text:
\begin{document}
\maketitle
\begin{abstract}
Let $G$ be a finite group, and $V$ a finite dimensional vector space over a field $\kk$ of characteristic dividing the order of $G$. Let $H \leq G$. The transfer map $\Tr_H^G: \kk[V]^H \rightarrow \kk[V]^G$ is an important feature of modular invariant theory. Its image is called a transfer ideal $I^G_H$ of $\kk[V]^G$, and this ideal, along with the quotients $\kk[V]^G/I^G_H$ are widely studied.

In this article we study $\kk[V]^G/I$, where $I$ is any sum of transfer ideals. Our main result gives an explicit regular sequence of length $\dim(V^G)$ in $\kk[V]^G/I$ when $G$ is a $p$-group. We identify situations where this is sufficient to compute the depth of $\kk[V]^G/I$, in particular recovering a result of Totaro. We also study the cases where $G$ is cyclic or isomorphic to the Klein 4 group in greater detail. In particular we use our results to compute the depth of $\kk[V]^G/I^G_{\{1\}}$ for an arbitrary indecomposable representation of the Klein 4 group.

\end{abstract}
\section{Introduction}

Let $\kk$ be an infinite field and $V$ a
finite-dimensional  $\kk$-vector space, and $G \leq \GL(V)$ a
finite group. Then the induced action  on $V^*$  extends to the
symmetric algebra $\kk[V]:=S(V^*)$ by the formula $$\sigma
(f)=f\circ \sigma^{-1}$$ for $\sigma \in G$ and $f\in \kk[V]$. The
algebra of fixed points $\kk[V]^G$ is called the \emph{ring of
invariants}, and is the central object of study in invariant
theory.

Now let $H$ be a subgroup of $G$. There is a well-defined $\kk$-linear map
\[\Tr^G_H: \kk[V]^H \rightarrow \kk[V]^G\]
\[f \mapsto \sum_{\sigma \in G/H} \sigma f\] which is called the {\it relative transfer} from $H$ to $G$. If $H$ is trivial we call this the {\it transfer} to $G$ and denote it by $\Tr^G$.

If $[G:H]$ is nonzero in $\kk$, then it is easy to show that $\Tr^G_H$ is surjective. In particular, $\Tr^G$ is a surjective map if $|G|$ is not divisible by the characteristic of $\kk$.

From now on suppose $\kk$ is a field of positive characteristic $p$ dividing $|G|$. Then if $H$ is a $p$-subgroup of $G$, but not a Sylow-$p$-subgroup, the transfer $\Tr^G_H$ is not surjective and its image is a proper ideal of $\kk[V]^G$. We denote this ideal by $I^G_H$. More generally, to any set $\mathcal{X}$ of subgroups of $G$ we associate the ideal

\begin{equation}\label{transferideal}
I^G_{\mathcal{X}} = \sum_{X \in \mathcal{X}} \Tr^G_X(\kk[V]^X).
\end{equation}

Transfer maps and their images are an important feature of modular invariant theory. For instance, a long-standing conjecture of Shank and Wehlau \cite{ShankWehlauTransfer} states that if $G$ is a $p$-group then $\kk[V]^G$ is polynomial if and only if $I^G_{\{1\}}$ is a principal ideal. Other investigations concern the transfer ideal $I:= I^G_{<P}$ where $P$ is a Sylow-$p$-subgroup and $<P$ denotes the set of all proper subgroups of $P$. This can be shown to be independent of the choice of Sylow-$p$-subgroup. Notable are a conjecture of Wehlau et al \cite{QueensReport} stating that $\kk[V]^G/I$ is always generated by invariants of degree $ \leq |G|$, and a recent result of Totaro \cite{Totaro} stating that $\kk[V]^G/I$ is always a Cohen-Macaulay ring. Both of these conform to a general philosophy that $\kk[V]^G/I$ behaves rather like a non-modular ring of invariants.

In the present article we consider quotient rings of the form $R^G_{\mathcal{X}}:= \kk[V]^G/I^G_{\mathcal{X}}$ where $G$ is a $p$-group and $\mathcal{X}$ is any set of proper subgroups of $G$. Our main result (Proposition \ref{regularsequence}) gives an explicit regular sequence of length $\dim(V^G)$ in $R^G_{\mathcal{X}}$. In particular, we recover Totaro's result in the special case of a $p$-group by elementary means. We also give some new results on the depth of $R^G_{\mathcal{X}}$ where $G$ is a cyclic $p$-group, and classify indecomposable representations $V$ of the Klein 4-group $G$ for which $\kk[V]^G/I_{\{1\}}^G$ is Cohen-Macaulay.

 \section{A regular sequence}
In this section let $\kk$ be a field and $A = \oplus_{i \geq 0}A_i$ a commutative graded $\kk$-algebra with $A_0=\kk$. Let $M$ be a graded $A$-module. An $M$-{\it regular element} is an element $a \in A$ such that the map $\phi_a: M \rightarrow M$ defined by
\[\phi_a(m) = am\] is injective. An $M$-{\it regular sequence} is a sequence $a_1,a_2, \ldots, a_r$ such that $a_1$ is $M$-regular and for all $i=2, \ldots, r$, $a_i$ is $M/(a_1, \ldots, a_{i-1})M$-regular.

An $M$-regular sequence is called {\it maximal} if it cannot be extended to a longer $M$-regular sequence. If $A$ is Noetherian and $M$ is finite, then it can be shown that all maximal $M$-regular sequences have the same length. Define $A_+  = \oplus_{i \geq 1}A_i$.  Then we call the length of a maximal $M$-regular sequence in  $A_+$ the {\it depth} of $M$. The depth of $A$ is the depth of $A$ as a module over itself.

For further results on depth and regular sequences we recommend \cite{BrunsHerzog}.

Now let $G$ be a $p$-group and adopt the notation introduced in the introduction. Let $v_1, \ldots, v_l$ be a basis of the fixed-point space $V^G$ and extend to a basis $v_1, \ldots, v_n$ of $V$. Let $x_1, x_2, \ldots, x_n$ be the corresponding dual basis of $V^*$. Then $\kk[V]$ is the polynomial ring generated by $x_1, x_2, \ldots, x_n$. For $i=1, \ldots, l$ we define
\begin{equation}\label{norms}
N_i = N^G(x_i) = \prod_{\sigma \in G} \sigma x_i.
\end{equation}

Notice that $N_i$ is a monic polynomial of  degree $|G|$ as a polynomial in $x_i$, and degree zero as a polynomial in $x_j$ for any $j=1, \ldots, i-1,i+1, \ldots, l$. To see the second statement, note that
\[N_i(v_j) = \prod_{\sigma \in G} (\sigma x_i)v_j = \prod_{\sigma \in G} x_i(\sigma^{-1}v_j) = \prod_{\sigma \in G} x_i(v_j) = 0.\]
Further, let $\mathcal{X}$ be a set of proper subgroups of $G$. Given $f \in \kk[V]^G$, we denote by $\overline{f}$ the image of $f$ in $R^G_{\mathcal{X}}$. With this notation we have the following which was proven for a cyclic group of prime order in \cite[Theorem~12]{MR3437279}.

\begin{prop}\label{regularsequence}
\[\overline{N_1}, \ldots, \overline{N_l}\] is a regular sequence in $R^G_{\mathcal{X}}$.
\end{prop}

\begin{proof} We first show that $\overline{N_i}$ is regular on $R^G_{\mathcal{X}}$ for any $i=1, \ldots, l$. Let $i \in \{1, \ldots, l\}$ and suppose there exists $f \in \kk[V]^G$ such that $\overline{f}\overline{N_i} = 0$ in $R^G_{\mathcal{X}}$. In turn this means for each $X \in \mathcal{X}$, there exists a polynomial $g_X$ such that
\begin{equation}
fN_i = \sum_{X \in \mathcal{X}} \Tr^G_X(g_X).
\end{equation}

We view $g_X$ as a polynomial in  $x_i$ with coefficients in $\kk[x_1, \ldots, x_{i-1},x_{i+1},x_n]$.   Since $N_i$ is a monic polynomial in $x_i$, we may divide $g_X$ by $N_i$, finding unique polynomials $q_X$ and $r_X$ such that
\[g_X = q_X N_i + r_X\] where the $x_i$-degree of $r_X$ is $<|G|$.  As $v_i \in V^G$, the action of $G$ on $\kk[V]$ does not increase $x_i$-degrees so the uniqueness of this decomposition implies that $q_X, r_X \in \kk[V]^X$ and so
\[fN_i = \sum_{X \in \mathcal{X}} \Tr_X^G(q_XN_i + r_X) = \sum_{X \in \mathcal{X}} N_i \Tr_X^G(q_X)+\Tr^G_X(r_X).\]
This can be rearranged as
\[0 = (f-\sum_{X \in \mathcal{X}}\Tr^G_X(q_X))N_i + \sum_{X \in \mathcal{X}}\Tr^G_X(r_X).\]
Since $\deg_{x_i}\Tr^G_X(r_X) < |G|$ for each $X$, comparing the $x_i$-degrees in the above equation gives
\[f-\sum_{X \in \mathcal{X}}\Tr^G_X(q_X)) = 0,\]
i.e. $f \in I^G_{\mathcal{X}}$, which shows that $\overline{f} =0$ as desired. Therefore, $\overline{N_i}$ is regular on $R^G_{\mathcal{X}}$.

Now assume $1<j \leq l$ and $\overline{N_1}, \ldots, \overline{N_{j-1}}$ is a regular sequence in $R^G_{\mathcal{X}}$. Suppose $\overline{N_j}$ is not regular on $R^G_{\mathcal{X}}/(\overline{N_1}, \ldots, \overline{N_{j-1}})R^G_{\mathcal{X}}$. This means there exist $f_1, \ldots,f_j \in \kk[V]^G$ such that
\[\overline{f_1}\overline{N_1} + \ldots + \overline{f_j}\overline{N_j} = 0\]
in $R^G_{\mathcal{X}}$. This in turn means there exists, for each  $X \in \mathcal{X}$, an element $g_X \in \kk[V]^X$ such that

\begin{equation}
f_1N_1+ \ldots+ f_jN_j = \sum_{X \in \mathcal{X}} \Tr^G_X(g_X).
\end{equation}

We now divide each $f_i$ ( for $i=2 \ldots j)$ by $N_1$ with remainder, writing
\[f_i = q_{i,1}N_1+r_{i,1}.\]
 We obtain for the right-hand side
\[f_1N_1+ ((q_{2,1})N_1+r_{2,1})N_2 + ((q_{3,1})N_1+r_{3,1})N_3 + \ldots + ((q_{j,1})N_1+r_{j,1})N_j.\]
By collecting together all terms divisible by $N_1$ to replace $f_1$, and replacing each $f_i$ by $r_{i,1}$, we can assume $f_i$ has degree $< |G|$ as a polynomial in $x_1$. Continuing this process, dividing by each $N_i$ $(i=2, \ldots, l)$ in turn and relabeling if necessary, allows us to assume $f_i$ has degree $<|G|$ in all variables $x_1, \ldots, x_{i-1}$. Further, for each $X \in \mathcal{X}$ we write
 \[g_X = q_{X,1} N_1 + q_{X,2}N_2 + \ldots q_{X,j}N_j + r_X\] where the degree of $q_{X,i}$ is $<|G|$ in each variable $x_1, \ldots, x_{i-1}$ and the degree of $r_X$ is $<|G|$ in each of $x_1, \ldots, x_j$. With further relabeling (bringing terms $\Tr^G_X(q_{X,i})N_i$ to the left hand side) we obtain the expression

\begin{equation}
f_1N_1+ \ldots+ f_jN_j = \sum_{X \in \mathcal{X}} \Tr^G_X(r_X)
\end{equation}
in which each $f_i$ has degree $<|G|$ as a polynomial in each variable $x_1, \ldots, x_{i-1}$ and each $r_X$ has degree $< |G|$ as a polynomial in each variable $x_1, \ldots, x_j$.

Now notice that that, as a polynomial in $x_1$, the term $f_1N_1$ has degree $\geq |G|$ unless $f_1=0$. All other terms in the expression have $x_1$-degree $< |G|$. This shows that $f_1=0$. Considering degrees in $x_2, \ldots, x_{j-1}$ now shows that $f_2 = f_3 = \ldots =f_{j-1} = 0$, too. So we're left with
\[f_jN_j = \sum_{X \in \mathcal{X}} \Tr^G_X(r_X).\]
Since we already know that $N_j$ is regular on $R^G_{\mathcal{X}}$, we must have $f_j = 0$. This completes the proof.

\end{proof}

\section{Upper Bounds}
Proposition \ref{regularsequence} implies that
\[\depth(R^G_{\mathcal{X}}) \geq \dim(V^G)\]
for any set $\mathcal{X}$ of proper subgroups of the $p$-group $G$. In order to compute the exact depth of $R^G_{\mathcal{X}}$, we need an upper bound for depth. It is well-known that, for any Noetherian $\kk$-algebra $A$, $\depth(A) \leq \dim(A)$, see e.g. \cite[Proposition~1.2.12]{BrunsHerzog}. With this in mind, we want to compute $\dim(R^G_{\mathcal{X}})$.

