TITLE: I wonder whether the system of equations and inequations below have a solution.
QUESTION [1 upvotes]: I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired.
$$\begin{cases}
\frac{c_1}{1-x_1}+\frac{c_2}{1-x_2}+\frac{c_3}{1-x_3}=0\\
\frac{c_1}{1-x_4}+\frac{c_2}{1-x_5}+\frac{c_3}{1-x_6}=0\\
c_1\ln{\frac{x_1}{1-x_1}}+c_2\ln{\frac{x_2}{1-x_2}} +c_3\ln{\frac{x_3}{1-x_3}}=0\\
c_1\ln{\frac{x_4}{1-x_4}}+c_2\ln{\frac{x_5}{1-x_5}} +c_3\ln{\frac{x_6}{1-x_6}}=0\\
\frac{x_1(1-x_1)}{x_4(1-x_4)}=\frac{x_2(1-x_2)}{x_5(1-x_5)}=\frac{x_3(1-x_3)}{x_6(1-x_6)}>1
\end{cases}$$
where $ c_1,c_2,c_3 $ are constants, and $x_i\in(0,1), i=1,2,3,4,5,6$.

REPLY [3 votes]: A simple set of solutions are any $(c_{1},c_{2},c_{3})$ such that $c_{1}+c_{2}+c_{3}=0$, $x_{1}=x_{2}=x_{3}=\frac{1}{2}$, $x_{4}=x_{5}=x_{6}<\frac{1}{2}$.


TITLE: Can a Poisson distribution be understood as a discrete Gaussian?
QUESTION [2 upvotes]: I'm coming to this from a Machine Learning perspective. In ML, we have to compute what we call the loss when our model makes a prediction. This loss gives us a number, allowing us to tell our far or close from the "right answer" (assumed to be provided in some form). Then it can be use to tweak the weights of our model to hopefully make better predictions next time similar inputs are provided.
Therefore one could see each inputs with its associated answer as a probabilistic distribution of events, hence a number of loss functions are probabilistic.
Since these tend to be discrete in nature, a Poisson distribution can be used for regression predictions.
I'm not familiar with Poisson's distribution, but I have extensively used Gaussian. My current grasp is to see Poisson's as essentially a discrete Gaussian. I understand there's a bit of a skew for extreme values (e.g. a true Gaussian has a pretty long tail, which real values may not allow, such as a price of -10$ for instance).
Is that idea far off from reality? If this is far off, any better mental model to get an intuitive grasp of the Poisson distribution?

REPLY [2 votes]: You are not very far off - in fact, for sufficiently large values of $\lambda$ it is common practice to approximate the Poisson with a normal distribution.


TITLE: The Ward identity, the Lorentz invariance
QUESTION [1 upvotes]: Outline - heuristic derivation of the Ward identity from the requirement of the Lorentz invariance
Suppose we have the free quantized gauge theory (with quanta called photons) with the Hilbert space restricted to transversal polarizations only. Next, suppose interactions with the matter, and assume the matrix element $M$ with (at least) one external photon line having the polarization vector $\epsilon^{\mu}_{\perp}(p)$. It has the form
$$
\tag 1 M = \epsilon^{\mu}_{\perp}(p)M_{\mu}
$$
Assume $M_{\mu} \sim p_{\mu}$. Next, perform the massless Lorentz orbit little group transformation $\Lambda$ defined by
$$
\tag 2 (\Lambda p) = p, \quad (\Lambda \epsilon_{\perp_{1}}) = \epsilon_{\perp_{1}} +cp, \quad (\Lambda \epsilon_{\perp_{2}}) = \epsilon_{\perp_{2}}
$$
and apply it to transformed squared averaged matrix element
$$
|\bar{M}|^{2} \equiv \sum_{\perp_{1,2}}|M|^{2}
$$
Since it must be invariant in the Lorentz-invariant gauge theory, we obtain
$$
|\bar{M}|^{2} = |\bar{M}^{2}| + c(p^{\mu}\epsilon_{\perp_{1}}^{*\nu} + p^{\nu}\epsilon_{\perp_{1}}^{\mu})M_{\mu}M^{*}_{\nu},
$$
from which follows the requirement
$$
\tag 3 p^{\mu}M_{\mu} = p^{\nu}M_{\nu}^{*} = 0,
$$
i.e., the Ward identity.
My question
1) What is the physical sense of the transformation $(2)$? Precisely, may I state that it generates the unphysical "longitudinal" state from the physical transverse one?
2) Next, does from the derivation above follow that the unitarity (a-la optical theorem) is broken as long as $(3)$ isn't valid? Precisely, being inserted as the sub-diagram the amplitude $M_{\mu}$ will distinguish physical and unphysical polarizations, and that's enough for proving the non-unitarity (see this related question)? For example, one may calculate the optical theorem (with $A$ denoting the state containing the photon)
$$
2\text{Im}M_{A \to A} = (2\pi)^{4}\sum_{n}|M_{A \to n}|^{2},
$$
where for the given order of perturbation theory $M_{A \to A}$ doesn't respect the Ward identity while $M_{A \to n}$ respects it. Then one may perform the transformation $(2)$ and deduce that there is indeed the violation of the unitarity.
3) Finally, does this relate directly the Lorentz invariance to the unitarity of the gauge theory?

REPLY [2 votes]: 1) see Weinberg's book.
2) You (and some introductory texts) have the logic inverted. Your "derivation" of $(3)$ is not a derivation of the Ward identity; it only serves to explain why the Ward identity is important. A quick proof of the Ward identity can be found in Srednicki's book.
3) If you use the Coulomb gauge then the Ward identity guarantees that the theory is covariant. If it is not satisfied, then covariance is broken but the theory is still unitary: there are no negative norm states.
If you use the $R_\xi$ gauge then the Ward identity guarantees that the theory is unitary. If it is not satisfied, then unitarity is broken but the theory is still covariant: the polarisation vectors are all true vectors, but the longitudinal ones are not decoupled.
This means that the Ward identity is related to both covariance and unitarity, but not at the same time. If the Ward identity becomes anomalous you either lose covariance or unitarity, but not both. And you must decide which one you want to give up (the choice is irrelevant anyway because if one of these properties isn't satisfied, then the whole theory becomes meaningless).


TITLE: Probability that a < b < c
QUESTION [0 upvotes]: There are 50 numbered objects from 1 to 50. We extract 3 of these objects one after the other without reinsertion. Calculate the probability P(a<b<c)
Given 3 objects, they can be ordered in $3!$ different ways. Of all these ways only 1 satisfies a<b<c so the probability should be:
$P(a<b<c) = \frac{1}{3!} = \frac{1}{6}$
Is it solved in the right way?

REPLY [0 votes]: [Completely basic probability point of view]
Given a set of $3$ numbers from the bag, what is the probability that the numbers are arranged the way you want? (Conditional)
Now use the theorem of total probability.
Same way, you can actually answer the question for drawing $k$ numbers out of the bag.


TITLE: Frequency and angular velocity of washing machines drum
QUESTION [0 upvotes]: I have the question "A washing machine runs its spin cycle at 800 rpm(revolutions per minute).
What is the frequency and the angular velocity  of the machines drum?"
I know that frequency is 1/T where T is the period.
So I multiplied 800 by 60 to get 48000 seconds.
So for frequency I got 2.1 x 10^-5 hz  and for angular velocity I got 1.3 x 10^-4 radians per second.
I am not sure if this is correct though.

REPLY [1 votes]: Frequency is nothing else but the same which is given in the question. It is $800$ rpm, but remember whenever you give you give your answer keep it in standard units. So, the frequency will be $800/60= (40/3) Hz$.
Now, the angular velocity is given by $w=2\pi v$ where v is frequency. Just put the value of frequency in  $w=2\pi v$ and get the angular velocity.


TITLE: Condition to obtain a not compact embedding
QUESTION [1 upvotes]: I have the two spaces $W_0^{1,p}$ with the norme $$||u||^p=||u||^p_{L^p}+||\nabla u||^p_{L^p}$$ and $$L^{p^*}_{\alpha}=\{ u~\text{measurable}, \int_{\Omega} (|x|^{\alpha} u(x)|)^{p^*} dx<\infty\}$$ equiped with the norm $$||u||_{L^{p^*}_{\alpha}}^{p^*}=\int_{\Omega} (|x|^{\alpha}|u(x)|)^{p^*} dx$$
How to find the condition on $\alpha>0$ such that $W^{1,p}_0$ do not be compactly embeded in $L^{p^*}_{\alpha}$ ?
Where $\Omega\subset\mathbb{R}^N$ is bounded,  $N>p$ and $p^*=\frac{Np}{N-p}$
Thank you

REPLY [1 votes]: The failure of the compactness of the embedding $W^{1, p} (\Omega) \subset L^{p^*} (\Omega)$ is local, that is, it can be exhibited in any ball.
An abstract way of seeing the failure of the embedding is to consider a ball $B$ such that $\overline{B} \subset \Omega \setminus \{0\}$. The compactness of the embedding of $W^{1, p} (\Omega)$ in $L^{p^*}_\alpha (\Omega)$ would then imply the compactness of the embedding of $W^{1, p} (B)$ in the unweighted space $L^{p^*} (B)$, in contradiction with the classical theory.
Alternatively, fix a point $a \in \Omega$ and a test function $\varphi \in C^1_c (\mathbb{R}^N)$, and define
$$
  \varphi_{\lambda} (x) = \lambda^{1-\frac{N}{p}} \varphi \Bigl(\frac{x - a}{\lambda}\Bigr).
$$
There exists $\lambda_0>0$ such that if $\lambda \in (0, \lambda_0)$, then $\varphi_\lambda \in C^1_c (\Omega) \subset W^{1,p} (\Omega)$. Moreover the family $(\varphi_\lambda)_{\lambda \in (0, \lambda_0)}$ is bounded in $W^{1, p} (\Omega)$ but is not relatively compact in $L^{p^*}_\alpha (\Omega)$.


TITLE: Finding mixed (and pure) Nash equilibria of a 2-players $n \times n$ game
QUESTION [1 upvotes]: To find the mixed NE for a 2-player 2x2 game, one can write the well-known formulation where each of the players makes another one indifference between its actions. But even in the case of a 2x2 game, this approach does not find all the equilibria of the game. For example, let's assume the actions of player 1 and 2 are $A_1=\{U,D\}$, $A_2=\{L,R\}$ and the utility matrix of the game is
$$\begin{array}{c|cc}
&\text{L}&\text{R} \\
\hline
\text{U}&A,E&B,F \\
\text{D}&C,G&D,H \\
\end{array}$$
Let's further assume that the mixed strategy of player 1 and 2 are $(q_1,1-q_1)$ and $(q_2,1-q_2)$, respectively. Now, player 2, to make player 1 indifference between its action writes $Aq_2 + B(1-q_2) = C(q_2) + D(1-q_2)$
and this gives:
$$ q_2 = \frac{D-B}{A-B-C+D} $$
Now, applying the similar procedure for the following matrix gives $q_1=q_2= 0$.
$$\begin{array}{c|cc}
&\text{L}&\text{R} \\
\hline
\text{U}&5,5&0,0 \\
\text{D}&0,0&0,0 \\
\end{array}$$
This means that it finds the NE $(D, R)$ and cannot find the NE $(U,L)$ which is a much better NE than the former one.
As far as I know, Lemke-Howson algorithm generalizes the above procedure for $n \times n$ game, but similar to the example above, it cannot find all the NEs.
I wondered if there is a mathematical program like the Lemke-Howson algorithm, regardless of its complexity, to find all the NEs of the game (considering pure NEs as special cases of mixed NEs)?
Many thanks.
P.S: An algorithm is available here: http://banach.lse.ac.uk/, based on the paper of Avis, et al., but I think implementing that algorithm takes very long time.

REPLY [1 votes]: The "well-known formulation" you describe is simply wrong. It is not sufficient for a player to be indifferent between some pure strategies to be able to mix between them in equilibrium; they must also be best response, i.e., give the best possible payoff among all pure strategies.
Finding all equilibria is harder for "degenerate games", where there be infinitely many equilibria, but if you are content with just finding the extreme equilibria (i.e., essentially equilibria which are not convex combinations of other equilibria) then the simplest approach is to do support enumeration.
Definition: the support of a mixed strategy is the set of pure strategies that it plays with positive probability.
Now given a pair of supports, you first check can each player make the other player indifferent between his support strategies using only her support strategies and, if this is possible, are the resulting mixed strategies that do this are best responses (i.e. give higher payoff than all things outside the support).
Pure equilibria have support size 1 for both players. When checking for the indifference part is trivial, since there is a single strategy, and you just need to check if the pure strategies are best responses against each other. For larger supports, you need to do both parts: see if indifference is even possible given the candidate support of the other player, and if it is possible for both players, see if the payoffs of the pure strategies in the supports (which will now all be the same for a given player since we achieved indifference) are the best possible.
In some sense, we are taking what you know about finding pure equilibria, and finding 2x2 mixed equilibria in 2x2 games, and combining them into a general algorithm.)
This is described as Algorithm 1 in the paper you refer to:
David Avis, Gabriel D. Rosenberg, Rahul Savani, and Bernhard von Stengel.
Enumeration of Nash equilibria for two-player games.
Economic Theory, 42(1):9–37.
It is not the algorithm used on banach.lse.ac.uk, which is indeed more complicated.
P.S. I wrote banach.lse.ac.uk and am a co-author on the Avis et al. paper.


text:
\section{Primitive of Inverse Hyperbolic Cosecant of x over a over x}
Tags: Primitives involving Inverse Hyperbolic Cosecant Function

\begin{theorem}
:$\displaystyle \int \frac {\operatorname{csch}^{-1} \dfrac x a \ \mathrm d x} x = \begin{cases}
\displaystyle \frac {\ln \left({\dfrac x a}\right) \ln \left({\dfrac {4 a} x}\right)} 2 - \sum_{k \mathop \ge 0} \frac {\left({2 k + 1}\right)!} {2^{2 k} \left({k!}\right)^2 \left({2 k + 1}\right)^3 \left({2 k}\right)^2} \left({\frac x a}\right)^{2 k} + C & : 0 < x < a \\
\displaystyle \frac {\ln \left({\dfrac {-x} a}\right) \ln \left({\dfrac {-x} {4 a} }\right)} 2 + \sum_{k \mathop \ge 0} \frac {\left({2 k + 1}\right)!} {2^{2 k} \left({k!}\right)^2 \left({2 k + 1}\right)^3 \left({2 k}\right)^2} \left({\frac x a}\right)^{2 k} + C & : -a < x < 0 \\
\displaystyle \sum_{k \mathop \ge 0} \frac {\left({-1}\right)^{k + 1} \left({2 k + 1}\right)!} {2^{2 k} \left({k!}\right)^2 \left({2 k + 1}\right)^3} \left({\frac a x}\right)^{2 k + 1} + C & : \left\vert{x}\right\vert > a \\
\end{cases}$
\end{theorem}

config: wiki


TITLE: A differentiable function is onto or not in higher dimension
QUESTION [0 upvotes]: Suppose we have a differentiable function $ F : \mathbb R^n \rightarrow \mathbb R^n$ and it's derivative 'function(linear function)' at each point always has non zero determinent then is the function onto? I know it is one-one and onto in small domain but can we extend it over all to whole domain?

REPLY [1 votes]: No. For instance, the function $f: \Bbb R \to \Bbb R$ defined by $$f(x) = e^x$$
has non-zero derivative but is not onto.

REPLY [0 votes]: No: a counterexample already for $n=1$ is $f(x) = \arctan x$.


