TITLE: Why can't my body draw a tiny piece of paper despite gravitation pull it exerts on the piece of paper?
QUESTION [2 upvotes]: We observe in static electricity experiments that a comb rubbed on dry hair lifts tiny pieces of paper.This is due to the attraction force exerted by the charge collecting on the surface of the comb on the opposite charges collecting on the hair.This is a very small amount of force.But our body is always exerting gravitational force on all objects around it.Still it cannot draw such tiny pieces of papers.Why?

REPLY [0 votes]: Let's say you have a mass of $64kg$ and a small bit of paper has a mass of $1g$ Let's also assume that you curl up into a ball and hover above the paper bits (you must be David Blaine) such that your center of mass is $1m$ above the paper. The gravitational force that your entire body exerts on that small bit of paper is then:
$$F_{you}=\frac{Gm_1m_2}{r^2}=\frac{G(64kg)(0.001kg)}{(1m)^2}=4.27\times10^{-12}N$$
The force of gravity that Earth uses to keep that paper in place is:
$$F_{Earth}=\frac{G(5.97\times10^{24}kg)(0.001kg)}{(6371000m)^2}=9.81\times10^{-3}N$$
So Earth is keeping the paper from moving toward you. But what if you were in space with nothing else keeping the paper from moving toward you? That $4.27\times10^{-12}N$ applied to the $1g$ paper bit means it will accelerate towards you at $4.27\times10^{-9}m/s^2$. Granted, that value will increase as the paper moves closer to you; by the time it reaches half a meter from you, it's accelerating four times that much. But even at that rate, it would take $t=\sqrt{2d/a}=\sqrt{2(0.5m)/(1.71\times10^{-8}m/s^2)}=2.12h$ for that paper bit to cover half a meter. Compare that to the $0.32s$ for Earth to make it fall half a meter. Not only that, but with such a weak force, friction is more than enough to keep that paper bit where it is on the table.
The reason electrostatics can move the paper is that electrostatic forces are much stronger than gravity. As mentioned, the electromagnetic force is $10^{36}$ times stronger than gravity, so it's easy to overcome gravity with even a small static charge.


TITLE: Find Three Mutual Friends in a Mathematical Society
QUESTION [2 upvotes]: I am having trouble with the following combinatorics/graph theory problem:

A mathematical society has three divisions (Pure, Applied, and Statistics), and exactly $n$ mathematicians in each division. Every mathematician has at least $n+1$ friends outside of their division. Prove there exist three mathematicians, one in each division, who are all friends.

This was a problem after the chapter about the Extreme Value Principle, which involves using the fact that every finite set of numbers has a minimum and maximum. Of course, there are lots of potential ways to apply this ("consider the mathematician with the most friends...", or "consider the longest chain of mathematicians, each friends with the next..."), but all the ways I've tried have been fruitless.
There is some sort of clever insight needed to solve this problem that I am missing. I would appreciate any hints in the right direction.

REPLY [4 votes]: Denote the divisions by $D_1,D_2,D_3$. Consider the mathematician $v\in D_1$ with the greatest difference between the number of friends in each division. Call this difference $k$. Denote the friends of $v$ in $D_2$ by $A$, and his friends in $D_3$ by $B$. WLOG let $|A|\geq|B|$. Then $|A| = \frac{n+1+k}{2}$ and $|B| = \frac{n+1-k}{2}$.
Let $u$ be in $B$. Denote the friends of $u$ in $D_2$ by $C$. By the maximality of $k$, the number of friends of $u$ in either division must be between $|B|$ and $|A|$ since the total number of friends is a constant $n+1$. In particular, $|B|\leq|C|\leq|A|$, so $|C| + |A| \geq n+1$. This means that $|A\cap C|\geq 1$. Let $w$ be in $A\cap C$, then $u, v, w$ is the required group of friends.


TITLE: Characters of a Group: two definitions
QUESTION [6 upvotes]: If $G$ is an abelian group, the characters associated to the rapresentations of $G$ over $\textrm{GL}_1(\mathbb C)=\mathbb C^\ast$ are simply the group homomorphisms:
$$\chi:G\longrightarrow\mathbb C^\ast$$
On the contrary if $G$ is a topological group (assume locally compact) then a character is a continous homomorphism:
$$\chi':G\longrightarrow\mathbb R/\mathbb Z\cong S^1$$
Why do we have two apparently different definitions? I know that the range of $\chi$ (first definition) is $S^1$ only when $G$ is finite.

REPLY [2 votes]: Agreeing with Adam in the comments: if the bigger codomain $\Bbb C^*$ is used we have what number theory would term a quasicharacter, at least from what I've read. True, from the viewpoint of representation theory, it's just a one-dimensional character. However for convenience and elegance we sometimes impose unitarity on our representations, and stipulating that the codomain be $S^1$ is essentially saying we want it to be a unitary one-dimensional representation.
KCd in another thread states why $U(1)$ is so useful: it is the "universal dualizer," which is where we get Pontryagin duality. Precisely: $G\cong\hom(\hom(G,A),A)$ for all loc. comp. $G$ iff $A=U(1)$.


TITLE: differential of a section, linear connections and Leibniz rule
QUESTION [2 upvotes]: Let $\pi:E \rightarrow M$ be a vector bundle and let $s$ $\in$ $\Gamma(E)$ (a section of the bundle). If we have a linear Ehresmann connection on the bundle, we can define a linear connection as $$\nabla s:= \rho_V \circ ds,$$ where $\rho_V$ is projection into $Ker(d\pi)$ (the vertical bundle). But I don't understand why it woks, because of $d\pi(ds(X))=X$, so $Im(ds)$ would be a subset of the horizontal bundle and then $\rho_V \circ ds=0$.
What's going on here?
Another question: With that definition of $\nabla$, how can I get the Leibniz rule  $\nabla(fs)=f\nabla(s)+df\otimes s$?. I mean, I understand why $\nabla$ is a map of $\Gamma(E)$ into $\Gamma(Hom(TM,E))=\Gamma(T^{\ast}M\otimes E)$ but I cannot see why $$\rho_V\circ d(fs)=f(\rho_V \circ ds)+(df)s.$$

REPLY [1 votes]: It is not true in general that the image of $ds$ is contained in the horizontal space. Let us consider a simple example. Let $E=\mathbb{R}^2$, $M=\mathbb{R}$, and $\pi$ be the projection on the first coordinate. In other words, $E$ is the trivial line bundle over the real line. Let $\nabla$ be the trivial connection, that is, the horizontal space at any point $(x,y)\in E$ is spanned by $\partial/\partial x$. In this setting, any section is given by$$x\mapsto(x,f(x)),$$and a section is horizontal (the image of its differential is contained in the horizontal space at any point) if and only if the function $f$ is constant.
Edit: A few words about the Leibniz rule. There are numerous ways to think about connections and the Leibniz rule, and I guess they are all the same in a way. Here is one of those ways.
Let $p\in M$ and $w\in T_pM$. We want to show that for a section $s$ and a function $f$ we have $$\nabla_wfs=df(w)s+f\nabla_ws.$$We proceed by the following steps.
1) Let $e_1,\ldots,e_k$ be a local trivialization of $E$ around $p$, and suppose first that $\nabla$ is the corresponding trivial covariant derivative. In this case, the Leibniz rule is just the usual one for paths in Euclidean space.
Note that step 1) is not enough, as most of the connections do not look locally like a trivial connection.
2)Let $\rho,\rho'$ be two different projections on the vertical bundle, each one of which respects the linear structure of $E$, and let $\nabla,\nabla'$ denote the corresponding covariant derivatives, respectively. We claim that the difference $\nabla-\nabla'=:T$ is a tensor. That is, the map$$(X,s)\mapsto T(X,s)=\nabla_Xs-\nabla_X's$$ is $C^\infty(M)$-linear in both parameters. To see this, let $\widetilde{X}$ and $\widetilde{X}'$ be  horizontal lifts of $X$ with respect to both connections. Then $$T(X,s)=\nabla_Xs-\nabla_X's=\left(ds(X)-\widetilde{X}\right)-\left(ds(X)-\widetilde{X}'\right)=\widetilde{X}-\widetilde{X}'.$$ In other words, $T(X,s)$ does not depend on the derivative of $s$ but only on the value of $s$ at a point, and the claim follows.
3) Finally, we claim that if $\nabla$ and $\nabla'$ are two covarivant derivatives, such that $\nabla-\nabla'$ is a tensor and $\nabla'$ satisfies the Leibniz rule, then so does $\nabla$. This is a nice exercise for anyone who is new to connections.
The full Leibniz rule follows from the above three steps.


text:
\subsection{The Index Computation}\label{ssec:index}
Fix two connected components of the respective critical sets $\Lambda_0 \subseteq \Crit (\mathcal{A}^{H_0})$ and $\Lambda \subseteq \operatorname{Crit}( \mathcal{A}^H)$. In this section we establish that the hybrid moduli problem is Fredholm, and compute the virtual dimension of the the moduli space $\mathcal{M}_{\operatorname{hyb}}(\Lambda_0 , \Lambda)$.

Let us introduce an anti-symplectic involution on $(T^*\R^k,\omega_k)$ by
\begin{equation}
\rho \colon  T^*\R^k \to T^*\R^k, \qquad (q,p)  \mapsto (q,-p). \label{rho}
\end{equation}
The standard symplectic structure $\mathbb{J}_k$ satisfies
\begin{equation}\label{JnRho}
- D\rho \circ \mathbb{J}_k \circ D\rho =\mathbb{J}_k.
\end{equation}
Therefore, we have the associated diffeomorphism:
\[
\mathcal{J}_\star^k \ni J \mapsto -D\rho \circ J \circ D\rho \in \mathcal{J}_\star^k.
\]
Finally, observe that $H_0\circ \rho = H_0$, which gives us
\begin{equation}
\label{XH0Rho}
X_{H_0}(x)= - D\rho[X_{H_0}(\rho (x))].
\end{equation}
As a result we obtain an automorphism of $\operatorname{Crit} (\mathcal{A}^{H_0})$ defined by:
\begin{equation}\label{Lambda0map}
\operatorname{Crit} (\mathcal{A}^{H_0}) \ni (v,\eta) \longmapsto (\rho \circ v, -\eta)\in \operatorname{Crit} (\mathcal{A}^{H_0}).
\end{equation}
Using the notation from \eqref{LambdaEta}, this automorphism maps $\Lambda_0^\eta$ to $\Lambda_0^{-\eta}$.
\begin{theorem}
\label{thm-index}
The hybrid moduli problem is Fredholm, and the space $ \mathcal{M}_{\operatorname{hyb}}(\Lambda_{0} , \Lambda)$ has virtual dimension
\[
\operatorname{vir\,dim} \mathcal{M}_{\operatorname{hyb}}(\Lambda_{0} , \Lambda) =  \tCZ(\Lambda) - \tCZ(\Lambda_0) + \frac{1}{2} \left( \dim \Lambda_0 + \dim \Lambda\right).
\]
\end{theorem}
\begin{proof}
We prove the theorem in five steps.\\

\paragraph{\textbf{1.}}
Denote by $\mathfrak{y}\coloneqq  \cA^{H_0}(\Lambda_0) \in \operatorname{CritVal}(\cA^{H_0})$, so that by Corollary \ref{cor:CritVal} we have $\Lambda_0=\Lambda_0^{\mathfrak{y}}$. In this first step, we show that $\mathcal{M}_{\operatorname{hyb}}(\Lambda_0^{\mathfrak{y}} , \Lambda)$ is a Fredholm problem. We begin by identifying $\mathcal{M}_{\operatorname{hyb}}(\Lambda_0^{\mathfrak{y}} , \Lambda)$ with a closely related set\footnote{The fact that ``hyb'' is occurs both as a subscript and a superscript is not a typo! They represent formally different spaces.} $\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ defined as follows:

Let $L \subseteq T^*\R^k\times T^*\R^n= \R^{2(n+k)}$ denote the set of elements of the form
\[
L = \left\{ (a,-b,a,b,c,0_{n-k})\ \Big|\ a,b \in \R^k, c \in \mathbb{R}^{n-k} \right\}.
\]
In other words, $L$ is the preimage of $\Delta_{T^*\R^k}\times \chi_0$ under $\rho \oplus \sigma^{-1}$, where $\rho$ is the anti-symplectic map defined in \eqref{rho},  $ \sigma$ is the symplectic splitting defined in \eqref{splitting}, $\Delta_{T^*\R^k}$ stands for the diagonal in $(T^*\R^k)^2$ and $\chi_0$ is the $0$-section of $T^*\R^{n-k}$. Then $L$ is a Lagrangian submanifold of $(T^*\R^k\times T^*\R^n,\omega_k\oplus \omega_n)$.

