.
Let $u \in \overline{B(p'\,;2\delta)}$.
Then
\[
\int_{[-2\delta,2\delta]^{k-1} \times [-\delta \, m \, M^{-1},0)} \Phi(u,v) \, dv
\leq (4\delta)^{k-1} \, \delta \, m
\leq \int_{[-2\delta,2\delta]^{k-1} \times (0,\delta]} \Phi(u,v) \, dv
\;\;\; .  \]
So there exists a $\beta(u) \in (0,\delta]$ such that
\[
\int_{[-2\delta,2\delta]^{k-1} \times [-\delta \, m \, M^{-1},0)} \Phi(u,v) \, dv
= \int_{[-2\delta,2\delta]^{k-1} \times (0,\beta(u)]} \Phi(u,v) \, dv
\;\;\; .  \]
Now choose
\[
A_1 = (y^{-1} \circ F^{-1}) \Big( \overline{B(p',;2\delta)} \times [-2\delta,2\delta]^{k-1} \times [-\delta \, m \, M^{-1},0) \Big)
\]
and
\[
A_2 = \{ (y^{-1} \circ F^{-1})(u,v) : u \in \overline{B(p'\,;2\delta)} \mbox{ and }
                 v \in [-2\delta,2\delta]^{k-1} \times (0,\beta(u)] \}
\;\;\; .  \]
Then $A_1 \cap A_2 = \emptyset$, $0 < |A_1|,|A_2| < \infty$ and
$\int_{A_1} U \varphi = \int_{A_1} U \varphi$ for all $\varphi \in C_c^\infty(M_1)$ by
(\ref{eldrum202.6;4}) and the choice of $\beta(u)$.
This contradicts \ref{ldrum202.6-3}.
So $d_1 = d_2$.
Thus all four conditions imply that the dimensions of $M_1$ and $M_2$ are equal.

`\ref{ldrum202.6-3}$\Rightarrow$\ref{ldrum202.6-4}'.
Clearly \ref{ldrum202.6-3} implies that $|M_2 \setminus M_2'| = 0$.
Suppose \ref{ldrum202.6-3} and $\tau$ is not injective.
Then there are $q_1,q_2 \in M_2'$ such that $\tau(q_1) = \tau(q_2)$ and $q_1 \neq q_2$.
Since $\dim M_1 = \dim M_2$ it follows from Proposition~\ref{pdrum202.3} that
$\tau|_{M_2'}$ is locally an isomorphism.
There are open connected relative compact $\Omega_1, \Omega_2 \subset M_2'$ such that
$\Omega_1 \cap \Omega_2 = \emptyset$ and for all $j \in \{ 1,2 \} $ one has
$q_j \in \Omega_j$ and $\tau_j = \tau|_{\Omega_j} \colon \Omega_j \to \tau(\Omega_j)$ is an
isometry.
Since $\Omega_j$ is connected there is a $c_j \in (0,\infty)$ such that
$h|_{\Omega_j} = c_j$.
Without loss of generality we may assume that $\tau_1(\Omega_1) = \tau_2(\Omega_2)$.
Then for all $\varphi \in C_c^\infty(M_1)$ one has
$(U \varphi)(q) = h(q) \, \varphi(\tau_1(q)) = c_1 \, \varphi(\tau_1(q))$
for all $q \in \Omega_1$.
So
\[
\int_{\Omega_1} U \varphi
= c_1 \int_{\Omega_1} \varphi \circ \tau_1
= c_1 \int_{\tau_1(\Omega_1)} \varphi
\]
where we used (\ref{eSdrum1;1}).
Similarly
\[
\int_{\Omega_1} U \varphi
= c_2 \int_{\tau_2(\Omega_2)} \varphi
\]
for all $\varphi \in C_c^\infty(M_1)$.
But $\tau_1(\Omega_1) = \tau_2(\Omega_2)$.
This contradicts the independence of the functionals.

`\ref{ldrum202.6-4}$\Rightarrow$\ref{ldrum202.6-2}'.
Since $|M_2 \setminus M_2'| = 0$ the space $C_c^\infty(M_2')$ is dense in $L_p(M_2)$.
Therefore it suffices to show that $C_c^\infty(M_2') \subset U C_c^\infty(M_1')$.