As a first step, we note that $I^G_{\mathcal{X}} = I^G_{\overline{\mathcal{X}}}$, where $\overline{\mathcal{X}}$ is obtained from $\mathcal{X}$ by including all subgroups of every $X \in \mathcal{X}$ and their $G$-conjugates. This follows easily from e.g. \cite[Equation~(1)]{FleischmannTransfer}. Now we have

\begin{prop}
\label{dim}
\[\dim(R^G_{\mathcal{X}}) = \max \{\dim(V^Q): Q \not \in \overline{\mathcal{X}}\}.\]
\end{prop}

\begin{proof} Note that $\kk[V]^G$ can be interpreted as the ring of regular functions on the quotient $V/G$. Then by definition we have
\[\dim(R^G_{\mathcal{X}}) = \dim(\mathcal{V}(I^G_{\mathcal{X}}))\] where, for an ideal $J$ of $\kk[V]^G$, $\mathcal{V}(J)$ denotes the set of $G$-orbits vanishing on $J$. By \cite[Proposition~12.4(iii)]{FleischmannTransfer} we have
\[\mathcal{V}(I^G_{\mathcal{X}}) = i_G^*(\bigcup_{Q \le G: Q \not \in \overline{\mathcal{X}}} V^Q)\] where $i_G$ is the inclusion map $\kk[V]^G \hookrightarrow \kk[V]$ and $i_G^*: V \rightarrow V/G$ is its dual. As the dimension of an algebraic variety is the largest dimension of an irreducible component, the result follows.
\end{proof}

Now in case $\mathcal{X} = \{X: X < G\}$ we get $$\dim(V^G) \leq \depth(R^G_{\mathcal{X}}) \leq  \dim(R^G_{\mathcal{X}}) = \dim(V^G).$$ Consequently we obtain

\begin{cor}[Totaro]\label{totaro}
$R^G_{<G}$ is a Cohen-Macaulay ring.
\end{cor}

For stronger upper bounds, we use the following, which is inspired by \cite[Theorem~3.17]{FleischmannCohomology}.

\begin{prop}\label{ub} Let $\mathcal{X}$ be a family of subgroups of $G$. Let $K\le H$ for all $H \in \mathcal{X}$. If $((\bigcap_{H \in \mathcal{X}} I_K^H)\setminus I_K^G)\cap \kk[V]^G\neq \emptyset$, then we have $$\depth(R^G_K)\le \dim(R^G_\mathcal{X}).$$
\end{prop}
\begin{proof} Let $H \in \mathcal{X}$ and let $f\in (I_K^H\setminus I_K^G)\cap \kk[V]^G$. Then $f=\Tr^H_K(g)$ for some $g\in \kk[V]^K$. For $h \in\kk[V]^H$  we have $$\Tr^G_K(gh)=\Tr^G_H(\Tr^H_K(gh))=\Tr^G_H(h\Tr^H_K(g))=\Tr^G_H(hf)=f\Tr^G_H(h).$$ This  shows that $I^G_H$ annihilates $f+I^G_K\in \kk[V]^G/I_K^G$. It follows that $I^G_{\mathcal{X}}$ annihilates $f+I^G_K\in \kk[V]^G/I_K^G$.
Therefore $I^G_{\mathcal{X}}$ is contained in one of the associated primes of $\kk[V]^G/I_K^G$ in $\kk[V]^G$. Since the depth of a module is smaller than the coheights of its associated primes, see \cite[Proposition~1.2.13]{BrunsHerzog}, the result follows.
\end{proof}

\section{Cyclic groups}
In this section $G$ denotes a cyclic group of order $p^r$. There are exactly $p^r$ indecomposable $\kk G$-modules $V_1, \dots , V_{p^r}$ and each indecomposable module $V_n$ is afforded by a Jordan block of size $n$.  We fix a generator $\sigma$ of $G$.
We choose a basis $e_1, e_2, \dots , e_n$ for $V_n$ such that the action of $\sigma$  is given by $\sigma (e_i)=e_i+e_{i-1}$ for $2\le i\le n$ and $\sigma (e_1)=e_1$.  Define $\Delta=\sigma-1$. We consider the subgroups $K\subseteq H$
of $G$ generated by $\sigma^{p^a}$ and $\sigma^{p^b}$, respectively. We identify a case when the premise of  Proposition \ref{ub} is attained.

\begin{Lemma}

Assume that    $\kk [V]$ contains a summand which is  isomorphic to $V_n$ with  $p^a-p^b+1 \le n\le p^a-1$. Then $(I^H_K\setminus I^G_K)\cap \kk [V]^G \neq \emptyset$.
\end{Lemma}
\begin{proof}
Clearly, $e_1\in V_n^G$. We demonstrate that $e_1\in (I^H_K\setminus I^G_K)$. Note that $\Tr^H_K$ applies to $K$-invariants and a basis element $e_i$ is in $V_n^K$ if and only if  $\sigma^{p^a}(e_i)=e_i$, or equivalently $\Delta^{p^a}(e_i)=0$. It follows that $V_n^K=V_n$ since $n\le p^a-1$.
The set $1, \sigma^{p^b}, \sigma^{2p^b}, \dots ,\sigma^{p^a-p^b}$ is a complete set of representatives for $H/K$. Therefore we have
\begin{alignat*}{2} \Tr^H_K=&(1+\sigma^{p^b}+\sigma^{2p^b}+ \cdots +\sigma^{p^a-p^b})\\
=&(1+\sigma+\sigma^2+ \cdots +\sigma^{p^{a-b}-1})^{p^b}\\
=&(\Delta^{p^{a-b}-1})^{p^b}=\Delta^{p^a-p^b}.
\end{alignat*}
 Then by assumption on $n$, we get $e_1\in I^H_K $ because $\Tr^H_K(e_{p^a-p^b+1})=e_1$ and $e_{p^a-p^b+1}\in V_n^K$. Similarly, $1, \sigma, \dots , \sigma^{p^a-1}$ is a complete set of representatives for $G/K$ and so  $\Tr^G_K=(1+\sigma+\sigma^2+ \cdots +\sigma^{p^a-1})=\Delta^{p^a-1}$. It follows that $e_1\notin I^G_K$ because $\Delta^{p^a-1}(V_n)=0$ by assumption on $n$.
\end{proof}

For subgroups  $K\subseteq H$ in $G$ we have the decomposition  $\Tr^G_{K}=\Tr^G_{H}  \circ\Tr^{H}_{K}$, so $I^G_{K}\subseteq I^G_{H}$. It follows that for a collection
$\mathcal{X}$ of subgroups, we have $I^G_{\mathcal{X}}=I^G_{H}$, where $H$ is the largest subgroup in $\mathcal{X}$. So we consider just subgroups of $G$, rather than collections of subgroups.

\begin{prop}
Let $K$ be a subgroup of $G$, generated by $\sigma^{p^a}$. If there is $0< b \le a$ such that there is an indecomposable summand $V_n$ of $\kk [V]$ with
$p^a-p^b+1 \le n\le p^a-1$, then $$\depth(R^G_K)\le \dim(V^{\sigma^{p^{b-1}}}) \le p^{b-1}\dim(V^G).$$
\end{prop}
\begin{proof}
Let $H$ denote the subgroup of $G$ generated by $\sigma^{p^{b}}$. Then from the  previous lemma we have that $(I^H_K\setminus I^G_K)\cap \kk [V]^G   \neq \emptyset$. So Proposition \ref{ub} applies for  $\mathcal{X}=\{H\}$ and we get that
$\depth(R^G_K)\le \dim (R^G/I^G_H)$. On the other hand $\dim (R^G/I^G_H)=\max \{\dim(V^Q): H\subsetneqq Q \}$ by Proposition \ref{dim}. Since $G$ is cyclic, we have that $\max \{\dim(V^Q): H\subsetneqq Q \}=\dim(V^{\sigma^{p^{b-1}}})$ as desired. For the second inequality, note that for each summand $V_m$ of $V$ with basis $e_1, e_2, \dots ,e_m$, $V_m^G$ is spanned by $e_1$ and $V_m^{\sigma^{p^{b-1}}}$ is spanned by $e_1, \dots , e_k$, where $k=\min (p^{b-1}, m)$.
\end{proof}

We note an application for cyclic $2$-groups.
\begin{cor}
Let $G$ be a cyclic $2$-group of order $2^r$ and let $V=V_t$ be the indecomposable module of dimension $t\le 2^r$, where $t$ is odd. Then $\depth (R^G_{\{1\}})=\dim (V^G)$.
\end{cor}
\begin{proof}
Note that the $d$-th symmetric power $S^d(V_t)$ is isomorphic to $\Omega^{-d} (\wedge^d(V_{2^r-t}))$ modulo induced modules, by \cite[Corollary~3.12]{SymondsCyclic}. Putting $d=2^r-t$ we get  $\wedge^d(V_{2^r-t}) = V_1$ and as $d$ is odd we have $\Omega^{-d}(V_1) = V_{2^r-1}$. Therefore, $V_{2^r-1}$ is a summand of the $d$-th homogeneous component of $\kk [V_t]$. So applying the previous proposition for $b=1$ and trivial $K$ yields  $\depth (R^G_{\{1\}})\le \dim (V^G)$. The reverse inequality follows from Proposition \ref{regularsequence}.
\end{proof}
\section{The Klein 4-group}

In this section we consider the action of the group $G = \langle \sigma_1,\sigma_2 \rangle \cong C_2 \times C_2$ on an indecomposable $\kk G$-module $V$, where $\kk$ is a field of characteristic 2. For each $V$, the depth of $\kk[V]^G$ was computed in \cite{ElmerFleischmann}. Further, if $\mathcal{X}$ contains all three proper subgroups of $G$, $R^G_{\mathcal{X}}$ is Cohen-Macaulay by Corollary \ref{totaro}. We will use the results of previous sections to compute $\depth(R^G_{\{1\}})$, for each indecomposable representation $V$ of $G$. Note that the results of Section 3 imply
\begin{equation}\label{ineq}
    \dim(V^G) \leq \depth(R^G_{\{1\}}) \leq \dim(R^G_{\{1\}})=\max_{1<Q<G} \dim(V^Q)
\end{equation}
and $R^G_{\{1\}}$ is Cohen-Macaulay if and only if the second inequality is an equality.

The classification of indecomposable modules for this group is given in \cite{Benson1}, and it is from this source we obtain our notation. To summarize, if $V$ is an indecomposable $\kk G$-module then $V$ is isomorphic to a module in the following list:

\begin{itemize}
    \item An even-dimensional representation of dimension $n$, denoted $V_{n, \lambda}$, where $\lambda \in \kk \cup \{\infty\}$;
    \item The representations $\Omega^m(\kk)$ where $m \in \Z$ and $\Omega$ represents the Heller shift operator. This representation has dimension $n = 2|m|+1$.
    \item The regular representation $\kk G$;
\end{itemize}

We analyse each case separately.

\subsection{Even dimensions, not projective}
First suppose $V \cong V_{n, \lambda}$ where $\lambda \in \kk$. Write $n=2m$. Then we can choose a basis $\{v_1,v_2, \ldots, v_m, w_1, \ldots, w_m\}$ with respect to which the action of $G$ is as follows:

\[\sigma_i(v_j)=v_j \ \qquad i=1,2\]
\[\sigma_1(w_i) = w_i+v_i\]
\[\sigma_2(w_i) = w_i+\lambda v_i+ v_{i-1}\]
where we set $v_0=0$. $V_{n, \infty}$ is obtained from $V_{n,0}$ by swapping the roles of $\sigma_1$ and $\sigma_2$. Note that, since the three modules $V_{n,0}$, $V_{n,1}$ and $V_{n,\infty}$ are linked by a group automorphism, they have isomorphic invariant rings, and in particular $\depth(R^G_{\{1\}})$ is the same for all three modules. So from now on we assume $\lambda \in \kk$ and $\lambda$ is invertible.

Now it is  clear that $\dim(V^G) = m$. Moreover, if $\lambda \neq 1$ then $\dim(V^Q)=m$ for each maximal subgroup of $G$. So we obtain immediately from Equation \ref{ineq} that $\depth(R^G_{\{1\}}) = m$ and this quotient ring is Cohen-Macaulay.

In case $\lambda=1$ we have $\dim(V^{\sigma_1\sigma_2}) =m+1$. So $\dim(R^G_{\{1\}}) =m+1$. We claim that $\depth(R^G_{\{1\}}) = m$, so this quotient is not Cohen-Macaulay. Note that $V_{2,1}$ is not faithful, so we may assume $n \geq 4$.

To see this, consider the basis $\{x_1, \ldots, x_m,y_1, \ldots, y_m\}$ of $V^*$, dual to the given basis of $V$. The action of $G$ on this basis is given by

\[\sigma_i(y_j)=y_j \ \qquad i=1,2\]
\[\sigma_1(x_i) = x_i+y_i\]
\[\sigma_2(x_i) = x_i+y_i+y_{i+1}\]
where $y_{m+1}=0$.

Now we have $$y_m = \Tr^{\langle\sigma_1\sigma_2\rangle}(x_{m-1}) \in I^{\langle\sigma_1\sigma_2\rangle}_{\{1\}}.$$
Moreover $y_m \in \kk[V]^G$ and $y_m \not \in I^G_{\{1\}}$ because $V$ and hence $V^*$ is not projective. Applying Proposition
\ref{ub} with $\mathcal{X} = \{\langle\sigma_1\sigma_2\rangle\} $ and $K$ trivial brings
\[\depth(R^G_{\{1\}}) \leq \dim(R^G_{\langle \sigma_1\sigma_2 \rangle}) = m.\]
Combining this with Equation \ref{ineq} proves the claim.

\subsection{Odd dimensions - negative Heller shift.}

Suppose $V$ is isomorphic to $\Omega^m(\kk)$ where $m<0$ (for $m=0$ this representation is not faithful). Then
we can choose a basis $\{v_1,v_2, \ldots, v_{m+1}, w_1, \ldots, w_m\}$ with respect to which the action of $G$ is as follows:

\[\sigma_i(v_j)=v_j \ \qquad i=1,2\]
\[\sigma_1(w_i) = w_i+v_i\]
\[\sigma_2(w_i) = w_i+v_{i+1}\]

Now its clear that $\dim(V^G) = m+1$. Moreover, $\dim(V^Q)=m+1$ for each maximal subgroup of $G$. So we obtain immediately from Equation \ref{ineq} that $\depth(R^G_{\{1\}}) = m+1$ and this quotient ring is Cohen-Macaulay.

\subsection{Odd dimensions - positive Heller shift.}

Suppose $V$ is isomorphic to $\Omega^m(\kk)$ where $m>0$. Then
we can choose a basis $\{v_1,v_2, \ldots, v_{m}, w_1, \ldots, w_{m+1}\}$ with respect to which the action of $G$ is as follows:

\[\sigma_i(v_j)=v_j \ \qquad i=1,2\]
\[\sigma_1(w_i) = w_i+v_i\]
\[\sigma_2(w_i) = w_i+v_{i-1}\] where we set $v_{m+1}=v_0=0$.