For $J_0\in \mathcal{J}_\star^k, J \in \mathcal{J}_\star^n$ we define, using \eqref{JnRho}, an element $\widehat{J}\in \mathcal{J}_\star^{n+k}$ by
\begin{equation}\label{J1}
\widehat{J} \coloneqq \begin{pmatrix}
-D\rho \circ J_0 \circ D\rho & \\
& \;J
\end{pmatrix}.
\end{equation}
An element of $\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ is a pair $(w,\xi)$
with
\begin{equation}\label{eq-w-first}
    w \colon [0,+\infty) \times S^1 \to T^*\R^k\times T^*\R^n, \qquad \xi \colon [0,+\infty) \to \R^2,
\end{equation}
satisfying the equations
\begin{align}
 \partial_s w + \widehat{J}\partial_t w - \xi \cdot X_1 (w)& = 0, \label{eq-w-third}\\
 \partial_s \xi -X_2(w)& = 0, \label{eq-xi-1st}
\end{align}
where
\begin{align}
  (\xi_0,\xi_1) \cdot X_1(w_0,w_1) & \coloneqq  \widehat{J}\begin{pmatrix} \xi_0 X_{H_0}(w_0) \\ \xi_1 X_H(w_1) \end{pmatrix}, \label{eq-w-fourth}\\
  X_2(w) & \coloneqq  \begin{pmatrix} -\int_{S^1}H_0(w_0) \\ -\int_{S^1} H (w_1)\end{pmatrix}, \label{eq-xi-2nd}
\end{align}
together with the limit conditions
\begin{equation}
\lim_{s\to + \infty}w(s)\in \Lambda_0^{-\mathfrak{y}}\times \Lambda, \label{limit2}
\end{equation}
and the coupling conditions
\begin{equation}
w(0,t) \in L, \qquad \forall \,t \in S^1,\qquad \textrm{and} \qquad \xi_1(0)=-\xi_0(0).\label{eq-w-second}
\end{equation}
There is a natural identification of elements $((v_0,\eta_0),(v,\eta))\in\mathcal{M}_{\operatorname{hyb}}(\Lambda_0^{\mathfrak{y}} , \Lambda)$ with elements $(w,\xi)\in\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ given by
\begin{equation}\label{bijection}
w(s,t)\coloneqq  \begin{pmatrix} \rho \circ v_0(-s,t) \\ v(s,t) \end{pmatrix}, \qquad
\xi(s)\coloneqq  \begin{pmatrix} -\eta_0(-s)\\ \eta(s) \end{pmatrix}.
\end{equation}
Indeed, under this identification conditions \eqref{limit1} and \eqref{coupling} become conditions \eqref{limit2} and \eqref{eq-w-second}.
Furthermore, we compute that
\[
    \partial_s w = \begin{pmatrix} - D \rho [\partial_s v_0] \\ \partial_s v \\ \end{pmatrix}, \qquad
    \partial_t w  = \begin{pmatrix} D\rho [\partial_t v_0 ]\\ \partial_t v  \\  \end{pmatrix}, \qquad
    \partial_s \xi = \begin{pmatrix} \partial_s \eta_0 \\ \partial_s \eta  \\  \end{pmatrix}.
\]
so that by \eqref{JnRho}, \eqref{XH0Rho}, \eqref{J1}, and \eqref{bijection} we have
\begin{align*}
 \partial_s w + \widehat{J}\partial_t w & = \begin{pmatrix} -D\rho (\partial_s v_0+J_0 \partial_t v_0) \\ \partial_s v+ J \partial_t v \end{pmatrix} =\begin{pmatrix} -\eta_0 D\rho \circ J_0 X_{H_0}(v_0) \\ \eta J X_H (v) \end{pmatrix}\\
& = \begin{pmatrix} \eta_0 D\rho \circ J_0 \circ D\rho [X_{H_0}(w_0)] \\ \eta J X_H (w_1) \end{pmatrix} = \widehat{J}\begin{pmatrix} \xi_0 X_{H_0}(w_0) \\ \xi_1 X_H(w_1) \end{pmatrix}.
\end{align*}
In this way the Rabinowitz Floer equations \eqref{Floer} become equivalent to \eqref{eq-w-third}--\eqref{eq-xi-2nd}.  So $ \mathcal{M}_{\operatorname{hyb}}(\Lambda_0^{\mathfrak{y}} , \Lambda)$ is a Fredholm problem if and only if $\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ is a Fredholm problem.

Fix $r >2$. Let $\mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)$ denote the Banach manifold of pairs $(w,\xi)$ as in \eqref{eq-w-first} satisfying \eqref{limit2} and \eqref{eq-w-second}. For $(w,\xi)\in \mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)$, we do not require equations \eqref{eq-w-third}--\eqref{eq-xi-2nd} to hold. Instead, $(w,\xi)$ should be locally of class $W^{1,r}$ and converge for $s\rightarrow +\infty$ exponentially to an element $(x_0,x_1) \in \Lambda_0^{-\mathfrak{y}}\times\Lambda$. The tangent space of $\mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)$ at $(w,\xi)$ can be identified with
\begin{align*}
T_{(w,\xi)} \mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)&\cong \mathcal{W}^{1,r}_{\delta,L}\oplus \mathcal{W}^{1,r}_{\delta,\Delta}\oplus  T_{ x_0}\Lambda_0\oplus T_{x_1}\Lambda,\\
\text{where }\qquad\qquad\mathcal{W}^{1,r}_{\delta,L} & \coloneqq \left\lbrace z \in W^{1,r}_{\delta} \left([0, \infty) \times S^1 , \mathbb{R}^{2k} \times \mathbb{R}^{2n} \right) \ \Big|\ z(0,t) \in L \right\rbrace,\\
\mathcal{W}^{1,r}_{\delta,\Delta} & \coloneqq \left\lbrace \zeta \in W^{1,r}_{\delta}\left([0,\infty), \mathbb{R}^2\right) \ \Big|\ \zeta_1(0)=-\zeta_0(0) \right\rbrace.
\end{align*}
Here, $\delta$ indicates that we are working with weighted Sobolev spaces with weight $\gamma(s)=e^{\delta s}$ (as our asymptotic operators will not be bijective).
 Explicitly, $f\in \mathcal{W}^{1,r}_\delta$ if and only if $f\cdot \gamma\in \mathcal{W}^{1,r}$. Let $\mathcal{E}$ be the Banach bundle over $\mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)$ whose fiber at $(w,\xi)$ is given by
 \[\mathcal{E}_{(w,\xi)}\coloneqq L^r_{\delta}\left( [0,\infty) \times S^1, \mathbb{R}^{2k} \times \mathbb{R}^{2n} \right) \times L^r_{ \delta}\left([0,\infty), \mathbb{R}^2\right).\]
 Define a section
 \begin{equation}
 \bar{\partial} \colon \mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)\to \mathcal{E}, \label{eq:Banach-sec}
 \end{equation}
 by the left-hand side of the equations \eqref{eq-w-third} and \eqref{eq-xi-1st}. Then $\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} ; \Lambda)=\bar{\partial}^{-1}(0)$. Since \eqref{eq-w-second} is a Lagrangian boundary condition and \eqref{eq-w-third} is a perturbed Cauchy-Riemann equation on a half-cylinder, coupled with a pair of ordinary differential equations \eqref{eq-xi-1st} for $\xi$, the proof that the linearisation
 \[
 D\bar{\partial} \colon T_{(w,\xi)}\mathcal{B}(\Lambda_0^{-\mathfrak{y}},\Lambda)\to \mathcal{E}_{(w,\xi)},
 \]
  is a Fredholm operator is a routine argument (see for instance \cite[Sec. 5.4]{AbbondandoloSchwarz2010}), and we will not go over the details here. \\

\paragraph{\textbf{2.}}
Computing the index however is somewhat less standard, for two reasons:
\begin{enumerate}[label=(\roman*)]
    \item In contrast to standard index computations in Rabinowitz Floer homology, here there are two Lagrange multipliers to worry about.
    \item We are on a half cylinder, and thus arguments using the spectral flow are not directly applicable.
\end{enumerate}
This computation is very similar to \cite[Theorem 4.12]{AbbonMerry2014}. In order to keep the exposition reasonably clean, we perform the calculation only in the case where both $ \mathcal{A}^{H_0}(\Lambda_0^{-\mathfrak{y}})$ and $ \mathcal{A}^H(\Lambda)$ are non-zero. The other cases are easier and are left to the reader.

In this step, we formulate an appropriate local model for the problem $ \mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ by linearising equations \eqref{eq-w-third}--\eqref{eq-xi-2nd} at a solution $(w,\xi) \in \mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$, and express the virtual dimension of  $\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)$ in terms of the index of this linearised operator.