Using again that $\dim M_1 = \dim M_2$ it follows from Proposition~\ref{pdrum202.3} that
$\tau|_{M_2'}$ is locally an isomorphism and $h_{M_2'}$ is locally constant.
If $\psi \in C_c^\infty(M_2')$ and there exists an open connected set $\Omega$ in
$M_2'$ such that $\supp \psi \subset \Omega$, then $h$ is constant on $\Omega$, say $c$,
and $\varphi = c^{-1} \psi \circ (\tau|_{M_2'})^{-1} \in C_c^\infty(M_1') \subset C_c^\infty(M_1)$
satisfies $U \varphi = \psi$.
Then the general case follows by a partition of the unity.\hfill$\Box$

\ruimte

For open subsets in $\Ri^d$ the surjectivity of $U$ follows from the fact that $U \neq 0$
(see \cite{Are4}, Theorem~2.1).
In general the condition $U \neq 0$ is not sufficient to
establish the surjectivity of $U$.

\begin{exam} \label{xdrum501}
Let $S^1 = \{ z \in \Ci : |z| = 1 \} $.
Let $g_1$ be the Riemannian metric on $M_1 = S^1$ such that
$(g_1)|_{e^{i \theta}}(\frac{\partial}{\partial x^1} |_{e^{i \theta}}, \frac{\partial}{\partial x^1}|_{e^{i \theta}}) = 1$
for each $\theta \in \Ri$, where $(V,x)$ is a chart on $S^1$ such that $V$ is an open neighbourhood
of $e^{i \theta}$, $\theta \in x(V)$ and $x^{-1}(\xi) = e^{i \xi}$ for all $\xi \in x(V)$.
Set $M_2 = S_1$ and choose the Riemannian metric $g_2$ on $M_2$
by $g_2 = 4 g_1$.

Define $U \colon L_2(M_1) \to L_2(M_2)$ by
\[
(U \varphi)(z) = \varphi(z^2)
\;\;\; .  \]
Then $U$ is a lattice homomorphism, $U \neq 0$ and
$U S^{(1)}_t = S^{(2)}_t U$ for all $t > 0$, where
$S^{(j)}$ is the semigroup
on $L_2(M_j)$ generated by the Dirichlet Laplace--Beltrami operator on $M_j$
for all $j \in \{ 1,2 \} $.
Moreover, $M_1$ and $M_2$ are regular in capacity.
But the Riemannian manifolds $(M_1,g_1)$ and $(M_2,g_2)$ are
not isomorphic.
\end{exam}

We combine the previous results.

\begin{prop} \label{pdrum203}
Let $(M_1,g_1)$ and $(M_2,g_2)$ be two connected Riemannian manifolds.
Let $p \in [1,\infty)$.
For all $j \in \{ 1,2 \} $ let $\Delta_j$ be the Dirichlet Laplace--Beltrami operator
on $M_j$ and let $S^{(j)}$ be the associated semigroup on $L_p(M_j)$.
Let  $U \colon L_p(M_1) \to L_p(M_2)$ be a lattice homomorphism such that
$U L_p(M_1)$ is dense in $L_p(M_2)$ and
\begin{equation}
U S^{(1)}_t = S^{(2)}_t U
\label{epdrum203;1}
\end{equation}
for all $t > 0$.

Then $U$ is an order isomorphism
and there exist open connected sets $M_1' \subset M_1$ and $M_2' \subset M_2$, a
map $\tau \colon M_2 \to M_1$ and a constant $c > 0$ such that $M_1' = \tau(M_2')$,
$\tau|_{M_2'} \colon M_2' \to M_1'$ is an isometry, and
\[
U \varphi = c \, \one_{M_2'} \cdot (\varphi \circ \tau)
\]
pointwise for all $\varphi \in C_c^\infty(M_1)$ and a.e.\ for all $\varphi \in L_p(M_1)$.
Moreover,
$\capp(M_1 \setminus M_1') = \capp(M_2 \setminus M_2') = 0$
and for all $\tilde p \in [1,\infty)$ there exists an order isomorphism $\widetilde U$ such that
such that
$\widetilde U \varphi = U \varphi$ for all $\varphi \in L_p(M_1) \cap L_{\tilde p}(M_1)$.
Finally, if $p = 2$ then $U$ maps $H^1_0(M_1)$ continuously into $H^1_0(M_2)$.
\end{prop}
\proof\
We use the notation as above.
Let $\tilde p \in [1,\infty)$.
Then the map
$\widetilde U \colon L_{\tilde p}(M_1) \to L_{\tilde p}(M_2)$
defined by $\widetilde U \varphi = h \cdot (\varphi \circ \tau)$ (a.e.) for all
$\varphi \in L_{\tilde p}(M_1)$ is well defined
since $h$ is bounded and $\tau^{-1}(N)$ is a null-set in $M_2'$ for every null-set $N$
in $M_1'$.
It is a lattice homomorphism and is consistent with $U$.
Moreover, $U L_{\tilde p}(M_1)$ is dense in $L_{\tilde p}(M_2)$ by
Proposition~\ref{ldrum202.6} \ref{ldrum202.6-4}$\Rightarrow$\ref{ldrum202.6-1}.