TITLE: Knowing how to order $\{a,b,c,0\}$ implies knowing how to order $\{a,b,c,0,-a,-b,-c\}$?
QUESTION [1 upvotes]: Suppose I have $4$ real numbers $\{a,b,c,0\}$ and I know

that they are all different
how to order them from smallest to largest, e.g., I know that $b<a<0<c$

Does this imply that I know how to order from smallest to largest $\{a,b,c,0,-a,-b,-c\}$?

REPLY [1 votes]: You can iff all are greater or smaller than zero.
Otherwise consider
$$-2<-1<0<c$$
If $c=1.5$ then $-c$ will be between $a$ and $b$; if $c=3$ then $-c$ will be on the far left...


TITLE: Understanding the Initial-value Theorem in the Laplace Transform Theory
QUESTION [1 upvotes]: I am an Electrical Engineering student and I study from the book "Engineering Circuit Analysis 8th Ed." authored by Hayt, Kemmerly and Durbin. On Chapter 14, page 562 dealing with an COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM, there is a subsection on the Initial-Value Theorem.

Now, what don't understand is the part where the initial integral is split into two:
$$
\lim_{s\to\infty}\left(\int_{0^{-}}^{\infty} e^{-st}\frac{df}{dt}dt\right) = \lim_{s\to\infty}\left(\int_{0^{-}}^{0^+} e^{0}\frac{df}{dt}dt + \int_{0^{+}}^{\infty} e^{-st}\frac{df}{dt}dt\right)
$$

Why did $e^{-st}$ become $e^{0}$ in the first part of the split integral when taking the limit to infinity? I have taken Calculus course before but I am just lost in here ($s$ is a complex number in this context).

I would be really grateful if someone answers my question.

REPLY [2 votes]: In a very hand wavy way, what is the only number between $0^+$ and $0^-$? $0$, of course. So the only number that gets plugged in for $t$ is $0$. However, this is a terrible way to prove the Initial Value Theorem. Here is a better way. We already have that
$$\int_0^\infty f'(t) e^{-st}dt = sF(s) -f(0)$$
Then taking $s\to\infty$, and assuming $f'(t)$ was bounded, we get that
$$\int_0^\infty f'(t)\cdot 0 dt = 0 = \lim_{s\to\infty} sF(s) -f(0)$$


TITLE: Is this partition problem strongly NP-complete?
QUESTION [4 upvotes]: Some computational problems have variants that appear to be harder. For instance, Graph Automorphism (GA) problem has quasi-polynomial time algorithm ( by Babai's Graph Isomorphism result) while the fixed-point free GA problem is NP-complete.
Partition problem is weakly NP-complete problem since it has pseudo-polynomial time algorithm. I am interested in variants that are strongly NP-complete.
Here is a variant of partition problem:
Restricted partition problem
Input: Set $S$ of $2N$ integers, and a collection of pairs $P$ from $S$
Query: Is there a partition of $S$ into two equal cardinality parts $A$ and $S-A$ such that both parts have the same sum and no pair in $P$ has both elements in one side of the partition?

Is this variant of partition problem NP-complete in the strong sense?

REPLY [1 votes]: It seems, that there exists a pseudo-polynomial time algorithm for your problem.
Denote by $N$ the sum of $S$. Add $N \cdot |S|$ to each element of $S$. This allows us to not carry about the cardinality restriction.
Consider a square boolean table $T$ of size $N^2 \cdot |S|$  with the following sense:
$T(X,Y)=1$ if there exist two  disjoint subsets $S_X$ and $S_Y$ of $S$ such that no pair in $P$ has both elements in $S_X$ or in $S_Y$ and $|P \cap (S_X \cup S_Y)| \not= 1 $.
Now use standard dynamic programming.
First fill all elements of $T$ by $0$.
Fill $T(0,0):=1$.
Assume we have filled a part of the table.
Let $a$ be an element of $S$ that we have not considered yet. If $a$ is not element of a pair  then
fill $T(X + a, Y) = T(X, Y+a)=1$ for all $X$ and $Y$ such that $T(X, Y)=1$.
If $(a,b)$ is a pair then $T(X + a, Y+b) = T(X+b, Y+a)=1$.
After  filling all table the answer is the element  $T(\frac{N\cdot|S|^2 + N}{2},\frac{N\cdot|S|^2 + N}{2})$
since the all sum now is $N\cdot|S|^2 + N$.


TITLE: Convergence of $ \sum_{n=1}^\infty \frac{\log n}{n^q+1} $
QUESTION [0 upvotes]: Can anyone give a me a hint on how to check if this series is convergent or not?
 $$ \sum_{n=1}^\infty \frac{\log n}{n^{q}+1} $$
Thank you :)

REPLY [0 votes]: Comparison test
We have that $\log n < n$. If $q > 2$ then
$\frac{\log n}{n^q + 1} < \frac{n}{n^q +1} = \frac{1}{n^{q-1} + \frac{1}{n}}< \frac{1}{n^{q-1}}$, then
since for $q> 2$ the series $\sum_{n\geq 1}\frac{1}{n^{q-1}}$ converges, then by comparison theorem so is $\sum_{n \geq 1} \frac{\log n }{n^q + 1}$.
Comparison test and Limit comparison test
Now let $q \in (0, 1]$
for $n \geq 3$
$\frac{1}{n^q+1}<\frac{\log{n}}{n^q+1}$
and the series $\sum\frac{1}{n^q+1}$ is divergent due to limit comparison test with the sum $\sum \frac{1}{n^q}$. Indeed $\frac{n^q+1}{n^q} \to 1$. Hence,
$\sum \frac{\log n }{n^q+1}$ is divergent when $q \in (0, 1]$.
The case which I'm leaving for you is when $q \in (1, 2]$.


TITLE: How find this limits $\lim_{n\to\infty}(\frac{2}{1^4}+1)(\frac{2}{2^4}+1)(\frac{2}{3^4}+1)\cdots(\frac{2}{n^4}+1)$
QUESTION [4 upvotes]: How to Find this limit
$$\lim_{n\to\infty}\left(\dfrac{2}{1^4}+1\right)\left(\dfrac{2}{2^4}+1\right)\left(\dfrac{2}{3^4}+1\right)\cdots\left(\dfrac{2}{n^4}+1\right)$$

see  1
I remeber in sack have solve this follow problem: But I can't find it
$$\prod_{n=1}^{\infty}\left(\dfrac{1}{n^4}+1\right)$$
May be we use same methods can solve it.

REPLY [10 votes]: (Note to the reader: If you find this answer rather terse, then, first, you are right and, second, I suggest that you read the comment thread.)
$$2\pi^2x^2\cdot\prod_{n=1}^{\infty}\left(1+\frac{x^4}{n^4}\right)=\cosh(\pi\sqrt2x)-\cos(\pi\sqrt2x)$$


TITLE: Eulerian graph theorem
QUESTION [1 upvotes]: How can I prove the following theorem:
For a connected multi-graph
G, G is Eulerian if and only if every vertex has even degree.

I found a proof here: in this PDF file, but, it merely consists of language that is very hard to follow and doesn't even give a conclusion that the theorem is proved. So, how can I prove this theorem?

REPLY [1 votes]: Def: An Eulerian cycle in a finite graph is a path which starts and ends at the same vertex and uses each edge exactly once.
Def: A finite Eulerian graph is a graph with finite vertices in which an Eulerian cycle exists.
Def: A graph is connected if for every pair of vertices there is a path connecting them.
Def: Degree of a vertex is the number of edges incident to it.
Claim: A finite connected graph is Eulerian iff all of its vertices are even degreed.
Pf:
($\Longrightarrow$)
Let $G=(V,E)$ be a connected Eulerian graph. By def. of being an Eulerian graph, there is an Eulerian cycle $Z$, starting and ending, say, at $u\in V$. Each visit of $Z$ to an intermediate vertex $v\in V\setminus\{u\}$ contributes 2 to the degree of $v$, so each $v\in V\setminus\{u\}$ has an even degree. As for $u$, each intermediate visit of $Z$ to $u$ contributes an even number, say $2k$ to its degree, and lastly, the initial and final edges of $Z$ contribute 1 each to the degree of $u$, making a total of $1+2k+1=2+2k=2(1+k)$ edges incident to it, which is an even number.
($\Longleftarrow$) (By Strong Induction on $|E|$)
B.S.: $|E|=0$. Since $G$ is connected, there must be only one vertex, which constitutes an Eulerian cycle of length zero.
I.H.: The claim holds for all graphs with $|E|<n$.
I.S.: Let $G$ be a graph with $|E|=n\in \mathbb{N}$.
Def: A tree is a graph which does not contain any cycles in it.
Def: A spanning tree of a graph $G$ is a subset tree of G, which covers all vertices of $G$ with minimum possible number of edges.
Def: A leaf is a vertex of degree 1.
Since $G$ is connected, there should be spanning tree $T=(V',E')$ of $G$.
Lemma: A tree on finite vertices has a leaf. Pf: Let $V=\{v_1,\ldots, v_n\}$. For a contradiction, let $deg(v)>1$ for each $v\in V$. Now start at a vertex, say $v_{i_1}$. Now 'walk' over one of the edges connected to $v_{i_1}$ to a vertex $v_{i_2}$. Since the degree of $v_{i_2}$ is 2, we can walk to a vertex $v_{i_3}\neq v_{i_2}$ and continue this process. Since $V$ is finite, at a given point, say $N$, we will have to connect $v_{i_N}$ to $v_{i_1}$, and have a cycle, $(v_{i_1}, \ldots, v_{i_N}, v_{i_1})$, contradicting the hypothesis that $G$ is a tree.
Hence our spanning tree $T$ has a leaf, $u\in T$. (It might help to start drawing figures from here onward.) Since $deg(u)$ is even, it has an incidental edge $e\in E\setminus E'$. Now consider the cycle, $C:=(V',E\cup\{u\})$. Let $G':=(V,E\setminus (E'\cup\{u\}))$.
Clearly, $deg_{G'}(v)= \left\{\begin{array}{lr}
        deg_G(v)-2, & \text{if } v\in C\\
        deg_G(v), & \text{if } v\notin C
        \end{array}\right.$
Suppose $G'$ consists of components $G_1,\ldots, G_k$ for $k\geq 1$. By Inductive Hypothesis, each component $G_i$ has an Eulerian cycle, $S_i$. Also each $G_i$ has at least one vertex in common with $C$. Let $x_i\in V(G_i)\cap V(C)$. By a renaming argument, we may assume that $S_i$ begins with $x_i$ and ends at $x_i$, since $S_i$ passes all edges in $G_i$ in a cyclic manner. Now, a traversal of $C$, interrupted at each $x_i$ to traverse $S_i$ gives an Eulerian cycle of $G$.


TITLE: What is the limit of this sequence created using Pascal's triangle?
QUESTION [2 upvotes]: Let use the Pascal Triangle to create a sequence. We sum from the first term on the left to the term in the middle
and then we take the inverse root times two. We get then that the terms are

\begin{eqnarray}
&&a_0 := 2 \\
&&a_1 = 2 \frac{1}{1^{1/1}} = 2\\
&&a_2 = 2 \frac{1}{(2+1)^{1/2}} = 1.154700\dots\\
&&a_3 = 2 \frac{1}{(3+1)^{1/3}} = 1.25992\dots\\
&&a_4 = 2 \frac{1}{(6+4+1)^{1/4}} = 1.098200\dots\\
&&a_5 = 2 \frac{1}{(10+5+1)^{1/5}} = 1.14869\dots\\
&&a_6 = 2 \frac{1}{(20+15+6+1)^{1/6}} = 1.0727\dots\\
&&a_7 = 2 \frac{1}{(35+21+7+1)^{1/7}} = 1.10408\dots\\
\end{eqnarray}

The first line of the triangle creates the $a_0$ the second creates the $a_1$ and so on. Then I have my conjecture:

This sequence converges to 1

But I couldn't prove it or disprove it. What I did was tried to relate to the Binomial expansion: Let $0\leq x \leq 1$ so we have that
$$(1+x)^2 = 1 + 2x^2 + x^4 \geq x^2 + 2x^2 = 3x^2$$
So we have that
$$\frac{x}{x+1} \leq \frac{1}{\sqrt{3}}$$
And I tried to translate this to some limit using this $x$. There is a way to solve this problem like this? This series converges?

REPLY [2 votes]: The sum of the elements in "half a row" of Pascal's triangle is either $2^{n-1}$ (if we are talking about the $n$-th row with $n$ being odd) or $2^{n-1}+\frac{1}{2}\binom{n}{n/2}$ if $n$ is even, due to $\binom{n}{k}=\binom{n}{n-k}$.
The central binomial coefficient $\binom{2n}{n}$ behaves in the following way
$$ \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}} $$
hence you are essentially asking about
$$ \lim_{n\to +\infty} 2\cdot\frac{1}{(2^{n-1}+E(n))^{1/n}} $$
that is clearly $1$.


TITLE: Can we exactly find where is instatenous axis of rotation is of a rod subjected to a constraint?
QUESTION [0 upvotes]: Suppose we have a rod which is connected with two strings with some angle theta eith vertical , suppose one string is cut out , is it possible to figure out where the IAOR of rod is located at just after the instant string is cut ? (My thought process was that we know point A acceleration direction very well for t= dt moment ,but we dont know centre of mass acceleration exact direction , is it still possible to figure out IAOR ?)

REPLY [0 votes]: From newtons laws and torque equation we can calculate both centre of mass acceleration and end point acceleration where string is connected , from there we can say in dt time speed would be v1 and v2 , now draw perpendiculars we will get the IAOR of the rod .


TITLE: Generalised quantifiers and Boolean Algebra
QUESTION [2 upvotes]: On page 4 of http://www.llf.cnrs.fr/sites/llf.cnrs.fr/files/biblio//the-same-ter.pdf
Richard Zuber writes:

"Given a fixed universe $E$, (where $|E| ≥ 2$), a type $n$ quantifier is a function from
  $n$-ary relations to truth values. A type $\langle 1 \rangle$ quantifier is a function from sets (sub-sets
  of $E$) to truth values, and thus it is a set of sub-sets of $E$. A type $\langle 1, 1 \rangle$ quantifier
  is a function from sets to type $\langle 1 \rangle$ quantifiers. In natural language semantics type
  $\langle 1 \rangle$ quantifiers are denotations of NPs and a type $\langle 1, 1 \rangle$ quantifiers are denotations
  of (unary nominal) determiners, that is expressions like $\textit{every, no, most, five}$, etc.
  Since both types of quantifiers form Boolean algebras they have Boolean complements (negations)."

In what sense do type $\langle 1 \rangle$ and $\langle 1, 1 \rangle$ quantifiers form Boolean algebras, as Zuber says? In what sense, for example, does the denotation of $\textit{every}$ form a boolean algebra?
Could it be that the Boolean algebra is formed by taking the boolean operations ON the set of subsets that is the denotation of a type $\langle 1, 1 \rangle$ quantifier? But then the boolean algebra would not be identical to the denotation of $every$ but a boolean algebra formed by operations performed on its denotation.

REPLY [2 votes]: It's not that a single quantifier forms a Boolean algebra; rather, the set of all quantifiers of a given type is a Boolean algebra (with the Boolean operations being performed pointwise, when you consider quantifiers as functions in the manner described).  So for instance, the Boolean complement of the quantifier "every" is the quantifier "not every" (i.e., the function that takes a unary relation to true iff there is some input that makes the unary relation become false).


text:
\section{Introduction}

Let us recall some standard notions from set-theoretic geology.
We generally assume $\ZFC$, though at times
(in particular in \S\ref{sec:grounds})
we will also consider a weaker theory $T_1$ (which includes $\AC$).