Write $m_0\coloneqq  k, m_1\coloneqq n$ and let
\[
    P_i \in W^{1, \infty}\left( [0,+\infty) \times S^1 , \operatorname{Mat}(2m_i)\right),
\]
for $i=0,1$ denote two paths of matrices that extend to the compactification $[0, +\infty] \times S^1$ in such a way that for each $i=0,1$
\[
\lim_{s_0 \rightarrow + \infty}\underset{(s,t)\in[s_0,\infty )\times S^1} {\operatorname{ess\, sup}}\Big(\left|\partial_s P_i(s,t)\right|+\left|\partial_t P_i(s,t)-\partial_t P_i(t,\infty)\right|\Big)=0.
\]
Assume in addition that the limit matrices $S_i(t) \coloneqq P_i(+ \infty, t)$ are symmetric:
\begin{equation}
    \label{eq-limit-S}
    S_0(t) \in \operatorname{Sym}( \mathbb{R}^{2k}), \qquad S_1(t) \in \operatorname{Sym}( \mathbb{R}^{2n}).
\end{equation}
Next, for $i=0,1$ let
\[
\beta_i \in W^{1, \infty}( [0,\infty) \times S^1 , \mathbb{R}^{2m_i}),\\
\]
denote two vector-valued paths which extend to the compactification $[0,+\infty] \times S^1$ in such a way that for each $i = 0,1$,
 \[
\lim_{s_0 \rightarrow + \infty}\underset{(s,t)\in (s_0,\infty)\times S^1} {\operatorname{ess\,sup}}\Big(\left|\partial_{s}\beta_i(s,t)\right|+\left|\partial_{t}\beta_i(s,t)-\partial_{t}\beta_i(t,+\infty)\right|\Big)=0.
\]
Abbreviate $b_i(t) \coloneqq \beta_i(+\infty,t)$, and assume that
\[
b_i  \in W^{1,2}(S^1, \mathbb{R}^{2m_i}) \cap \operatorname{range}(\mathbb{J}_{m_i}\partial_t +S_i), \qquad i = 0,1.
\]
We are now ready to introduce an appropriate local model. Consider the operator $\mathbf{D}  \colon \mathcal{W}^{1,r}_{ \delta, L} \times \mathcal{W}^{1,r}_{\delta,\Delta}   \to \mathcal{E}$ defined by\footnote{We somewhat ambiguously switch between row and column notation, favouring whichever is cleaner for any given equation.}
\begin{equation}\label{eq-D}
    \mathbf{D} \begin{pmatrix}
z_0 \\ z_1 \\ \zeta_0 \\ \zeta_1
\end{pmatrix}
\coloneqq
\begin{pmatrix}
\partial_s z_0 + \mathbb{J}_k \partial_t z_0 + P_0 z_0 + \zeta_0 \beta_0 \\
\partial_s z_1 + \mathbb{J}_n \partial_t z_1 + P_1 z_1 + \zeta_1 \beta_1 \\
\zeta_0' + \int_{S^1} \langle z_0, \beta_0  \rangle \,dt\\
\zeta_1' + \int_{S^1} \langle z_1, \beta_1  \rangle \,dt
\end{pmatrix}
\end{equation}
Note that $\mathbf{D}$ is the restriction to $\mathcal{W}^{1,r}_{\delta,L}\oplus\mathcal{W}^{1,r}_{\delta,\Delta}$ of the linearisation $D\bar{\partial}$ of the problem \eqref{eq:Banach-sec} at a solution $(w, \xi)\in\mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} ; \Lambda)$ of \eqref{eq-w-third}--\eqref{eq-xi-2nd}, when viewed in a suitable symplectic trivialisation.
Explicitly, for $H=H_0+H_1$ satisfying \eqref{defH} we have
\begin{align}
S_0 & =  -\eta_0 \mathbb{J}_k D_{v_0}X_{H_0}, && b_0 (t)  = -\mathbb{J}_k [ X_{H_0}] (v_0(t)),\label{Sb0}\\
S_1 & = -\eta \mathbb{J}_n D_v X_H, && b_1(t)  = - \mathbb{J}_n [X_H] (v(t)),\label{Sb1}
\end{align}
for some $(v_0,\eta_0)\in \Lambda_0^{-\mathfrak{y}}, (v,\eta)\in \Lambda$.
Consequently, we have
\begin{equation}\label{dimM}
 \operatorname{vir\,dim} \mathcal{M}^{\operatorname{hyb}}(\Lambda_0^{-\mathfrak{y}} , \Lambda)  =  \operatorname{ind} D\bar{\partial}= \operatorname{ind} \mathbf{D} + \dim \Lambda_0 + \dim \Lambda.
\end{equation}
$\ $
\paragraph{\textbf{3.}}
Our aim is to homotope the operator $\mathbf{D}$ from \eqref{eq-D} through Fredholm operators into a new operator of product form
\begin{equation}\label{Dnew}
\mathbf{D}^{\operatorname{new}} (z, \zeta) = \left( \mathbf{D}_1(z), \mathbf{D}_2( \zeta) \right).
\end{equation}
The operator $\mathbf{D}^{\operatorname{new}}$ decouples the perturbed Cauchy-Riemann equation satisfied by $z$ and the ordinary differential equation satisfied by $ \zeta$. This will allow us to compute the index of $\mathbf{D}$:
\begin{align}
    \operatorname{ind} \mathbf{D}& = \operatorname{ind} \mathbf{D}^{\operatorname{new}}  & &\text{since the index is constant along deformations,} \nonumber \\
  &  = \operatorname{ind} \mathbf{D}_1 + \operatorname{ind}\mathbf{D}_2 & & \text{by additivity.} \label{eq-index-sum}
\end{align}
The operator $\mathbf{D}^{\operatorname{new}}$ is constructed using methods from \cite[App.\ C]{CieliebakFrauenfelder2009}.
The index of $ \mathbf{D}_1$ can be computed using the methods from \cite{AbbondandoloSchwarz2010}. Meanwhile the index of $ \mathbf{D}_2$ can be computed by hand.

In order to construct the homotopy we first need to compute the perturbation term defined in \cite[Def.\ C.3]{CieliebakFrauenfelder2009} as follows: for $i=0,1$ and $\delta>0$ choose $z_i \in W^{1,2}(S^1, \mathbb{R}^{2m_i})$ such that
\begin{equation}\label{Av=b}
(\mathbb{J}_{m_i}\partial_t +S_i+\delta \operatorname{Id})z_i = b_i,
\end{equation}
and define
\begin{equation}
\label{tau1}
\tau_i \coloneqq  \int_{S^1} \langle z_i(t), b_i(t) \rangle \,dt.
\end{equation}
Since the operator $\mathbb{J}_{m_i}\partial_t +S_i$ is self-adjoint, the number $\tau_i$ does not depend on the choice of $z_i$. We will show that in our setting, for $\delta>0$ small enough
\begin{equation}\label{tau2}
\tau_0, \tau_1>0.
\end{equation}
First observe, that by Lemma \ref{lem:orbits} every periodic orbit $(v,\eta)\in \Crit(\cA^H), \eta\neq 0$ is of the form $( \sigma^{-1}(v_1,0),\eta)$ with $(v_1,\eta)\in \Crit(\cA^{H_0})$. As a result
\[
dH(v)=dH \circ \sigma^{-1} (v_1,0)=dH_0(v_1).
\]
Therefore, if $(z_0,z)$ is a solution to \eqref{Av=b} and
 $ (z_1,z_2) \coloneqq \sigma(z) \in W^{1,2}(S^1,T^*\R^k)\times  W^{1,2}(S^1,T^*\R^{n-k})$, then by \eqref{Sb1} we get
\[
\tau_1 = \int_{S^1}d_v H (z)\,dt =\int_{S^1}d_{v_1}H_0 (z_1)\,dt.
\]
Therefore, without loss of generality we can assume that $z_2=0$. Consequently, by \eqref{CritEta}, \eqref{Sb0} and \eqref{Sb1}, equation \eqref{Av=b} translates to
\begin{equation}\label{Av=b2}
(\mathbb{J}_k\partial_t +\eta_i A_0 +\delta \operatorname{Id})z_i = A_0 \exp[t \eta_i \mathbb{J}_k A_0] v_i(0),
\end{equation}
where $(v_0,\eta_0)\in \Lambda_0^{-\mathfrak{y}}, ( \sigma^{-1}(v_1,0),\eta_1)\in \Lambda \subseteq \Crit (\cA^H)$.

It is easy to check that
\[
z_i(t)\coloneqq \frac{1}{\delta} A_0 \exp[t \eta_i \mathbb{J}_k A_0] v_i(0),
\]
solves \eqref{Av=b2} and as a result for $i=0,1$ we have
\[
\tau_i = \frac{1}{\delta}\int_{S^1} \left\|A_0  \exp[t \eta_i \mathbb{J}_k A_0] v_i(0)\right\|^2dt>0,
\]
which proves \eqref{tau2}.\\

\paragraph{\textbf{4.}}
In this step we will construct the homotopy between the operator $\mathbf{D}$ from \eqref{eq-D} and a new operator $\mathbf{D}^{\operatorname{new}}$ of the form as in \eqref{Dnew}.

If we set
\[
Q = \begin{pmatrix}
\mathbb{J}_k \partial_t + P_0  &  \\
& \mathbb{J}_n \partial_t + P_1
\end{pmatrix}, \qquad  \beta = \begin{pmatrix}
\beta_0 \\
\beta_1
\end{pmatrix},
\qquad
B \begin{pmatrix}
z_0 \\ z_1
\end{pmatrix}
=
\begin{pmatrix}
\int_{S^1} \langle z_0, \beta_0  \rangle \,dt\\
\int_{S^1} \langle z_1, \beta_1  \rangle \,dt
\end{pmatrix},
\]
then the operator $ \mathbf{D}$ can be written as
\[
\mathbf{D} = \partial_s + \begin{pmatrix}
Q & \beta \\
B & 0
\end{pmatrix}.
\]
We now consider the homotopy $\{ \mathbf{D}^{\theta }\}_{ \theta \in [0,1]}$ of operators given by
\[
\mathbf{D}^{\theta} = \partial_s +  \begin{pmatrix}
Q & (1 - \theta)  \beta \\
(1 - \theta) B & c( \theta)
\end{pmatrix},
\]
where $c  \colon  [0, 1] \to \mathbb{R}^2$ is defined by
\begin{equation}
\label{eq-c}
c( \theta)  \coloneqq
\begin{pmatrix}
\theta \tau_0 \\
\theta \tau_1
\end{pmatrix}.
\end{equation}
By \cite[Thm. C.5]{CieliebakFrauenfelder2009} the operators $\mathbf{D}^{\theta}$ are all Fredholm of the same index. We define
\begin{align*}
 \mathbf{D}^{\operatorname{new}} & \coloneqq \mathbf{D}^1=(\mathbf{D}_1,\mathbf{D}_2), & & \textrm{where} \\
 \mathbf{D}_1 & \colon \mathcal{W}^{1,r}_{\delta, L}  \to L^r_{\delta} \left( [0,\infty) \times S^1, \mathbb{R}^{2n+2k}\right),
  & z & \mapsto   \partial_s z + Q z,\\
\mathbf{D}_2 & \colon \mathcal{W}^{1,r}_{\delta, \Delta}  \to L^r_{ \delta} \left( [0,\infty), \mathbb{R}^2\right), &
 \begin{pmatrix}
 \zeta_0 \\ \zeta_1
 \end{pmatrix}  &\mapsto \begin{pmatrix}
 \zeta_0' + \tau_0 \zeta_0 \\ \zeta_1' + \tau_1\zeta_1
 \end{pmatrix} .
\end{align*}