Therefore, for the remainder of the proof we may assume that $p = 2$.
Then one deduces from (\ref{epdrum203;1}) that
$(I + \Delta_2)^{-1/2} U = U (I + \Delta_1)^{-1/2}$.
Hence
\begin{eqnarray*}
U H^1_0(M_1)
& = & U (I + \Delta_1)^{-1/2} L_2(M_1)  \\
& = & (I + \Delta_2)^{-1/2} U L_2(M_1)
\subset (I + \Delta_2)^{-1/2} L_2(M_2)
= H^1_0(M_2)
\;\;\; .
\end{eqnarray*}
Then by the closed graph theorem the restriction of $U$
to $H^1_0(M_1)$ is a continuous map from $H^1_0(M_1)$ into $H^1_0(M_2)$.
Next $U L_2(M_1)$ is dense in $L_2(M_2)$ and
$(I + \Delta_2)^{-1/2}$ is continuous from $L_2(M_2)$ onto $H^1_0(M_2)$.
So $U H^1_0(M_1)$ is dense in $H^1_0(M_2)$.
Therefore $U C_c^\infty(M_1)$ is dense in $H^1_0(M_2)$.

Now suppose $\capp(M_2 \setminus M_2') > 0$.
There exist compact subsets $K_1 \subset K_2 \subset \ldots$ of $M_2$
such that $M_2 = \bigcup_{n=1}^\infty K_n$.
Then $\capp(M_2 \setminus M_2') = \lim_{n \to \infty} \capp(K_n \setminus M_2')$
by \cite{BH} Proposition~8.1.3c.
Hence there exists an $n \in \Ni$ such that $\capp(K_n \setminus M_2') > 0$.
Let $\psi \in C_c^\infty(M_2)$ be such that
$\psi|_{K_n} = 1$.
Since $U C_c^\infty(M_1)$ is dense in $H^1_0(M_2)$ there exists a
$\varphi \in C_c^\infty(M_1)$ such that
$\|\psi - U \varphi\|_{H^1(M_2)}^2 < \capp(K_n \setminus M_2')$.
Then $\psi - U \varphi = 1$ on $K_n \setminus M_2'$ by definition of $M_2'$.
Moreover, $\psi - U \varphi$ is continuous and
$\psi - U \varphi \in H^1_0(M_2) \subset H^1(M_2)$.
Hence $\capp(K_n \setminus M_2') \leq \|\psi - U \varphi\|_{H^1(M_2)}^2$.
This is a contradiction.
Hence $\capp(M_2 \setminus M_2') = 0$.

This allows to apply Theorem~\ref{tdrum302} to deduce that $M_2'$ is connected.
Then $h|_{M_2'}$ is constant, say $c > 0$.
It follows from (\ref{eSdrum1;1}) that (for any $p \in [1,\infty)$) the map
$U$ is an isometry between $L_p$-spaces and since the range is dense, it is surjective.

Finally, $H^1_0(M_2') = \{ \psi|_{M_2'} : \psi \in H^1_0(M_2) \} $
by Lemma~\ref{ldrum310}, since $\capp(M_2 \setminus M_2') = 0$.
But $H^1_0(M_2) = U H^1_0(M_1)$.
In addition, $U H^1_0(M_1') = H^1_0(M_2')$ since
$\tau|_{M_2'} \colon M_2' \to M_1'$ is an isometry.
Therefore
\[
U H^1_0(M_1')
= \{ (U \varphi)|_{M_2'} : \varphi \in H^1_0(M_1) \}
= \{ U (\varphi|_{M_1'}) : \varphi \in H^1_0(M_1) \}
\;\;\; .  \]
Hence $H^1_0(M_1') = \{ \varphi|_{M_1'} : \varphi \in H^1_0(M_1) \} $.
Using Lemma~\ref{ldrum310} again one deduces that
$\capp(M_1 \setminus M_1') = 0$.\hfill$\Box$

\ruimte

Now we are able to prove the main theorem of this paper.

\ruimte

\noindent
{\bf Proof of Theorem~\ref{tdrum101} \hspace{5pt}}\
The implication \ref{tdrum101-1}$\Rightarrow$\ref{tdrum101-3} is trivial.
But if Condition~\ref{tdrum101-3} is valid then Proposition~\ref{pdrum203}
implies that $M_1 \simh M_2$.
Hence the Riemannian manifolds $(M_1,g_1)$ and $(M_2,g_2)$ are isomorphic by Theorem~\ref{tdrum324}.
Moreover, there exist $c > 0$ and a isometry $\tau \colon M_2 \to M_1$ such that
$U \varphi = c \, \varphi \circ \tau$ for all $\varphi \in L_p(M_1)$.
\hfill$\Box$

\ruimte

In fact, it follows from the proof that under
Condition~\ref{tdrum101-3} it follows that $M_2' = M_2$ and $M_1' = M_1$.
Therefore $U \varphi = c \, \varphi \circ \tau$ for all $\varphi \in L_p(M_1)$.


text:
\section*{Appendix} \label{sec:appendix}
Here, we prove Lemmas~\ref{lem:1}-\ref{lem:ordering}, and Propositions~\ref{prop:1} \& \ref{prop:2}.