Given a transitive model $W$\footnote{Here we are not specific
about exactly what formalization of classes we use. We could
work in some  class set theory,
which allows quantification over such classes $W$, or we could
with more care restrict to classes definable from parameters;
for us a class must have the property
that the structure $(V,\in,W)$ satisfies $\ZFC$
in the language with symbols $\dot{\in},\dot{W}$ which interpret
$\in$ and $W$.} of $\ZFC$ and a
forcing $\PP\in W$,
a \emph{$(W,\PP)$-generic} is a filter $G\sub\PP$
which is generic with respect to $W$.
For a cardinal $\kappa$, a \emph{${<\kappa}$-ground}
of $V$ is a transitive proper class $W\sats\ZFC$ such that
there is $\PP\in W$ with $\PP$ of cardinality ${<\kappa}$ (with cardinality
as computed in $W$, or equivalently, in $V$) and a $(W,\PP)$-generic filter $G$
such that $V=W[G]$. A \emph{ground} is a $<\kappa$-ground
for some cardinal $\kappa$.\footnote{Throughout,
we consider only set-forcing, no class-forcing.} The \emph{mantle} $\Mmm$ is
the intersection of all
grounds.
The \emph{$\kappa$-mantle} $\Mmm_\kappa$ is the intersection of all
${<\kappa}$-grounds.

By \cite{laver_vlc}, as refined in \cite{stgeol},
there is a formula $\varphi(x,y)$ in two free variables
such that (i) for all $r$,
$W_r=\{x\bigm|\varphi(r,x)\}$
is a ground (possibly $W_r=V$), and (ii) for every ground
$W$ there is $r$ such that $W=W_r$.
Therefore we can discuss grounds uniformly,
and $\Mmm$ and $\Mmm_\kappa$ are transitive classes
which are definable ($\Mmm_\kappa$ from parameter $\kappa$).

In \S\ref{sec:grounds} we will give the proof of ground definability,
but from somewhat less than $\ZFC$: we show that it holds
under a certain theory $T_1$ (see \ref{dfn:T_1}),
which is true in $\her_\kappa$ whenever $\kappa$ is a strong
limit cardinal (assuming $\ZFC$). The proof is essentially
the usual $\ZFC$ proof, however.

From now on, we take $W_r$ to be defined as in
\S\ref{sec:grounds}, by which $r=(\her_{\gamma^+})^W$
for some $\gamma\geq\om$ for which there is $\PP\in r$
and a $(W,\PP)$-generic $G$ with $W[G]=V$.

Let $\theta$ be a strong limit cardinal.
By Usuba \cite{usuba_ddg},
the grounds
are set-directed. By \cite{usuba_ddg} and \cite{schindler_fsttimtg},
this is moreover
reasonably local,
and in particular if $X\in\her_\theta$, then there is $s\in\her_\theta$
with
$W_s\sub\bigcap_{r\in X}W_r$. (For following Usuba's proof of \cite[Proposition
5.1]{usuba_ddg},  note that we can take the regular
cardinal $\kappa$ of that proof with $\kappa<\theta$,
and then the model $W$ constructed there satisfies the $\kappa^{++}$-uniform
covering property for $V$. Usuba then uses Bukovsky's theorem, \cite[Fact
3.9]{usuba_ddg}, to deduce that $W$ is a ground of $V$. But
by \cite[Theorem 3.11]{schindler_fsttimtg}, the forcing for this can be taken
of size
$2^{\kappa^{++}}$ in $W[g]=V$.)

Also by \cite{usuba_ddg}, $\Mmm\sats\ZFC$,
and by \cite[\S2]{usuba_extendible},
 $\Mmm_\theta\sats\ZF$
(so note $\Mmm_\theta\sats$``$\theta$ is a strong limit cardinal'',
in the $\ZF$ sense that $\Mmm_\theta$ has no surjection
$\pi:V_\alpha\to\theta$ with $\alpha<\theta$).
If
$V_\theta\elem_n V$ with $n$ large enough, then
$V_\theta^{\Mmm_\theta}=V_\theta^{\Mmm}$, and hence
$V_\theta^{\Mmm_\theta}\sats\AC$.\footnote{\label{ftn:false_claim}An earlier
draft asserted here that
(\cite{usuba_ddg}  shows) $\her_\theta^{\Mmm_\theta}=\her_\theta^{\Mmm}$
for every strong limit cardinal $\theta$. This
was however not used anywhere, and it is false, at least if $M_1$
exists; see \cite{local_mantles_in_Lx}.}
 Usuba  showed in \cite{usuba_extendible} that
if $\kappa$ is an extendible
cardinal then $\Mmm_\kappa=\Mmm$, so in this case,
$\Mmm_\kappa\sats\ZFC$.
Hence Usuba asked in \cite{usuba_extendible} about whether
$\Mmm_\kappa\sats\ZFC$ in general. We consider
related questions in this paper. Let us first sketch some further history.

Suppose now $\kappa$ is inaccessible.
Then $V_\kappa^{\Mmm_\kappa}\sats\ZFC$.
For note that by inaccessibility and the remarks above,
for each $\alpha<\kappa$ there is some $r\in V_\kappa$
such that $V_\alpha^{W_r}=V_\alpha^{\Mmm_\kappa}$.
Since each $W_r\sats\ZFC$, it follows that
$V_\kappa^{\Mmm_\kappa}\sats\ZFC$.
Clearly $\Mmm_\kappa\sats$``$\kappa$ is
inaccessible'', and if $\kappa$ is Mahlo
then $\Mmm_\kappa\sats$``$\kappa$ is Mahlo''.

However, A. Lietz (\cite{lietz}) answered Usuba's question above negatively
(assuming large cardinals),
showing that in fact it is consistent
relative to a Mahlo cardinal that  $\kappa$ is Mahlo but
$\Mmm_\kappa\sats$``$\kappa$-$\AC$ fails''. In fact, Lietz constructs
a forcing extension $L[G]$ of $L$ in which $\kappa$ is Mahlo and
$\Mmm_\kappa^{L[G]}$ satisfies ``there is a function
$f:\kappa\to\her_{\kappa^+}$ for which there is no choice function''.
He also proved other related things.

In the last few years, the theory of Varsovian models
has also been developed by Fuchs, Schindler, Sargsyan and more recently the
author.
Here, among other things, full mantles $\Mmm$ of certain fully
iterable mice  have
been analyzed,
and shown to be strategy mice, hence satisfying $\ZFC$.
Analysis of natural $\kappa$-mantles of those mice was, however,  missing.
But using Varsovian model techniques,
the author then analyzed the $\kappa_0$-mantle of the mouse
$M_{\swsw}$ (Definition \ref{dfn:Mswsw}),
showing that it is a strategy mouse, modelling $\ZFC+\GCH$.
A very brief outline is given in \S\ref{sec:choice}
(but the other results in the note do not rely on this,
and no inner model theory
appears elsewhere in the paper).
The argument
has elements in common with
 Usuba's extendibility
 proof.

 Schindler then
showed that if $\kappa$ is measurable then $\Mmm_\kappa\sats\AC$,
hence $\ZFC$; see \cite{schindler_fsttimtg}.
In this note we adapt this
argument,
deducing that fragments of choice hold in $\Mmm_\kappa$
from the weak compactness and inaccessibility  of $\kappa$ respectively.

\begin{dfn}\label{dfn:(alpha,X)-Choice}
 Given an ordinal $\alpha$ and set $X$,
let \emph{$(\alpha,X)$-Choice} be the assertion
that for every function $f:\alpha\to X$,
there is a choice function for $f$.
And \emph{$({<\alpha},X)$-Choice}
is the assertion that $(\beta,X)$-Choice
holds for all $\beta<\alpha$.
\end{dfn}

Part \ref{item:4} of the following theorem applies to the kind
of functions involved in the failure of $\kappa$-$\AC$ in Lietz' example,
but now with domain ${<\kappa}$.
Note that we assume $\ZFC$ except where otherwise stated;
 \emph{$\kappa$-amenable-closure}
is defined in \ref{dfn:amenable-closure}.

\begin{tm*}[\ref{tm:inaccessible_tm}]
\inacctmwithout
\end{tm*}

\begin{rem}
In part \ref{item:<kappa-Choice},
the ``$\kappa^+$'' and ``$\her_{\kappa^+}$''
are both in the sense of $\Mmm_\kappa$.
However, it can be that $\kappa$ is Mahlo
 and $\Mmm_\kappa\sats$``$(\kappa,\her_{\kappa^+})$-Choice
 fails, and $(\her_{\kappa^+})^\kappa\not\sub\her_{\kappa^+}$'';
 indeed, note that this occurs in Lietz' example $L[G]$ mentioned above.
\end{rem}

In the following theorem, the initial observation
that $\Mmm_\kappa\sats$``$\her_\kappa$ is
wellordered'' was due to Lietz:

\begin{tm*}[\ref{tm:weak_compact_tm}] \footnote{Regarding
part \ref{item:<kappa+-DC},
the author initially
 observed that a variant
 of Schindler's argument gives that $\Mmm_\kappa\sats\kappa$-$\DC$,
 and then  Lietz and the author independently noticed
 that the argument can be adjusted to
 show that every
 set in $\her_{\kappa^+}\inter\Mmm_\kappa$ is wellordered
 in $\Mmm_\kappa$.}
\weakcompacttmwithout
\end{tm*}

As a corollary to Schindler's proof, one easily gets:

\begin{fact*}[\ref{cor:meas}]
\textthmmeas
\end{fact*}

As mentioned above, Usuba showed that $\Mmm=\Mmm_\kappa$ assuming $\kappa$ is
extendible.
The next result indicates that there are signs of this
in the leadup to an extendible cardinal
(for the definition of a \emph{$\Sigma_2$-strong} cardinal,
see \ref{dfn:Sigma_2-strong}):

\begin{tm*}[\ref{tm:Sigma_2-strong}]
\Sigmatwostrong
\end{tm*}

Analogously, down lower:

\begin{tm*}[\ref{tm:sharp}]
\textthmsharp
\end{tm*}

For further related results, which involve
some inner model theory, see  \cite{local_mantles_in_Lx}.

Before beginning our discussion of these results,
we go through some background set-theoretic geology,
including a proof of the the definability of grounds
from a theory modelled by $\her_\kappa$ whenever
$\kappa$ is a strong limit cardinal.

\section{Grounds and mantles}\label{sec:grounds}
We discuss here some background, starting with the key
fact of the definability of set-forcing grounds under $\ZFC$,
proved by some combination of Laver, Woodin and Hamkins:

\begin{fact}
Let $M,N$ be proper class
transitive inner models of $\ZFC$
and $\gamma\in\OR$ with $\pow(\gamma)\inter M=\pow(\gamma)\inter
N$.
Let $\PP\in M$ and $\QQ\in N$, with $\PP,\QQ\sub\gamma$,
and let $G$ be $(M,\PP)$-generic and $H$ be $(N,\QQ)$-generic
and suppose $M[G]=N[H]=V$. Then $M=N$.
\end{fact}

We will discuss the proof of the result above, for two purposes.
The fact is central to our concerns, and the proof
contains elements which will come up in various places later,
so it is natural to collect all these things together.
Second, we wish to prove a version which assumes less background theory
(than $\ZFC$). The authors of \cite{bhtu} make use of an analysis
of the complexity of the definability of grounds.
As shown there, each ground $W$ is, in particular, $\Sigma_2$ in a parameter
$r$. However, the $\Sigma_2$ definition given there is not particularly
local: to compute $V_\alpha^W$, they work in $V_\beta$,
for a significantly larger ordinal $\beta$. So for  \cite[Theorem
4]{bhtu}, they adopt the background theory  $\ZFC_\delta$.
We show here that the ground definability can be done much more locally
(though still requiring $\Sigma_2$ complexity), hence requiring significantly
less than $\ZFC_\delta$.

\begin{dfn}\label{dfn:T_1}
Let $T_1^-$
 be the following theory in the language of set theory.
The axioms are Extensionality, Foundation, Pairing, Union,
Infinity, ``Every set is bijectable with an ordinal'',
$\Sigma_1$-Separation
and $\Sigma_1$-Collection. Now let
\[ T_1=T_1^-+\text{ Powerset}.\qedhere\]
\end{dfn}

Note that $T_1^-\sats\AC$. We will show that models of $T_1$ can uniformly
define
their grounds from parameters. First we give some lemmas.

\begin{lem}
 Assume $\ZFC$. Then for every cardinal $\kappa\geq\om$,
(i) $\her_\kappa\sats T_1^-$, and
(ii) $\her_\kappa\sats T_1$ iff $\kappa$ is a strong limit cardinal.
\end{lem}

The usual proofs from $\ZFC$ easily adapt to give:
\begin{lem}
Assume $T_1$. Then  (i) for each ordinal $\xi$,
 $\her_\xi$ exists, (ii) $V=\bigcup_{\xi\in\OR}\her_\xi$,
  (iii) $\her_\xi\elem_1 V$, (iv) $\her_\xi\sats T_1^-$,
  (iv) the Lowenheim-Skolem theorem holds.
\end{lem}

In the following lemma, the forcing relations $\forces_{\Sigma_i}$
for $i\in\{0,1\}$, and
$\forces_{\Pi_1}$, are the relations defined in a first-order
manner over $M$ in the usual
manner, and the strong-$\Sigma_{i+1}$-forcing relation
$\forces^*_{\Sigma_{i+1}}$ is the relation for which,
given a $\Pi_i$ formula $\psi(\vec{x},\vec{y})$
with free variables $\vec{x},\vec{y}$, and
given $\vec{\tau}\in(M^\PP)^{<\om}$,
we say $p\forces^*_{\Sigma_{i+1}}\exists\vec{y}\ \psi(\vec{\tau},\vec{y})$
iff there is $\vec{\sigma}\in(M^\PP)^{<\om}$
such that $p\forces_{\Pi_i}\psi(\vec{\tau},\vec{\sigma})$.

\begin{lem}[Forcing over $T_1^-$ and $T_1$]\label{lem:T_1_forcing}
Let $M\sats T_1^-$. Let $\PP\in M$ be a poset with $\PP\sub\gamma\in\OR^M$
and $G$ be $(M,\PP)$-generic.
Then:
\begin{enumerate}
 \item\label{item:Sigma_0_fr} We have:
 \begin{enumerate}
 \item The $\Sigma_0$-forcing relation
$\forces_{\Sigma_0}$ for $(M,\PP)$ is
$\Delta_1^M(\{\PP\})$, uniformly.
\item The $\Sigma_1$-forcing relation $\forces_{\Sigma_1}$ for
$(M,\PP)$ is
$\Sigma_1^M(\{\PP\})$, uniformly.\footnote{In a previous
draft of this document,
it mistakenly said that the $\Sigma_1$-forcing relation
is $\Delta_1^M$-definable, which is clearly false,
since in the case of trivial forcing, it would imply
that $\Sigma_1^M=\Delta_1^M$.}
\item The $\Pi_1$-forcing relation $\forces_{\Pi_1}$ for $(M,\PP)$ is
$\Pi_1^M(\{\PP\})$, uniformly.
\end{enumerate}
Hence, $\forces_{\Sigma_0}$, $\forces_{\Sigma_1}$ and $\forces_{\Pi_1}$ are
 absolute to $\her_{\kappa}^M$, for $M$-cardinals $\kappa>\gamma$.
\item The strong-$\Sigma_2$-forcing relation $\forces^*_{\Sigma_2}$
for $(M,\PP)$ is $\Sigma_2^M(\{\PP\})$, uniformly.
 \item\label{item:Sigma_1_ft} The forcing theorem for
$\Sigma_0$, $\Sigma_1$, $\Pi_1$ formulas
 holds for  $M[G]$, with respect to $\forces_{\Sigma_0}$, $\forces_{\Sigma_1}$,
 $\forces_{\Pi_1}$; likewise for $\Sigma_2$ and $\forces^*_{\Sigma_2}$.
That is, if $\varphi$ is $\Sigma_i$, where $i\in\{0,1\}$, and
$\vec{\tau}\in(M^\PP)^{<\om}$,
then
\[ M[G]\sats\varphi(\vec{\tau}_G)\iff \exists p\in G\ \Big[
M\sats\text{``}p\forces_{\Sigma_i}\varphi(\vec{\tau})\text{''}\Big].\]
Likewise for $\Pi_1$ with $\forces_{\Pi_1}$, and for $\Sigma_2$ with
$\forces^*_{\Sigma_2}$.