\paragraph{\textbf{5.}}
In this final step we compute the index of $ \mathbf{D}_1$ and $ \mathbf{D}_2$. For this denote by $ \Psi_i$ for $i=0,1$ the following paths of symplectic matrices:
\begin{align*}
    \Psi_0 & \colon [0,1] \to \operatorname{Symp}( \mathbb{R}^{2k}, \omega_k), && \begin{cases}
    \Psi'_0(t) = \mathbb{J}_k S_0(t) \Psi_0(t),\\
    \Psi_0(0) = \operatorname{Id}_{2k},
    \end{cases} \\
        \Psi_1 & \colon [0,1] \to \operatorname{Symp}( \mathbb{R}^{2n}, \omega_n), && \begin{cases}
    \Psi'_1(t) = \mathbb{J}_n S_1(t) \Psi_1(t),\\
    \Psi_1(0) = \operatorname{Id}_{2n},
    \end{cases}
\end{align*}
Denote by $\CZ(\Psi_i) \in \frac{1}{2} \mathbb{Z}$ the (full) Conley-Zehnder index of these paths, and let $ \operatorname{null}( \Psi_i) \coloneqq \dim \left( \ker \Psi_i(1) - \operatorname{Id} \right) $ denote the nullity. Then the Fredholm index of $\mathbf{D}_1$ is given by \cite[Theorem 5.25]{AbbondandoloSchwarz2010} as
\begin{equation}
\label{eq-index-D1}
\operatorname{ind} \mathbf{D}_1 = \sum_{i=0}^1 \left( \CZ(\Psi_i) - \frac{1}{2}\operatorname{null}(\Psi_i) \right)+m(L),
\end{equation}
where $m(L)$ stands for the correction term coming from the boundary conditions.
Actually \eqref{eq-index-D1} is much simpler than the general statement in \cite[Thm. 5.25]{AbbondandoloSchwarz2010}, since we only have a single boundary condition (rather than a set of jumping boundary conditions). The correction term $m(L)$ is computed by \cite[Thm. 5.25]{AbbondandoloSchwarz2010} to be
\[
m(L)\coloneqq \frac{\dim T^*\R^{n+k}}{2} -\frac{1}{2}\dim \Delta_{T^*\R^{n+k}} - \dim L\\
 + \dim \left(\Delta_{T^*\R^{n+k}}\cap(L\times L) \right),
\]
where $\Delta_{T^*\R^{n+k}}$ is the diagonal in $(T^*\R^{n+k})^2$. In this case,
\begin{align*}
\dim \Delta_{T^*\R^{n+k}} & =2(n+k), \\
\dim \left(\Delta_{T^*\R^{n+k}}\cap(L\times L) \right) & =\dim L=n+k,
\end{align*}
hence\footnote{Sanity check: The same computation works for any Lagrangian $L$. Thus the correction term is always $0$ when there is but a single Lagrangian boundary condition. This can also be proved directly, without appealing to the machinery of \cite{AbbondandoloSchwarz2010}.} $m(L)=0$. On the other hand, by Proposition \ref{prop:muCZ} we have
\begin{align*}
\CZ(\Psi_0) & =\CZ(\Lambda_0^{-\mathfrak{y}})= - \CZ(\Lambda_0^\mathfrak{y})=-\tCZ(\Lambda_0),\\
\CZ(\Psi_1) & = \CZ(\Lambda)= \tCZ(\Lambda).
\end{align*}
Moreover, by Lemma \ref{lem:orbits} and \eqref{CritEta} we have
\begin{align*}
\dim \Lambda_0 & = \dim \Lambda_0^{-\mathfrak{y}}= \dim \left( \ker \left(\Psi_0(1) - \operatorname{Id}\right) \cap \Sigma_0\right)  =\operatorname{null}( \Psi_0)-1,\\
\dim \Lambda & = \dim \left(\ker \left( \Psi_1(1) - \operatorname{Id}\right) \cap \left(\Sigma_0\times \{0\}\right)\right) =\operatorname{null}( \Psi_1)-1,
\end{align*}
which gives
\begin{equation}\label{D1}
\operatorname{ind} \mathbf{D}_1 =   \tCZ(\Lambda) - \tCZ(\Lambda_0) -\frac{1}{2}\left( \dim \Lambda_0 + \dim \Lambda \right)-1.
\end{equation}
Now we would like to compute $\operatorname{ind}  \mathbf{D}_2$. Note that
\[
\operatorname{ind} \mathbf{D}_2 = 2 \operatorname{ind} \widetilde{\mathbf{D}}_2- 1,
\]
where the $-1$ comes from the boundary condition $\zeta_1(0)=-\zeta_0(0)$ and $\widetilde{\mathbf{D}}_2$ is an operator of the form
\begin{gather*}
\widetilde{\mathbf{D}}_2 \colon  W^{1,r}_{\delta}\left([0,+\infty), \R\right)  \to  L^r_{ \delta}\left([0,+\infty), \R\right),\\
f \mapsto f'+\tau f.
\end{gather*}
Note that $\dim \operatorname{coker} \widetilde{\mathbf{D}}_2 =0$. However, since  we are working with weighted Sobolev spaces on a positive half-cylinder, the dimension of $\ker \widetilde{\mathbf{D}}_2=1$, since in both cases $\tau>0$ by \eqref{tau1}. Consequently, $\operatorname{ind} \mathbf{D}_2 = 1$. Combining this with \eqref{dimM} and \eqref{D1}
we have
\begin{align*}
   \operatorname{vir\,dim} \mathcal{M}_{\operatorname{hyb}}(\Lambda_0 , \Lambda) & = \operatorname{ind} \mathbf{D}^{\operatorname{new}}  + \dim \Lambda_0 + \dim \Lambda \\
   & = \tCZ(\Lambda) - \tCZ(\Lambda_0) + \frac{1}{2} \left( \dim \Lambda_0 + \dim \Lambda\right).
\end{align*}
This completes the proof of Theorem \ref{thm-index}.
\end{proof}

config: arxiv


text:
\section{Computing the Brauer-Picard Ring of $\vvec{\ZZ{p}}$}\label{S:Zp}

\begin{table}[t]
	\resizebox{\linewidth}{!}{

		\begin{tabular}{!{\vrule width 1pt}>{\columncolor[gray]{.9}[\tabcolsep]}c!{\vrule width 1pt}c !{\color[gray]{.8}\vrule} c !{\color[gray]{.8}\vrule} c !{\color[gray]{.8}\vrule} c !{\color[gray]{.8}\vrule} c !{\color[gray]{.8}\vrule} c!{\vrule width 1pt}}
			\toprule[1pt]
			\rowcolor[gray]{.9}[\tabcolsep]Bimodule label & Subgroup & \# objects & Cosets $\to$ object label & Left action & Right action & Associator \\
			\toprule[1pt]
			$T$ & $\{(0,0)\}$ & $p^2$ & $\{(g,h)\}\to(g,h)$&
			$
			\begin{array}{c}
			\includeTikz{Zp_T_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$(a,b)$][$(a+g,b)$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_T_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$(a,b)$][$(a,b+g)$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_T_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$(a,b)$][$(a+g,b+h)$];
				\end{tikzpicture}
			}
			\end{array}
			=
			\begin{array}{c}
			\includeTikz{Zp_T_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$(a,b)$][$(a+g,b+h)$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\greycline{2-7}
			$L$ & $\langle (1,0) \rangle\cong\ZZ{p}$ & p & $\{(h,g)\,|\,h\in \ZZ{p}\}\to g$&
			$
			\begin{array}{c}
			\includeTikz{Zp_L_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$a$][$a$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_L_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$a$][$a+g$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_L_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$a$][$a+h$];
				\end{tikzpicture}
			}
			\end{array}
			=
			\begin{array}{c}
			\includeTikz{Zp_L_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$a$][$a+h$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\greycline{2-7}
			$R$ & $\langle (0,1) \rangle\cong\ZZ{p}$ & p & $\{(g,h)\,|\,h\in \ZZ{p}\}\to g$&
			$
			\begin{array}{c}
			\includeTikz{Zp_R_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$a$][$g+a$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_R_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$a$][$a$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_R_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$a$][$g+a$];
				\end{tikzpicture}
			}
			\end{array}
			=
			\begin{array}{c}
			\includeTikz{Zp_R_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$a$][$g+a$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\greycline{2-7}
			$F_0$ & $\langle (0,1), (1,0) \rangle\cong\ZZ{p}\times\ZZ{p}$ & 1 & $\ZZ{p}\times \ZZ{p}\to *$&
			$
			\begin{array}{c}
			\includeTikz{Zp_F0_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_F0_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_F0_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			=
			\begin{array}{c}
			\includeTikz{Zp_F0_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\toprule[1pt]
			$X_k$ & $\langle (-k,1) \rangle\cong\ZZ{p}$ & p & $\{n(-k,1)+(h,0)\,|\,n\in \ZZ{p}\}\to h$&
			$
			\begin{array}{c}
			\includeTikz{Zp_X_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$a$][$a+g$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_X_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$a$][$a+kg$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_X_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$a$][$g+a+kh$];
				\end{tikzpicture}
			}
			\end{array}
			=
			\begin{array}{c}
			\includeTikz{Zp_X_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$a$][$g+a+kh$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\greycline{2-7}
			$F_q$& $\langle (0,1), (1,0) \rangle\cong\ZZ{p}\times\ZZ{p}$ & 1 & $\ZZ{p}\times \ZZ{p}\to *$&
			$
			\begin{array}{c}
			\includeTikz{Zp_Fq_Left}
			{
				\begin{tikzpicture}
				\Laction[$g$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$
			&
			$
			\begin{array}{c}
			\includeTikz{Zp_Fq_Right}
			{
				\begin{tikzpicture}
				\Raction[$g$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$&
			$
			\begin{array}{c}
			\includeTikz{Zp_Fq_AssL}
			{
				\begin{tikzpicture}
				\Lassociator[$g$][$h$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			=
			e^{\frac{2\pi i}{p} q g h}
			\begin{array}{c}
			\includeTikz{Zp_Fq_AssR}
			{
				\begin{tikzpicture}
				\Rassociator[$g$][$h$][$*$][$*$];
				\end{tikzpicture}
			}
			\end{array}
			$
			\\
			\hline
		\end{tabular}
	}
	\caption{Data for all $\vvec{\ZZ{p}}-\vvec{\ZZ{p}}$ bimodules. $q \in H^2(\ZZ{p},U(1)) \cong \ZZ{p}$. Bimodules below the thick line are invertible.}\label{tab:zpdata}
\end{table}

In this section, we compute the Brauer-Picard ring for $\vvec{\ZZ{p}}$ with $p$ prime. We remark that the ring for the twisted case $\vvectwist{\ZZ{p}}{\phi}$ can be similarly computed. In that case, there are additional consistency conditions between the associators and cocycle twist that restrict the bimodules compared to $\vvec{\ZZ{p}}$.

Recall from Chapter~7 of \onlinecite{MR3242743} that each $\vvec{G}\text{ -}\vvec{G}$ bimodule $\mathcal{M}$ is labelled by a conjugacy class of subgroups $H\subseteq G\times G$ and a 2-cocycle $\omega\in H^2(H,U(1))$. The simple objects in $\mathcal{M}$ are labelled by cosets of $H$. In Table~\ref{tab:zpdata}, we list the bimodules, and the associated functors ($\vartriangleright,\, \vartriangleleft$ in Definition~\ref{bimodule_definition}). The associators are obtained from the cocycle $\omega$ using the formula explained in \onlinecite{288303}. We now provide some example calculations. The full multiplication table is presented in Table~\ref{tab:zptable}.

\begin{exmp}[$T \otimes_{\vvec{\ZZ{p}}} T$]
 We begin by computing $T\otimes_{\vvec{\ZZ{p}}} T$. We proceed in two steps: First we construct the \emph{ladder category} $\lad{T,T}{}$, then we construct its \emph{Karoubi envelope} $\kar{\protect\lad{T,T}{}}$ and identify the left and right actions and associators. With this data, we can identify the resulting bimodule. This example is unnaturally simple because the ladder category is already idempotent complete. The (simple) objects in $\lad{T, T}{}$ are given by pairs of objects from $T$, written $(g_0,g_1)(g_2,g_2)$ with $g_i\in \ZZ{p}$. Therefore, $\lad{T,T}{}$ has $p^4$ simple objects. Given two objects $(a,b)(c,d)$ and $(a,b-g)(c+g,d)$, the full set of morphisms between them are
\begin{align}
\begin{array}{c}
\includeTikz{TxT_ladder_mor}{
	\begin{tikzpicture}[scale=0.7, every node/.style={scale=0.7}]
	\clip (-4,-3) rectangle (8,3);
	\coordinate (A) at (0,0);
	\coordinate (B) at (4,0);
	\coordinate (C) at (-4,0);
	\coordinate (D) at (3,2.5);
	\coordinate (E) at (9.5,0);
	\coordinate (F) at (3,-3);

	\fill (A) circle(2pt) node[below,inline text] {\footnotesize$(a,b)(c,d)$};

	\fill (B) circle(2pt) node[below,inline text] {\footnotesize$(a,b-g)(g+c,d)$};