\textbf{Lemma~\ref{lem:1}.}
\begin{proof}
  We can compute:
  \begin{align}
    \T_\tau^c &= \left(\D_\tau ^T \D_\tau\right)^{-1} \D_\tau^T\D_c\\

              &=\left( \left( D_\tau^TD_\tau \right)^{-1} D_\tau D_c\right) \otimes I_k
              = T_\tau^C\otimes I_k,
  \end{align}
  and similarly,
  $\D = D\otimes I_k = \left(D_\tau \otimes I_k \right)\left( R\otimes I_k \right) = \D_\tau \mathbf{R}$.

\end{proof}

\textbf{Lemma~\ref{lem:2}.}
\begin{proof}
      Denote the product $S_v^{-1}S_v$ in block form:
      \begin{align}
        S_v^{-1}S_v =
        \begin{bmatrix}
          \mathcal{A}&\mathcal{B}\\\mathcal{C}&\mathcal{D}
        \end{bmatrix}.
      \end{align}
      First, note that
      \begin{align}
        F\E^{-1} =
        \begin{bmatrix}
          E_1&\cdots& E_n
        \end{bmatrix} \E^{-1} =
                      \begin{bmatrix}
                        I&\cdots & I
                      \end{bmatrix}= \mathbf{1}^T_n \otimes I_k.
      \end{align}
      Then, we can compute each term:
      \begin{align}
        \mathcal{A}&= \D_\tau ^T \E^{-1} \D_\tau \left(D_\tau^T \E^{-1} \D_\tau\right)^{-1} = I\\
        \mathcal{B}&= \D_\tau^T \mathbb{1}

         = \left(D_\tau^T\mathbf{1}_n \right) \otimes I_k = 0 \otimes I_k = \mathbf{0}.\\
        \mathcal{C}&= \Xi^{-1} F \E^{-1} \D_\tau \left(\D_\tau^T \E^{-1} \D_\tau\right)^{-1}\\

         &= \Xi^{-1} \left(\mathbf{1}^T_n \D_\tau \otimes I_k\right) \left(\D_\tau^T \E^{-1} \D_\tau\right)^{-1}= \mathbf{0}.
      \end{align}
      Lastly, we have
      \begin{align}
        \mathcal{D} &= \Xi^{-1} F \mathbb{1}\\
          &=
            \begin{bmatrix}
              \epsilon_{s,1}^{-1} & & \\
              & \ddots & \\
              & & \epsilon_{s,k}^{-1}
            \end{bmatrix}
                  \begin{bmatrix}
                    E_1 & \cdots & E_n
                  \end{bmatrix} \left( \mathbf{1}_n \otimes I_k\right)\\
          &=
            \begin{bmatrix}
              \epsilon_{s,1}^{-1} & & \\
              & \ddots & \\
              & & \epsilon_{s,k}^{-1}
            \end{bmatrix}
                  \begin{bmatrix}
                    \sum_{j=1}^n\epsilon_{j,1} & & \\
              & \ddots & \\
              & & \sum_{j=1}^n\epsilon_{j,k}
            \end{bmatrix}

= I_k.
      \end{align}
    \end{proof}

\textbf{Lemma~\ref{lem:ordering}.}
    \begin{proof}
For simplicity, we adopt the following notation the state and input matrix of~\eqref{eq:treemodel}, $A := -\lap_{e,s} \R \W \R^T$, and $B := B_{\tau} \Omega \Omega^T B_{\tau}^T + B_c \Gamma \Gamma^T B_c^T$. Due to the stability of $\lap(\graph)$, we know the solution to~\eqref{eq:lyapunov} is positive definite, and can be written as in integral form~\cite{skogestad2001},

\[
X = \int_0^\infty e^{A t} B_{\tau} \Omega \Omega^T B_{\tau}^T e^{A^T t} \diff  + \int_0^\infty e^{A t} B_c \Gamma \Gamma^T B_c^T e^{A^T t} \diff
\]