Now it's clear that $\dim(V^G) = m$. Further, $\dim(V^{\sigma_1\sigma_2})= m$ but $\dim(V^{\sigma_i}) = m+1$ for $i=1,2$. So $\dim(R^G_{\{1\}}) = m+1$.

Assume $m \geq 2$. We claim that $\depth(R^G_{\{1\}})=m$ and this quotient ring is not Cohen-Macaulay.

To see this, consider the basis $\{x_1, \ldots, x_m,y_1, \ldots, y_{m+1}\}$ of $V^*$, dual to the given basis of $V$. The action of $G$ on this basis is given by

\[\sigma_i(y_j)=y_j \ \qquad i=1,2\]
\[\sigma_1(x_i) = x_i+y_i\]
\[\sigma_2(x_i) = x_i+y_{i+1}.\]

 Then we have
\[\Tr^{\langle \sigma_1 \rangle}(x_m) = \Tr^{\langle \sigma_2 \rangle}(x_{m-1})=y_m.\] Moreover $y_m \in \kk[V]^G$ and $y_m \not \in I^G_{\{1\}}$ because $V$ and hence $V^*$ is not projective. Applying Proposition
\ref{ub} with $\mathcal{X} = \{\langle\sigma_1\rangle, \langle \sigma_2\rangle\} $ and $K$ trivial brings
\[\depth(R^G_{\{1\}}) \leq \dim(R^G_{\mathcal{X}}) = \dim(V^{\sigma_1\sigma_2})= m.\]
Combining this with Equation \ref{ineq} proves the claim.

For $m=1$ the ring of invariants is easy to compute: Let $N(x_1) = \prod_{g \in G}gx_1$. Note that $N(x_1)$ is $x_1^4$ modulo the ideal generated by $y_1, y_2$ in $\kk[V]$. So the common zero set of the invariants $y_1, y_2, N(x_1)$ contains only the point zero, so this set is a homogeneous system of parameters for $\kk[V]^G$. But since the product of their degrees is the group order, from a standard fact in invariant theory  we get that
$\kk[V]^G = \kk[y_1,y_2,N(x_1)]$. In particular, $\kk[V]^G$ is a polynomial ring. Note that the ideal $\kk[V]_+^G \kk[V]$ is generated by $y_1, y_2, x_1^4$. Therefore $1, x_1, x_1^2, x_1^3$ forms a  $\kk $-basis for $\kk [V]/\kk[V]_+^G \kk[V]$.  Since $\Tr^G$ is a $\kk[V]^G$-linear map, it follows that $I^G_{\{1\}}$ is generated by $\Tr^G(1), \Tr^G(x_1), \Tr^G(x_1^2), \Tr^G(x_1^3)$. But $\Tr^G(1)=\Tr^G(x_1)= \Tr^G(x_1^2)=0$ and so $I^G_{\{1\}}$ is a principal ideal, generated by $\Tr^G(x_1^3) = y_1^2y_2+y_1y_2$. Clearly this is a regular element of $\kk[V]^G$.  Therefore by \cite[Theorem~2.1.3]{BrunsHerzog}, $R^G_{\{1\}}$ is Cohen-Macaulay.

\subsection{The regular representation}
Let $V_R$ denote the regular  $\kk G$-module with a  basis $v_1, v_2, v_3,v_4$. The action of $\sigma_1, \sigma_2$ on $V_R$ is given by the permutations  $(1,2)(3,4)$ and $(1,3)(2,4)$, respectively. That is, we have $\sigma_i (v_j)=v_{\sigma_i (j)}$ for $i=1,2$ and $1\le j\le 4$. Note that  $\dim(V^{\sigma_1})=\dim(V^{\sigma_2})=\dim(V^{\sigma_1\sigma_2})=2$ and so by Proposition \ref{dim} we have $\dim(R^G_{\{1\}})=2$. We demonstrate that $\depth (R^G_{\{1\}})=2$ as well. Consider the  basis $x_1, x_2, x_3, x_4$ of $V_R^*$ dual to the given basis of $V_R$. Since the inverse transpose of a matrix representing a product of disjoint tranpositions is equal to itself, the action of $G$ on $V_R^*$ is again given by
$$\sigma_i (x_j)=x_{\sigma_i (j)} \text { for } i=1,2 \text { and } 1\le j\le 4.$$
Since  $G$ permutes the variables,  $\kk [V_R]^G=\kk [x_1, x_2, x_3, x_4]^G$ is generated as a vector space by orbit sums $o(m)$ of monomials in $\kk [x_1, x_2, x_3, x_4]$. Note that $o(m)\in I^G_{\{1\}}$ for a monomial $m$ if the orbit sum contains four monomials. Otherwise the monomial $m$ has a non-trivial stabilizer and so $o(m)\notin I^G_{\{1\}}$. In this case the orbit sum $o(m)$ contains one or two monomials. We show that $N:=x_1x_2x_3x_4$ and $H:=o(x_1x_2)+o(x_1x_3)+o(x_1x_4)=x_1x_2+x_3x_4+x_1x_3+x_2x_4+x_1x_4+x_2x_3$ form a regular sequence in $R^G_{\{1\}}$. Assume that $N(\sum_t c_to(m_t))\in I^G_{\{1\}}$, where $c_t\in \kk $ and  $ o(m_t)\notin I^G_{\{1\}}$. So we may assume that each $o(m_t)$ contains one or two monomials. Note that the number of monomials in the orbits of $m_t$ and $Nm_t$ are the same and $N\sigma_i (m_t)=\sigma_i(Nm_t)$. Therefore $N(\sum_t c_to(m_t))=\sum_t c_to(Nm_t)$ which is a sum of orbits with one or two monomials. It follows that  $N(\sum_t c_to(m_t))\notin I^G_{\{1\}}$.

Next we show that $H$ is a non-zero divisor in the quotient ring  $R^G_{\{1\}}/NR^G_{\{1\}}=\kk [x_1, x_2, x_3, x_4]^G/(I^G_{\{1\}}, N)$. Note that the ideal $(I^G_{\{1\}}, N)$ is spanned by orbit sums $o(m)$ of monomials where $o(m)$ has 4 monomials or $m$ is divisible by $N$. Assume that $H(\sum_t c_to(m_t))\in (I^G_{\{1\}}, N)$ with $\sum_t c_to(m_t)\notin (I^G_{\{1\}}, N)$. Since $H$ is a homogeneous polynomial and  $(I^G_{\{1\}}, N)$ is a homogeneous ideal, we may assume $\sum_t c_to(m_t)$ is homogeneous as well. Each $m_t$ has two monomials in its orbit (if it has one, then $m_t$ is divisible by $N$), so each $m_t$  is fixed by one of the $\sigma_1, \sigma_2, \sigma_1\sigma_2$. It follows that $x_1, x_2, x_3, x_4$ group in two pairs and the  variables in the same pair appear with the same multiplicity in $m_t$. But $m_t$ is not divisible by $N$ so variables in one of the pairs do not appear at all in $m_t$.   Therefore we may assume that $m_t$ is equal to either $x_1^dx_2^d$, $x_1^dx_3^d$ or $x_1^dx_4^d$ for some positive integer $d$. So we have $H(c_1o(x_1^dx_2^d)+c_2o(x_1^dx_3^d)+c_3o(x_1^dx_4^d))\in (I^G_{\{1\}}, N)$. Note that if $c_1\neq 0$, then $x_1^{d+1}x_2^{d+1}$ appears in $Ho(x_1^dx_2^d)$. But  $x_1^{d+1}x_2^{d+1}$ does not appear in $Ho(x_1^dx_3^d)$ or in $Ho(x_1^dx_4^d)$. So  if $c_1\neq 0$, then $x_1^{d+1}x_2^{d+1}$ appears in
  $H(c_1o(x_1^dx_2^d)+c_2o(x_1^dx_3^d)+c_3o(x_1^dx_4^d))$. But $x_1^{d+1}x_2^{d+1}$ is fixed by $\sigma_1$ and it so it has two monomials in its orbit and we get that it does not appear in a polynomial in $(I^G_{\{1\}}, N)$. This gives $c_1=0$. Along the same lines one sees that $c_2=0$ and $c_3=0$.
  \subsection{Decomposable Representations}
   We let $V_1 \oplus V_2$ denote the direct sum of the
 $\kk G$-modules $V_1$ and $V_2$,  and let $kV$ denote the direct sum of $k$ copies of a $\kk G$-module $V$.  Assume that $V$ is isomorphic to $c_{n,\lambda}V_{n, \lambda}+d_m \Omega^m(\kk)$. Set $c=\sum c_{n, 1}$ and $d=\sum_{m>0}d_m$. We prove that the Cohen-Macaulay defect of $R^G_{\{1\}}$ is bounded above by $\max \{c, d\}$.
 \begin{prop}
 Assume the notation of the previous paragraph. Then we have $\dim (R^G_{\{1\}})- \depth (R^G_{\{1\}})\le \max \{c, d\}$. In particular, if $V$ is a direct sum of modules of the form $V_{n,\lambda}$ with $\lambda\neq 1$ and  $\Omega^m(\kk)$ with $m<0$, then $R^G_{\{1\}}$ is Cohen-Macaulay.
 \end{prop}
 \begin{proof}
 Note that the dimensions of the fixed point spaces
 of $\sigma_i$ and $G$ are equal in all summands except $\Omega^m(\kk)$ for $m>0$ and $i=1,2$. Furthermore $\dim (\Omega^m(\kk)^{\sigma_i})=\dim (\Omega^m(\kk)^{G})+1$ for  $m>0$ and $i=1,2$. It follows that $\dim (V^{\sigma_i})=\dim (V^G)+d$ for $i=1,2$. On the other hand the dimensions of the fixed point spaces
 of $\sigma_1\sigma_2$ and $G$ are equal in all summands except $V_{n,1}$, and $\dim (V_{n,1}^{\sigma_1 \sigma_2})=\dim (V_{n,1}^G)+1$. Therefore $\dim (V^{\sigma_1\sigma_2})=\dim (V^G)+c$. So from Proposition \ref{dim} we get $\dim (R^G_{\{1\}})=\dim (V^G)+\max \{c, d\}$. Since the depth of  $R^G_{\{1\}}$ is at least $\dim (V^G)$ by Proposition \ref{regularsequence}, the result follows.
\end{proof}
\bibliographystyle{plain}
\bibliography{MyBib}

\end{document}

config: arxiv


TITLE: Cofinality of $\omega _1$
QUESTION [4 upvotes]: I need a little help regarding one of the basic properties of $\omega _1$.
Namely, I am supposed to show that for any $\alpha \in \omega_1$ there exists $\beta \in \omega_1\cap A$ such that $\alpha < \beta$.
Here $A$ stands for a set $A:=$ { $\alpha$ $ |$ $ \alpha \in Ord$ $\wedge$ $cof(\alpha)=\omega$ }.
What I am trying to show is that for any $\alpha \in \omega_1$ $\exists \beta \in \omega_1 \cap Lim$, where $Lim$ is the class of all limit ordinals such that $\alpha < \beta$.
This will be enough since every countable limit ordinal has cofinality $\omega$ which is not so hard to show.
My problem is, how to show that this defined $\beta$ indeed exists, i.e. $\beta$ being $Lim$.

REPLY [5 votes]: Define $\beta = \sup \{ \alpha + n \mid n < \omega \}$. Then $\beta > \alpha$ and $\beta$ has cofinality $\omega$; moreover, it is less than $\omega_1$ since it is a countable union of countable sets, so is countable.


TITLE: Is this assumption reasonable?
QUESTION [0 upvotes]: In physics book which I use right now proposes a simple deduction for drag equation. Almost every part of it is nice. But when deducting, it says: Whenever object hits an air molecule (which we assume to be stationary before the collision) it accelerates it to velocity $v$. In here, $v$ is velocity of a falling object through air.
The thing made me confused is part that says implicitly, "air molecules accelerate into velocity $v$" It's like author assumes it. (Like an assumption) But shouldn't, since molecules' mass is so less than free falling object, molecules must accelerate into much higher velocities compared to $v$?
I mean, electrons are very sensitive, and also, molecules are sensitive like electrons -but not exactly as electrons are-. Electrons go frenzy when they are interacted ,and with that, shouldn't molecules go frenzy; not as the same degree as electrons but you know, much higher velocities than just $v$? Is this a reasonable assumption -that molecules of air accelerate to $v$- when deriving drag equation, or I am missing a point? Can you help me? Thanks from now.

REPLY [0 votes]: yes molecules are a little frenzy / vibrate, but their average velocity is 0 (for example the air in a room doesn't move from a macroscopic view). Likely the author is assuming that the falling object is dragging with him(they don't bounce) the air molecules , that start with an average velocity v = 0 and then accelerate at the same speed of the falling object.


TITLE: Can the square root of certain quantities be turned into another square root plus an arbitrary constant
QUESTION [0 upvotes]: So I was sitting in math class and this hit me, can the square root of a quantity with a constant in it be equivalent to another square root plus an arbitrary constant?
What I did was something like $\frac{\sqrt{9x^2+5}}{2x^3}$ and turned it into $\frac{\sqrt{9x^2}+C}{2x^3}$
I was thinking this could work because each case is only in incremented by a constant, even though the first one is under a square root
I plugged in different numbers for x and then solved for $C$, assuming they would be equal at all values, and what I kept getting was close(especially when the value I chose was around 1) but not exactly the same(except at that one value of x).
Some simple algebra got me a general solution for this approximation as$$\frac{\sqrt{a(x)}+\sqrt{a(k)+C_a}-\sqrt{a(k)}}{b(x)}$$ where a(x) is the original numerator function($9x^2$), b(x) is the denominator function($2x^3$), $C_a$ is the original constant under the radical(5), and k is the chosen point to approximate off of. This approximation does change whether or not the approximation function would look like $\frac{\sqrt{a(x)}+C}{b(x)}$ or $\frac{\sqrt{a(x)}}{b(x)}+C$, this is the approximation for the former
Would it be possible to get an exact other equation, with keeping the square root term the same as before and have to modify the constant with a quantity that changes with x, or would the previous square root also have to change to accommodate the constant being removed?

REPLY [0 votes]: ...[C]an the square root of a quantity with a constant in it be equivalent to another square root plus a ... constant?