 \item\label{item:M[G]_sats_T_1} $M[G]\sats T_1^-$,
 and if $M\sats T_1$ then $M[G]\sats T_1$.
 \item $M$ and $M[G]$ have the same
 cardinals $\kappa>\gamma$,
 \item for each $M$-cardinal $\kappa>\gamma$,
we have $\her_\kappa^{M[G]}=\her_\kappa^M[G]$.
\end{enumerate}
\end{lem}

Such
local forcing calculations are very common in the literature,
in particular in fine structure theory, where much more local calculations are
often used. But we include a proof in case the reader has
not seen these before.
\begin{proof}
Parts \ref{item:Sigma_0_fr}, \ref{item:Sigma_1_ft} for $\Sigma_0$:
The usual internal
definition of the $\Sigma_0$-forcing relation $\forces_0$
works locally; in fact,
for each $\xi\in\OR^M$ with $\xi\geq\gamma$,
the $\Sigma_0$-forcing
relation for names in $\her_\xi$, is $\Delta_1^{\her_\xi}(\{\PP\})$,
uniformly in $\xi$. This gives the Forcing Theorem
for $\Sigma_0$ formulas in the usual manner.

Parts  \ref{item:Sigma_0_fr}, \ref{item:Sigma_1_ft} for $\Sigma_1$:
We defined the strong-$\Sigma_1$-forcing relation $\forces^*_1$ over $M$
above. Using the $\Sigma_0$-Forcing Theorem, note that
$M[G]\sats\exists y\ \varphi(y,\tau_G)$
iff there is $p\in G$ such that
$M\sats$``$p\forces^*_1\exists y\ \varphi(y,\tau)$''.
Moreover, $\forces^*_1$ is uniformly $\Sigma_1^M(\{\PP\})$-definable.

Note that we take $\forces_1$ defined over $M$ as follows:
Working in
$M$, for $\varphi$ being $\Sigma_1$
and $\tau\in M^\PP$, set
\[ p\forces_1\varphi(\tau) \iff \all q\leq p\ \exists r\leq q\
\Big[r\forces^*_1\varphi(\tau)\Big].\]

We claim that $p\forces_1\varphi(\tau)$
iff $p\forces_1^*\varphi(\tau)$.
For the non-trivial direction,
suppose $p\forces_1\varphi(\tau)$.
Then working in $M$, using
$\Sigma_1$-Collection and $\AC$,
we can put together a name $\sigma\in M^\PP$
showing that $p\forces^*_1\varphi(\tau)$.
This completes the calculation for $\Sigma_1$.

Parts  \ref{item:Sigma_0_fr}, \ref{item:Sigma_1_ft} for $\Pi_1$:
$\forces_{\Pi_1}$ is defined as usual: Working in $M$,
for $\varphi$ being $\Pi_1$ and $\tau\in M^\PP$,
say $p\forces_{\Pi_1}\varphi(\tau)$
iff there is no $q\leq p$ such that
$q\forces_{\Sigma_1}\neg\varphi(\tau)$.
So $\forces_{\Pi_1}$ is $\Pi_1^M(\{\PP\})$.
If $p\in G$ and $p\forces_{\Pi_1}\varphi(\tau)$,
then clearly $M[G]\sats\varphi(\tau_G)$.
So suppose $M[G]\sats\varphi(\tau_G)$ where $\varphi$ is $\Pi_1$.
Let
\[ D=\{p\in\PP\bigm|p\forces_{\Sigma_1}\neg\varphi(\tau)\}.\]
By $\Sigma_1$-Separation, $D\in M$.
Let $D'=D\cup\{p\in\PP\bigm|\neg\exists q\in D\ [q\leq p]\}$,
then $D'\in M$, and since $D'$ is dense, this easily suffices.

Parts  \ref{item:Sigma_0_fr}, \ref{item:Sigma_1_ft} for $\Sigma_2$:
Here we only consider the strong-$\Sigma_2$ forcing relation
$\forces^*_{\Sigma_2}$, and the claims regarding this follow immediately
just like for $\forces^*_{\Sigma_1}$.

Part \ref{item:M[G]_sats_T_1}: Most of the axioms are routine consequences
of the previous parts.
Let us verify that $M[G]\sats\Sigma_1$-Collection.
Fix a $\Sigma_0$ formula $\varphi$ and $\sigma,\tau\in M^\PP$.
Let $t\in M$ be the transitive closure of $\{\sigma,\tau\}$.
Then there is $w\in M$ such that for all $p\in\PP$
and $\varrho\in t$,
if
\[ p\forces_{\Sigma_1}\text{``}\varrho\in\sigma\text{ and }\exists
y\varphi(\varrho,\tau,y)\text{''},\]
then there is $y\in M^\PP\inter w$ such that
$p\forces_{\Sigma_0}\text{``}\varrho\in\sigma\text{ and
}\varphi(\varrho,\tau,y)\text{''}$.
But then using $w$, we easily get a bound on witnesses in $M[G]$,
as desired. This and the $\Sigma_0$-Forcing Theorem
easily yields $\Sigma_1$-Separation.

The remaining parts follow from routine calculations with nice names.
\end{proof}

\begin{dfn}
 Let $(M,E)\sats T_1^-$. A \emph{ground} of $M$ is a
 $W\sub M$ such that:
 \begin{enumerate}
  \item $(W,E\rest W)$ is \emph{$M$-transitive};
 that is, for all $x\in W$ and all $y\in M$, if $yEx$ then $y\in W$,
  \item $W\sats T_1^-$,
  \item there is $\PP\in W$ and a $(W,\PP)$-generic
 $G\in M$ such that $M=W[G]$.
 \item If $(M,E)\sats T_1$ then $(W,E\rest W)\sats T_1$.\qedhere
 \end{enumerate}
\end{dfn}

We now prove that $T_1$ suffices for the definability of grounds
(in the sense of the definition above).
The proof is essentially that due to some combination of Laver, Woodin and
Hamkins.
In the proof we make implicit use of
Lemma \ref{lem:T_1_forcing}, to allow the forcing calculations:

\begin{tm}[Ground definability under $T_1$]
Assume $T_1$.  Let
$\gamma\in\OR$,
$H\sub\her_{\gamma^+}$ and $\kappa\geq\gamma^+$ a cardinal.
Then there is at most
one transitive $M\sub\her_{\kappa}$ such that
$M\sats T_1^-$, $(\her_{\gamma^+})^M=H$, and
$M$ is a ground for $\her_{\kappa}$ via some  $\PP\in H$.
\end{tm}
\begin{proof}
We proceed by induction on $\kappa$.
For $\kappa=\gamma^+$ it is trivial.

Suppose $\kappa$ is a limit cardinal,
and that for each cardinal $\theta\in[\gamma^+,\kappa)$,
there is a (unique) model $M_\theta$ of ordinal height $\theta$
with the stated
properties. Then clearly $M=\bigcup_{\theta<\kappa}M_\theta$
is the unique candidate at $\kappa$. To see that $M$ works,
we just need to verify that $M$ is indeed a set-ground
of $\her_\kappa$ via some $\PP\in H$;
i.e. there is $\PP\in H$ and an $(M,\PP)$-generic
$G\sub\PP$ such that $M[G]=\her_\kappa$. But we can use any $(\PP,G)$
which worked at some earlier $\theta$.
For let $\theta_0\leq\theta_1<\kappa$, and let
$(\PP_0,G_0),(\PP_1,G_1)$ work for $M_0=M_{\theta_0}$ and $M_1=M_{\theta_1}$.
So
$G_0$ is also $(M_1,\PP_0)$-generic, and vice versa.
And since
$\her_{\gamma^+}^{M_0}=H=\her_{\gamma^+}^{M_1}$,
and $H[G_0]=\her_{\gamma^+}=H[G_1]$, it follows
that $\her_\kappa=M_0[G_0]=M_0[G_1]$ and $M_1[G_0]=M_1[G_1]=\her_\kappa$,
so the specific choice of $(\PP,G)$ is irrelevant.

So consider $\kappa=\theta^+>\gamma^+$. Let $M,N$
be grounds of $\her_\kappa$ with the stated properties. By induction,
$M\inter\her_\theta=N\inter\her_\theta$. It just
remains to verify that
$\pow(\theta)\inter M=\pow(\theta)\inter N$.
The proof is, however, not by contradiction;
we will not assume that $M\neq N$.
Fix $(\PP,G)$ such that $\PP\in H$ and $G$ is $(M,\PP)$-generic
and $M[G]=\her_\kappa$.

Suppose first that $\cof(\theta)>\gamma$, as this case is easier;
however, it is in the end subsumed into the general case.
Let $A\sub\theta$. Then:

\begin{clm} $A\in M$ iff $A\inter\alpha\in M$ for all $\alpha<\theta$.\end{clm}
\begin{proof}
For the non-trivial direction,
suppose $A\inter\alpha\in M$ for every $\alpha<\theta$.
Let $f:\theta\to M$ be
$f(\alpha)=A\inter\alpha$.
Then $f\in\her_\kappa$. So
there is a $\PP$-name $\dot{f}\in M$ with $\dot{f}_G=f$.
Working in $M$, for $p\in\PP$, compute
\[ D_p=\{\alpha<\theta\bigm|\exists x\
[p\forces\dot{f}(\check{\alpha})=\check{x}]\},\]
and let
$f_p:D_p\to\theta$ be the function
\[ f_p(\alpha)=\text{ unique }x\text{ such that
}p\forces\dot{f}(\check{\alpha})=\check{x}.\]
So $\left<D_p,f_p\right>_{p\in\PP}\in M$, and because $\cof(\theta)>\gamma$,
there is $p\in G$ such that $D_p$ is cofinal in $\theta$.
Then $f=\left(\bigcup_{\alpha\in D_p}f_p(\alpha)\right)\in M$.
\end{proof}

We now argue in general.

\begin{clm}
 Let $A\sub\theta$. Then $A\in M$ iff
 for every $X\in\pow(\theta)\inter M$
 such that $\card(X)<(\gamma^+)=(\gamma^+)^M$,
we have $A\inter X\in M$.
\end{clm}
\begin{proof}
The forward direction is trivial. So let $A\sub\theta$ with $A\notin M$.
Let $\dot{A}\in M$ be a $\PP$-name
and $p_0\in G$ such that
$p_0\forces\dot{A}\sub\check{\theta}$.
For each $q\leq p_0$, \emph{if}
there is $\alpha<\theta$ such that
\[ q\not\forces\check{\alpha}\in\dot{A}\text{ and
}q\not\forces\check{\alpha}\notin\dot{A},\]
then let $\alpha_q$ be the least such $\alpha$;
otherwise $\alpha_q$ is undefined.
Let $D$ be the set of all $q\leq p_0$
such that $\alpha_q$ exists.
Then $G\sub D$, because
otherwise $q$ decides all elements of $\dot{A}$,
so $A\in M$.

In $M$, let
 $X=\{\alpha_q\bigm|q\in D\}$.
Then  $X\in M$, $\card^M(X)\leq\gamma$ and
$X\inter A\notin M$, as desired. For given $Y\in\pow(X)\inter M$,
an easy density argument shows that $Y\neq X\inter A$.
\end{proof}

\begin{clm}Let $X\sub\theta$ with
$\card(X)<\gamma^+$.
Then $X\in M$ iff $X\in N$.
\end{clm}
\begin{proof}
Suppose $X_0=X\in N$.
Let $\dot{X}\in M$ be a $\PP$-name for $X$.
Using the forcing relation and $\dot{X}$,
there is a set $X_1\in\pow(\theta)\inter M$
with $X_0\sub X_1$ and $\card(X_1)<(\gamma^+)^V$.
Proceeding back-and-forth, construct (in $V$)
a continuous sequence of sets $\left<X_\alpha\right>_{\alpha<\gamma^+}$ such
that
(i) $X_0=X$,
(ii) $X_{\om\alpha+2n+1}\in M$ and $X_{\om\alpha+2n+2}\in N$, and
(iii) $\card(X_\alpha)<(\gamma^+)^V$.

Now $\gamma^+<\kappa$, so
$\left<X_\alpha\right>_{\alpha<\gamma^+}\in\her_\kappa$,
so $M,N$ have names for this sequence.
So as in the $\cof(\theta)>\gamma$ case,
we get a cofinal set $D_M\sub\gamma^+$
such that $D_M\in M$ and $\left<X_\alpha\right>_{\alpha\in D_M}\in M$.
Likewise with a cofinal set $D_N\in N$.
Let $D'_M$ be the set of limit points of $D_M$,
and $D'_N$ likewise. So these are club in $\gamma^+$.
Let $\alpha\in D'_M\inter D'_N$. Then note
that
\[ X_\alpha=\left(\bigcup_{\beta\in
D_M\inter\alpha}X_\beta\right)=
\left(\bigcup_{\beta\in D_N\inter\alpha}X_\beta\right)\in
M\inter N.\]
Let $\pi:\xi\to X_\alpha$ be the increasing enumeration
of $X_\alpha$. Then $\xi<\gamma^+$
and  $\pi\in M\inter N$.
We have $X\sub\rg(\pi)$.
Let $\bar{X}=\pi^{-1}(X)$.
Then $\bar{X}\in N$. But $\her_{\gamma^+}^M=H=\her_{\gamma^+}^N$,
so $\bar{X}\in M$. So $\pi``\bar{X}=X\in M$, as desired.
\end{proof}

This completes the proof of  ground definability under $T_1$.
\end{proof}

\begin{rem}
If $M\sats T_1^-$+``there is a largest cardinal
 $\kappa$, and $\kappa$ is regular'', then grounds of
 $M$ via forcings $\PP$ of $M$-cardinality ${<\kappa}$
are also definable from parameters over $M$,
by arguing much as above.
\end{rem}

\begin{dfn}Assume $T_1$.
 Let $\varphi_{\mathrm{grd}}(r,x)$ be the formula ``
 there are $\gamma$, $\PP$, $G$, $M$, $\kappa$ such that
 $\gamma<\kappa$ are cardinals,
$M\sub\her_\kappa$ is transitive, $M\sats T_1^-$,
$\PP\in r=(\her_{\gamma^+})^M$,
$G$ is $(M,\PP)$-generic, $\her_\kappa=M[G]$ and $x\in M$''.