Consider either term in the above solution, and generalize as,
\[
X_p = \int_0^\infty e^{A t} B_p Z Z^T B_p^T e^{A^T t} \diff
\]
\noindent where $Z$ and $B_p$ are placeholders for a covariance matrix and input matrix, respectively. From this, consider the quadratic form with any $u$,
\[
u^T X_p u = \int_0^\infty \| Z^T B_p^T e^{A^T t} u \|_2^2 \diff
\]
\noindent This form is usually employed to show the positive definiteness of $X_p$ based on the stability of $A$, but here it can be used to order $X_p$'s based on the ordering of $Z$:
\begin{align*}
u^T(X_p - \bar{X}_p)u & = \int_0^\infty \| Z^T B_p^T e^{A^T t} u \|_2^2 - \| \bar{Z}^T B_p^T e^{A^T t} u \|_2^2 \diff  \\
& \leq \int_0^\infty  \left(\|Z^T\|_2^2 - \|\bar{Z}^T\|_2^2\right) \| B_p^T e^{A^T t} u \|_2^2 \diff
\end{align*}
\noindent From $Z \preceq \bar{Z}$, it follows that $\|Z\|_2^2 \leq\|\bar{Z}\|_2^2$. Thus, this integral is always negative, which then implies $X_p - \bar{X}_p \preceq 0 \Rightarrow X_P \succeq \bar{X}_p$. By identical argument,
\[
u^T(\underline{X}_p - X_p)u \leq  \int_0^\infty  \left(\|\underline{Z}^T\|_2^2 - \|Z^T\|_2^2\right) \| B_p^T e^{A^T t} u \|_2^2 \diff
\]
\noindent implies that $\underline{X}_p \preceq X$. This shows the ordering for the solutions of~\eqref{eq:lyapunov} for ordered covariances.

Now, consider the ordering of the performance, which is given by $\mathbf{tr}(R X R^T)$. From the ordering and positive definiteness of $\underline{X},X,\bar{X}$, we have, $u^T \underline{X} u \leq u^T X u \leq u^T \bar{X} u$, for all $u$. Letting $u = R x$ gives,
\[
x^T R^T \underline{X} R x \leq x^T R^T X R x \leq x^T R^T \bar{X} R x.
\]
Now, note that the trace of a matrix $M$ can be written as $\sum_i e_i^T M e_i$. Taking $M = R^T X R$ and $x = e_i$ in the above equation gives the desired ordering on the performances.

\end{proof}

\textbf{Proposition~\ref{prop:2}.}
\begin{proof}

We first identify the gradient of the cost function with respect to $W_c$, the weight on the $c$th edge in $\Edges$.
To this end, consider the functions
\begin{align}
  f(X) &= \R \left[ \mathbf{e}_c X \mathbf{e}_c^T + \sum_{l \in \Edges\setminus c} \mathbf{e}_l  W_l \mathbf{e}_l^T \right] \R^T\\
       &= \R \left[ \mathrm{blk}_c^k (X) + \sum_{l\in\Edges\setminus c} \mathrm{blk}_c^k(W_l) \right] \R^T\\
  g(X) &= X^{-1},~ h(X) = \mathbf{tr}\left[ \R^T X \R\right],
\end{align}
where
\begin{align}
  \mathrm{blk}_l^k(W_l) =  \mathbf{e}_l  W_l \mathbf{e}_l^T = \left( e_l\otimes I_k\right) W_l \left( e_l^T \otimes I_k\right)
\end{align}
denotes the $kn\times kn$ matrix with $W_l$ on the $l$th $k\times k$ diagonal block, with zeros otherwise.
Then, the cost function with $W_c$ as the argument, is given by
\begin{align}
  \mathrm{tr}\left[ \R^T \left(\R\W\R^T\right)^{-1} \R\right] = \left(h\circ g \circ f\right)(W_c).\label{eq:1}
\end{align}
These functions have differentials
\begin{align}
  df_X[H] &= \R \left[ \mathrm{blk}_c^k(H) \right] \R^T,~  dg_X[H] = - X^{-1} H X^{-1}\\
  dh_X[H] &= \mathrm{tr}\left[ \R^TH\R \right].
\end{align}
By the chain rule, we have that
\begin{align}
  d(g\circ f)_X[H] &= - Y^{-1} \R \mathrm{blk}_c^k (H) \R^T Y^{-1}\\
  Y \triangleq f[H] &= \R \left[ \mathrm{blk}_c^k (H) + \sum_{l\in\Edges\setminus c} \mathrm{blk}_l^k(W_l) \right] \R^T\\
                   &\triangleq \R \W_H^c \R^T,
\end{align}
and so
\begin{align}
      \resizebox{\columnwidth}{!}
    {$
  d(g\circ f)_X[H] = - \left(\R \W_H^c \R^T\right)^{-1} \R \mathrm{blk}_c^k (H) \R^T \left(\R \W_H^c \R^T\right)^{-1}
    $}\\\vspace{-0.5cm}
        \resizebox{\columnwidth}{!}
    {$
  d(h\circ g\circ f)_X[H]= -\mathrm{tr} \left[Q \mathrm{blk}_k^c(H) Q^T \right],~
  Q^T \triangleq \R^T\left( \R \W_H^c \R^T \right)^{-1} \R.
      $}
\end{align}
Hence, we can write
\begin{align}
   &d(h\circ g\circ f)_X[H]= -\mathrm{tr} \left[Q \mathrm{blk}_k^c(H) Q^T \right]\\
                          &= -\mathrm{tr} \left[Q \mathbf{e}_c  H \mathbf{e}_c^T Q^T \right]

                          = \left \langle -\mathbf{e}_c^T Q^TQ \mathbf{e}_c, H \right \rangle,
\end{align}
and so the gradient of \eqref{eq:1} with respect to the $c$th weight $W_c$ is identified as the $c$th $k\times k$ diagonal block of $-Q^TQ$.