	\draw[-stealth,red] ($(A)+(0,.15)$) arc (20:330:.7);
	\draw[-stealth,red] ($(B)+(0,.15)$) arc (330:20:-.7);
	\draw[-stealth,red] ($(A)+(.15,.15)$) to[out=20,in=180-20] ($(B)+(-.15,.15)$);
	\draw[-stealth,red] ($(B)+(-.15,-.45)$) to[out=180+20,in=-20] ($(A)+(.15,-.45)$);

	\ladder[C][$(a,b)$][$(c,d)$][][][];
	\ladder[E][$(a,b-g)$][$(g+c,d)$][][][];
	\ladder[D][$(a,b)$][$(c,d)$][$g$][$(a,b-g)$][$(g+c,d)$];
	\ladder[F][$(a,b-g)$][$(c+g,d)$][$-g$][$(a,b)$][$(c,d)$];
	\end{tikzpicture}
}
\end{array}.\label{eq:TTmorph}
\end{align}
For all other pairs of objects, only the self-morphisms exist. Since all morphisms in Equation~\ref{eq:TTmorph} are invertible, the objects $(a,b)(c,d)$ and $(a,b-g)(c+g,d)$ are isomorphic in $\lad{T,T}{}$. The objects in the Karoubi envelope of the ladder category are labelled by pairs $(A,e)$, where $A$ is an object from $\lad{T,T}{}$ and $e:A\to A$ and idempotent. The $p^3$ non-isomorphic objects of the Karoubi envelope can be represented by
\begin{align}
\begin{array}{c}
\includeTikz{KarTTobj}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder [A][$(a,b)$][$(0,c)$][][][];
	\end{tikzpicture}
}
\end{array}.
\end{align}

\begin{table}[t]
	\begin{tabular}{!{\vrule width 1pt}>{\columncolor[gray]{.9}[\tabcolsep]}c!{\vrule width 1pt}c c c c!{\vrule width 1pt}c c!{\vrule width 1pt}}
		\toprule[1pt]
		\rowcolor[gray]{.9}[\tabcolsep]$\otimes_{\vvec{\ZZ{p}}}$&$T$&$L$&$R$&$F_0$&$X_l$&$F_r$\\
		\toprule[1pt]
		$T$&$p\cdot T$&$T$&$p\cdot R$&$R$&$T$&$R$\\

		$L$&$p\cdot L$&$L$&$p\cdot F_0$&$F_0$&$L$&$F_0$\\

		$R$&$T$&$p\cdot T$&$R$&$p\cdot R$&$R$&$T$\\

		$F_0$&$L$&$p\cdot L$&$F_0$&$p\cdot F_0$&$F_0$&$L$\\
		\toprule[1pt]
		$X_k$&$T$&$L$&$R$&$F_0$&$X_{kl}$&$F_{k^{-1}r}$\\

		$F_q$&$L$&$T$&$F_0$&$R$&$F_{ql}$&$X_{q^{-1}r}$\\
		\toprule[1pt]
	\end{tabular}
	\caption{Multiplication table for $\protect\bpr{\protect\vvec{\ZZ{p}}}$.}\label{tab:zptable}
\end{table}

We now need to identify the left action of $\vvec{\ZZ{p}}$. A morphism $f:(A,e)\to(A^\prime,e^\prime)$ is a morphism $f\in \lad{T,T}{}$ which obeys $ef=f=fe^\prime$. Such a morphism allows us to identify the left action of $g$
\begin{align}
\begin{array}{c}
\includeTikz{TTiso}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lmorphism[A][$(a,b)$][$(0,c)$][][$(g+a,b)$][$(0,c)$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\begin{array}{c}
\includeTikz{TTisoLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lladder[A][$(a,b)$][$(0,c)$][][][][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\begin{array}{c}
\includeTikz{TTisoRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$(g+a,b)$][$(0,c)$][][][];
	\end{tikzpicture}
}
\end{array}.\label{eqn:TTleftiso}
\end{align}
One can readily check that this morphism ``absorbs'' the appropriate idempotents and therefore establishes an isomorphism. The left action is therefore $g \otimes (a,b)(0,c)=(g+a,b)(0,c)$. A similar isomorphism
\begin{align}
\begin{array}{c}
\includeTikz{TTisoR}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rmorphism[A][$(a,b)$][$(0,c)$][][$(g,b)$][$(0,c+g)$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\begin{array}{c}
\includeTikz{TTisoRLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rladder[A][$(a,b)$][$(0,c)$][][][][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\begin{array}{c}
\includeTikz{TTisoRRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$(a,b)$][$(0,c+g)$][][][];
	\end{tikzpicture}
}
\end{array},\label{eqn:TTrightiso}
\end{align}
which identifies the right action as $(a,b)(0,c)\otimes g=(a,b)(0,c+g)$. These actions are consistent with the bimodule $T$. Since there is no way to change the label $b$, it labels $p$ different blocks with no morphisms between them. It is straightforward to check that the morphisms in Equation~\ref{eqn:TTleftiso} and Equation~\ref{eqn:TTrightiso} commute, meaning that the associator is trivial. It follows that
\begin{align}
	T\otimes_{\vvec{\ZZ{p}}}T =p T.
\end{align}
\end{exmp}
\begin{center}\rule{.8\textwidth}{1pt}\end{center}

\begin{exmp}[$R \otimes_{\vvec{\ZZ{p}}} F_0$]
The simple objects in $\lad{R,F_0}{}$ are labelled by $a$, with $a\in \ZZ{p}$, so there are $p$ simple objects. Given an object $a\in \lad{R,F_0}{}$, there is a $p$ dimensional algebra of morphisms
\begin{align}
\begin{array}{c}
\includeTikz{RF0Morph1}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$a$][$*$][$g$][$a$][$*$];
	\end{tikzpicture}
}
\end{array}&,\\
\text{with }
\begin{array}{c}
\includeTikz{RF0Morph2}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);\coordinate (B) at (0,.5);
	\ladder[A][$a$][$*$][$g$][][];
	\ladder[B][][][$h$][$a$][$*$];
	\end{tikzpicture}
}
\end{array}
&=\begin{array}{c}
\includeTikz{RF0Morph3}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$a$][$*$][$g+h$][$a$][$*$];
	\end{tikzpicture}
}
\end{array}
\end{align}
which form the group algebra $\mathbb{C}\left[\ZZ{p}\right]$. The $p$ indecomposable idempotents $a\to a$ are given by
\begin{align}
I_{k}&=\frac{1}{p}\sum_{g\in\ZZ{p}}
e^{\frac{2\pi i}{p}kg}
\begin{array}{c}
\includeTikz{RF0Morph4}
{
	\begin{tikzpicture}
	\coordinate (A) at (0,0);
	\ladder[A][$a$][$*$][$g$][$a$][$*$];
	\end{tikzpicture}
}
\end{array}.
\end{align}
The $p^2$ simple objects of $\kar{\lad{R,F_0}{}}$ can be labelled by $(a,k)$. To compute the action of $g$ on these objects, we again need to find a morphism $f:g\otimes(a,k)\to(a^\prime,k^\prime)$ in $\kar{\lad{R,F_0}{}}$. In this case, we can take $k' = k$ and $a' = g + a$. Recall that this needs to absorb the idempotents $I_k$ and $I_k^\prime$ on the bottom and top respectively. Such a morphism is given by
\begin{align}
\frac{1}{p}\sum_{x\in\ZZ{p}}e^{\frac{2\pi i}{p}kx}
\begin{array}{c}
\includeTikz{RF0iso}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lmorphism[A][$a$][$*$][$x$][$g+a$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}ky}
\begin{array}{c}
\includeTikz{FR0isoLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lladder[A][$a$][$*$][$y$][$a$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}ky}
\begin{array}{c}
\includeTikz{RF0isoRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$g+a$][$*$][$y$][$g+a$][$*$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:RF0leftiso}
\end{align}
One can check that the idempotents are absorbed:
\begin{align}
	I_kf&=
	\frac{1}{p^2}
	\sum_{x,y\in \ZZ{p}}
	e^{\frac{2\pi i}{p}k(x+y)}
	\begin{array}{c}
		\includeTikz{RF0LeftAbsorbBelow1}
		{
			\begin{tikzpicture}
				\coordinate (A) at (0,0);
				\coordinate (B) at (0,-1);
				\Lmorphism[A][][][$x$][$g+a$][$*$][];
				\Lladder[B][$a$][$*$][$y$][][][$g$];
			\end{tikzpicture}
		}
	\end{array}
	=
	\frac{1}{p^2}
	\sum_{x,y\in \ZZ{p}}
	e^{\frac{2\pi i}{p}k(x+y)}
	\begin{array}{c}
	\includeTikz{RF0LeftAbsorbBelow2}
	{
		\begin{tikzpicture}
		\coordinate (A) at (0,0);
		\coordinate (B) at (0,-1);
		\Lmorphism[A][][][$x+y$][$g+a$][$*$][];
		\Lladder[B][$a$][$*$][][][][$g$];
		\end{tikzpicture}
	}
	\end{array}\\
	&=
	\frac{1}{p^2}
	\sum_{x,z\in \ZZ{p}}
	e^{\frac{2\pi i}{p}kz}
	\begin{array}{c}
	\includeTikz{RF0LeftAbsorbBelow3}
	{
		\begin{tikzpicture}
		\coordinate (A) at (0,0);
		\coordinate (B) at (0,-1);
		\Lmorphism[A][][][$z$][$g+a$][$*$][];
		\Lladder[B][$a$][$*$][][][][$g$];
		\end{tikzpicture}
	}
	\end{array}
	=
	\frac{1}{p}
	\sum_{z\in \ZZ{p}}
	e^{\frac{2\pi i}{p}kz}
	\begin{array}{c}
	\includeTikz{RF0LeftAbsorbBelow4}
	{
		\begin{tikzpicture}
		\coordinate (A) at (0,0);
		\coordinate (B) at (0,-1);
		\Lmorphism[A][][][$z$][$g+a$][$*$][];
		\Lladder[B][$a$][$*$][][][][$g$];
		\end{tikzpicture}
	}
	\end{array}
	\\
	&=f.
\end{align}
Absorption from above can be checked similarly, recalling that the associator on the left strand is trivial
\begin{align}
	fI_{k+g}&=
	\frac{1}{p^2}
	\sum_{x,y\in \ZZ{p}}
	e^{\frac{2\pi i}{p}k(x+y)}
	\begin{array}{c}
		\includeTikz{RF0LeftAbsorbAbove1}
		{
			\begin{tikzpicture}
				\coordinate (A) at (0,0);
				\coordinate (B) at (0,-1);
				\Lmorphism[B][$a$][$*$][$x$][][][$g$];
				\ladder[A][][][$y$][$g+a$][$*$];
			\end{tikzpicture}
		}
	\end{array}
	=
	\frac{1}{p^2}
	\sum_{x,y\in \ZZ{p}}
	e^{\frac{2\pi i}{p}k(x+y)}
	\begin{array}{c}
		\includeTikz{RF0LeftAbsorbAbove2}
		{
			\begin{tikzpicture}
			\coordinate (A) at (0,0);
			\coordinate (B) at (0,-1);
			\Lmorphism[A][][][$x+y$][$g+a$][$*$][];
			\Lladder[B][$a$][$*$][][][][$g$];
			\end{tikzpicture}
		}
	\end{array}
	=f.
\end{align}