We write $W'_r=\{x\bigm|\varphi_{\mathrm{grd}}(r,x)\}$.
We say $r$ is a \emph{true index} iff $W'_r$ is proper class.
We write $W_r=W'_r$ for true indices $r$,
and $W_r=V$ otherwise.
\end{dfn}

\begin{cor}
 Assume $\ZFC+\GCH$ and let $\lambda$ be a limit cardinal.
 Then the grounds of $\her_\lambda$ are definable from parameters
 over $\her_\lambda$.
\end{cor}

\begin{rem}
 Assume $\ZFC+\GCH$. Then for each limit ordinal $\xi$,
$V_{\om+\xi}$ is equivalent in the codes to
 $\her_{\aleph_{\xi}}$.
 So one can correctly formulate ``grounds''
 of $V_{\om+\xi}$, and they are definable over $V_{\om+\xi}$ from parameters.
\end{rem}

So we have the standard uniform definability of grounds, just assuming $T_1$:
\begin{lem}
Let $M\sats T_1$. Then
$\{W_r^M\bigm|r\in M\}$
enumerates exactly the grounds of $M$ (with repetitions, including
$M$ itself).\end{lem}

\begin{rem}Assume $T_1$. Note that $\varphi_{\grd}$ is $\Sigma_2$,
and the assertion ``$r$ is a true index'' is $\Pi_2$.
(In fact, there are fixed $\Sigma_2$ and $\Pi_2$ formulas,
such that $T_1$ proves that these fixed formulas always work.)
Moreover, letting $\xi=\card(\trcl(\{r,x\}))$, note that
$\varphi_{\mathrm{grd}}(r,x)$ is absolute between $V$ and
$\her_{(2^{\xi})^+}$. (It is witnessed
by some $(\her_{\xi^+},M)$,  a structure
of size $2^{\xi}$.) Therefore:\end{rem}

\begin{fact}[Local definability of grounds]\label{fact:local_ground_def}
Assume $T_1$+``There is a proper class of strong limit cardinals''.
Let $\lambda$ be a strong limit cardinal.
Let $r\in\her_\lambda$ be a true index.
Then $\her_\lambda\sats$``$r$ is a true index''
 and
 $W_r^{\her_\lambda}=W_r\inter\her_\lambda=\her_\lambda^{W_r}$.
\end{fact}

It seems it might be possible, however, that $\her_\lambda\sats$``$r$ is a true
index'' while $r$ fails to be a true index in $V$.

The remaining facts in this section, and the rest of the paper,
have a background theory of $\ZFC$. We have not investigated
to what extent things go through under $T_1$.
By \cite[Proposition 5.1]{usuba_ddg} and an
examination of its proof, we have:

\begin{fact}[Local set-directedness of
grounds (Usuba)]\label{fact:local_ground_directedness}
(Assume $\ZFC$.) Let $\theta$ be a strong limit cardinal and $R\in\her_\theta$.
 Then there is
 $t\in\her_\theta$ with
 $t\in\bigcap_{r\in R}W_r$
 and $W_t\sub W_r$ and $W_t=W_t^{W_r}$ for each
 $r\in R$. In particular, $W_t\sub\bigcap_{r\in R}W_r$.
\end{fact}

\begin{proof}
We refer here to the \emph{$\lambda$-uniform covering property for $V$};
see  \cite[Definition
4.2]{usuba_ddg} or \cite[Definition 2.1]{schindler_fsttimtg}.
Let us set up some of the notation from the proof of \cite[Proposition
5.1]{usuba_ddg}. Let $X=R$ (following the notation
from \cite{usuba_ddg}).\footnote{We wrote $R$ in the
statement of the fact for consistency with later notation.}We may assume
that $X$ is a set of true indices $r$.
For $r\in X$ let $\PP_r\in W_r$ be a forcing witnessing that $r$ is a true
index.
Let $\kappa$ be a regular cardinal with $\kappa>\card(X)$
and $\kappa>\card(\PP_r)$ for each $r$ (so it suffices
if $\kappa>\card(\trcl(X))$). Then the proof of \cite[Proposition
5.1]{usuba_ddg} constructs a ground $W\sub\bigcap_{r\in X}W_r$
with the $\lambda=\kappa^{++}$-uniform covering property for $V$.
Therefore by \cite[Theorem 3.3]{schindler_buk},
there is $\PP\in W$ such that $W\sats$``$\card(\PP)=2^{2^{<\lambda}}$''
and $W$ is a ground of $V$ via $\PP$. Let
$\gamma_0=\card^W(\PP)$ and $t_0=(\her_{\gamma_0^+})^W$.
So $\gamma_0<\theta$, $t_0$ is a true index and $W=W_{t_0}$.
Let $\BB\in W$ be such that $W\sats$``$\BB$ is the complete Boolean algebra
determined by $\PP$''
(so $\PP$ is a dense sub-order of $\BB$).
So $\card^W(\BB)\leq(2^{\gamma_0})^W<\theta$.
Then by \cite[Lemma 15.43]{jech} (or \cite[Fact
3.1]{usuba_ddg}) for each $r\in X$ there is some $\BB_r\in W$
with $\BB_r\sub\BB$ and there is a $(W,\BB_r)$-generic
$G_r$ such that $W[G_r]=W_r$.
So letting $\gamma=(2^{\gamma_0})^W$, then  $t=(\her_{\gamma^+})^W$
is as desired.
\end{proof}

An easy corollary of local set-directedness is:
\begin{fact}[Invariance of $\Mmm_\kappa$]\label{fact:mantle_invariance}
Let $\kappa$ be a strong limit cardinal
and $r\in\her_\kappa$.
 Then $\Mmm_\kappa^{W_r}=\Mmm_\kappa$.
\end{fact}

\begin{lem}[Absoluteness of $\Mmm_\kappa$]\label{lem:ground_Sigma_2_elem}
Let $\kappa<\lambda$ be strong limit cardinals and suppose
$\her_\lambda=V_\lambda\elem_{2}V$. Then for each $r\in\her_\kappa$, we
have:
\begin{enumerate}[label=\tu{(}\roman*\tu{)}]
\item\label{item:grounds_mantle_absolute} ${<\kappa}$-grounds and $\Mmm_\kappa$
are
absolute to $V_\lambda$:
\[ W_r^{V_\lambda}=W_r\inter V_\lambda=V_\lambda^{W_r}\text{
and }\Mmm_\kappa^{V_\lambda}=\Mmm_\kappa\inter
V_\lambda=V_\lambda^{\Mmm_\kappa},\]
\item\label{item:elem_ground} $V_\lambda^{W_r}\elem_2 W_r$,
\item\label{item:local_mantle_invariance}
$\Mmm_\kappa^{V_\lambda^{W_r}}=\Mmm_\kappa^{W_r}\inter
V_\lambda^{W_r}=\Mmm_\kappa\inter V_\lambda=\Mmm_\kappa^{V_\lambda}$.
\end{enumerate}
\end{lem}
\begin{proof}
 Part \ref{item:grounds_mantle_absolute}:
 The absoluteness of $W_r$ is because the class true indices
 $r$  is $\Pi_2$, and each $W_r$ is
  $\Sigma_2(\{r\})$. But then clearly
 \[ \Mmm_\kappa^{V_\lambda}=\bigcap_{r\in
V_\kappa}W_r^{V_\lambda}=\bigcap_{r\in
V_\kappa}V_\lambda^{W_r}=V_\lambda^{\Mmm_\kappa}.\]

 Part \ref{item:elem_ground}:
 If $W_r=V$ then this is trivial.
 Suppose $W_r\psub V$ and let $\varphi$ be $\Sigma_2$
 and $x\in W_r\inter V_\lambda$ and suppose that
 $W_r\sats\varphi(x)$. Then by Fact \ref{fact:local_ground_def},
 $V\sats\psi(x)$ where $\psi$
 asserts ``There is a strong limit cardinal $\xi$
 such that $W_r^{\her_\xi}\sats\varphi(x)$'',
 but this is also $\Sigma_2$, so $V_\lambda\sats\psi(x)$,
 so letting $\xi<\lambda$ witness this,
 again by Fact \ref{fact:local_ground_def}, we get $W_r\inter
\her_\xi\sats\varphi(x)$,
 so $W_r\inter V_\lambda\sats\varphi(x)$.

 Part \ref{item:local_mantle_invariance}: This follows from the previous parts
 and Fact \ref{fact:mantle_invariance}.
\end{proof}

\begin{dfn}\label{dfn:amenable-closure}
Let $N$ be an inner model. Let $f:\kappa\to N$.
Say that $f$
is \emph{amenable to $N$} iff $f\rest\alpha\in N$ for every
$\alpha<\kappa$. Say that $N$ is \emph{$\kappa$-amenably-closed}
 iff for every $f:\kappa\to N$, if $f$ is amenable to $N$ then $f\in N$.
Say that $N$ is \emph{$\kappa$-stationarily-computing}
(\emph{$\kappa$-unboundedly-computing})
iff for every $f:\kappa\to N$,
there is a stationary (unbounded) $A\sub\kappa$ such that $f\rest A\in N$.
\end{dfn}

\begin{lem}
Let $N$ be an inner model of $\ZF$ and $\kappa>\om$ be regular.
If $N$ is $\kappa$-stationarily-computing then $N$ is
$\kappa$-unboundedly-computing.
If $N$ is $\kappa$-unboundedly-computing then $N$ is $\kappa$-amenably-closed.
\end{lem}
\begin{lem}\label{lem:ground_amenably-closed}
 Let $W$ be a $<\kappa$-ground of $V$,
 where $\kappa>\om$ is regular.
 Then $W$ is $\kappa$-stationarily-computing.
\end{lem}
\begin{lem}The intersection of any family of $\kappa$-amenably-closed
structures is  $\kappa$-amenably-closed.\end{lem}

\begin{lem}\label{lem:kappa-amenably-closed}
 If $\kappa$ is inaccessible
 then  $\Mmm_\kappa$ is $\kappa$-amenably-closed.
\end{lem}

\section{Choice principles in the $\kappa$-mantle}\label{sec:choice}

As mentioned above, from now on we have $\ZFC$ as background theory.

The first positive results along the lines of what we
will prove here (regarding about $\kappa$-mantles
when $\kappa<\infty$), consists in Usuba's work,
including his extendibility
result. This
was followed by Lietz' negative results \cite{lietz}.
Some time after
this, using the general theory
of \cite{vm2},
the author showed that the $\kappa_0$-mantle
$\Mmm^M_{\kappa_0}$ of $M=M_{\swsw}$ (see below) is a strategy mouse.
We
give an outline of this argument, but it is primarily intended
for the reader familiar with inner model theory,
and can be safely skipped over, as the remainder
of the paper does not depend on it.
We omit all
specifics to do with Varsovian models, just mentioning
enough to indicate what is relevant here.
The full proof will appear in \cite{vm2}.

\begin{dfn}\label{dfn:Mswsw}
$M_{\swsw}$ denotes the least iterable proper class
mouse with ordinals
$\delta_0<\kappa_0<\delta_1<\kappa_1$
satisfying ``each $\delta_i$ is Woodin and each $\kappa_i$ is strong''.
\end{dfn}

The Varsovian model analysis produces a mouse $M_\infty$,
which is the direct limit of (pseudo-)iterates $P$ of $M$
 via correct iteration trees $\Tt$ on $M$, with
$\Tt\in M|\kappa_0$, and which are based on $M|\delta_0$.
It also defines a certain fragment $\Sigma$ of the iteration strategy
 for $M_\infty$, yielding a strategy mouse
$M_\infty[\Sigma]$. It turns out that
$M_\infty[\Sigma]$ has universe $\Mmm^M_{\kappa_0}$.

What is relevant here is the proof that $\Mmm^M_{\kappa_0}\sub
M_\infty[\Sigma]$.
 Let $X\in\Mmm^M_{\kappa_0}$ be a set of ordinals.
 We must see that $X\in M_\infty[\Sigma]$.
\footnote{What blocks the more obvious attempt to prove this
is that it is not clear
that the iteration maps $i_{PQ}$ between the iterates $P,Q$
of the direct limit system eventually fix $X$.}
Now $\kappa_0$ is measurable in $M$.
Let $E$ be a normal measure on $\kappa_0$,
in the extender sequence of $M$,
and let
\[ j:M\to U=\Ult(M,E) \]
be the ultrapower map. By elementarity,
$j(X)\in\Mmm_{j(\kappa_0)}^U$.
With methods from the Varsovian model analysis,
one can then construct a specific ${<j(\kappa_0)}$-ground $W$ of $U$,
with $W\sub M_\infty[\Sigma]$. So
\[ j(X)\in\Mmm_{j(\kappa_0)}^U\sub W\sub M_\infty[\Sigma].\]
Other facts from Varsovian model analysis give
$j\rest\alpha\in M_\infty[\Sigma]$
for each $\alpha\in\OR$.
But then  $X\in M_\infty[\Sigma]$, as desired, since
\[ \beta\in X\iff j(\beta)\in j(X).\]

The preceding argument has structural similarities to
Usuba's extendibility proof (see
\cite{usuba_extendible}). Schindler then
found the following result (see \cite{schindler_fsttimtg}). We will use an
adaptation of the proof for Theorem
\ref{tm:weak_compact_tm} later, so
we present this one first as a warmup, and in order to note
a simple corollary. We give
essentially Schindler's proof, although the
precise implementation might differ
slightly.

\begin{fact}[Schindler]\label{fact:meas} Let $\kappa$ be measurable.
 Then $\Mmm_\kappa\sats\AC$, so $\Mmm_\kappa\sats\ZFC$.
\end{fact}

\begin{proof}
 Let $A\in\Mmm_\kappa$. We will find a wellorder
${<_A}$ of $A$ with ${<_A}\in\Mmm_\kappa$.

Let $\mu$ be a normal measure on $\kappa$, $M=\Ult(V,\mu)$ and
$j=i^V_\mu:V\to M$
the ultrapower map.
So $\kappa=\crit(j)$
and $j(A)\in\Mmm_{j(\kappa)}^M$.

\begin{clmtwo}\label{clm:first_meas}
We have:
\begin{enumerate}
 \item\label{item:Mmm_sub} $\Mmm^M_{j(\kappa)}
\sub\Mmm_\kappa^M\sub\Mmm_\kappa$, and
\item\label{item:j_amenable} $j\rest\Mmm_\kappa$  is amenable to
$\Mmm_\kappa$.
\end{enumerate}
\end{clmtwo}
\begin{proof}
Part \ref{item:Mmm_sub}: The first $\sub$ is immediate.
For the second,
we
have
\[ \Mmm_\kappa=\bigcap_{r\in V_\kappa}W_r \text{\ \ and\ \ }
\Mmm_\kappa^M=\bigcap_{r\in V_\kappa}W^M_r. \]
Let $\mu_r=\mu\inter W_r$. Then by standard
forcing calculations and elementarity,
we get $\mu_r\in W_r$ and
\[ W^M_r=j(W_r)=\Ult(W_r,\mu)^V=\Ult(W_r,\mu_r)^{W_r}, \]
so $W^M_r\sub W_r$, so
$\Mmm_\kappa^M\sub\Mmm_\kappa$
as desired.

Part \ref{item:j_amenable}:
Let $r\in V_\kappa$. Then calculations as above
give  $i^{W_r}_{\mu_r}\rest W_r\sub j$.
But $\Mmm_\kappa\sub W_r$, and so $j\rest\Mmm_\kappa$
is amenable to $W_r$. Therefore $j\rest\Mmm_\kappa$
is amenable to $\Mmm_\kappa$, as desired.
\end{proof}

Since $\kappa$ is a strong limit, Fact
\ref{fact:local_ground_directedness} gives $s\in V_{j(\kappa)}^M$ such that
\[ \Mmm_{j(\kappa)}^M\sub W=W_s^M\sub\Mmm_\kappa^M.\]
So $j(A)\in W\sats\ZFC$, so there is a wellorder ${<^*}$ of $j(A)$
with ${<^*}\in W$.
But $W\sub\Mmm_\kappa^M$, so ${<^*}\in\Mmm_\kappa^M\sub\Mmm_\kappa$.