Absolutely yes. $\sqrt{7} = 2 + c$ for some constant $c$.
The problem here is not that we cannot perform some kind of substitution $\sqrt{a^2 + b} \rightarrow a+c$. Such a step is reasonable. The problem is that $c$ is not actually a constant independent of $a$ or $b$ which you can see directly by trying to get an expression for it. Because your constant is not independent from the variable in question, you're not actually eliminating the square root you want to eliminate but hiding it in some function $C(x)$. If there are other constraints on the problem such that the values of $a$ and $b$ are already fixed then $c$ is indeed a constant. But that's basically not saying much, since if $a$ and $b$ are fixed the entire expression is a constant. Oh well!


TITLE: Free algebras on sets of different cardinality – for what theories are they non-isomorphic?
QUESTION [4 upvotes]: Following the case of groups, I asked in this MSE question for a quick proof that given a free-forgetful adjunction $F\dashv U$ for some algebraic theory, we have $X\not\cong Y\implies FX\not\cong FY$. I was really surprised to find out this isn't true at all.
On the other hand, it seems that for the usual theories, i.e of (abelian) groups, (commutative) rings, modules, monoids, this is true, usually by finding a canonical way to vector spaces and using the dimension theorem. What is the underlying property of such theories that makes this theorem true? Can it be stated without reference to vector spaces, using only the axiom of choice? Is there a nice sufficient condition?

REPLY [3 votes]: Jónsson and Tarski (Math. Scand. 9 (1961), 95-101, link) proved the following:

Consider a variety $\mathcal{V}$ of algebras of a certain signature. If $\mathcal{V}$ contains a finite algebra of cardinal $\ge 2$, then the free algebra $F_n(\mathcal{V})$ in $\mathcal{V}$ on $n$ elements cannot be generated by less than $n$ elements. In particular, $F_n(\mathcal{V})$ and $F_m(\mathcal{V})$ are not isomorphic for any $0\le n<m$ for $n$ finite.

This applies in many cases, for which the result itself is trivial or not: set with no law, free lattices, free magmas, free semigroups, free groups, free associative commutative unital ring, free associative ring, etc.
They also provide a counterexample without the assumption that $\mathcal{V}$ contains a finite algebra of cardinal $\ge 2$. Namely, consider the variety $\mathcal{V}_2$ with signature $(2,1,1)$, encoding a set $M$ with a bijection $M^2\to M$. Then the free algebras $F_n(\mathcal{V}_2)$ for finite $n\ge 1$ are all isomorphic. (This is called the variety of Jónsson-Tarski algebras, or pairing functions.)
(It was later checked by Higman that if we consider the variety $\mathcal{V}_k$ encoding a bijection $M^k\to M$, then $F_n(\mathcal{V}_k)$ and $F_m(\mathcal{V}_k)$, for finite $n,m\ge 1$, are isomorphic if and only if $k-1$ divides $m-n$.)
They also mention the variety of $M$-sets, where $M$ is the monoid $\langle u,v\mid uv=1\rangle$: although it has the given property, the free algebra on 1 generator in this algebra can be non-freely generated by a single element. (They also give a sufficient condition in variety ensuring that every generating $n$-tuple in the free $n$-generated algebra is free, namely residual finiteness of free algebras.)


TITLE: Nets and directed sets problem
QUESTION [2 upvotes]: I am trying to solve the following exercise:
Let $\Lambda$ be a directed set, and for each $\alpha \in \Lambda$ let $\Gamma_\alpha$ be a directed set. Suppose that for each $\alpha \in \Lambda$ there is a net $({x_k}^{\alpha})_{k \in \Gamma_\alpha}$ that converges to $x^{\alpha} \in X$, and such that $(x^{\alpha})_{\alpha \in \Lambda}$ converges to $x \in X$. Consider $\Phi=\Lambda \times \prod_{\alpha \in \Lambda} \Gamma_{\alpha}$ ordered by the product, which means
$$ (\alpha, (k_\beta)_{\beta \in \Lambda}) \geq  (\alpha', (k_\beta')_{\beta \in \Lambda}) \space \text{if and only if} \space \alpha \geq \alpha' \space \text{and} \space k_{\beta}\geq k_{\beta}' \space \forall \beta \in \Lambda$$
Then the net $(\alpha, (k_\beta)_{\beta \in \Lambda}) \rightarrow {x_{k_{\alpha}}}^{\alpha}$ converges to $x$.
$$$$$$$$ I am trying to understand the net they ask me to prove it converges, $\alpha$ determines that $(\alpha, (k_\beta)_{\beta \in \Lambda})$ will be send to an element of the net $\Gamma_{\alpha}$. Now, I am not so sure what the subscript $k_{\alpha}$ determines. I don't completely get how the map $(\alpha, (k_\beta)_{\beta \in \Lambda}) \rightarrow {x_{k_{\alpha}}}^{\alpha}$ works, I would appreciate if someone could explain me that and a hint to show the statement is true. Thanks in advance

REPLY [2 votes]: Assuming you already understand that the set $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$ with the given ordering is directed, then the proposed net (in the space $X$) works like this.  To be a net in $X$, it has to associate,  to each element of the index set, some point in $X$. So look at an element of the index set $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$; it's a pair consisting of an element $\alpha\in\Lambda$ and an element of $\prod_{\alpha\in\Lambda}\Gamma_\alpha$.  The latter is, by definition of products, a system $(k_\beta)_{\beta\in\Lambda}$ indexed by $\Lambda$, such that, for each $\beta\in\Lambda$, the element $k_\beta$ with index $\beta$ is in $\Gamma_\beta$. Given such a pair $(\alpha,(k_\beta)_{\beta\in\Lambda})$, the net ignores most of the system $(k_\beta)_{\beta\in\Lambda}$, and pays attention only to the one element in this system with index $\alpha$, namely $k_\alpha$. In other words, it uses the first component $\alpha$ of the pair $(\alpha,(k_\beta)_{\beta\in\Lambda})$ to select one element $k_\alpha$ from the second component.  Then it goes back to the originally given system of nets $(x_k^\alpha)_{k\in\Gamma_\alpha}$, looks at the $\alpha$-th of these nets, and picks out the $k_\alpha$-term from that net.
To show that this complicated-looking net converges to $x$, I'd consider an arbitrary open neighborhood $U$ of $x$ and see what I can prove to be in $U$. Since the net $(x^\alpha)_{\alpha\in\Lambda}$ converges to $x$, there must be some $\alpha'\in\Lambda$ such that $x^\alpha\in U$ for all $\alpha\geq\alpha'$. Then for each such $\alpha$, the fact that $(x^\alpha_k)_{k\in\Gamma_\alpha}$ converges to a point $x^\alpha$ in the open set $U$ means that there must be some $k'_\alpha$ such that $x^\alpha_k\in U$ for all $k\geq k'_\alpha$ in $\Gamma_\alpha$.  I've almost defined an element $(\alpha',(k'_\alpha)_{\alpha\in\Lambda})$ of $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$, for which I hope that all larger elements are mapped into $U$ by the complicated-looking net.  There are two catches: (1) I've defined $k'_\alpha$ only for $\alpha\geq\alpha'$; what about other $\alpha$'s in $\Lambda$? (2) I've only hoped, not proved, that all larger elements of $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$ get mapped into $U$.  My answer to (1) is extremely cheap: choose the remaining $k'_\alpha$'s, for $\alpha\not\geq\alpha'$, arbitrarily in the corresponding $\Gamma_\alpha$'s. My answer to (2) is even cheaper: I'll leave that to you because I've already done too much work.


TITLE: Complex numbers question
QUESTION [6 upvotes]: Consider 3 nonzero complex numbers $z_1,z_2,z_3$ each satisfying $z^2=i \bar{z}$. We are supposed to find $z_1+z_2+z_3, z_1z_2z_3,  z_1z_2+z_2z_3+z_3z_1$.
The answers- $0$, purely imaginary , purely real respectively.
I have no idea how to proceed. I tried to use the expansion for $(a+b+c)^2$ for them, but I am not getting anywhere. Please help. Thanks in advanced!

REPLY [11 votes]: Taking the modulus gives $|z|^2=|i\overline{z}|=|z|$ hence $|z|=1$ since $z\not=0$. Multiplying both sides of the original equation by $z$ then gives:
$$z^3=i\overline{z}z=i|z|^2=i$$
that is $z^3-i=0$.
If you call the solutions of that $z_1$, $z_2$ and $z_3$, how can you find the numbers $z_1+z_2+z_3$, $z_1z_2z_3$ and $z_1z_2+z_2z_3+z_3z_1$ straight from the equation?

REPLY [3 votes]: Try writing $z = a + ib$ for $a,b \in \mathbb{R}$ and expand $z^2$ and $iz^-$. Then compare real and imaginary coefficients in the two expansions.


text:
\section{Icosahedral algebra: invariants, and equivariants}

An account of the algebraic objects that emerge from the icosahedral action on \CP{1} appears in other places (\cite{klein}, \cite{dm}, and \cite{soccer}).  Results relevant to the task at hand appear without discussion.

Denote by \I\ the \alt{5}-isomorphic group of $60$ rotational symmetries of the regular icosahedron as a graph structure on the sphere.  (See Figure~\ref{fig:5tet}.)  Three polynomials generate the ring of \I-invariants:
$$\C[x,y]^\I=\left<F(x,y), H(x,y), T(x,y)\right>$$
where $(x,y)$ are homogeneous coordinates on \CP{1}.  The forms $F$, $H$, and $T$ vanish at the special \I-orbits: the $12$ vertices $v_k$, $20$ face-centers $f_k$, and $30$ edge-midpoints $e_k$ respectively.  For ease of reference, call the members of these sets ``$12$-points," etc.  We can express the generating invariants as products:
\begin{align*}
F=&\ \prod_{k=1}^{12} (x-v_k y)= x y \left(x^{10}-11x^5 y^5-y^{10}\right)\\
H=&\ \prod_{k=1}^{20} (x-f_k y)= x^{20}+228x^{15} y^5+494x^{10} y^{10}-228x^5
y^{15}+y^{20} \\
T=&\ \prod_{k=1}^{30} (x-e_k y)=  x^{30}-522x^{25} y^5-10005x^{20} y^{10}-10005x^{10} y^{20}+522x^5 y^{25}+y^{30}.
\end{align*}

Accordingly, $F$ and $H$  are algebraically independent, while we can arrange for an algebraic combination of the two generators in degree $60$ to vanish (with multiplicity two) at the $30$-points:
$$T^2=H^3-1728\,F^5.$$

We also need the system of invariants for each of the five tetrahedral subgroups $\T{1},\dots,\T{5}$ of \I.  Each \T{k} acts as an alternating group \alt{4} on a set of four $20$-points.  Overall, these disjoint sets occupy the vertices of five regular tetrahedra. Figure~\ref{fig:5tet} shows the icosahedral net and the vertical decomposition into tetrahedral sets.  Taking $k=5$, there are two degree-four relative \T{5} invariants: one, $q_5$, given by the product that involves the tetrahedral vertices and the other, $\Hat{q}_5$, given by the product that uses the tetrahedral face-centers (antipodal to the vertices).  The results are
\begin{align*}
q_5=&\  \frac{1}{4} \bigl(4 x^4+(2+2 i \sqrt{15}) x^3 y+(6-2 i \sqrt{15}) x^2 y^2+(-2-2 i \sqrt{15}) x y^3+4 y^4\bigr)\\
\Hat{q}_5=&\ \frac{1}{2} \bigl(2 x^4+ (1-i \sqrt{15})x^3 y)+(3+i \sqrt{15}) x^2 y^2-
(1-i\sqrt{15}) x y^3+2 y^4\bigr).
\end{align*}
\begin{figure}[ht]
	\centering

	\includegraphics[width=4in]{5tet.png}
	\caption{Configuration of five tetrahedral vertices}

	\label{fig:5tet}

\end{figure}
A \emph{relative invariant} is a form for which a non-trivial multiplicative character appears under the group's action.  For instance, for some $A\in \mathcal{T}_5$,
$$q_5(A(x,y))=\lambda_A\, q_5(x,y)\qquad \lambda_A\neq 1.$$

Using in a product the six-point tetrahedral orbit associated with edges  gives an absolute \T{5} invariant:
$$t_5= x^6-2 x^5 y-5 x^4 y^2-5 x^2 y^4+2 x y^5+y^6.$$
In this case, $t_5(A(x,y))=t_5(x,y)$ for all $A\in \mathcal{T}_5$.
The product of the degree-four forms yields a degree-eight \T{5} invariant:
$$
u_5=q_5 \widehat{q}_5=x^8+x^7 y+7 x^6 y^2-7 x^5 y^3+7 x^3 y^5+7 x^2 y^6-x y^7+y^8.
$$
Since the eight zeroes of $u_5$ have order-three symmetry, they are also face-centers of the icosahedron.  Hence, $H$ is divisible by $u_5$ and the quotient is a \T{5} invariant of degree $12=20-8$:
\begin{align*}
m_5=\frac{H}{u_5}=&\ x^{12}-x^{11} y-6 x^{10} y^2+20 x^9 y^3+15 x^8
y^4+24 x^7 y^5+11 x^6 y^6-24 x^5 y^7+15 x^4
y^8\\
&\ -20 x^3 y^9-6 x^2 y^{10}+x y^{11}+y^{12}.
\end{align*}
The tetrahedral invariants satisfy a  degree-$24$ relation
$$
m_5^2=\frac{1}{64} \left(95 t_5^2 m_5-40 t_5^4+9 u_5^3\right).
$$
Applying powers of an order-$5$ element $P\in \I$ manufactures the remaining \T{k} invariants:
$$
t_k=t_5\circ P^k  \qquad u_k=u_5\circ P^k \qquad m_k=m_5\circ P^k \qquad  k=1,\dots,4.
$$
In the chosen coordinates, we can take $P(x,y)=(\epsilon^3 x,\epsilon^2 y)$ where $\epsilon=e^{2 \pi i/5}$.