\end{proof}

\textbf{Proposition~\ref{prop:2}.}
\begin{proof}
Consider the cost function (denoted by $f(\epsilon_{i,j}^{-1})$) without the box constraint, and note that the $\Htwo$ portion can be rewritten as a double sum of the same form as the regularization term,

\[
\mathbf{tr}\left(\R^T \lap_{e,s}^\tau  \R\right) = \mathbf{tr}\left(\E^{-1} \lap \right) = \sum_{i=1}^n \sum_{j=1}^k \epsilon_{i,j}^{-1} \text{deg}(\nu_i),
\]

resulting in,

\[
f(\epsilon_{i,j}^{-1}) = \frac{1}{2} \sum_{i=1}^n \sum_{j=1}^k \epsilon_{i,j}^{-1} \text{deg}(\nu_i) + \frac{h}{2}\sum_{i=1}^n \sum_{j=1}^k (\epsilon_{i,j}^{-1})^{-r}.
\]

Minimizing this cost alone can be achieved by setting its gradient equal to zero,
\begin{align*}
\frac{\partial f}{\partial \epsilon_{i,j}^{-1}} & = \frac{\text{deg}(\nu_i)}{2} - \frac{h r}{2} (\epsilon_{i,j}^{-1})^{-(r+1)}=0,
\end{align*}
implying that,
$\epsilon_i^\ast  = ({\text{deg}(\nu_i)}/{h r})^{{1}/{r+1}}.$
Projecting this result onto the constraint set gives the result.
\end{proof}


TITLE: What is the complex partial fraction decomposition of $\frac{1}{z^2 - z + 1}$
QUESTION [1 upvotes]: I need this result to compute a residue. I haven't been successful so far.
What I have tried:
I have tried decomposing $\frac{1}{z^2 - z + 1} = \frac{A + Bi}{z - \omega} + \frac{C + Di}{z + \omega}$ where $\omega$ is the cube root of unity. I didn't get anything from this method.

REPLY [0 votes]: An alternate starting point:
$$
1-z+z^2 = \frac{z^3+1}{z+1}.
$$


TITLE: A question about behaviour of prime numbers in totally real field..
QUESTION [0 upvotes]: Let $F$ be a number field with class number 1. Consider its ring of integers $O_F$. Let $p$ ba a prime number. Is there a totally real field $F$ such that for each prime number $p$, $pO_F$ is also a prime ideal of $O_F$?
We know that this is not true if $F$ is not totally real. Take $F= \mathbb{Q}(i)$. $O_F=\mathbb{Z}[i]$. Take $p=2$. Then $2O_F = (1-i)^2$ and hance 2 is not prime in $O_F=\mathbb{Z}[i]$.
But what if $F$ is totally real??
Any help would be appreciated.

REPLY [2 votes]: For a number field $F$,

The primitive element theorem gives $F=\Bbb{Q}(a)$, we can take $a$ to be an algebraic integer with monic minimal polynomial $f\in \Bbb{Z}[x]$, if $p \nmid Disc(f)$ and $p | f(n)$ then $f$ isn't irreducible in $\Bbb{F}_p[x]$ thus $p$ isn't a prime ideal of $\Bbb{Z}[x]/(f(x))$ thus it is not a prime ideal of $O_F$. Whence there are infinitely many non-inert primes.
If finitely many primes of $\Bbb{Q}$ split completely in $F$ then $$\zeta_F(s) = \prod_p\prod_{P \ prime \ \in O_F,P \ni p}\frac1{1-N(P)^{-s}}$$
converges for $\Re(s) > 1/2$.

It is known to be untrue because for $F$ totally real and $s > 1$

$$\Gamma(s)\zeta_F(s)\ge \Gamma(s)\sum_{a\in (O_F-0)/O_F^\times} N(a)^{-s}=\int_{\Bbb{R}_{>0}^n / U} f(x)(\prod_{j=1}^n x_j)^{s-1}dx_j$$ where  $n=[F:\Bbb{Q}]$ $$f(x_1,\ldots,x_n)=\sum_{a\in O_F} \exp(-\sum_{j=1}^n x_j |\sigma_j(a)| x_j)$$ and $$U = \{ (|\sigma_1(u)|,\ldots,|\sigma_n(u)|,u\in O_F^\times\}$$
A more careful analysis shows that $\zeta_F(s)$ has a simple pole at $s=1$ which means that a positive density of the primes split completely in $F$.