For the right action, we find
\begin{align}
\frac{1}{p}\sum_{x\in\ZZ{p}}e^{\frac{2\pi i}{p}kx}
\begin{array}{c}
\includeTikz{RF0isoR}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rmorphism[A][$a$][$*$][$x$][$a$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}ky}
\begin{array}{c}
\includeTikz{FR0isoRLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rladder[A][$a$][$*$][$y$][$a$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}ky}
\begin{array}{c}
\includeTikz{RF0isoRRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$a$][$*$][$y$][$a$][$*$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:RF0rightiso}
\end{align}
The left action is therefore $g\otimes (a,k)=(g+a,k)$, with trivial right action. As before, the morphisms Equation~\ref{eqn:RF0leftiso} and Equation.~\ref{eqn:RF0rightiso} commute so the associator is trivial. The label $k$ is unaffected by the group action, so labels $p$ distinct blocks with no morphisms between them. This gives
\begin{align}
R\otimes_{\vvec{\ZZ{p}}} F_0=pR.
\end{align}

\end{exmp}
\begin{center}\rule{.8\textwidth}{1pt}\end{center}

\begin{exmp}[$X_k\otimes_{\vvec{\ZZ{p}}} X_l$]
The simple objects in $\lad{X_k,X_l}{}$ are labeled by $(a,b)$, with $a,b\in \ZZ{p}$. Given a pair of objects $(a,b)$ and $(a-kg,b+g)$ the full set of morphisms between them are
\begin{align}
	\begin{array}{c}
		\includeTikz{XxX_ladder_mor}{
			\begin{tikzpicture}[scale=0.7, every node/.style={scale=0.7}]
			\clip (-4,-3) rectangle (8,3);
			\coordinate (A) at (0,0);
			\coordinate (B) at (4,0);
			\coordinate (C) at (-4,0);
			\coordinate (D) at (3,2.5);
			\coordinate (E) at (9.5,0);
			\coordinate (F) at (3,-3);

			\fill (A) circle(2pt) node[below,inline text] {\footnotesize$(a,b)$};

			\fill (B) circle(2pt) node[below,inline text] {\footnotesize$(a-kg,b+g)$};

			\draw[-stealth,red] ($(A)+(0,.15)$) arc (20:330:.7);
			\draw[-stealth,red] ($(B)+(0,.15)$) arc (330:20:-.7);
			\draw[-stealth,red] ($(A)+(.15,.15)$) to[out=20,in=180-20] ($(B)+(-.15,.15)$);
			\draw[-stealth,red] ($(B)+(-.15,-.45)$) to[out=180+20,in=-20] ($(A)+(.15,-.45)$);

			\ladder[C][$a$][$b$][][][];
			\ladder[E][$a-kg$][$b+g$][][][];
			\ladder[D][$a$][$b$][$g$][$a-kg$][$b+g$];
			\ladder[F][$a-kg$][$b+g$][$-g$][$a$][$b$];
			\end{tikzpicture}
		}
	\end{array}.\label{eq:XXmorph}
\end{align}
For all other pairs of objects, only the self-morphisms exist. Since all morphisms in Equation~\ref{eq:XXmorph} are invertible, the objects $(a,b)$ and $(a-kg,b+g)$ are isomorphic. The $p$ non-isomorphic objects of $\kar{\lad{X_k,X_l}{}}$ can be represented by
\begin{align}
	\begin{array}{c}
		\includeTikz{KarXXobj}
		{
			\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
			\coordinate (A) at (0,0);
			\ladder [A][$a$][$0$][][][];
			\end{tikzpicture}
		}
	\end{array}.
\end{align}
The morphism associated to the left $g$ action is
\begin{align}
	\begin{array}{c}
		\includeTikz{XXiso}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\Lmorphism[A][$a$][$0$][][$g+a$][$0$][$g$];
			\end{tikzpicture}
		}
	\end{array}
	:
	\begin{array}{c}
		\includeTikz{XXisoLHS}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\Lladder[A][$a$][$0$][][$a$][$0$][$g$];
			\end{tikzpicture}
		}
	\end{array}
	\tilde{\to}
	\begin{array}{c}
		\includeTikz{XXisoRHS}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\ladder[A][$g+a$][$0$][][$g+a$][$0$];
			\end{tikzpicture}
		}
	\end{array},\label{eqn:XXleftiso}
\end{align}
so $g\otimes(a,0)=(a+g,0)$. The right action is more complicated. Consider the morphism
\begin{align}
	\begin{array}{c}
		\includeTikz{XXisoR1}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\Rmorphism[A][$a$][$0$][][$a$][$lg$][$g$];
			\end{tikzpicture}
		}
	\end{array}
	:
	\begin{array}{c}
		\includeTikz{XXisoRLHS1}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\Rladder[A][$a$][$0$][][$a$][$0$][$g$];
			\end{tikzpicture}
		}
	\end{array}
	\tilde{\to}
	\begin{array}{c}
		\includeTikz{XXisoRRHS1}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\ladder[A][$a$][$lg$][][$a$][$lg$];
			\end{tikzpicture}
		}
	\end{array}.\label{eqn:RF0rightiso1}
\end{align}
The codomain is not the chosen representative of the isomorphism class. Instead, we need to append the morphism
\begin{align}
	\begin{array}{c}
		\includeTikz{KarXXcorrection}
		{
			\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
			\coordinate (A) at (0,0);
			\ladder [A][$a$][$lg$][$-lg$][$a+lkg$][$0$];
			\end{tikzpicture}
		}
	\end{array}
\end{align}
to identify the correct block. The right action is then identified from
\begin{align}
	\begin{array}{c}
		\includeTikz{XXisoR2}
		{
			\begin{tikzpicture}[xscale=.75,yscale=.6]
			\coordinate (A) at (0,0);
			\coordinate (B) at (0,1);
			\Rmorphism[A][$a$][$0$][][][][$g$];
			\ladder[B][][][$-lg$][$a+lkg$][$0$]
			\end{tikzpicture}
		}
	\end{array}
	:
	\begin{array}{c}
		\includeTikz{XXisoRLHS2}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\Rladder[A][$a$][$0$][][$a$][$0$][$g$];
			\end{tikzpicture}
		}
	\end{array}
	\tilde{\to}
	\begin{array}{c}
		\includeTikz{XXisoRRHS21}
		{
			\begin{tikzpicture}[xscale=.75]
			\coordinate (A) at (0,0);
			\ladder[A][$a+lkg$][$0$][][$a+lkg$][$0$];
			\end{tikzpicture}
		}
	\end{array}.\label{eqn:RF0rightiso2}
\end{align}
The right action is therefore $(a,0)\otimes g=(a+klg,0)$. Since the associators on $X_i$ are trivial, the morphisms commute and the resulting associator is trivial. These $\vvec{\ZZ{p}}$ actions and associator show that
\begin{align}
	X_k\otimes_{\vvec{\ZZ{p}}} X_l=X_{kl},
\end{align}
where $kl$ is taken modulo $p$. This is the first instance of an invertible bimodule we have seen. The bimodules $X_k$ therefore form a subgroup of the \emph{Brauer-Picard group} of $\vvec{\ZZ{p}}$.

\end{exmp}
\begin{center}\rule{.8\textwidth}{1pt}\end{center}

\begin{exmp}[$F_q\otimes_{\vvec{\ZZ{p}}} F_r$]
There is a single simple object in $\lad{F_q,F_r}{}$, with a $p$ dimensional algebra of morphisms
\begin{align}
\begin{array}{c}
\includeTikz{FFMorph1}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$*$][$g$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}&,\\\text{with }
\begin{array}{c}
\includeTikz{FFMorph2}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);\coordinate (B) at (0,.5);
	\ladder[A][$*$][$*$][$g$][][];
	\ladder[B][][][$h$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}
&=\begin{array}{c}
\includeTikz{FFMorph3}
{
	\begin{tikzpicture}[scale=0.8, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$*$][$g+h$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}
\end{align}
which form the group algebra $\mathbb{C}\left[\ZZ{p}\right]$. The $p$ indecomposable idempotents $**\to **$ are given by
\begin{align}
I_{qk}&=\frac{1}{p}\sum_{g\in\ZZ{p}}
e^{\frac{2\pi i}{p}qkg}
\begin{array}{c}
\includeTikz{FFMorph4}
{
	\begin{tikzpicture}
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$*$][$g$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}.
\end{align}
The $p$ simple objects of $\kar{\lad{F_q,F_r}{}}$ can be labelled by $k$. The morphism associated to the left $g$ action is
\begin{align}
\frac{1}{p}\sum_{x\in\ZZ{p}}e^{\frac{2\pi i}{p}qkx}
\begin{array}{c}
\includeTikz{FFiso}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lmorphism[A][$*$][$*$][$x$][$*$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}qky}
\begin{array}{c}
\includeTikz{FFisoLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Lladder[A][$*$][$*$][$y$][$*$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}q(k+g)y}
\begin{array}{c}
\includeTikz{FFisoRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$*$][$y$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:FFleftiso}
\end{align}
One can check that $I_{k+g}$ is absorbed from above
\begin{align}
\frac{1}{p^2}\sum_{x,y\in \ZZ{p}}
e^{\frac{2\pi i}{p}q(kx+(k+g)y)}
\begin{array}{c}
\includeTikz{FFleftabsorbtop1}
{
	\begin{tikzpicture}
	\coordinate (A) at (0,0);
	\coordinate (B) at (0,1);
	\ladder[B][][][$y$][$*$][$*$];
	\Lmorphism[A][$*$][$*$][$x$][][][$g$];
	\end{tikzpicture}
}
\end{array}
&=
\frac{1}{p^2}\sum_{x,y\in \ZZ{p}}
e^{\frac{2\pi i}{p}q(kx+(k+g)y)}e^{-\frac{2\pi i}{p}qgy}
\begin{array}{c}
\includeTikz{FFleftabsorbtop2}
{
	\begin{tikzpicture}
	\coordinate (A) at (0,0);
	\coordinate (B) at (0,-1);
	\Lladder[B][$*$][$*$][$x$][][][$g$];
	\Lmorphism[A][][][$y$][$*$][$*$][];
	\end{tikzpicture}
}
\end{array}\\
&=\frac{1}{p}\sum_{x\in \ZZ{p}}
e^{\frac{2\pi i}{p}qkx}
\begin{array}{c}
\includeTikz{FFleftabsorbtop3}
{
	\begin{tikzpicture}
	\coordinate (A) at (0,0);
	\Lmorphism[A][$*$][$*$][$x$][$*$][$*$][$g$];
	\end{tikzpicture}
}
\end{array},
\end{align}
where the first equality uses the associator in Table~\ref{tab:zpdata}, where $y$ outgoing contributes $-y$. A similar computation on the right gives us
\begin{align}
\frac{1}{p}\sum_{x\in\ZZ{p}}e^{\frac{2\pi i}{p}qkx}
\begin{array}{c}
\includeTikz{FFisoR}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rmorphism[A][$*$][$*$][$x$][$*$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}qky}
\begin{array}{c}
\includeTikz{FFisoRLHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\Rladder[A][$*$][$*$][$y$][$*$][$*$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\frac{1}{p}\sum_{y\in\ZZ{p}}e^{\frac{2\pi i}{p}q(k+q^{-1}rg)y}
\begin{array}{c}
\includeTikz{FFisoRRHS}
{
	\begin{tikzpicture}[xscale=.75]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$*$][$y$][$*$][$*$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:FFrightiso}
\end{align}
The left and right actions are therefore $g\otimes k=k+g$ and $k\otimes g=k+q^{-1}rg$. One can again check that the associator is trivial, giving us
\begin{align}
	F_q\otimes_{\vvec{\ZZ{p}}} F_r=X_{q^{-1}r},
\end{align}
where $k^{-1}$ is the multiplicative inverse of $k$ taken modulo $p$. The bimodules $F_q$ are therefore invertible.