From a generating \I-invariant, we can construct an \I-equivariant (or \I-map) of one less degree using a ``cross" operator $\times$:
\begin{align*}
\phi=&\ \times F=(-\partial_y F,\partial_x F)= \bigl(-x^{11}+66 x^6 y^5+11 x y^{10},11 x^{10} y-66 x^5 y^6-y^{11}\bigr)\\
\eta=&\ \times H=(-\partial_y H,\partial_x H)\\
=&\ 20 \bigl(-57 x^{15} y^4-247 x^{10} y^9+171 x^5 y^{14}-y^{19} ,
x^{19}+171 x^{14} y^5+247 x^9 y^{10}-57 x^4 y^{15}\bigr).
\end{align*}
These maps behave in an elegant manner: $\phi$ twists and wraps a dodecahedral  face $\mathcal{F}$ onto the $11$ faces that comprise the complement of the face antipodal to $\mathcal{F}$ while $\eta$ does the analogous twisting and wrapping for an icosahedral face.  For edges of the respective polyhedra we can take great circle arcs between vertices to obtain sets that are forward invariant under the respective map.  Call this structure a \emph{dynamical polyhedron}.  Moreover, each map expands the internal angle of a face in its dynamical polyhedron onto an external angle of the antipodal face.  The vertices are thereby periodic critical points and their superattracting basins are full-measure subsets of \CP{1}.  The Doyle-McMullen iteration uses $\phi$ whose  attracting set is a special orbit.  Hence, \alt{5} symmetry is partially broken allowing for the extraction of two roots.

To break \alt{5} symmetry fully, we look for a map $g$ whose critical set \CC{g} is a generic $60$-point \I-orbit that is permuted under the action of $g$.  All maps of this sort have degree $31$ and were found in \cite{icos31}.  Excepting two cases, the dynamical polyhedra associated with these special ``$31$-maps" are derived from the icosahedral structure; they consist of twelve pentagons, twenty triangles, and thirty quadrilaterals.  The resulting configuration is called a $B_{62}$.  (It also goes by the awkward name rhombicosidodecahedron.).


TITLE: Does a lattice in $SL_n(\mathbb R)$ which is contained in $SL_n(\mathbb Z)$ have finite index in $SL_n(\mathbb Z)$?
QUESTION [6 upvotes]: A lattice $H$ in a locally compact group $G$ is a discrete subgroup such that the coset space $G/H$ admits a finite $G$-invariant measure.
I have read several places that any lattice H in $\operatorname{SL}_n(\mathbb{R})$ which is contained in $\operatorname{SL}_n(\mathbb{Z})$ must have finite index in $\operatorname{SL}_n(\mathbb{Z})$. But I have been unable to prove this.
I have tried using the correspondence between the Haar measure on $\operatorname{SL}_n(\mathbb{R})$ and the counting measure on $\operatorname{SL}_n(\mathbb{Z})$, where we can partition $\operatorname{SL}_n(\mathbb{R})$ into sets each containing one element of $\operatorname{SL}_n(\mathbb{Z})$, and then normalizing s.t. each of these has measure one. But this seemed to lead nowhere. Also just restricting the measure on $\operatorname{SL}_n(\mathbb{R})/H$ to $\operatorname{SL}_n(\mathbb{Z})/H$ does not work either since the latter has measure zero.
Thanks a lot to the ones who will answer.

REPLY [3 votes]: First, for $H\subset \Gamma\subset G$ with $G$ unimodular, $\Gamma$ discrete, fixing a Haar measure on $G$, there is a unique $G$-invariant measure on $\Gamma\backslash G$ such that
$$
\int_{\Gamma\backslash G} \sum_{\gamma\in \Gamma} \varphi(\gamma\cdot g)\;dg \;=;\ \int_G \varphi(g)\;dg
$$
for all $\varphi\in C^o_c(G)$. Suppose $\Gamma\backslash G$ has finite measure. Similarly, by the same general uniqueness results, there is a unique measure on $H\backslash G$ such that
$$
\int_{\Gamma\backslash G} \sum_{\gamma\in H\backslash \Gamma} \varphi(hg)\;dg
\;=\; \int_{H\backslash G} \varphi(g)\;dg
$$
for all $\varphi\in C^o_c(H\backslash G)$. This set-up answers most questions about the trio $H\subset \Gamma \subset G$. For example, yes, if $H\backslash G$ has finite volume, then $H$ must be of finite index in $\Gamma$, or else the sum over $H\backslash \Gamma$ is infinite...


TITLE: Calculating electric field in a dielectric
QUESTION [0 upvotes]: Why the electric field calculated in a dielectric is equal to the uniform charge density per unit area divided by epsilon-0 (which is the permitivity in space) shouldn't it be divided by another constant?

REPLY [0 votes]: Yes. It should be divided by the relative permittivity.


text:
\subsection{The general framework}

The general online optimization framework that we consider can be represented as the following sequence of events (presented for the moment from the viewpoint of a single, focal agent):
\begin{itemize}
\item
At each stage $\run=\running$, of the process, the agent picks an \emph{action} $\state_{\run}$ from a compact convex subset $\points$ of a $\vdim$-dimensional real space $\vecspace$.
\item
The choice of action generates a \emph{reward} $\est\pay_{\run} = \pay_{\run}(\state_{\run})$ based on a concave function $\pay_{\run}\from\points\to\R$ (assumed unknown to the player at stage $\run$).
\item
Simultaneously, $\state_{\run}$ triggers a \emph{delay} $\delay_{\run} \geq 0$ which determines the round $\run + \delay_{\run}$ at which the generated reward $\est\pay_{\run}$ will be received.
\item
The agent receives the rewards from all previous rounds $\info_{\run} = \setdef{\runalt}{\runalt + \delay_{\runalt} = \run}$, and the process repeats.
\end{itemize}

The above model has been stated in an abstract way that focuses on a single agent so as to provide the basis for the analysis to come.
The setting where there are no assumptions on the process generating the agent's payoff functions will be referred to as the \emph{unilateral} setting;
by contrast, in the multi-agent, \emph{game-theoretic} setting, the payoff functions of the focal agent will be determined by the stream of actions of the other players (see below for the details).
In the latter case, all variables other than the running counter $\run$ will be indexed by $\play$ to indicate their dependence on the $\play$-th player;
for example,
the action space of the $\play$-th player will be written $\points^{\play}$,
the corresponding action chosen by at stage $\run$ will be denoted $\state_{\run}^{\play}$,
etc.
For concreteness, we provide a diagrammatic illustration in \cref{fig:delays} above.

\begin{figure}[tbp]
\begin{center}
\small
\input{Figures/Delays.tex}
\end{center}
\caption{Schematic illustration of the delayed feedback framework considered in the paper.
Arrows illustrate the round to which the payoff is deferred.}
\label{fig:delays}
\end{figure}

In both cases, our blanket assumptions for the stream of payoffs and the delays encountered by the players will be as follows:

\begin{assumption}
\label{asm:pay}
For each $\run=\running$, $\pay_{\run}$ is
concave in $\point$,
$\vbound$-Lipschitz continuous,
and
$\smooth$-Lipschitz smooth.
Specifically, the gradient $\payv_{\run}(\point) \equiv \nabla_{\point} \pay_{\run}(\point)$ of $\pay_{\run}$ is bounded by $\vbound$ and satisfies $\norm{\payv_{\run}(\pointalt) - \payv_{\run}(\point)} \leq \smooth \norm{\pointalt - \point}$ for all $\point,\pointalt\in\points$.
\end{assumption}

\begin{assumption}
\label{asm:delay}
The delays $\delay_{\run}$ grow asymptotically as $\delay_{\run} = o(\run^{\alpha})$ for some $\alpha<1$.
\end{assumption}

Regarding the delay assumption above, large-scale analytic studies have shown that long delays \emph{are} observed in practice:
in a study by \citet{Chapelle} with data from the real-time bidding company Criteo, it was found that more than $10\%$ of the conversions were more than two weeks old.
Moreover, the conclusion of the same study was that the distribution of delays in online advertising can be fitted reasonably well by long-tailed distributions,

especially when conditioning on context and feature variables available to the advertiser, thus justifying the assumption of a possibly unbounded delay between choosing an action and receiving a reward.

We also note here that we are making no further assumptions on the way the sequence of delays is generated:
conceivably, delays could even be determined adversarially, as in \citet{QK15}.

\subsection{Multi-agent considerations}

For the multi-agent case, suppose there is a finite set of players $\players = \{1,\dotsc,\nPlayers\}$, each with their own action space $\points^{\play} \subseteq \R^{\vdim^{\play}}$ (always assumed convex and compact).
In this case, it will be convenient to encode the players' joint action profile $\point = (\point^{\play})_{\play\in\players} \in \points \equiv \prod_{\play\in\players} \points^{\play}$ by means of the shorthand $(\point^{\play};\point^{-\play}) \equiv (\point^{1},\dotsc,\point^{\play},\dotsc,\point^{\nPlayers})$ which highlights the action $\point^{\play}\in\points^{\play}$ of the $\play$-th player against the action profile $\point^{-\play} \in \points^{-\play} \equiv \prod_{\playalt\neq\play} \points^{\playalt}$ of $\play$'s opponents.
The payoff to each player $\play\in\players$ for a given action profile $\point \in \points$ will then be determined by an associated \emph{payoff \textpar{\emph{or} utility} function} $\pay^{\play}\from\points\to\R$, assumed here and throughout to be concave in the action variable $\point^{\play}$ of the player in question.
We will refer to the tuple $\game \equiv \game(\players,\points,\pay)$ as an \emph{$\nPlayers$-player continuous game} \citep{Deb52,Ros65,FT91}.

In this context, if $\state_{\run} = (\state_{\run}^{1},\dotsc,\state_{\run}^{\nPlayers}) \in \points$ is a sequence of joint actions, the payoff function encountered by the $\play$-th player at stage $\run$ will be given by
\begin{equation}
\label{eq:pay-play}
\pay_{\run}^{\play}(\point^{\play})
	\equiv \pay^{\play}(\point^{\play};\state_{\run}^{-\play})
	\quad
	\text{for all $\point^{\play} \in \points ^{\play}$},
\end{equation}
leading to the gradient expression
\begin{align}
\label{eq:payv-t}
\payv_{\run}^{\play}(\point^{\play})
	&\equiv \nabla_{\point^{\play}} \pay_{\run}^{\play}(\point^{\play};\state_{\run}^{-\play})
	= \payv^{\play}(\point^{\play};\state_{\run}^{-\play})
\intertext{where}
\label{eq:payv}
\payv^{\play}(\point)
	&= \nabla_{\point^{\play}} \pay^{\play}(\point^{\play};\point^{-\play}).
\end{align}
denotes the individual payoff gradient of the $\play$-th player at the action profile $\point\in\points$.
In the rest of our paper, we will assume that $\pay^{\play}$ is Lipschitz continuous and Lipschitz smooth, so \cref{asm:pay} is satisfied by default in this case.

\subsection{Regret and equilibrium}

With these preliminaries at hand, our principal performance indicators will be the minimization of \emph{regret} and the notion of a \acli{NE}.
Starting with the former, the regret of an agent in the unilateral setting is defined over a horizon of $\nRuns$ stages as
\begin{equation}
\label{eq:regret}
\reg(\nRuns)
	= \max_{\point\in\points} \sum_{\run=1}^{\nRuns} \bracks{\pay_{\run}(\point) - \pay_{\run}(\state_{\run})}.
\end{equation}
and, in the presence of randomness, we similarly introduce the agent's mean (or pseudo-) regret as
\begin{equation}
\label{eq:reg-mean}
\overline\reg(\nRuns)
	= \max_{\point\in\points}
		\exof{\sum_{\run=1}^{\nRuns}
			\bracks{\pay_{\run}(\point) - \pay_{\run}(\state_{\run})}}.
\end{equation}
Accordingly, we will say that a sequence of actions $\state_{\run}\in\points$, $\run=\running$, leads to \emph{no regret} if $\reg(\nRuns) = o(\nRuns)$.

On the other hand, the notion of a \acdef{NE} is a purely game-theoretic concept which characterizes those action profiles that are resilient to unilateral deviations.
In more detail, we say that $\eq \in \points$ is a \acl{NE} of $\game$ when
\begin{equation}
\label{eq:NE}
\tag{NE}
\pay^{\play}(\eq)
	\geq \pay^{\play}(\point^{\play};\eq^{-\play})

\end{equation}
for all $\point^{\play}\in\points^{\play}$ and all $\play\in\players$.
In full generality, the relation between \aclp{NE} and regret minimization is feeble at best:
if all players play a \acl{NE} for all $\run=\running$, they will trivially have no regret;
the converse however fails by a longshot, see \eg \citet{VZ13} and references therein.
\footnote{Specifically, \citet{VZ13} show that ther are games whose set of coarse correlated equilibria contain strategies that assign positive probability \emph{only} to strictly dominated strategies.}

In the game-theoretic literature, existence and uniqueness of equilibrium points has been mainly studied under a condition known as \acdef{DSC} \citep{Ros65}, which we define here as:
\begin{equation}
\label{eq:DSC}
\tag{DSC}
\sum_{\play\in\players}
	\lambda^{\play} \braket{\payv^{\play}(\pointalt) - \payv^{\point}(\point)}{\pointalt^{\play} - \point^{\play}}
	< 0

\end{equation}
for some $\lambda^{\play}>0$ and all $\point,\pointalt\in\points$ with $\pointalt\neq\point$.