TITLE: Counting Problems with n-Bit Strings
QUESTION [0 upvotes]: I'm a little confused with problems regarding n-bit strings- for instance if we are considering the number of n-bit strings of size n with a 1001 pattern, why is it alright to consider only adding things onto one side when generating recurrence relations? Is this an obvious fact, or is it something that generally confuses people? I think I understand that if we consider adding to one side, adding to the other side should be handled when we talk about the number of n-1 bit strings, but it still feels like a leap of faith for me.
For instance, the number of n bit strings with 1001 would be something like the number of n-1 bit strings with 1001 patterns, times 2 by adding a 1 or 0 on the end, plus some other things, but that already confuses me. Initially, I feel as if adding on a 1 or a 0 should be consider for both sides, so I feel like I would want to multiply by 4. Is there a better way of understanding this?

REPLY [0 votes]: You're asking about converse. We should check that both directions work out: going from smaller strings to larger strings, and from larger string to smaller strings.  It sometimes goes without mention from these proofs.
The simplest non-trivial case is $n$-bit strings without $11$ (so let's use this as an example).  Other cases are similar in flavor, but the method of counting might change.

One direction: Given an $n$-bit binary string without $11$:

If it ends with $1$, then it ends in $01$, so we can delete the last two bits to obtain a unique $(n-2)$-bit binary string without $11$.
If it ends with $0$, then we can delete the last bit to obtain a unique $(n-1)$-bit binary string without $11$.

The other direction:

Given a $(n-2)$-bit binary string without $11$, we can append $01$ to the end to get a unique $n$-bit binary string without $11$.
Given a $(n-1)$-bit binary string without $11$, we can append $0$ to the end to get a unique $n$-bit binary string without $11$.

This gives a bijection between the $n$-bit strings without $11$ and the $(n-1)$-bit strings without $11$ $\bigcup$ the $(n-2)$-bit strings without $11$.  This is illustrated below:

We could proceed similarly and add bits at either end.  It doesn't make the proof any easier though.  If we did that in the above case:

Every $n$-bit string without $11$ would be paired up with exactly $2$ smaller strings without $11$ (either $(n-1)$-bit or $(n-2)$-bit).
Every $(n-1)$-bit and $(n-2)$-bit string without $11$ would be paired up with exactly $2$ $n$-bit strings.

So the number of edges in the figure would double, but the result would still hold:  The number of edges going out of the left is $2$ times the number of $(n-1)$-bit strings without $11$ plus the number of $(n-2)$-bit strings without $11$.  This is equal to the number of edges going into the right, which is $2$ times the nmber of $n$-bit strings without $11$.  So the $2$'s cancel out and we get the same recurrence.


TITLE: Continuity of composite functions
QUESTION [1 upvotes]: The continuity theorem for composite functions states that if $f(x)$ is continuous at $x = a$ and $g(x)$ is continuous at $x = a$ , then the composite function $f\circ g$ and $g\circ f$ are also necessarily continuous at $x = a$. My question is why doesn't it work for the functions $\sin x$ and $[x]$. To be specific, let $f(x) = \sin x$ ; $g(x) = [x]$ G.I.F.   At $x = \pi/2$ both $f(x)$ and $g(x)$ are continuous but the composite function $g\circ f$ is discontinuous. Doesn't this violate the above theorem ?
Any help will be much appreciated. Thanks

REPLY [2 votes]: The theorem, as you stated it, is not true! Notice that for the function $f \circ g$ to be defined, the image of $g$ must be in the domain of $f$, and if $a$ is in the domain of $f$, it's not necessarly true that it' is in the domain of $g$. However, what is true is that if $f$ is continuous at $a$, and $g$ is continuous at $f(a)$, then $g \circ f$ is continuous at a.


TITLE: How to use dimensional analysis in an integral
QUESTION [0 upvotes]: $\int \frac{d x}{\sqrt{2 a x-x^{2}}}=a^{n} \sin ^{-1}\left[\frac{x}{a}-1\right]$
The value of n is
(a) 0
(b) -1
(c) 1
(d) none of these.
How would you find the answer using dimensional analysis and what is it?

REPLY [3 votes]: Check out the denominator of the integrated function. Since we have $2ax-x^2$ there and since one cannot add apples and oranges, we can immediately see that the unit of $a$ is the same as the unit of $x$. Let us say it is meters just for the sake of definiteness. The function you are integrating is $f(x)=\frac{1}{\sqrt{2ax-x^2}}$, which is meters$^{-1}$, and since integrating wrt $x$ is akin to finding the area below $f(x)$, the integral should then be meters$^{-1}\times$meters, i.e. non-dimensional.
The argument of $\sin^{-1}$ is clearly non-dimensinal, so is the value of $\sin^{-1}(x/a-1)$. Thus, the only way for the answer to be non-dimensional is if $n=0$.