Now working in $\Mmm_\kappa$, where we have $k=j\rest A$
and $j(A)$
and ${<^*}$, we can define a wellorder $<_A$ of $A$ by setting,
for $x,y\in A$:
\[ x<_Ay\iff k(x)<^*k(y).\]
This completes the proof.
\end{proof}

As a corollary to the proof above, we observe:

\begin{cor}\label{cor:meas}
 \textthmmeas
\end{cor}
\begin{proof}
Continue with the notation from the proof of Fact \ref{fact:meas}.
We show $\Mmm_\kappa^M\sats$``$V_{\kappa+1}$ is wellorderable''.

\begin{clm*}
$V_{\kappa+1}\inter\Mmm^M_{j(\kappa)}=V_{\kappa+1}\inter
\Mmm_{\kappa}^M=V_{\kappa+1}\inter\Mmm_\kappa$.
\end{clm*}
\begin{proof} We have
$V_{\kappa+1}\inter\Mmm_\kappa\sub V_{\kappa+1}\inter\Mmm_{j(\kappa)}^M$
since
$j\rest\Mmm_\kappa:\Mmm_\kappa\to\Mmm_{j(\kappa)}^M$
is elementary and $\kappa=\crit(j)$. By Claim \ref{clm:first_meas}
of the proof of Fact \ref{fact:meas}, this
suffices.
\end{proof}

By Fact \ref{fact:meas},
$\Mmm_\kappa\sats\AC$, so $\Mmm_{j(\kappa)}^M\sats\AC$ also.
Let ${<^*}\in\Mmm_{j(\kappa)}^M$ be a wellorder
of $V_{\kappa+1}\inter\Mmm_{j(\kappa)}^M$. Then
${<^*}\in\Mmm_\kappa^M$ and ${<^*}$ is a wellorder of
$V_{\kappa+1}\inter\Mmm_\kappa^M$.
\end{proof}

We next use the simple idea above to prove
that certain cardinals are ``stable'' with respect to the mantle.
The first observation is:

\begin{tm}\label{tm:sharp}
\textthmsharp
\end{tm}
\begin{proof}
Let $j:L(A)\to L(A)$ be elementary with $\crit(j)=\kappa$.
We write $\Mmm_\kappa$ for $\Mmm_\kappa^{L(A)}$;
likewise $\Mmm_{j(\kappa)}$. Now
$j\rest\Mmm_\kappa:\Mmm_\kappa\to\Mmm_{j(\kappa)}$
is elementary.
Clearly $\Mmm_{j(\kappa)}\sub\Mmm_\kappa$.
But also, $B=V_{\kappa+1}^{\Mmm_\kappa}\sub
V_{\kappa+1}^{\Mmm_{j(\kappa)}}$
as in the previous proof. So $V_{\kappa+1}^{\Mmm_{j(\kappa)}}=B$.
But
$V_{j(\kappa)}^{\Mmm_{j(\kappa)}}\sats\ZFC$,
so there is a wellorder of $B$ in
$\Mmm_{j(\kappa)}\sub\Mmm_\kappa$.

It now follows that $V_{\kappa+1}^{\Mmm_\kappa}=V_{\kappa+1}^{\Mmm}$,
because we can take $j(\kappa)$ as large as we like,
hence past any true index.
\end{proof}

\begin{dfn}\label{dfn:Sigma_2-strong}
A cardinal $\kappa$ is \emph{$\Sigma_2$-strong} iff
 for every $\alpha\in\OR$ there is an elementary
 embedding $j:V\to M$ with $\alpha<j(\kappa)$ and $V_\alpha\sub M$
 and $\Th_{\Sigma_2}^M(V_\alpha)=\Th_{\Sigma_2}^V(V_\alpha)$.\footnote{That is,
for each $\Sigma_2$ formula $\varphi$ and all $\vec{x}\in(V_\alpha)^{<\om}$,
we have $M\sats\varphi(\vec{x})$ iff $V\sats\varphi(\vec{x})$.}

An embedding $j:V\to M$ is \emph{superstrong}
iff $V_{j(\kappa)}\sub M$. A cardinal $\kappa$ is \emph{$\infty$-superstrong}
iff for every $\alpha\in\OR$ there is a superstrong embedding
$j$ with $\crit(j)=\kappa$ and $j(\kappa)>\alpha$.

A \emph{superstrong extender} is the $V_\beta$-extender
derived from a superstrong embedding $j:V\to M$
where $\beta=j(\kappa)$ and $\kappa=\crit(j)$.
\end{dfn}
Note that:
\begin{lem}
 If $E$ is a superstrong extender and $W\sats\ZFC$ is a transitive proper class
with $E\in W$, then $W\sats$``$E$ is a superstrong extender''.
\end{lem}

\begin{rem}
 Say that a cardinal $\kappa$ is \emph{$\infty$-$1$-extendible}
 iff for every $\alpha\in\OR$ there is $\beta\in\OR$ with $\beta\geq\alpha$ and
 and an elementary
 $j:V_{\kappa+1}\to V_{\beta+1}$
 (hence $j(\kappa)=\beta$) with $\crit(j)=\kappa$.
\end{rem}

\begin{tm}
We have:
\begin{enumerate}
 \item\label{item:extendible_is_limit} Every extendible cardinal
 is $\infty$-$1$-extendible and carries a normal measure
 concentrating on $\infty$-$1$-extendible cardinals.
 \item\label{item:1-extendible_is_limit} Every $\infty$-$1$-extendible cardinal
is
 $\infty$-superstrong and carries a normal
measure concentrating on $\infty$-superstrong cardinals.

\item\label{item:infty-ss_is_limit} Every $\infty$-superstrong cardinal is
$\Sigma_2$-strong and carries a normal measure concentrating on
$\Sigma_2$-strong cardinals.
\end{enumerate}
\end{tm}
\begin{proof}
Part \ref{item:extendible_is_limit}:
This is routine and left to the reader.

Part \ref{item:1-extendible_is_limit}:
Let $\kappa$ be $\infty$-$1$-extendible.
Let $j:V_{\kappa+1}\to V_{\beta+1}$ be
elementary
with $\crit(j)=\kappa$. Let $E$ be the extender
derived from $j$ with support $V_\beta$.
Let $M=\Ult(V,E)$ and $k:V\to M$ be the ultrapower map.
Then one can show that $k$ is a superstrong embedding
with $k(\kappa)=\beta$ and that
$M\sats$``$\kappa$ is $\infty$-superstrong''.
(For the last clause, consider ultrapowers $\Ult(M,E\rest\alpha)$
where $\alpha<\beta$, and show that unboundedly many of these
produce superstrong embeddings in $M$,
and also use that $\beta$ is $1$-$\infty$-extendible in $M$.)

Part \ref{item:infty-ss_is_limit}: Let $\kappa$ be $\infty$-superstrong.
We show first that $\kappa$ is $\Sigma_2$-strong.
So let $\alpha\in\OR$. We may assume that $V_\alpha\elem_2 V$.
Let $j:V\to M$ be any superstrong embedding with $\crit(j)=\kappa$ and
$\alpha<j(\kappa)$. It suffices to verify:
\begin{clmthree}
$\Th_{\Sigma_2}^M(V_\alpha)=\Th_{\Sigma_2}^V(V_\alpha)$.
\end{clmthree}
\begin{proof}
Let $\varphi$ be $\Sigma_2$ and $\vec{x}\in(V_\alpha)^{<\om}$.
If $V\sats\varphi(\vec{x})$ then $V_\alpha\sats\varphi(\vec{x})$,
which implies $M\sats\varphi(\vec{x})$. Conversely,
suppose $M\sats\varphi(\vec{x})$. Because $\kappa$
is $\infty$-superstrong, it is clearly strong,
which implies that $V_\kappa\elem_2 V$. Therefore $V_{j(\kappa)}^M\elem_2 M$.
Therefore $V_{j(\kappa)}^M\sats\varphi(\vec{x})$.
But $V_{j(\kappa)}^M=V_{j(\kappa)}$, so $V_{j(\kappa)}\sats\varphi(\vec{x})$,
so $V\sats\varphi(\vec{x})$, as desired.\end{proof}

Now let $j:V\to M$ be a superstrong embedding with $\crit(j)=\kappa$. We will
show that
$M\sats$``$\kappa$ is $\Sigma_2$-strong'', which completes the proof.

\begin{clmthree} $M\sats$``$\kappa$ is ${<\beta}$-$\Sigma_2$-strong'',
where $\beta=j(\kappa)$. That is, for each $\alpha<\beta$,
$M$ has an elementary $k:M\to N$ with $\crit(k)=\kappa$
and $V_\alpha\sub N$ and
$\Th_{\Sigma_2}^N(V_\alpha)=\Th_{\Sigma_2}^M(V_\alpha)$.\end{clmthree}
\begin{proof}Since $M\sats$``$\beta$ is strong'',
$V_\beta^M\elem_2 M$ and there are club many $\alpha<\beta$
such that $V_\alpha^M=V_\alpha\elem_2M$. Fix some such $\alpha$.
Let $E_\alpha$ be the extender derived from $j$
with support $V_\alpha$.
Then $E_\alpha\in V_\beta\sub M$, and $M\sats$``$E_\alpha$ is an extender''.
Moreover, letting $N_\alpha=\Ult(M,E_\alpha)$, we have $V_\alpha\sub
N_\alpha$
and
\[
\Th_{\Sigma_2}^{N_\alpha}(V_\alpha)=\Th_{\Sigma_2}^M(V_\alpha).\]
For let
$t=\Th_{\Sigma_2}^V(V_\kappa)=\Th_{\Sigma_2}^M(V_\kappa)$.
Then letting $k_\alpha:M\to N_\alpha$ be the ultrapower map,
\[ j(t)=\Th_{\Sigma_2}^M(V_\beta)\text{ and
}k_\alpha(t)=\Th_{\Sigma_2}^N(V_{k_\alpha(\kappa)}^N). \]
So
$\Th_{\Sigma_2}^M(V_\alpha)=j(t)\inter V_\alpha=k_\alpha(t)\inter
V_\alpha=\Th_{\Sigma_2}^N(V_\alpha)$.
\end{proof}

Now since $\kappa$ is $\Sigma_2$-strong, $M\sats$``$\beta=j(\kappa)$ is
$\Sigma_2$-strong''. So let $\alpha\in\OR$ be a strong limit cardinal. Then $M$
has an embedding $\ell:M\to N$ with $\crit(\ell)=\beta$
and $V_\alpha^M=V_\alpha^N$ and
$\Th_{\Sigma_2}^M(V_\alpha^M)=\Th_{\Sigma_2}^N(V_\alpha^M)$.
By the claim and elementarity, $N\sats$``$\kappa$ is
${<\ell(\beta)}$-$\Sigma_2$-strong''. But then extenders
in $N$ which witness ${<\alpha}$-$\Sigma_2$-strength
in $N$ also witness this in $M$. Since $\alpha$ was arbitrary, we are done.
\end{proof}

We now prove an analogue of Usuba's extendibility result down lower:

\begin{tm}\label{tm:Sigma_2-strong}
\Sigmatwostrong
\end{tm}
\begin{proof}
 Suppose not and let $r$ be such that $V_{\kappa+1}^{W_r}\psub
V_{\kappa+1}^{\Mmm_\kappa}$. Let $\lambda\in\OR$ be such
that $\beth_\lambda=\lambda$
and $r\in V_\lambda$. Let $j:V\to M$
witness $\Sigma_2$-strength
with respect to $\lambda$.

Since the class of true indices is $\Pi_2$,
$M\sats$``$r$ is a true index''.
Also, by the local definability of grounds,
\[W_r^M\inter V_\lambda=W_r^{V_\lambda^M}=W_r^{V_\lambda}=W_r\inter V_\lambda.\]
In particular, $V_{\kappa+1}^{W_r^M}=V_{\kappa+1}^{W_r}\psub
V_{\kappa+1}^{\Mmm_\kappa}$.

Since $r\in V_\lambda\sub V_{j(\kappa)}^M$, therefore
$\Mmm_{j(\kappa)}^M\inter
V_{\kappa+1}\psub\Mmm_\kappa\inter V_{\kappa+1}$.
But since $\crit(j)=\kappa$,
as in the proof of Theorem \ref{fact:meas}, we have
\[ \Mmm_\kappa\inter
V_{\kappa+1}\sub\Mmm_{j(\kappa)}^M\inter V_{\kappa+1},\]
a contradiction.
\end{proof}

\begin{ques} Suppose $\kappa$ is strong. Is
$V_{\kappa+1}^{\Mmm}=V_{\kappa+1}^{\Mmm_\kappa}$?\end{ques}

We now move toward the positive results
in the cases that $\kappa$ is inaccessible and/or weakly compact.
Toward these we first prove a couple of lemmas.

\begin{lem}[$\kappa$-uniform hulls]\label{lem:kappa-uniform_hull}
Let $\kappa$ be inaccessible. For true indices
$r\in V_\kappa$, let $(\PP_r,G_r)$ witness this,
and otherwise let $\PP_r=G_r=\emptyset$.
Let $\lambda=\beth_\lambda$ with $\cof(\lambda)>\kappa$ and $V_\lambda\elem_2
V$.
Let $S\in V_\lambda$. Then there is $\widetilde{X}$ such that,
letting $\widetilde{X}_r=\widetilde{X}\inter V_\lambda^{W_r}$ for $r\in
V_\kappa$, we have:
\begin{enumerate}
\item $V_\kappa\cup\{S,\kappa\}\sub \widetilde{X}\elem V_\lambda$
and $\widetilde{X}^{<\kappa}\sub \widetilde{X}$
and $|\widetilde{X}|=\kappa$,
\item $\widetilde{X}_r\in W_r$ and $\widetilde{X}_r\elem V_\lambda^{W_r}\elem_2
W_r$,
\end{enumerate}
and letting $X$ be the transitive collapse
of $\widetilde{X}$ and $\sigma:X\to \widetilde{X}$ the
uncollapse
and $X_r,\sigma_r$ likewise, then:
\begin{enumerate}[resume*]
 \item $X_r\sub X$ and in fact,
$X_r=W_r^{X}$,
 \item $\sigma:X\to V_\lambda$ is fully elementary with
  $\crit(\sigma)>\kappa$,
   \item $\sigma_r:X_r\to V_\lambda^{W_r}$ is fully elementary
   and $\sigma_r\sub\sigma$,
 \item $G_r$ is $(X_r,\PP_r)$-generic and
$X=X_r[G_r]$,
 \item\label{item:hull_mantle_invariance}
$\Mmm_\kappa^{X}=\Mmm_\kappa^{X_r}=\bigcap_{
s\in
V_\kappa}X_s$; hence
$\Mmm_\kappa^{X}\in\Mmm_\kappa$,
\item
\label{item:<kappa-closure}
$X^{<\kappa}\sub X$ and
$X_r^{<\kappa}\inter W_r\sub X_r$
and $(\Mmm_\kappa^{X})^{<\kappa}
\inter\Mmm_\kappa\sub
\Mmm_\kappa^{X}$,
\item
$\sigma\rest\Mmm_\kappa^{X}=\sigma_r\rest\Mmm_\kappa^{\bar{
\widetilde{X}} _r}$;
hence $\sigma\rest\Mmm_\kappa^{X}\in\Mmm_\kappa$,
\item
$\sigma\rest\Mmm_\kappa^{X}:\Mmm_\kappa^{X}
\to\Mmm_\kappa^{V_\lambda
}$ is fully elementary.
\item\label{item:statements_in_X}
$V_\lambda,\widetilde{X},X,\widetilde{X}_r,X
_r$
each satisfy $T_1$ and the following
statements:
\begin{enumerate}
\item ``There are
unboundedly many $\eta$
such that $\eta=\beth_\eta$'',
\item ``Fact \ref{fact:local_ground_directedness}'',
\item
``There is
$\xi=\beth_\xi$ such that for each $r\in V_\kappa$
and $s\in V_\kappa^{W_r}$,
we have $W_r\sats$``$s$ is a  index'' iff $V_\xi^{W_r}\sats$``$s$ is a true
index''.
\end{enumerate}
\end{enumerate}
\end{lem}

\begin{proof}The fact that $V_\lambda^{W_r}\elem_2 W_r$
is by Lemma \ref{lem:ground_Sigma_2_elem}.