\end{exmp}

\begin{center}\rule{.8\textwidth}{1pt}\end{center}

\begin{exmp}[$F_q\otimes_{\vvec{\ZZ{p}}} X_l$]

The $p$ simple objects of $\lad{F_q,X_l}{}$ are labeled by $x\in\ZZ{p}$. Given any two objects $x$ and $y$, there is an isomorphism in $\lad{F_q,X_l}{}$, given by
\begin{align}
\begin{array}{c}
\includeTikz{FXIso1}
{
	\begin{tikzpicture}
	\coordinate (A) at(0,0);
	\ladder[A][$*$][$x$][$y-x$][$*$][$y$];
	\end{tikzpicture}
}
\end{array}.
\end{align}
The single isomorphism class in $\kar{\lad{F_q,X_l}{}}$ can be represented using
\begin{align}
\begin{array}{c}
\includeTikz{FXKarObj1}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at(0,0);
	\ladder[A][$*$][$0$][][$*$][$0$];
	\end{tikzpicture}
}
\end{array}.
\end{align}
Since there is a single object, we already know that $F_q\otimes_{\vvec{\ZZ{p}}}X_l=F_r$ for some $r$. To identify the value of $r$, we need to compute the associator. The left action is given by
\begin{align}
\begin{array}{c}
\includeTikz{FXiso}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\Lmorphism[A][$*$][$0$][][$*$][$0$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\begin{array}{c}
\includeTikz{FXisoLHS}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\Lladder[A][$*$][$0$][][$*$][$0$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\begin{array}{c}
\includeTikz{FXisoRHS}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$0$][][$*$][$0$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:FXleftiso}
\end{align}
On the right, we identify the isomorphism
\begin{align}
\begin{array}{c}
\includeTikz{FXisoR1}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\Rmorphism[A][$*$][$0$][][$*$][$lg$][$g$];
	\end{tikzpicture}
}
\end{array}
:
\begin{array}{c}
\includeTikz{FXisoR1LHS}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\Rladder[A][$*$][$0$][][$*$][$0$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\begin{array}{c}
\includeTikz{FXisoR1RHS}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$lg$][][$*$][$lg$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:FXrightiso}
\end{align}
The codomain is not the chosen representative of the isomorphism class. Instead, we need to append the morphism
\begin{align}
\begin{array}{c}
\includeTikz{FXCap}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at(0,0);
	\ladder[A][$*$][$lg$][$-lg$][$*$][$0$];
	\end{tikzpicture}
}
\end{array}
\end{align}
to correctly absorb the $g$. The complete isomorphism is given by
\begin{align}
\begin{array}{c}
\includeTikz{FXisoR2}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\coordinate (B) at (0,1);
	\Rmorphism[A][$*$][$0$][][][][$g$];
	\ladder[B][][][$-lg$][$*$][$0$]
	\end{tikzpicture}
}
\end{array}
:
\begin{array}{c}
\includeTikz{FXisoR2LHS}
{
	\begin{tikzpicture}[xscale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\Rladder[A][$*$][$0$][][$*$][$0$][$g$];
	\end{tikzpicture}
}
\end{array}
\tilde{\to}
\begin{array}{c}
\includeTikz{FXisoR2RHS}
{
	\begin{tikzpicture}[xscale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,0);
	\ladder[A][$*$][$0$][][$*$][$0$];
	\end{tikzpicture}
}
\end{array}.\label{eqn:FXrightiso_full}
\end{align}
The associator $\omega$ can be identified by composing Equation~\ref{eqn:FXleftiso} and Equation~\ref{eqn:FXrightiso_full} in both orders
\begin{align}
\omega^{-1}(g,h)
\begin{array}{c}
\includeTikz{FX_ass1}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
		\coordinate (A) at (0,-1);
		\coordinate (B) at (0,0);
		\coordinate (C) at (0,1);
		\Lmorphism[A][$*$][$0$][][][][$g$];
		\Rmorphism[B][][][][][][];
		\ladder[C][][][$-lh$][$*$][$0$];
		\draw(1.8,-.5)--(1.8,-1.1);
		\node[below,inline text] at (1.8,-1.1) {\footnotesize$h$};
	\end{tikzpicture}
}
\end{array}
&=
\begin{array}{c}
\includeTikz{FX_ass2}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,-1);
	\coordinate (B) at (0,0);
	\coordinate (C) at (0,1);
	\Lmorphism[C][][][][$*$][$0$][];
	\Rmorphism[A][$*$][$0$][][][][$h$];
	\ladder[B][][][$-lh$][][];
	\draw(-1.8,.1)--(-1.8,-1.1);
	\node[below,inline text] at (-1.8,-1.1) {\footnotesize$g$};
	\end{tikzpicture}
}
\end{array}\\
&=e^{-\frac{2\pi i}{p}ql gh}
\begin{array}{c}
\includeTikz{FX_ass3}
{
	\begin{tikzpicture}[scale=0.6, every node/.style={scale=0.8}]
	\coordinate (A) at (0,-1);
	\coordinate (B) at (0,0);
	\coordinate (C) at (0,1);
	\Lmorphism[A][$*$][$0$][][][][$g$];
	\Rmorphism[B][][][][][][];
	\ladder[C][][][$-lh$][$*$][$0$];
	\draw(1.8,-.5)--(1.8,-1.1);
	\node[below,inline text] at (1.8,-1.1) {\footnotesize$h$};
	\end{tikzpicture}
}
\end{array},
\end{align}
where the final equality is obtained using the associator on the left strand. This gives us
\begin{align}
F_q\otimes_{\vvec{\ZZ{p}}} X_l=F_{ql}.
\end{align}

\end{exmp}

\begin{center}\rule{.8\textwidth}{1pt}\end{center}

\section{The topological phase $\vvec{\ZZ{p}}$}
Given a fusion category $\mathcal{C}$, the associated (nonchiral) topological phase is described by $Z(\mathcal{C})$, the \emph{Drinfeld center} of $\mathcal{C}$. The particles of the theory correspond to the simple objects of $Z(\mathcal{C})$, whilst the morphisms describe braiding processes. Given a modular category $Z(\mathcal{C})$, one can use the Levin-Wen procedure\cite{Levin2005} to construct a lattice model that realizes the topological phase in its low energy space.

Bimodule categories $\mathcal{C}\curvearrowright\mathcal{M}\curvearrowleft\mathcal{D}$ are associated to domain walls between the topological phases $Z(\mathcal{C})$ and $Z(\mathcal{D})$. In this section, we discuss a physical interpretation of the bimodule categories in Table~\ref{tab:zpdata}. We also show how the wall fusion rules, including the multiplicities, in Table~\ref{tab:zptable} can be obtained from the physical theory.

The lattice model associated to $Z(\vvec{\ZZ{p}})$ has $p^2$ particle-like excitations, usually labelled $e^am^b$, with $a,b\in\ZZ{p}$. The fusion rules are $e^a\times e^b=e^{a+b}$ and $m^a\times m^b=m^{a+b}$ and the braiding is defined by $e^a m^b=e^{\frac{2\pi i}{p}ab}m^be^a$. On a torus, this model has $p^2$ degenerate ground states which can be used to encode a pair of $p$-dimensional logical spaces (referred to a qu$p$its). The logical Pauli gates are implemented by moving a particle around one of the cycles of the torus, so the encoded information is protected against local noise.

There are two possible boundary conditions that can be imposed on the model, commonly referred to as rough and smooth due to their lattice implementation\cite{MR2942952}. At a rough boundary, a particle $e^a$ can be `condensed' into the vacuum. In other words, fusing an $e$ particle into the boundary leaves the boundary unaffected. The smooth boundaries behave similarly for the $m$ sector. A single qu$p$it can be encoded into a strip of the phase with smooth (or rough) boundaries on both sides. By creating a hole or puncture to vacuum in the middle of the code, additional qu$p$its can be encoded\cite{0610082}. By moving these punctures around each other, logical gates can be applied to the encoded information\cite{0610082,Bombin2007a,Brown2016}.

The possible invertible domain walls for this theory are well understood\cite{Bridgeman2017}. These walls form the group $\dih{p-1}$, the dihedral group of order $2p-2$. By sweeping these walls across the lattice, additional logical gates can be applied to the encoded information.

By allowing interfaces between distinct domain walls, such as corners between rough and smooth boundaries, or terminations of invertible domain walls, additional qu$p$its can be encoded and manipulated\cite{Bombin2010,1606.07116,Brown2016}. A comprehensive study of the possible \emph{binary interface defects} is the subject of the companion paper \onlinecite{DefectsPaper}.

The group of invertible domain walls, along with the boundaries can be used to realize all bimodules. A domain wall can be built from a pair of boundaries by simply stacking them, for example a smooth and rough boundary
\begin{align}
	\begin{array}{c}
		\includeTikz{domainfromboundary1}
		{
			\begin{tikzpicture}[scale=.75]
				\fill[red!10](-1,-1) rectangle (-.25,1);
				\draw[thick](-.25,-1)--(-.25,1);
				\fill[red!10](1,-1) rectangle (.25,1);
				\draw[thick](.25,-1)--(.25,1);
				\foreach \x in {0,1,...,10}
				{
					\draw[shift={(0,1-.2*\x)}](0,0)--(.25,0);
				}
				\node at (-.625,0) {$\mathcal{C}$};\node at (.625,0) {$\mathcal{C}$};
			\end{tikzpicture}
		}
	\end{array}.
\end{align}

\begin{table}
\begin{tabular}{!{\vrule width 1pt}c!{\vrule width 1pt}c|c!{\vrule width 1pt}}
	\toprule[1pt]
	\rowcolor[gray]{.9}[\tabcolsep]Bimodule label & Domain wall & Action on particles\\
	\toprule[1pt]
	$T$&$\begin{array}{c}\includeTikz{T}{
		\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (-.25,1);
		\fill[red!10](1,-1) rectangle (.25,1);
		\draw[thick](-.25,-1)--(-.25,1);
		\draw[thick](.25,-1)--(.25,1);
		\foreach \x in {0,...,9}{\draw (.1,-.9+.2*\x)--(.25,-.9+.2*\x);};
		\foreach \x in {0,...,9}{\draw (-.1,-.9+.2*\x)--(-.25,-.9+.2*\x);};
		\end{tikzpicture}}
	\end{array}$&Condenses $e$ on both sides\\

	$L$&$\begin{array}{c}\includeTikz{L}{\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (-.25,1);
		\fill[red!10](1,-1) rectangle (.25,1);
		\draw[thick](-.25,-1)--(-.25,1);
		\draw[thick](.25,-1)--(.25,1);
		\foreach \x in {0,...,9}{\draw (.1,-.9+.2*\x)--(.25,-.9+.2*\x);};
		\end{tikzpicture}}\end{array}$&Condenses $m$ on left and $e$ on right\\

	$R$&$\begin{array}{c}\includeTikz{R}{\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (-.25,1);
		\fill[red!10](1,-1) rectangle (.25,1);
		\draw[thick](-.25,-1)--(-.25,1);
		\draw[thick](.25,-1)--(.25,1);
		\foreach \x in {0,...,9}{\draw (-.1,-.9+.2*\x)--(-.25,-.9+.2*\x);};
		\end{tikzpicture}}\end{array}$&Condenses $e$ on left and $m$ on right\\