In optimization, this condition is known as \emph{monotonicity} \citep{BC17}, so we will interchange the terms ``diagonally strictly concave'' and ``\emph{monotone}''for games that satisfy \eqref{eq:DSC}.
Under \eqref{eq:DSC}, \citet{Ros65} showed the existence of a unique \acl{NE};
this is of particular importance to online advertising because of the following auction mechanism that can be seen as a monotone game:

\medskip
\begin{example}[Kelly auctions]
\label{ex:Kelly}
Consider a provider with a splittable commodity (such as advertising time or website traffic to which a given banner will be displayed).
Any fraction of this commodity can be auctioned off to a set of $\nPlayers$ bidders (players) who can place monetary bids $\point^{\play}\geq0$ up to each player's total budget $b^{\play}$ to acquire it.
Once all players have placed their respective bids, the commodity is split among the bidders proportionally to each player's bid;
specifically, the $\play$-th player gets a fraction

\(
\rho^{\play}
	= \point^{\play} \big/ \parens{c + \sum_{\playalt\in\players} \point_{\playalt}}
\)

of the auctioned commodity (where $c\geq0$ is an ``entry barrier'' for bidding on the resource).
A simple model for the utility of player $\play$ is then given by the \emph{Kelly auction mechanism} \citep{KMT98}:
\begin{equation}
\label{eq:pay-auction}
\pay^{\play}(\point^{\play};\point^{-\play})
	= g^{\play} \rho^{\play} - \point^{\play},
\end{equation}
where
$g^{\play}$ represents the marginal gain of player $\play$ from a unit of the commodity.
Using standard arguments, it is easy to show that the resulting game satisfies \eqref{eq:DSC}, see \eg \citet{Goo80}.
\end{example}

Other example of games satisfying \eqref{eq:DSC} are (strictly) convex-concave zero-sum games \citep{JNT11}, routing games \citep{NRTV07}, Cournot oligopolies \citep{MZ19}, power control \citep{SFPP10,MBNS17}, etc.
For an extensive discussion of monotonicity in game theory, see \citet{FK07,PSPF10,San15,LRS19} and references therein.
In the rest of our paper, we will assume that all games under consideration satisfy \eqref{eq:DSC}.


TITLE: Closed non-intersecting geodesics on a compact hyperbolic surface are finite
QUESTION [2 upvotes]: I recently came across this exercise.
Let $S$ be a closed orientable surface of genus strictly greater than one and let $g$ be a Riemannian metric on S with constant negative curvature. Then, given any infinite family of simple closed geodesics on S, there are at least two geodesics in this family that intersect in a point (in other words, a family of closed simple geodesics that are disjoint two by two must be finite).
So far, I only managed to prove that, if $(\gamma_n)_{n\in\mathbb{N}}$ is a sequence of simple closed geodesics, $\gamma_n : S^1 \to S  $, then
$$
\inf_{n,m\in\mathbb{N}}\mathrm{d}(\gamma_n,\gamma_m)=0,
$$
where $\mathrm{d}$ is the distance induced by $g$ on $S$. This infimum must be zero because, for any $t\in S^1$, the sequence $(\gamma_n(t))_{n\in\mathbb{N}}$ is contained in $S$, which is compact, so it admits a converging subsequence $(\gamma_{n_k}(t))_{k\in\mathbb{N}}$, which is a Cauchy sequence, and thus the equality above holds. Now, my (qualitative) idea is that the limit point is very near to all the geodesics in the sequence (for $k$ sufficiently big) and, since on a surface there is not much room, the geodesics are forced to meet in a point. Anyway, I do not how to make it rigorous nor if my idea is correct.
I am also aware of some basic results concerning hyperbolic manifolds, for example Cartan-Hadamard Theorem or the one to one corrispondence between closed geodesics and free homotopy loop classes, but I am not really familiar with them, so I do not know how to apply these theorems.
Any hint will be really appreciated.

REPLY [3 votes]: Suppose that $S$ is a closed connected oriented surface of genus $g\ge 2$ and $L$ is a 1-dimensional submanifold in $S$ such that the complement $S-L$ contains no components which are annuli or disks. Prove (using the Euler characteristic) that the number of components of $L$ is $\le 3g-3$.
Now, if $S$ is equipped with a Riemannian metric of negative curvature, then a closed geodesic in $S$  cannot be null-homotopic and two such geodesics cannot be homotopic unless their images are the same. Use these observation to conclude that if $L\subset S$ is a submanifold each component of which is geodesic then the number of components is at most $3g-3$.


TITLE: $f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$, $\exists x_n\ne0$ s.t. $x_n\to0$, $f(x_n)=0$ $\forall n$, then $c_k=0$ $\forall k$.
QUESTION [0 upvotes]: I meet a problem:
The power series $f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$. If  there exists a sequence $x_n\ne0$ such that $x_n\to0$ and $f(x_n)=0$ for all $n$, then $c_k=0$ for all $k\in\mathbb{N}$.
I have no clue at all. Any hint or help will be appreciated.

REPLY [0 votes]: Completion of Hagen von Eitzen's method:
Since $f(x)=\sum_{k=0}^\infty c_kx^k$ converges for $|x|<R$ with $R>0$, $f$ is continuous and infinitely differentiable in $(-R,R)$, and for any $j\in\mathbb{N}$, $$
f^{(j)}(x)=\sum_{k=j}^\infty \frac{k!}{(k-j)!}c_kx^{k-j}=\sum_{k=0}^\infty \frac{(k+j)!}{k!}c_{k+j}x^k
$$
has the same radius with $f$, and thereby is also continuous in $(-R,R)$.
By the continuity, $c_0=f(0)=\lim_{n\to\infty}f(x_n)=0$.
Notice that $f(0)=f(x_1)=0$. By Rolle's mean value theorem, there exist a point $\xi_1\in(0,x_1)$ such that $f'(\xi_1)=0$. Choose $x_{n_2}\in(0,\xi_1)$ in the sequence $\{x_n\}$, then $f(0)=f(x_{n_2})=0$. Also by Rolle's theorem, there exist a point $\xi_2\in(0,x_{n_2})$ such that $f'(\xi_2)=0$. Obviously $\xi_2<\xi_1$. Keep on this way, we can get a strictly decreasing sequence $\{\xi_n\}$ with $\xi_n\ne0$ $(\forall n)$, $\xi_n\to0$ $(n\to\infty)$, and $f'(\xi_n)=0$ $(\forall n)$. By the continuity, $c_1=f'(0)=\lim_{n\to\infty}f'(\xi_n)=0$.
By the same argument and the induction, one can show that for any $j\in\mathbb{N}$,
$$
c_j=\frac{1}{j!}f^{(j)}(0)=0.
$$
QED.


TITLE: Linear transformation + Rank & Nullity
QUESTION [1 upvotes]: If $$T : \mathbb{R}^4 \to \mathbb{R}^3$$ is a linear transformation which is defined by $$T(a,b,c,d) = (a-b+c+d, a+2c-d, a+b+3c-3d)$$ then verify $$Rank(T) + Nullity(T) = dim \mathbb{R}^4.$$

REPLY [1 votes]: $T(a,b,c,d)=0$ implies $a (1,2,0,1)=(a,b,c,d) $ for all  $(a,b,c,d)\in N(T) $.Those two vectors are linearly independent and generates  $N (T) $.Hence has dimension $1$.Consider the standard ordered basis ${e_1,e_2,e_3,e_4} $ ... (we could choose any basis,just for simplicity I choose it). $T(e_1)=(1,1,1);T(e_2)=(-1,0,1);T(e_3)=(1,2,3),T(e_4)=(1,-1,-3)$ generates image of $T $. $rank (T)=3$ .
Hence...$rank T + nullity T=4=dimR^4$


TITLE: Finding a differentiable function with a particular property
QUESTION [1 upvotes]: If $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable and satisfies $$2 \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=0,$$ show that there exists a differentiable function $\widetilde{f}:\mathbb{R} \to \mathbb{R}$ such that $f(x,y)=\widetilde{f}(x-2y)$ for every $(x,y) \in \mathbb{R}^2$.
A hint to the problem says to show that for suitable $a,b,c,d \in \mathbb{R}$, the function $F(u,v)=f(au+bv,cu+dv)$ is independent of $u$.  By the chain rule, we have that $\frac{\partial F}{\partial u}=a\frac{\partial f}{\partial x}+c \frac{\partial f}{\partial y}$, so by the hypotheses, $\frac{\partial F}{\partial u}=(a-2c)\frac{\partial f}{\partial x}$.  Hence, $F$ as defined is independent of $u$ if $a=2c$.
My question is, how can I use this information to obtain the function $\widetilde{f}$ with the required property?

REPLY [1 votes]: Hint: As you observed, we may write $f(2u + v, u + v) = g(v)$ for some differentiable function $g$ (because it is independent of $u$). Now if you find $u,v$ such that $2u + v = x$ and $u + v = y$, then you can write $f(x,y) = g(v)$.


TITLE: Why is this assumption correct in Laguerre's method?
QUESTION [0 upvotes]: From Wikipedia:

We then make what Acton calls a 'drastic set of assumptions', that the root we are looking for, say, $x_1$ is a certain distance away from our guess $x$, and all the other roots are clustered together some distance away. If we denote these distances by

I don't understand how such an assumption can be correct. I don't really know what a drastic set of assumptions mean but how can we just write it as $\frac{n-1}b$? They are all different values and we need to expand the expression and do the operations later, isn't it? $x_i$ has actually $n-1$ values.

REPLY [1 votes]: They are just using the approximation $x-x_i \approx b$ for $i=2,3,\dots,n$, which is basically saying that those roots are relatively clustered together, at least enough that the "true values" $G$ and $H$ aren't modified too much by assuming they're at the same point. The validity of that depends on some things, in particular it depends on $x$ being significantly closer to $x_1$ than to the other roots.
This is structurally necessary for the method since you are basically trying to (perhaps crudely) guess where all the other roots are at once using only three scalar properties of $p$: $p(x_k),p'(x_k),p''(x_k)$.


TITLE: Rothberger game and Meager in itself sets
QUESTION [2 upvotes]: On $(\mathbb{R}, \tau)$ the euclidean space of real numbers, we define a new topology by letting $\tau^{*}=\{X\subseteq \mathbb{R}: X=\emptyset \hspace{0.1cm}\mbox{or}\hspace{0.1cm}\mathbb{R}\setminus X\hspace{0.1cm}\mbox{is}\hspace{0.1cm} \mbox{compact} \hspace{0.1cm} \mbox{in}\hspace{0.1cm} (\mathbb{R}, \tau) \}$, it is known that $(\mathbb{R}, \tau^{*})$ is Lindelöf, meager in itself, my question is if Player II has a winning strategy in the Rothberger game played in  $(\mathbb{R}, \tau^{*})$.
Remember that:
The Rothberger game on a topological space $X$ is played according to the following rules:
In each inning $n\in\omega$, Player I chooses an open cover $\mathcal{U_n}$ of $X$, and then Player II picks an open set $U_{n}\in\mathcal{U}_{n}$. At the end of the play $\langle \mathcal{U}_{0}, U_{0}, \mathcal{U}_{1}, U_{1}, ..., \mathcal{U}_{n}, U_{n}, ....  \rangle $. The winner is Player II if $X\subseteq\bigcup_{n\in\omega}U_n$, and Player I otherwise.
Thank you

REPLY [2 votes]: The game is a win for Player I, not Player II. A winning strategy for Player I is to choose, in inning $n$, the open cover
$$\mathcal U_n=\{(a,b)\cup(-\infty,-10)\cup(10,\infty):a,b\in\mathbb R,\ 0\lt b-a\lt2^{-n}\}.$$
If more details are wanted, see the answer I posted to the same question at math.SE:
https://math.stackexchange.com/questions/3200075/compact-complement-topology-and-rothberger-game/3200150#3200150


TITLE: Proving that if $E\subseteq D$ is relatively open in $D$ then $f(E)$ is relatively open in $f(D)$
QUESTION [1 upvotes]: I am not confident in my proof for the problem below. Can someone please help solve this? Thanks!
$\def\x{{\mathbf x}}
\def\C{{\mathcal C}}
\def\R{{\mathbb R}}
\def\f{{\mathbf f}}$
Let $\f\colon D\to \R^m$ where $D$ is compact be one-to-one, and assume that $\f\in\C(D)$.  Prove that if $E\subseteq D$ is
relatively open in $D$ then $\f(E)$ is relatively open in $\f(D)$. The result will follow from the proof of the Open Mapping Theorem,
$\textbf{Solution:}$ Let $\f\colon D\to \R^m$, where $D$ is compact, be one-to-one. Since $\f \in \C(D)$, i.e. $\f$ is continuous and we know that the continuous image of a compact set is compact. Then it follows that $\f(D)$ is compact set as $\f$ is one-to-one which means $\f$ is a bijective map. Thus, $\f$ is invertible so $\f^{-1}$ is also continuous and a bijective map on the compact set $\f(D)\subseteq \R^m$. Since, for every continuous function, the inverse image of open sets is open. So, we define a map $g$ and show that $g=\f^{-1}\colon \f(D)\to D.$ Now, let $E\subseteq D$ be an open set. So $g^{-1}(E) = (\f^{-1})^{-1}E = \f(E)$ as $g$ and $\f$ are bijective. Then $g^{-1}(E)$ is open which means $\f(E)$ is open in $\f(D).$ Thus, $\f$ is an open map.

REPLY [1 votes]: Your proof is correct, but we can simplify the proof once more. I assume that $D$ is a subset of $\mathbb{R}^m$.
If $E$ is open in $D$, then $D-E$ is closed, so $D-E$ is compact. As you mentioned, $\mathbf{f}[D-E]$ is compact, so it is closed in $\mathbb{R}^m$ and in $\mathbf{f}[D]$.
As $\mathbf{f}$ is one-to-one, $\mathbf{f}[D-E]=\mathbf{f}[D]-\mathbf{f}[E]$. Therefore, $\mathbf{f}[E]$ is open in $\mathbf{f}[D]$.


TITLE: Is this property of parabola true?
QUESTION [0 upvotes]: Attached text (taken from "Pre Calculus Mathematics in a Nutshell bu George F Simmons) defines a parabola as containing a "property that each of its points is equidistant from a given fixed point and a given fixed line" :

But it appears that the fixed point(F) and directrix distance varies at each point. As each point on the parabola is at a different distance in relation to the fixed point ? Or am I not interpreting the statement correctly ?

REPLY [1 votes]: Let $P_1,P_2,P_3,\ldots,P_n$ be points along the parabola. Let $F_1,F_2,F_3,\ldots,F_n$ be the distances from each such point to the focus F, and $D_1,D_2,D_3,\ldots,D_n$ be the distances from each such point to the directrix D. What the book is saying is that $F_k=D_k$ for each $1\le k\le n$, not that $F_k=F_j$ and $D_k=D_j$ for all j and k between $1$ and n.