Construct an increasing sequence
$\left<\widetilde{X}_\alpha\right>_{\alpha<\kappa}$ such that
 $\widetilde{X}_\alpha\elem V_\lambda$
and $V_\kappa\cup\{x\}\sub \widetilde{X}_\alpha$
and $\widetilde{X}_\alpha^{<\kappa}\sub \widetilde{X}_\alpha$
and $|\widetilde{X}_\alpha|=\kappa$,
and such that for each $r\in V_\kappa$
there are cofinally many $\alpha<\kappa$
such that $\widetilde{X}_\alpha\inter W_r\in W_r$.

To construct this sequence, suppose we have
constructed $\widetilde{X}_\alpha$,
and let $r\in
V_\kappa$.
Let $X=\widetilde{X}_\alpha\inter W_r$.
By elementarity, $X\elem V_\lambda^{W_r}$ and
\[
\widetilde{X}_\alpha=X[G_r]=\{\tau_{G_r}\bigm|\tau\in\bar{
\widetilde{X}}\} .\]
Since $|X|=\kappa$, there is some $\widetilde{X}'\in
W_r$ with $|\widetilde{X}'|=\kappa$ (hence
$W_r\sats$``$|\widetilde{X}'|=\kappa$''),
and $X\sub \widetilde{X}'$, so there is also
$\widetilde{X}''\in W_r$
with $\widetilde{X}''\elem V_\lambda^{W_r}$
and $\widetilde{X}'\sub \widetilde{X}''$ and $|\widetilde{X}''|=\kappa$ (in $V$
and $W_r$)
and such that $W_r\sats$``$(\widetilde{X}'')^{<\kappa}\sub (\widetilde{X}'')$''.
It easily follows that
\[ \widetilde{X}_\alpha\sub \widetilde{X}''[G_r]=\{\tau_{G_r}\bigm|\tau\in
\widetilde{X}''\}\elem
V_\lambda \]
and
$\widetilde{X}''[G_r]\inter W_r=\widetilde{X}''$.
We set $\widetilde{X}_{\alpha+1}=\widetilde{X}''[G_r]$.
Then everything is clear except for the requirement
that $\widetilde{X}_{\alpha+1}^{<\kappa}\sub \widetilde{X}_{\alpha+1}$.
So let $f:\gamma\to \widetilde{X}_{\alpha+1}$ where $\gamma<\kappa$ (with $f\in
V$);
we claim that $f\in \widetilde{X}_{\alpha+1}$.
Let $g:\gamma\to \widetilde{X}''$ be such that $g(\alpha)_{G_r}=f(\alpha)$ for
each
$\alpha<\gamma$.
So $g\in V$, but we don't know that $g\in W_r$.
But there is a $\PP_r$-name $\dot{g}\in V_\lambda^{W_r}$
such that $\dot{g}_{G_r}=g$. And $\widetilde{X}''\in W_r$, so there is $p_0\in
G_r$
forcing that $\rg(\dot{g})\sub \widetilde{X}''$. Working in $W_r$ then,
we may fix for each $\alpha<\gamma$ an antichain
$A_\alpha\sub\PP_r$ maximal below $p_0$ and for each $p\in A_\alpha$
some $\tau_{\alpha p}\in \widetilde{X}''$ such that $p$ forces
that $\dot{g}(\alpha)=\tau_{\alpha p}$.
Then the sequence $\left<\tau_{\alpha p}\right>_{(\alpha,p)\in I}$,
where
\[ I=\{(\alpha,p)\bigm|\alpha<\gamma\text{ and }p\in A_\alpha\},\]
is $\sub \widetilde{X}''$, and hence in $\widetilde{X}''$. Since
$W_r\sats$``$(\widetilde{X}'')^{<\kappa}\sub (\widetilde{X}'')$'',
this gives
a name $\dot{g}''\in \widetilde{X}''$ such that
$p_0$ forces $\dot{g}''=\dot{g}$, and therefore
\[ g=\dot{g}_{G_r}=\dot{g}''_{G_r}\in
\widetilde{X}''[G_r]=\widetilde{X}_{\alpha+1}. \]
But since $G_r\in \widetilde{X}_{\alpha+1}$, therefore $f\in
\widetilde{X}_{\alpha+1}$,
so $\widetilde{X}_{\alpha+1}^{<\kappa}\sub \widetilde{X}_{\alpha+1}$ as desired.
With some simple bookkeeping then, we get an appropriate sequence.

Let now
$\widetilde{X}=\bigcup_{\alpha<\kappa}\widetilde{X}_\alpha$.
We claim that $\widetilde{X}$ is as desired. The only thing we need to verify
is that for each $r\in V_\kappa$, we have
\[ \widetilde{X}_r=\widetilde{X}\inter W_r\in W_r. \]

Fix $r$.  There is a $\PP_r$-name $\tau\in W_r$
such that $\tau_{G_r}=\left<\widetilde{X}_\alpha\right>_{\alpha<\kappa}$, and
for cofinally
many $\alpha<\kappa$
there is $p_\alpha\in G_r$ and $\widetilde{X}^r_\alpha\in W_r$ such that
\[ p_\alpha\forces\tau_\alpha\inter W_r=\check{\widetilde{X}^r_\alpha} \]
(hence $\widetilde{X}^r_\alpha=\widetilde{X}_\alpha\inter W_r$).
Since $\PP_r\in V_\kappa$, there is therefore a fixed $p\in\PP_r$
such that $p_\alpha=p$ for cofinally many $\alpha$. So
$\widetilde{X}_r=\bigcup_{\alpha\in I}\widetilde{X}^r_\alpha $
where
\[ I=\{\alpha<\kappa\bigm|\exists x\ [p\forces\tau_\alpha=\check{x}]\},\]
so $\widetilde{X}_r\in W_r$.

This completes the construction. The verification of the remaining
properties is now straightforward.
We omit discussing them, other than two remarks.
In part \ref{item:<kappa-closure},
the third statement follows directly from the first two
together with part \ref{item:hull_mantle_invariance};
the first two follow readily from the construction.
And in part \ref{item:statements_in_X}, note that $\xi$ exists
because $\cof(\lambda)>\kappa=|V_\kappa|$.\end{proof}

\begin{fact}
 Let $\kappa$ be weakly compact.
 Then $X$ be transitive with $\kappa\in X$ and $X^{<\kappa}\sub X$
 and $|X|=\kappa$.\footnote{In an earlier draft,
 the hypothesis ``$|X|=\kappa$'' was accidentally omitted,
 which obviously makes the statement equivalent
 to measurability.}
Then there is a non-principal $X$-$\kappa$-complete $X$-normal\footnote{That
is, $\kappa$-completeness and normality with respect to
sequences in $X$.}
ultrafilter
$\mu$ over $\kappa$ such that letting $Y=\Ult(X,\mu)$
and $i^X_\mu$ the ultrapower embedding, then $Y$ is wellfounded. Moreover,
$i^X_\mu$ is $\Sigma_1$-elementary and cofinal and $\crit(i^X_\mu)=\kappa$.
\end{fact}
\begin{proof}

 Let $\pi:X\to Z$ be any elementary embedding with $Z$ transitive
 and $\crit(\pi)=\kappa$. Let $\mu$ be the normal measure derived from $\pi$.
Note that $\mu$ works.
\end{proof}

We now extend the situation above, adding the assumption that
$\kappa$ is weakly compact.

\begin{lem}[$\kappa$-uniform weak compactness
embedding]\label{lem:kappa-uniform_wc}
Adopt the assumptions and notation from the statement
and proof of Lemma \ref{lem:kappa-uniform_hull}.
Assume further that $\kappa$ is weakly compact.
Let $\pi:X\to Y$
witness the weak compactness of $\kappa$ in $V$,
with $Y=\Ult(X,\mu)$ for an $X$-$\kappa$-complete $X$-normal ultrafilter
$\mu$ over $\kappa$, and $\pi=i^X_\mu$.
For $r\in V_\kappa$,
let $\mu_r=\mu\inter X_r$.
Then:
\begin{enumerate}
\item\label{item:mu_r_in_W_r} $\mu_r\in W_r$ and $\mu_r$ is an
$X_r$-$\kappa$-complete ultrafilter over $\kappa$;
let
\[ Y_r=\Ult(X_r,\mu_r)\text{ and }\pi_r:X_r\to Y_r \]
the ultrapower map; so $Y_r,\pi_r\in W_r$,
\item\label{item:mu_is_upward_closure} $\mu$ is the $X$-ultrafilter
generated by $\mu_r$ ($\mu_r$ is dense in $\mu$).
\item\label{item:function_representation}
For each $f:\kappa\to X_r$ with $f\in X$, there
is
$f_r\in X_r$ with $f_r:\kappa\to X_r$
and $f_r(\alpha)=f(\alpha)$ for $\mu$-measure one many
$\alpha<\kappa$.\footnote{A draft
assumed only $f:\kappa\to X$, not $f:\kappa\to X_r$,
which obviously makes the statement false when $X\neq X_r$.}
 \item\label{item:Los_theorem} The ultrapowers satisfy
 Los' theorem
for $\Sigma_1$ formulas, and $\pi_r,\pi$ are $\Sigma_2$-elementary.
 \item\label{item:Y_r_is_ground} $Y,Y_r\sats T_1$ and
 $Y_r$ is transitive,
$Y_r=W_r^{Y}$, and $Y=Y_r[G_r]$.
\item\label{item:pi_r_sub_pi}$\pi_r\sub\pi$.
\item\label{item:pi(kappa)_mantles_invariant}
$\Mmm_{\pi(\kappa)}^Y=\Mmm_{\pi_r(\kappa)}^{Y_r}\in W_r$;
hence this belongs to $\Mmm_\kappa$.
\item\label{item:pi_rest_in_mantle}
$\pi\rest\Mmm_\kappa^X:\Mmm_\kappa^X\to\Mmm_{\pi(\kappa)}^Y$ is cofinal
$\Sigma_1$-elementary; this map belongs to $\Mmm_\kappa$.
\item\label{item:kappa_mantles_invariant} $\Mmm_\kappa^Y=\bigcap_{s\in
V_\kappa}W^Y_s=\Mmm_\kappa^{Y_r}\in W_r$; hence this belongs to $\Mmm_\kappa$.
\item\label{item:ground_between_kappa_and_pi(kappa)}
$Y,Y_r$ each satisfy $T_1$ and the following statements:
\begin{enumerate}
\item\label{item:ground_between_a} ``There are
unboundedly many $\eta$
such that $\eta=\beth_\eta$'',
\item\label{item:ground_between_b} ``Fact \ref{fact:local_ground_directedness}
holds at $\theta=\pi(\kappa)=\beth_{\pi(\kappa)}$'',
\item\label{item:ground_between_c}
``There is
$\xi=\beth_\xi$ such that for each $r\in V_{\pi(\kappa)}$ and $s\in
V_{\pi(\kappa)}^{W_r}$,
we have $W_r\sats$``$s$ is true'' iff $V_\xi^{W_r}\sats$``$s$
is true''.
\end{enumerate}
Therefore there is $t\in V_{\pi(\kappa)}^Y$ with $W_t^Y\sub\Mmm_\kappa^Y$.
\end{enumerate}
\end{lem}

\begin{proof}
Parts \ref{item:mu_r_in_W_r}--\ref{item:function_representation}:
These are simple variants of the version for measurable cardinals $\kappa$
of $V[G]$ via  small forcing $\PP\in V_\kappa$;
one uses especially, however, the fact that $X_r$ is ${<\kappa}$-closed in
$W_r$.
We leave the details to the reader.

Part \ref{item:Los_theorem}: Note that $V_\lambda$
satisfies $\Sigma_1$-Collection and ``For all $\alpha\in\OR$,
$V_\alpha$ exists and $\beth_\alpha\in\OR$ exists, and $\OR=\beth_\OR$'',
so $X_r,X$ do also.
Therefore if $\varphi$ is $\Sigma_0$ and $x\in X$
and
\[ X\sats\all\alpha<\kappa\ \exists y\ \varphi(x,y,\alpha) \]
then some $V_\xi^X\in X$ satisfies the same statement,
and hence there is $f\in X$ picking witnesses $y$.
This gives Los' theorem for $\Sigma_1$ formulas.
The $\Sigma_2$-elementarity of $\pi:X\to Y$ follows.
Likewise for $X_r,\pi_r$.

Parts \ref{item:Y_r_is_ground}, \ref{item:pi_r_sub_pi}:
The fact that $Y,Y_r\sats T_1$ follows from $\Sigma_2$-elementarity
and cofinality of $\pi,\pi_r$, and (for $\Sigma_1$-Collection)
that for each $\xi\in\OR^X$,
we have $\her_\xi^X\elem_1 X$ and $\her_\xi^{X_r}\elem_1 X_r$.
The rest follows as usual from the fact that
functions in $X$ with codomain $X_r$ are
represented in $X_r$ (part \ref{item:function_representation}),
and again the $\Sigma_2$-elementarity of $\pi,\pi_r$.

Parts \ref{item:pi(kappa)_mantles_invariant}, \ref{item:pi_rest_in_mantle}:
Basically by invariance of $\Mmm_\kappa$  (Fact \ref{fact:mantle_invariance}),
we have $\Mmm_\kappa^X=\Mmm_\kappa^{X_r}$, and by part
\ref{item:statements_in_X} of \ref{lem:kappa-uniform_hull},
 there is $\xi<\OR^X$ such that for each $r\in V_\kappa$
 and $s\in V_\kappa^{W_r}$, we have
$X_r\sats$``$s$ is true'' iff $V_\xi^{X_r}\sats$``$s$ is true''. Let
\[ T_r=\{s\in V_\kappa^{W_r}\bigm|W_r\sats\text{``}s\text{
 is true''}\}.\]
So $T_r\in X_r$ and has the same definition there;
likewise for $T_r\in Y_r$, since $\pi_r$ is $\Sigma_2$-elementary.
And because of the existence of $\xi$,
\[ \pi(T_r)=\{s\in V_{\pi_r(\kappa)}^{Y_r}\bigm|Y_r\sats\text{``}s\text{
is true''}\},\]
and it follows (in the case of $r=\emptyset$, but similarly in general),
\[ \Mmm_{\pi(\kappa)}^Y=\left(\bigcap_{s\in V_{\pi(\kappa)}^Y}W^Y_s\right)=
 \left(\bigcup_{\zeta\in I}
 \pi(\Mmm_\kappa^{V_\zeta})\right)\]
where $I$ is the set of all $\zeta\in[\xi,\OR^X)$
such that $\beth_\zeta^X=\zeta$.
But $\Mmm_\kappa^X=\Mmm_\kappa^{X_r}$
and $\pi_r\sub\pi$, so
$\Mmm_{\pi(\kappa)}^Y=\Mmm_{\pi_r(\kappa)}^{Y_r}$.
The calculations above also show that
\[ \pi\rest\Mmm_\kappa^X:\Mmm_\kappa^X\to\Mmm_{\pi(\kappa)}^Y \]
is cofinal $\Sigma_1$-elementary, and likewise for $\pi_r\sub\pi$.