TITLE: Lagrangian method for an $LC$-Circuit
QUESTION [9 upvotes]: In the paper http://physics.unipune.ernet.in/~phyed/26.2/File5.pdf, the author solves the LC-circuit using Euler-Lagrange equation. She assumes that the Lagrangian function for the circuit is
$$L=T-V$$
where $T=L\dot q^2 / 2$ is the kinetic energy part and $V=q^2 / 2C$ is the potential energy.
She then goes on to say that
$$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q} \right ) - \frac{\partial L}{\partial q} = 0$$ is the equation of motion.
I have a couple of questions:

Is Hamilton's principle applicable here? If yes, why is the kinetic energy part $\dfrac{{L\dot q^2}}{2}$?
Inductors also store potential energy temporarily in the magnetic field. Why don't we consider that term?

REPLY [10 votes]: In this answer, I'm going to use $Q$ instead of $q$ for the capacitor charge.

There is only one certain rule for finding Lagrangians: The Lagrangian is chosen such as to get the correct equations of motion.
Never forget that.
In the case of a circuit problem, the most sure way to know you got the right Lagrangian is to see if it gives you the right equations of motion, i.e. the equations you get from Kirchhoff's laws.
Hamilton's principle

Is Hamilton's principle applicable here?

Let's start out by writing down Kirchhoff's laws for the LC circuit.
Choose the orientation of current so that it's flowing into the inductor as shown.
Then Kirchhoff's current rule and the definition of an inductor says
$$
I = -\dot{Q} \qquad \dot{I} = V/L$$
where $Q$ is the charge on the capacitor and $V$ is the voltage on the capacitor and inductor.
The rule for a capacitor is $CV = Q$, so we can rewrite the second equation as
$$\dot{I} = Q / LC \, .$$
Differentiating the first equation gives
$$ \dot{I} = - \ddot{Q}$$
which plugs into the line above to get
$$\ddot{Q} = - Q/LC \, .$$
Defining $\omega_0^2 = 1/LC$ we now have the usual equation for a harmonic oscillator
$$\ddot{Q} = -\omega_0^2 Q \, . $$
That's the equation of motion for the charge in an LC oscillator.
Now let's see if the Lagrangian proposed by the author gives the same equation if we use the Euler Lagrange equations (a.k.a. Hamilton's principle).
The Lagrangian is
$$
\mathcal{L} = \frac{L \dot{Q}^2}{2} - \frac{Q^2}{2C}$$
so the Euler Lagrange equations are
\begin{align}
\frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{Q}} \right) - \frac{\partial \mathcal{L}}{\partial Q} =& 0 \\
\frac{d}{dt} \left(L \dot{Q} \right) + \frac{Q}{C} =& 0 \\
L \ddot{Q} + \frac{Q}{C} =& = 0 \\
\ddot{Q} = - \frac{Q}{LC} = - \omega_0^2 Q \, .
\end{align}
The result from Hamilton's principle matches the result from Kirchhoff's laws, so yes, Hamilton's principle is applicable here.
Kinetic and potential energy

If yes, why is the kinetic energy part $L \dot{Q}^2 / 2$?

In the strictest sense, the kinetic energy part of the Lagrangian is $L \dot{Q}^2 / 2$ because it works to give the right equations of motion and that's the only thing that really matters in Lagrangian mechanics.
However, another way to look at the Lagrangian of the LC oscillator is to simply note that each term is the energy of the capacitor and inductor respectively.
The energy of a capacitor with charge $Q$ and capacitance $C$ is $Q^2 / 2C$.
The author chose to call that the potential energy.
The energy of an inductor with current $I$ and inductance $L$ is $LI^2/2$.
Using $I = \dot{Q}$ we get that the energy of the inductor is $L\dot{Q}^2/2$, which is what the author used for the kinetic energy.
Therefore, you can reason that the author simply identified the two types of energy in the circuit, called one "kinetic" and the other "potential" and guessed the Lagrangian that way.
More general Lagrangians in the context of circuits
A lot of us first learn about Lagrangians in the context of a physical particle moving around in a potential field.
In that particular case, the Lagrangian is always $\mathcal{L} = T-V$, but that's far from the most general form of a Lagrangian.
What happens, for example, if we have two systems connected to each other, such as two particles connected by a spring or two LC circuits connected through a capacitor?
It turns out there is actually a recipe for finding more complex Lagrangians without just guessing.
That recipe is described for example in this article by Michel Devoret.
The article is a rewrite of a famous set of course notes from the Les Houches summer school(pdf) done to correct a number of errors and clarify some points.
Do not be discouraged by the word "quantum" in the title.
Much of the discussion is purely classical physics.

Inductors also store potential energy temporarily in the magnetic field. Why don't we consider that term?

As just explained, we do, and it's what the author called the "kinetic" energy.