TITLE: How many possibilities are there for two full houses to be dealt to two players in one game?
QUESTION [1 upvotes]: Imagine dealing cards from a classic 52 card deck to two poker players.
How many possibilities are there for both of them to be dealt a full house(three cards in same rank and two cards of another rank) in same round?
As I know totally ${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}$ possibilities for one player (AAABB). According to that, I tried to solve my problem as shown below: $${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 1}{4 \choose 2} + {13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 3}{11 \choose 1}{4 \choose 2}$$ for two players ((AAABB)(AAACC) or (AAABB)(CCCBB)). Is my solution correct or am I missing something?

REPLY [2 votes]: My approach would be
For the first hand

Choose a rank to be in first hand as the three-card rank: $13\choose1$, $12$ of full ranks remaining
Choose $3$ cards of that rank: $4\choose3$.
Choose a rank to be in first hand as the two-card rank: $12\choose1$, $12$ of $2^+$-card ranks remaining, $11$ of $4$-card ranks remaining.
Choose $2$ cards of that rank: $4\choose2$.

First hand total: ${13\choose1}\cdot{4\choose3}\cdot{12\choose1}\cdot{4\choose2}$
Here comes the tricky part:

Choose a rank to be in second hand as the three-card rank: $11\choose1$, 10 of $4$-card ranks remaining and $11$ of $2^+$-card ranks remaining.
Choose $3$ cards of that rank: $4\choose3$.

Case 1: The two-card rank that appears in first hand is in the second hand

Choose that rank as the two-card rank: $1$.
Choose the remaining $2$ cards of that rank: $1$.

Case 2: The two-card rank that appears in first hand is not in the second hand

Choose a rank to be in first hand as the two-card rank:$10\choose1$.
Choose $2$ cards of that rank:$4\choose2$.

Second hand total: ${11\choose1}\cdot{4\choose3}\cdot\left[1+{10\choose1}\cdot{4\choose2}\right]$
Grand total: ${13\choose1}\cdot{4\choose3}\cdot{12\choose1}\cdot{4\choose2}\cdot{11\choose1}\cdot{4\choose3}\cdot\left[1+{10\choose1}\cdot{4\choose2}\right]=10 \ 048 \ 896
$

REPLY [0 votes]: There are two possibilities.  Either both players receive two cards of the same rank or they do not.
Both players receive two cards of the same rank:  There are $\binom{13}{1}$ ways to choose the rank from which player A receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank.  That leaves $\binom{12}{1}$ possible ways to choose the rank from which player B receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank.  There are $\binom{11}{1}$ ways to choose the rank from which both players receive two cards.  There are $\binom{4}{2}$ ways to select which two of the four cards are received by player A.  Hence, there are
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}$$
such deals.
The players receive two cards from different ranks:  The number of ways to select three cards of one rank for each player is the same as above.  There are $\binom{11}{1}$ ways to select the rank from which player A receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank.  There are $\binom{10}{1}$ ways to select the rank from which player B receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank.  Hence, there are
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}\binom{10}{1}\binom{4}{2}$$
such deals.
Since the two cases are mutually exclusive and exhaustive, the number of deals in which both players receive a full house is found by adding the results for the above cases.


TITLE: Universal cover of $(S^1\times S^2)\vee S^1$
QUESTION [2 upvotes]: How does the universal cover $X$ of $(S^1\times S^2)\vee S^1$ looks like? I know that the universal cover $Y$ of $S^1\vee S^1$ is as in the picture. First I thought that  $X$ would be the space obtained from $Y$ by attachin $S^2$ at  each vertex, but I soon found that it would be the universal cover of $S^1\vee S^1\vee S^2$. Then I thought that since $\Bbb R \times S^2$ is the universal cover of $S^1\times S^2$, $X$ would be obtained from $\Bbb R$ by attaching $\Bbb R \times S^2$ at each integer point, but I think this is also wrong, because the preimage of the wedge point will be not only the integers, but also infinitely many in each $\Bbb R\times S^2$. Any ideas?

REPLY [2 votes]: Choose a point $p \in S^1$. Consider the subset embedding
$$(S^1 \times \{p\}) \vee S^1 \hookrightarrow (S^1 \times S^2) \vee S^1
$$
On the left we have a wedge of two circles, with fundamental group free of rank 2. The induced homomorphism on fundamental groups that is induced by this embedding is an isomorphism between two rank 2 free groups:
$$\pi_1((S^1 \times \{p\}) \vee S^1) \approx \pi_1(S^1 \times S^2) \vee S^1)
$$
It follows that universal covering space of $(S^1 \times \{p\}) \vee S^1$, which is depicted in your post, embeds (equivariantly) into the universal covering space of $(S^1 \times S^2) \vee S^1$.
There are now two key questions to ask yourself.
First, how does the universal cover of $S^1 \times \{p\}$ embed into the universal cover of $S^1 \times S^2$?
Answer: by including $\mathbb R \times \{p\}$ into $\mathbb R \times S^2$.
Second, how does one visualize this inclusion in the figure in your post?
Answer: Think of each vertical line in that figure as a copy of $\mathbb R \times \{p\}$. And now, to each of those lines, attach a separate copy of $\mathbb R \times S^2$.


TITLE: Without directly evaluting, show that the determinant of $A = 0$
QUESTION [1 upvotes]: Without directly evaluting, show that
  $det \left[
       \begin{array}{ccc}
         b + c & c + a & b + a \\
           a & b & c \\
           1 & 1 & 1
       \end{array}
     \right]
=0$

I am stuck on this one. I can only do this by evuating.
Things that I know:
1) Square matrix $A$ with two proportional rows or columns as a $det(A) = 0$
2) Square matrix $A$ has $det(A)=0$ if it has a row or column of zeros.
Can anyone help?
Thanks.

REPLY [5 votes]: The row vectors are linearly dependent.
Specifically, denoting the row vectors as $$\vec r_1=(b+c,a+c,a+b)\quad \vec r_2=(a,b,c)\quad \vec r_3=(1,1,1)$$  then we have $$\vec r_1+\vec r_2=(a+b+c)\,\vec r_3$$


TITLE: Endomorphism on polynomial vector space $\mathbb{R}_3[x]$
QUESTION [0 upvotes]: Let $ \mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ \le 3$. Given the endomorphism on this vector space, $$ T:\mathbb{R}_3[x] \to \mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ \{{1,x,x^2, x^3} \} $ of this space.
So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = \lambda x$. This resulted in an equation depending on $\lambda : f(x+1)-f(x)(\lambda-1)=0$. What should I do next?
Thank you!

REPLY [1 votes]: There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :

Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ \operatorname{deg} R < k $. We have
$$ \begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \\
&= a(X+1 -X)\sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\\
&= a\sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) \end{align}$$
which is of degree $ < k $, and thus cannot be equal to $ P $
Consider your polynomial $ P $ in $ \mathbb{C}[X] $ instead, then suppose $ T(P) = \lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.

Calculation of $ T $ on constant polynomials then gives you the desired conclusion.
As for the matrix representation, just use the identity $ (a + b)^n = \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}.$


TITLE: Question about the purpose of a transistor
QUESTION [0 upvotes]: I've been studying about diodes and transistors and something made me wonder about the purpose of transistors.
In the examples we were talking about increasing current by using two batteries with a transistor and I understood how the transistor transfers the current from one side to another while increasing it with another battery connected to both sides(probably I still have gaps with my knowledge).
But the problem is: I didn't understand why we need transistor to increase current, can't we do the same thing by just connecting two batteries together or just using a transformator to increase voltage to get more current or just simply using a bigger battery?
Maybe I'm thinking so straight logic now but examples did not look satisfying to me to understand the usefulness of transistors as I know how widely they are used especially in computer science (which is I'm interested and planing to study in university.) so I would like to hear better examples with explanations.

REPLY [2 votes]: Although there can be a current gain (and voltage gain) between the input and output of a transistor the important thing is that the (large) output current (and voltage) can be controlled by (small) input currents (and voltages).
This control of the output by the input can even be such that the transistor acts like a switch being turned on and off by the input current/voltage without using very much electrical power and this is the basis of the operation of a computer


text:
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\section{Main Results \label{sec:main}}
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\paragraph{Policy Benchmark and Assumptions}
Our goal is be to have total cost which competes with stabilizing, linear controllers. Formally, a linear dynamical controller, or \emph{LDC}, is a is a linear dynamical system $(\Api,\Bpi,\Cpi,\Dpi) $, with internal state $\sopenpi_t \in \R^{\dimpi}$,  equipped with the internal dynamical equations $\sopenpi_{t+1} = \Api \sopenpi_t + \Bpi \yopenpi_t \quad \text{and} \quad \uopenpi_t := \Cpi \sopenpi_t + \Dpi \yopenpi_t$.
We let $\Pildc$ denote the set of all LDC's $\pi$.  These policies include static fedback laws  $\uopenpi_t = K \yopenpi_t$, but are considerably more general due to the internal state. The \emph{closed loop} sequence $(\matypi_t,\matupi_t,\matxpi_t,\matspi_t)$ denotes the unique sequence consistent with  \Cref{eq:LDS_system}, the above internal dynamics, and the equalities $\uopenpi_t = \matu_t$, $\yopenpi_t = \maty_t$.\footnote{The sequence $(\yk_t,\uk_t)$ is a special case with $\Dpi = K$ and $\Cpi =0$.} We say consider \emph{stabilizing} $\pi$ meaning that, for all bounded disturbances sequences $\max_{t \ge 1} \|\matw_t\|,\|\mate_t\| <\infty$, it holds that $\max_{t \ge 1} \|\matypi_t\|,\|\matupi_t\| < \infty$. Since stabilizing policies enjoy geometric decay, our policy benchmark is pararametrized as follows:
\begin{definition}[Policy Benchmark]\label{defn:benchmark} Fix parameters $\rhocomp \in (0,1)$ and $\constcomp > 0$. Our regret benchmark competes LDC's $\pi \in \Pistar := \Pistab(\constcomp,\rhocomp)$, where we define
\iftoggle{nips}
{
	$\Pistab(c,\rho) := \{\pi \in \Pildc : (\|\Gpicl^{[i]}\|_{\op} \le c\rho^n, \forall n \ge 0\}$,
}
{
	\begin{align*}
\Pistab(c,\rho) := \{\pi \in \Pildc : \|\Gpicl^{[i]}\|_{\op} \le c\rho^n, \forall n \ge 0\},
\end{align*}
}
where the Markov operator $\Gpicl$ is formally defined in \Cref{defn:gpicl}.
\end{definition}

Next, we turn to the necessary assumptions under which our bounds hold:
\begin{comment}
\begin{assumption}\label{asm:stab}For some $\constk > 0$ and $\rhok \in (0,1)$ and all $n \ge 0$, $\|\Gk^{[i]}\|_{\op} \le \constk\,\rhok^n$.
\end{assumption}
\begin{assumption}\label{asm:bound}
We assume that $(\matw_t,\mate_t)$ are bounded such that, for all $t \ge 1$, $\|(\yk_t,\uk_t)\|_{2} \le \radnat$, where we recall that $(\yk_t,\uk_t)$ are interates under $\yalg_t = K\ualg_t$. For simplicity, we take $\matx_1 = 0$.
\end{assumption}
\Cref{asm:loss_reg} captures general smoothness and growth bounds for quadratic functions. \Cref{asm:stab} is analogous to ``strong stability'' \cite{cohen2018online}, and holds for some $\rhok \in (0,1),\constk > 0$ for any stabilizing $K$. Regarding \Cref{asm:bound}: since $K$ is stabilizing, any bounded sequence of disturbances will yield a uniform upper bound on $\|(\yk_t,\uk_t)\|_{2}$, and the assumption can be restated to depend directly on $\max_{t}\|\matw_t,\mate_t\|_2$.

Moreover, one may allow $\radnat$ to grow logarithmically in $T$ (e.g. $\radnat = \mathcal{O}(\log^{1/2} T)$ for subguassian noise), by inflating logarithmic factors in our bounds.
\end{comment}

Complete descriptions of \Cref{alg:nfc,alg:estimation_static} referenced above are detailed in \Cref{app:general}. The proof of the above theorems rests on generic policy regret bounds for the \semions{} algorithm, which are detailed in the following subsection. The proof of the above theorems is then completed in \Cref{sec:control_proofs_ful}, along with complete statements which specify the parameter choices \Cref{thm:drconsstat_known_granular,thm:drcons_unknown_specific}. In addition, \Cref{app:general} describes  generalizations which replace \emph{static} $K$ in the \drc{} algorithm with a dynamic nominal controller $\pinot$,

and \Cref{sec:control_proofs_ful} states and proves regret bounds for this more general algorithm as well.
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\subsection{Policy Regret for \semions\label{ssec:semi_ons_guarantees_known}}
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Our main results for control derive from general-purpose policy regret bounds for the \semions{} algorithm. Specifically,  we consider the $\PolicyRegret$ benchmark \Cref{eq:pol_reg}, evaluated on losses of the form \iftoggle{nips}{$F_t(z_{t:t-h}) = \cost_t(\matv_t + \sum_{i=0}^h G^{[i]}\matY_{t-i}z_{t-i})$, and $f_t(z) = F_t(z,\dots,z) = \cost_t(\matv_ t + \matH_t z) $,}{
	\begin{align*}
	F_t(z_{t:t-h}) = \cost_t(\matv_t + \sum_{i=0}^h G^{[i]}\matY_{t-i}z_{t-i}), \quad f_t(z) = F_t(z,\dots,z) = \cost_t(\matv_ t + \matH_t z),
	\end{align*}
}where $\matH_t := \sum_{i=0}^h G^{[i]}\matY_{t-i}$. To address the known system case, we consider generic bounds on policy regret when running $\semions$ on losses $(f_t,\matH_t)$.
To address unknown systems, we bound the policy regret when $\semions$ with losses $(\fhat_t,\matHhat_t)$, where
\iftoggle{nips}{ $\fhat_t(z) := \cost_t(\matvhat_t + \matHhat_t z)$, and $\matHhat_t = \sum_{i=0}^h \Ghat^{[i]}\matY_{t-i}$.
}
{
\begin{align*}
 \fhat_t(z) := \cost_t(\matvhat_t + \matHhat_t z), \quad \text{ and } \matHhat_t = \sum_{i=0}^h \Ghat^{[i]}\matY_{t-i}.
\end{align*}
}
We assume the following bounds:

\begin{restatable}{definition}{defnPolPars}\label{defn:pol_reg_pars} Set $\diamz := \max\{\|z-z'\|:z,z' \in \calC\}$, $\radY := \max_{t} \|\matY_t\|_{\op}$, and  $\radyc := \max_{t}\max_{z \in \calC}\|\matY_t z\|$. Further take $\radv := \max_{t} \max\{\|\matv_t\|_2, \|\matvhat_t\|_2\}$ and we define $\radG := \max\{1,\|G\|_{\loneop},\|\Ghat\|_{\loneop}\}$, where $\|G\|_{\loneop} := \sum_{i \ge 0}\|G^{[i]}\|_{\op}$ Finally, we set $\radH = \radG \radY$, and define the the effective Lipschitz constant  $\Leff := L\max\{1,\radv + \radGst \radyc\}$.
\end{restatable}
\iftoggle{nips}{}{Note that we let $\matz_{t},\matY_t$ be arbitrary even for $t \le 1$, provided the above bounds hold.}
In addition, our analysis hinges on $G$ having the following `invertibility' property: that  its smallest singular value, viewed as an operator on sequences $(u_{1},u_{2},\dots)$, is bounded below:
\begin{definition}\label{defn:input_recov}  Let $G = (G^{[i]})_{i \ge 0}$ be a Markov operator. We define its \emph{invertibility-modulus} as
\iftoggle{nips}{
$\kappa(G) := 1 \wedge \inf_{(u_{0},u_{1},\dots)} \{\sum_{n \ge 0}\|\sum_{i=0}^n G^{[i]}u_{n-i}\|_2^2 : \sum_{t}\|u_t\|_2^2 = 1 \}$.
}
{
	\begin{align*}
\kappa(G) := 1 \wedge \inf_{(u_{0},u_{1},\dots)} \left\{\sum_{n \ge 0}\left\|\sum_{i=0}^n G^{[i]}\matu_{n-i}\right\|_2^2 : \sum_{t\ge 0}\|\matu_t\|_2^2 = 1 \right\}
\end{align*}
}
We also define its \emph{decay-function} $\psiG(n) := \sum_{i \ge n} \|G\|^{[i]}$, which is exponentially decay in the control setting.
\end{definition}
Our subsequent policy regret bounds will rely on $\kappa := \kappa(G) > 0$. Importantly, this assumption holds for our control setting (see \Cref{app:invertibility} for a proof):

Our subsequent analysis centers around the fact that, for $\kappa(G) > 0$, the covariance of the $\matH_t$ sequence dominates that of the $\matY_t$ sequence, up to lower order terms:

This implies roughly that the precondition dominates the $\matY$-convariance: $\Lambda_t \succsim  \lambda I + \sum_{s=1-h}^{t} \matY_{s}^{\top}\matY_{s}$ which is at the heart of the following regret bound. With this insight, we prove our first theorem, yielding logarithmic regret for $\kappa > 0$:

\begin{proof}[Proof Sketch] Since $f_t(z) = F_t(z,\dots,z) = \ell_t(\matv_t + \matH_t v)$, the preconditioner increments lie in a certain ``sweet spot'' $\Omega(1)\cdot \nabla f_t(\nabla f_t)^\top  \preceq  \matH_t^\top \matH \preceq \frac{1}{\alpha}\nablatwo f_t$ (\Cref{lem:ft_facts,lem:quad_lb}). This lets us to replicate the standard analysis of Online Newton Step \cite{hazan2007logarithmic} to bound $f$-regret (\Cref{thm:semions}).  From \Cref{lem:Ft_diff}, it remains to analyze the geometry-adaptive movement cost (\Cref{eq:geo_adaptive}). We use here that for $\kappa(G) > 0$,  invertibility of $G$ implies that $\Lambda_t$ roughly dominates the covariance of the $\matY_t$-sequence $\sum_{s=1-h}^t \matY_t^\top \matY_t$ (\Cref{prop:covariance_lb}). Thus, we can analyze \Cref{eq:geo_adaptive} as if $\Lambda_t$ were based on the $\matY_t$-geometry, which yields logarithmic geometry-aware movement by an application of the log-det potential lemma (\Cref{lem:log_potential}). The full proof is detailed in \Cref{sec:known}.
\end{proof}

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Lastly, we state the guarantee pertinent for \semions{} on approximate losses, bounding the policy regret on the $f_t,F_t$ sequence, but using updates from the estimates $\fhat_t$ defined above:

\begin{proof}[Proof Sketch] Using the geometry of the errors and AM-GM, the error introduced by the gradients of inexact $\fhat_t \approx f_t$ scales as  $\tau^{-1} T \epsG^2 + \tau \sum_{t=1}^T \|\matY_t(\matz_t - \zst)\|_2^2$, where $\tau$ is a free parameter, and where the second term measures distance between iterates $\matz_t$ and a comparator $\zst$ in $\matY_t$-geometry. Fortunately, our regret bound also contains a \emph{negative term} scaling as $- \sum_{t=1}^T \|\matX_t(\matz_t - \zst)\|_2^2$, which we use to offset the previous term. \iftoggle{nips}{}
{

}
By making $\tau$ small, we would like to show roughly that $\tau \sum_{t=1}^T \|\matY_t(\matz_t - \zst)\|_2^2 - \sum_{t=1}^T \|\matX_t(\matz_t - \zst)\|_2^2$ is non-positive. But this does not directly follow in the PSD-order comparison from \Cref{prop:covariance_lb}, because the vectors $\matz_t - \zst$ vary with $t$. Instead, we \emph{block} times $t$ into epochs of size roughly $\sqrt{T}$. We then argue that for large enough regularizer $\lambda \sim \sqrt{T}$, the iterates do not move too much within the blocks, allowing us to use \Cref{prop:covariance_lb}. The argument is quite careful, and a full proof is given in \Cref{sec:unknown}.
\end{proof}
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In fact, the above technique yields a stronger version of the theorem, which also incorporates a \emph{negative term} in the regret scaling with  $\sum_{t=1}^T \|\matY_{t}(\matz_t - \zst)\|^2 $ (\Cref{thm:semions_unknown_clam}); this offset term is essential in obtaining fast rates for the control problem by accounting for the mismatch between $\matY_t = \embed(\yk_{t:t\shortminus m +1})$ based on the estimates of $\yk$, and the true $\yk_{t:t\shortminus m +1}$ (see \Cref{sssec:prop_redux_unknown}).  More generally, we remark that negative regret is in fact a cornerstone of the study of online learning with curvature (e.g., \citep{rakhlin2014online}). In addition, we prove a more general bound, \Cref{thm:unknown_granular}, which accomodates $\epsG^2 \ll \sqrt{T}$, showing a $(T\epsG)^{2/3}$ regret scaling for very small $\epsG$.


TITLE: Number of paths between two points on a directed grid
QUESTION [1 upvotes]: Let $G$ be a simple directed graph with vertex set $\{(i,j) \in \mathbb{Z}^2 : 15 \leq i \leq 21 , 21 \leq j \leq 25\}$ and an arc from $(i,j)$ to $(i',j')$ if and only if $(i',j')$ is obtained from $(i,j)$ by adding 1 to one coordinate. Then the number of directed paths from $(15,21)$ to $(21, 25)$ is
(a) $1260$
(b) $1420$
(c) $980$
(d) None of these

My answer: The number of paths from $(15,21)$ to $(21, 25)$ would be $${(21-15)+(25-21) \choose (25-21)} = {10 \choose 4} = 210.$$
Is this correct? Just want to get a second opinion.

REPLY [2 votes]: Yes. Your answer is correct. The solution to the general problem is that you must take X right steps, and Y down steps then the number of paths is simply the ways of choosing where to take the up (or right) step. i.e.
((X+Y),X)=((X+Y),Y)
For example, you are traversing graph points of equal distance i.e. squares then there are 6 right steps and 4 up step :
((6+4),4) = ((6+4),6) = (10,4) = (10,6) = 210
P.S (a,b) -> a! / b!


TITLE: Why does the Redfield equation model thermal relaxation while the Lindblad equation does not?
QUESTION [3 upvotes]: In open quantum systems, we model a process known as thermal relaxation. What is this process, and why is it that only the Redfield equation models this process, and the Lindblad equation doesn't?

REPLY [1 votes]: In open quantum systems, the evolution of the total System is unitary. The freedom degrees of the environment is large, so we always think the environment will not evolve and that is the Born approximation.$$\rho(t) = \rho_{s}(t)\otimes\rho_{E}$$ The relaxation time means the time what the environment need back to equilibrium again. Redfield equation is always the Non-Markovian master equation, so we have to consider the relaxation time, the Lindblad equation is always Markovian, so we don't consider the relaxation time here.


TITLE: How to Prove that if $f(z)$ is entire, and $f(z+i) = f(z), f(z+1) = f(z)$, then $f(z)$ is constant?
QUESTION [11 upvotes]: So the problem states that if $f(z)$ is entire, and satisfies the relation $f(z+i) = f(z)$ and $f(z+1) = f(z)$, show that $f(z)$ is constant. So I was thinking that since any point in $\mathbb{C}$ can be written as $\alpha * 1 + \beta * i $ we can say that $f(z + z_0) = f(z) $ in which case it is constant, but I'm having trouble breaking down the steps and using the fact that $f$ is entire, which makes me feel like I'm missing something. What should I review to figure this out?

REPLY [16 votes]: I think you can just use Liouville's theorem. The thing is that $f$ is bounded on compact subsets of the complex plane, so in particular it is bounded in the unit square $S = \{ a + bi \in \mathbb{C} \mid a, b \in [0, 1] \}$.
So then you can use the two relations $f(z+1) = f(z)$ and $f(z+i) = f(z)$ to prove that the values of $f$ are determined by its values in the unit square because you can write $z = x + iy$ and then
$$f(z) = f(x + iy) = f(a + bi + (n + mi)) = f(a + bi)$$
where $0 \leq a, b \leq 1$ and $n, m \in \mathbb{Z}$. In this case $n = \lfloor x \rfloor$ and $m = \lfloor y \rfloor$. Then of course $f$ is bounded in the whole complex plane so Liouville's theorem applies, and $f$ has to be constant.

REPLY [3 votes]: Here is a Hint: Liouville Theorem.


TITLE: Finding cos(a-B) using tan a = 4/3 in quad III and cos B = 12/13 in quad IV
QUESTION [1 upvotes]: Really stuck with this problem, I need help with this asap.
To find cos(a-B), I'm supposed to use:
tan a = 4/3, with a in quadrant III
cos B = 12/13, with B in quadrant IV

REPLY [0 votes]: I have shown the relevent identities . I hope this helps answer your question.


TITLE: Is this a black hole?
QUESTION [3 upvotes]: I came across this metric definition:
$ds^2=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2+r^2d\Omega^2$
I was trying to figure out if it describes a black hole, but I was unable to find the proper definition.
My hints so far:

The time coordinate does not change its sign when r crosses $r_s$
There are two singularities, one at $r=r_s$ which I guess arises from coordinate singularity and another at $r=0$, which looks like the center of a black hole?

What do you think?

REPLY [3 votes]: We need to be careful about terminology because the phrase black hole tends to be used to mean a vacuum solution. The only two vacuum solutions are the Schwarzschild and Kerr metrics, so in everyday use black hole means either the Schwarzschild or Kerr metric.
But if we relax the requirement for the geometry to be a vacuum solution then we can get other forms of black hole. For example if we add a cosmological constant we get the de Sitter-Schwarzschild solution and if we add null dust we get the Vaidya metric. Your geometry is also a non-vacuum geometry.
In principle we can tell it's not a vacuum solution by calculating the Einstein tensor from your metric, because this gives us the stress-energy tensor and that will turn out not to be zero. However the GR Mathematica notebook I have does not include a routine for calculating the Einstein tensor and calculating it by hand would be something of an ordeal. I did calculate the Ricci tensor and it seems fairly obvious from a glance that this is going to give a non-zero Einstein tensor.
But I would guess your question is really whether this geometry has a horizon. In general finding horizons is a surprisingly hard thing to do because we can only find them by considering the entire spacetime. However in a simple, static, geometry like this one we can simply look at the coordinate speed of light and see if it's zero. We do this by noting that for a light ray $ds = 0$. If we choose a radial trajectory, so $d\Omega=0$ and make this substitution we get:
$$ 0=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2 $$
Giving us:
$$ \frac{dr}{dt} = \pm\left(1-\frac{r_s}{r}\right)^2  $$
And since this is zero at $r=r_s$ this is indeed an event horizon.
Courtesy of AccidentalFourierTransform we can calculate the Ricci and Kretschmann scalars. The Ricci scalar turns out to be zero everywhere, which isn't much help, but the Kretschmann scalar is:
$$ K = \frac{8r_s^2 ( 6 r^2 - 12 r r_s + 7 r_s^2)}{r^8} $$
This remains finite at $r=r_s$, but goes to infinity as $r \to 0$. So the former is a coordinate singularity while the latter is a true curvature singularity.


TITLE: volume of the region bounded by the surfaces $x = 0, y = x, y = 2 - x^2, z = 0$ and $z = x^2$
QUESTION [1 upvotes]: Find the volume of the region in the first octant bounded by the surfaces $x = 0, y = x, y = 2 - x^2, z = 0$ and $z = x^2.$
It is difficult for me recognize the region geometrically. Can someone please give some idea about the sketch of the region?
I am confused with the limits of integration.
Required volume = $\int_{z = ?}^{?}\int_{y = ?}^{?}\int_{x = ?}^{?}dxdydz.$

REPLY [0 votes]: Firstly, consider the region in the $x,y$ plane bounded by $y=x$, $x=0$ and $y=2-x^2$. We see that, for a given $x$, that $y$ ranges from $x$ to $2-x^2$. If we make a similar diagram for $x,z$, we find that $z$ ranges from $0$ to $x^2$. We have that $x$ ranges from $0$ to $1$, as $y=x$ intersects $y=2-x^2$ at $x=1$. This gives the integral as
$$\int_{x=0}^{x=1} \int_{y=x}^{y=2-x^2} \int_{z=0}^{z=x^2} \mathrm{d}z\mathrm{d}y\mathrm{d}x$$