	$F_0$&$\begin{array}{c}\includeTikz{F_0}{\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (-.25,1);
		\fill[red!10](1,-1) rectangle (.25,1);
		\draw[thick](-.25,-1)--(-.25,1);
		\draw[thick](.25,-1)--(.25,1);
		\end{tikzpicture}}\end{array}$&Condenses $m$ on both sides\\
	\toprule[1pt]
	$X_k$&$\begin{array}{c}\includeTikz{Xk}{\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (1,1);
		\draw[ultra thick,dashed] (0,-1)--(0,1);
		\end{tikzpicture}}\end{array}$&$X_k:e^am^b\mapsto e^{ka}m^{k^{-1} b}$ (moving left to right), where $k^{-1}$ is taken multiplicatively modulo $p$\\

	$F_{q}=F_1 X_q$&$\begin{array}{c}\includeTikz{Fq}{\begin{tikzpicture}[yscale=.3]
		\draw[white](0,-1.1)--(0,1.21);
		\fill[red!10](-1,-1) rectangle (1,1);
		\draw[ultra thick,dotted] (0,-1)--(0,1);\end{tikzpicture}}\end{array}$&
	$F_1:e^am^b\mapsto e^{b}m^{a}$\\
	\toprule[1pt]
\end{tabular}
	\caption{Domain walls on the lattice corresponding to bimodules}\label{tab:zpinterp}
\end{table}

Since there are 2 types of boundary, there are 4 non-invertible domain walls in $\vvec{\ZZ{p}}$. In Table~\ref{tab:zpinterp}, we associate a domain wall to each bimodule. The fusion of bimodules can be understood via this association. For example,
\begin{align}
	T\otimes_{\vvec{\ZZ{p}}} T\to
	\begin{array}{c}
		\includeTikz{TxTpic1}
		{
			\begin{tikzpicture}[scale=.75]
				\fill[red!10] (-2,-1)rectangle(-1.25,1);
				\draw[thick] (-1.25,-1)--(-1.25,1);
				\fill[red!10] (-.75,-1)rectangle(.75,1);
				\draw[thick,fill=none] (.75,-1)--(.75,1) (-.75,1)--(-.75,-1);
				\fill[red!10] (2,-1)rectangle(1.25,1);
				\draw[thick] (1.25,-1)--(1.25,1);
				\foreach \x in {0,...,9}
				{
					\draw[shift={(0,.1+.2*\x)}] (-1.25,-1)--(-1.1,-1);
					\draw[shift={(0,.1+.2*\x)}] (-.75,-1)--(-.9,-1);
					\draw[shift={(0,.1+.2*\x)}] (1.25,-1)--(1.1,-1);
					\draw[shift={(0,.1+.2*\x)}] (.75,-1)--(.9,-1);
				}
			\end{tikzpicture}
		}
	\end{array}
	&\propto
	\begin{array}{c}
		\includeTikz{TxTpic2}
		{
			\begin{tikzpicture}[scale=.75]
			\fill[red!10] (-1,-1)rectangle(-.25,1);
			\draw[thick] (-.25,-1)--(-.25,1);
			\fill[red!10] (1,-1)rectangle(.25,1);
			\draw[thick] (.25,-1)--(.25,1);
			\foreach \x in {0,...,9}
				{
					\draw[shift={(0,.1+.2*\x)}] (-.25,-1)--(-.1,-1);
					\draw[shift={(0,.1+.2*\x)}] (.25,-1)--(.1,-1);
				}
			\end{tikzpicture}
		}
	\end{array},
\end{align}
by discarding the isolated strip. The theory $Z(\vvec{\ZZ{p}})$ on a strip or tube with both rough (or both smooth) boundaries supports a $p$ dimensional vector space. These $p$ distinct states for the central strip give the coefficient $p$. On the other hand, if the opposite boundaries are not the same, the state is unique and the coefficient is 1
\begin{align}
	T\otimes_{\vvec{\ZZ{p}}} L\to
	\begin{array}{c}
		\includeTikz{TxLpic1}
		{
			\begin{tikzpicture}[scale=.75]
			\fill[red!10] (-2,-1)rectangle(-1.25,1);
			\draw[thick] (-1.25,-1)--(-1.25,1);
			\fill[red!10] (-.75,-1)rectangle(.75,1);
			\draw[thick,fill=none] (.75,-1)--(.75,1) (-.75,1)--(-.75,-1);
			\fill[red!10] (2,-1)rectangle(1.25,1);
			\draw[thick] (1.25,-1)--(1.25,1);
			\foreach \x in {0,...,9}
			{
				\draw[shift={(0,.1+.2*\x)}] (-1.25,-1)--(-1.1,-1);
				\draw[shift={(0,.1+.2*\x)}] (-.75,-1)--(-.9,-1);
				\draw[shift={(0,.1+.2*\x)}] (1.25,-1)--(1.1,-1);
			}
			\end{tikzpicture}
		}
	\end{array}
	&=
	\begin{array}{c}
		\includeTikz{TxTpic2}
		{
			\begin{tikzpicture}[scale=.75]
			\fill[red!10] (-1,-1)rectangle(-.25,1);
			\draw[thick] (-.25,-1)--(-.25,1);
			\fill[red!10] (1,-1)rectangle(.25,1);
			\draw[thick] (.25,-1)--(.25,1);
			\foreach \x in {0,...,9}
			{
				\draw[shift={(0,.1+.2*\x)}] (-.25,-1)--(-.1,-1);
				\draw[shift={(0,.1+.2*\x)}] (.25,-1)--(.1,-1);
			}
			\end{tikzpicture}
		}
	\end{array}.
\end{align}

The action of the invertible domain walls corresponding to $F_q$ swap the $e$ and $m$ sectors of the theory. The effect on the boundaries is to switch rough and smooth. From this, we obtain the same multiplication Table~\ref{tab:zptable}.

It follows from our computations that every domain wall between the $\protect\vvec{\ZZ{p}}$ theory and itself are either invertible, or have a strip of vacuum separating the two bulks. If $p \neq q$, then all the domain walls between the $\protect\vvec{\ZZ{p}}$ and $\protect\vvec{\ZZ{q}}$ theories have a strip of vacuum separating the two bulks. This is consistent with the computation using ladder string diagrams.


TITLE: Taking a limit of $f(x,y)$ using the definition
QUESTION [0 upvotes]: I am reviewing for a test in my Calculus III class and I ran into a problem. I need to prove that $\lim_{(x,y)\to(a,b)} x+y=a+b$ using the formal definition of a limit.
The definition given is:
Let $f$ be defined on the interior of a circle centered at the point $(a, b)$, except possibly
at $(a, b)$ itself. We say that $lim_{
(x,y)→(a,b)}f (x, y) = L$ if for every $ε > 0$ there exists a
$δ > 0$ such that $| f (x, y) − L| < ε$ whenever $0 <
\sqrt{(x − a)^2 + (y − b)^2} < δ$.
Can someone please help me? Thank you.

REPLY [1 votes]: Assume that $ε > 0$ is given. Note that $0 <
\sqrt{(x − a)^2 + (y − b)^2} < δ$ implies $|x-a| < \delta$ and $|y-b|<\delta$. So if we take $\delta=\frac{\epsilon}{2}$, then $|(x+y) - (a+b)| \leq | x-a| + |y-b| < 2\delta=\epsilon$.


TITLE: Eigenvalues of $A=\mathbf{u}_1\mathbf{u}_1^T + \mathbf{u}_2\mathbf{u}_2^T$
QUESTION [0 upvotes]: Let $\mathbf{u}_1, \mathbf{u}_2\in \mathbb{R}^{N\times 1}$ be real-valued vectors. Assume that $\mathbf{u}_1$ is orthogonal to $\mathbf{u}_2$ (i.e., $\mathbf{u}_1^T\mathbf{u}_2=0$).
Let us consider the following matrix $$A= \mathbf{u}_1\mathbf{u}_1^T + \mathbf{u}_2\mathbf{u}_2^T.$$
Could you please prove that $A$ has two non-zero eigenvalues: $\lambda_1=\|\mathbf{u}_1\|^2$ and $\lambda_2=\|\mathbf{u}_2\|^2$?

REPLY [2 votes]: What is $A\textbf{u}_1$? What is $A\textbf{u}_2$? If you have a vector $\textbf{v}$ that is orthogonal to both $\textbf{u}_1$ and $\textbf{u}_2$, what is $A\textbf{v}$?

REPLY [1 votes]: For example:
$$Au_1=\left(u_1u_1^t+u_2u_2^t\right)u_1=\overbrace{u_1u_1^t}^{=\left\|u_1\right\|^2}u_1+u_2\;\overbrace{u_2^tu_1}^{=0}=\left\|u_1\right\|^2\,u_1\implies\left\|u_1\right\|^2$$
is an eigenvalue with eigenvector $\;u_1\;$


TITLE: How to find the initial value of a solution to a differential equation in order to comply with certain limits
QUESTION [2 upvotes]: For the IVP, $y'+\frac{2x^2-4xy-y^2}{3x^2}=0, x>0, y(1)=y_0$
I got this, once all the calculations have been done:
$$y'=\frac{y_0^2(-2x^2+4x+1)+4y_0(x^2+x+1)-2(x^2+4x-2)}{(2+y_0-x(y_0-1))^2}$$
I am confident with this calculation, and I also verified with Wolfram Alpha. Then the question asked there is exactly one value of $y_0$ such that the IVP satisfied $\lim_{x\to 0}y'(x)\neq 1$, while $\lim_ {x\to0}y'(x)=1$ for all other values of $y_0$. What is this value of $y_0$ corresponding to the different limits?
So I took the limit
$$\lim_{x\to0}y'(x)=\frac{y_0^2+4y_0+4}{(2+y_0)^2}=1$$
So no matter what value of $y_0$ I have, the limit will always be 1. Where did I misunderstand or did wrong? Any help will be great, stuck about 2 days...

REPLY [2 votes]: This is, among other things, a Riccati equation. Set thus $y=-3x^2\frac{u'}{u}$ to find
$$
0=y'+\frac{2x^2-4xy-y^2}{3x^2}
=\left[-3x^2\frac{u''}{u}+3x^2\frac{u'^2}{u^2}-6x\frac{u'}{u}\right]
+\left[\frac23+4x\frac{u'}{u}-3x^2\frac{u'^2}{u^2}\right]
\\~\\
0=3x^2u''+2xu'-\frac23u
$$
This now is an Euler-Cauchy DE with characteristic polynomial
$$
0=3m(m-1)+2m-\tfrac23=\tfrac13((3m-\tfrac12)^2-2-\tfrac14)=\tfrac13(3m-2)(3m+1)
$$
So
\begin{align}
u&=Ax^{-1/3}+Bx^{2/3},~~~|A|+|B|\ne 0
\\~\\
\implies
y(x)&=-3x^2\frac{-\frac13Ax^{-4/2}+\frac23Bx^{-1/3}}{Ax^{-1/3}+Bx^{2/3}}
\\
&=x\frac{A-2Bx}{A+Bx}\\&=x-\frac{3Bx^2}{A+Bx}
\\
y'(x)&=1-\frac{6ABx+3B^2x^2}{(A+Bx)^2}
\end{align}
For $A\ne 0$ the solution has slope 1 in $x=0$, only for $A=0$ one gets a deviating result with $y(x)=-2x$, $y'(0)=-2$. Then $y_0=y(1)=-2$.

You excluded this case from consideration in the limit $\lim_{x\to 0}y'(x)$, but did not treat this case separately. The quotient law in this limit can only be applied for $y_0+2\ne 0$.

REPLY [2 votes]: Observe that for $y_0=-2$ the solution $y$ of the IVP has the property that $y'$ is constant:
$$y'(x)=-2.$$