Part \ref{item:kappa_mantles_invariant}: By part \ref{item:Y_r_is_ground},
$W^Y_s=Y_s$, so $\Mmm_\kappa^Y=\bigcap_{s\in V_\kappa}W^Y_s$.
And note that the density of the grounds of $X_r$
in the grounds of $X$ is lifted
to that for those of $Y_r$ in those of $Y$.
(That is, for example, if $r,s$ are such that $X_r\sub X_s$,
then $Y_r\sub Y_s$, as this is preserved by $\pi$.)
So $\Mmm_\kappa^{Y_r}=\Mmm_\kappa^Y$, as desired.

Part \ref{item:ground_between_a}: For each $\zeta\in X$ with
$\zeta=\beth_\zeta^X$,
we have $\pi(\zeta)=\beth_{\pi(\zeta)}^Y$.

Part \ref{item:ground_between_c}: If $\xi$ witnesses the
corresponding statement in $X$, note that
$\pi(\xi)$ works in $Y$.

Part \ref{item:ground_between_b}:
We consider literally $Y$, but the same proof works for $Y_r$.
Note that there is a function $f:V_\kappa\to V_\kappa$
with $f\in X$, such that for each $R\in V_\kappa$,
$X\sats$``$t=f(R)$ is a true index
and $t$ witnesses Fact
\ref{fact:local_ground_directedness}
for $R$'' ($f$ exists by the elementarity of $\sigma$).
We claim that $\pi(f)$ has the same property for $Y$.
For by $\Pi_2$-elementarity, $Y\sats$``Every $t\in\rg(\pi(f))$
is a true index''. Moreover, let $\xi$ be as before.
Then for each $\zeta$ such that $\xi<\zeta<\OR^X$ and $\zeta=\beth_\zeta^X$,
$V_\zeta^X$ satisfies ``$W_{f(R)}\sub W_r$ for each $R\in V_\kappa$
and $r\in R$''.
This lifts to $Y$ under $\pi$, and since $\pi$ is cofinal, this suffices.
\end{proof}

We are now ready to prove the main theorem for weakly compact $\kappa$.
The first proof that, under this assumption, $\Mmm_\kappa\sats$``$V_\kappa$ is
wellordered''
is due to Lietz:

\begin{tm}\label{tm:weak_compact_tm}
\weakcompacttm
\end{tm}

\begin{proof}
 Part \ref{item:ult_emb_elem} follows directly from part
\ref{item:kappa-choice}, as the wellfoundedness of $\Ult(\Mmm_\kappa,\mu)$
requires only $\om$-DC, and the
proof of Los' theorem here
only uses $\kappa$-choice.
The conclusion that $x^\#$ exists in part \ref{item:small_power}
follows easily from the rest,
using the elementarity of $i_\mu$ and that $\Ult(\Mmm_\kappa,\mu)$ is
wellfounded.
To see that $\Mmm_\kappa\sats$``$\kappa$ is weakly compact'',
let $T\sub{^{<\kappa}2}$ be a tree in $\Mmm_\kappa$.
Then $T$ has a cofinal branch $b$ in $V$, by weak compactness in $V$.
But $b\inter V_\alpha\in\Mmm_\kappa$ for each $\alpha<\kappa$.
Therefore by \ref{lem:ground_amenably-closed}, $b\in\Mmm_\kappa$.

Here is Lietz' argument that
$\Mmm_\kappa\sats$``$V_\kappa$ is wellordered'':\footnote{The author first
mistakenly
thought that a similar argument worked with $\kappa$ only inaccessible,
but Lietz noted that one seems to need weak compactness for this.}
Working
in $\Mmm_\kappa$, let $T$ be the tree
of all attempts to build a wellorder of $V_\kappa$.
(For example, let $T\sub{^{<\kappa}V_\kappa}$ be the set of all functions
$f:\alpha\to V_\kappa$ where $\alpha<\kappa$,
such that for each $\beta<\alpha$,
$f(\beta)$ is a wellorder of $V_\beta$,
and for all $\beta_1<\beta_2<\alpha$, $f(\beta_2)$ is an end extension of
$f(\beta_1)$.) Since $V_\kappa^{\Mmm_\kappa}\sats\ZFC$, $T$ is unbounded in
$V_\kappa$, and clearly $T\rest\alpha\in V_\kappa$ for each $\alpha<\kappa$.
Therefore by weakly compactness in $\Mmm_\kappa$,  $\Mmm_\kappa$ has a
$T$-cofinal
branch, and clearly this gives a wellorder of $V_\kappa\inter\Mmm_\kappa$.

We proceed now to the proof that $\Mmm_\kappa\sats\kappa$-$\DC$,
and that every set $A\in\Mmm_\kappa\inter\her_{\kappa^+}$ is wellordered in
$\Mmm_\kappa$.
Let $\mathscr{T}\in\Mmm$ be a $\kappa$-$\DC$-tree,\footnote{
That is, a set $\mathscr{F}$ of functions $f$
such that $\dom(f)<\kappa$, with $\mathscr{F}$ closed under initial segment,
and no maximal elements; that is, for every $f\in\mathscr{F}$
there is $g\in\mathscr{F}$ with $\dom(f)<\dom(g)$ and $f=g\rest\dom(f)$.
Note that $\kappa$-$\DC$ is just the assertion
that for every $\kappa$-$\DC$ tree $\mathscr{T}$,
there is a $\mathscr{T}$-maximal branch;
that is, a function $f\notin\mathscr{T}$ such that $f\rest\alpha\in\mathscr{T}$
for all $\alpha<\dom(f)$.} and let
$A\in\Mmm_\kappa\inter\her_{\kappa^+}$.
Let $S=(\mathscr{T},A)\in V_\lambda$ and $X$ be a $\kappa$-uniform
hull, etc, with $S\in X$ and everything as in Lemma
\ref{lem:kappa-uniform_hull},
and let $\pi:X\to Y$, etc, be as in Lemma \ref{lem:kappa-uniform_wc}.
So $\sigma:X\to V_\lambda$ is fully elementary with
$\kappa<\crit(\sigma)$. Let $\sigma(\bar{\mathscr{T}})=\mathscr{T}$
and $\sigma(A)=A$.

By \ref{lem:kappa-uniform_wc},
$\pi'=\pi\rest\Mmm^X_\kappa:\Mmm^X_\kappa\to\Mmm^Y_{\pi(\kappa)}$ is cofinal
$\Sigma_1$-elementary, and these models and map
belong to $\Mmm_\kappa$. We have $A,\bar{\mathscr{T}}\in\Mmm^X_\kappa$.

We first find a wellorder of $A$ in $\Mmm_\kappa$,
by arguing as in Schindler's proof of Fact \ref{fact:meas}, but using the weak
compactness
embedding.
We have $\pi'(A)\in\Mmm_{\pi(\kappa)}^Y$.
By \ref{lem:kappa-uniform_wc}, there is a ground $W$ of $\Mmm^Y_{\pi(\kappa)}$
such that
\[ \Mmm_{\pi(\kappa)}^Y\sub W\sub\Mmm_{\kappa}^Y\in\Mmm_\kappa.\]
So $W\sats\AC$ and $\pi'(A)\in W$.
Let ${<^*}\in W$ be a wellorder of $\pi'(A)$.
So ${<^*}\in\Mmm_\kappa$. Working in $\Mmm_\kappa$,
we can therefore wellorder $A$ by setting, for $x,y\in A$:
\[ x<_Ay\iff\pi'(x)<^*\pi'(y).\]

We now find a branch through $\bar{\mathscr{T}}$ in $\Mmm_\kappa$,
with length $\kappa$.
Let $B\in\Mmm_\kappa^X$ be the field of $\bar{\mathscr{T}}$.
As above, there is a wellorder
$<^*$ of $B$ in $\Mmm_\kappa$.
Working in $\Mmm_\kappa$,
we  recursively construct a sequence $\left<x_\alpha\right>_{\alpha<\kappa}$
constituting a branch through $\bar{\mathscr{T}}$,
using $<^*$ to pick next elements,
and noting that at limit stages $\eta<\kappa$,
we get $\left<x_\alpha\right>_{\alpha<\eta}\in\Mmm_\kappa^X$,
because
by \ref{lem:kappa-uniform_wc} part \ref{item:<kappa-closure}
we have $(\Mmm_\kappa^X)^{<\kappa}
\inter\Mmm_\kappa\sub\Mmm_\kappa^X$. By \ref{lem:kappa-uniform_hull},
$\sigma'=\sigma\rest\Mmm_\kappa^X\in\Mmm_\kappa$,
and note that
 $\left<\sigma'(x_\alpha)\right>_{\alpha<\kappa}$
is a cofinal branch through $\mathscr{T}$, as desired.

 Part \ref{item:small_power}: Now suppose
$\pow(\kappa)\inter\Mmm_\kappa\in\her_{\kappa^+}$.
Then we may assume that $A=\pow(\kappa)\inter\Mmm_\kappa$ above.
Therefore $\pi':\Mmm_\kappa^X\to\Mmm_{\pi(\kappa)}^Y$
is $\Mmm_\kappa$-total. Therefore $\kappa$ is measurable in $\Mmm_\kappa$.
Since  $\Mmm_\kappa\sats\kappa$-$\DC$,
the rest now follows, as discussed in the first paragraph of the proof.
\end{proof}

Recall  $(\alpha,X)$-Choice
from Definition \ref{dfn:(alpha,X)-Choice}:

\begin{tm}\label{tm:inaccessible_tm}
\inacctm\end{tm}

\begin{rem}
Note that in part \ref{item:<kappa-Choice},
the ``$\kappa^+$'' and ``$\her_{\kappa^+}$''
are both in the sense of $\Mmm_\kappa$.
Note that also, as $\kappa$ is inaccessible,
$V_\kappa^{\Mmm_\kappa}\sats\ZFC$,
$\Mmm_\kappa\sats$``$\kappa$ is inaccessible'',
and $\Mmm_\kappa$ is $\kappa$-amenable closed,
by Lemma \ref{lem:kappa-amenably-closed}.
\end{rem}

\begin{proof}
Part \ref{item:Mmm_kappa_kappa-amenc} was Lemma \ref{lem:kappa-amenably-closed},
and since
$V_\kappa^{\Mmm_\kappa}=\her_\kappa^{\Mmm_\kappa}\sats\ZFC$,
part  \ref{item:V_kappa_wo} is easy.

Part \ref{item:<kappa-Choice}: Let
$\gamma<\kappa$ and $f\in\Mmm_\kappa$ with
$f:\gamma\to(\her_{\kappa^+})^{\Mmm_\kappa}$.
We find a choice function for $f$ in $\Mmm_\kappa$.
Write $f_\alpha=f(\alpha)$.
Fix a function $g:\gamma\to\Mmm_\kappa$ with
\[ g_\alpha=g(\alpha):\kappa\to
\mathrm{trancl}(f_\alpha) \]
surjective
for each $\alpha<\gamma$.
Let $c:\gamma\to\Mmm_\kappa$ be $c_\alpha=c(\alpha)\sub\kappa$
the
induced code for $g_\alpha$
(so $c_\alpha,g_\alpha\in\Mmm_\kappa$,
but note we don't know that $c,g\in\Mmm_\kappa$).
Fix $\lambda$ and a $\kappa$-uniform hull
$\widetilde{X}\elem V_\lambda$ with
$f,c,g\in\widetilde{X}$
and everything else as
in \ref{lem:kappa-uniform_hull}.
So
$\sigma(f,c,g)=(f,c,g)$.
Fix a club $C$ of $\bar{\kappa}<\kappa$
such that $\gamma<\bar{\kappa}$ and $V_{\bar{\kappa}}\elem V_\kappa$
and such that we get a corresponding system of
structures $X^{\bar{\kappa}}_{r}$ and elementary embeddings
$\pi^{\bar{\kappa}}_r:X^{\bar{\kappa}}_r\to X_r$,
for $r\in V_{\bar{\kappa}}$,
with $X^{\bar{\kappa}}_r,\pi^{\bar{\kappa}}_r\in W_r$, $X^{\bar{\kappa}}_r$ of
cardinality
$\bar{\kappa}$ in $W_r$,
 $\crit(\pi^{\bar{\kappa}}_r)=\bar{\kappa}$
and $\pi^{\bar{\kappa}}_r(\bar{\kappa})=\kappa$, and each
$X^{\bar{\kappa}}_r[G_r]=X_{\emptyset\bar{\kappa}}$
and $\pi^{\bar{\kappa}}_r\sub\pi_{\emptyset\bar{\kappa}}$,
and with $f,c_\alpha,g_\alpha\in\rg(\pi^{\bar{\kappa}}_r)$
for each $\alpha<\gamma$.
Write
$\pi^{\bar{\kappa}}_r(f^{\bar{\kappa}},
c^{\bar{\kappa}}_\alpha,g^{\bar{\kappa}}_\alpha)=
(f,c_\alpha,g_\alpha)$.
So $c^{\bar{\kappa}}_\alpha=c_\alpha\inter\bar{\kappa}$,
so
$c^{\bar{\kappa}}_\alpha,g^{\bar{\kappa}}_\alpha\in
(\her_{\bar{\kappa}^+})^{\Mmm_\kappa}$,
and $f^{\bar{\kappa}}:\gamma\to(\her_{\bar{\kappa}^+})^{\Mmm_\kappa}$
with $f^{\bar{\kappa}}_\alpha\sub\rg(g^{\bar{\kappa}}_\alpha)$.
Let $c^{\bar{\kappa}}:\gamma\to\Mmm_\kappa$
be $c^{\bar{\kappa}}(\alpha)=c^{\bar{\kappa}}_\alpha$
and likewise for $g^{\bar{\kappa}}$.

In $V$,
pick a sequence $\left<{<_{\bar{\kappa}}}\right>_{\bar{\kappa}\in C}$
of wellorders $<_{\bar{\kappa}}$
of $(\her_{\bar{\kappa}^+})^{\Mmm_\kappa}$
with ${<_{\bar{\kappa}}}\in\Mmm_\kappa$.
Let  $z^{\bar{\kappa}}_\alpha$ be the $<_{\bar{\kappa}}$-least
element of $f^{\bar{\kappa}}_\alpha$,
and let $\xi^{\bar{\kappa}}_\alpha<\bar{\kappa}$ be the
least $\xi$ with $g^{\bar{\kappa}}_\alpha(\xi)=z_{\alpha}^{\bar{\kappa}}$.

Let $S$ be the stationary set of all strong limit cardinals
$\bar{\kappa}\in C$ of cofinality $\gamma^+$.
Enumerate ${^\gamma}\kappa$ as $\{s_\beta\}_{\beta<\kappa}$,
with ${^\gamma}\bar{\kappa}=\{s_\beta\}_{\beta<\bar{\kappa}}$
for each $\bar{\kappa}\in S$.
For $\bar{\kappa}\in S$, let $\beta_{\bar{\kappa}}$
be the $\beta<\bar{\kappa}$ such that
$s_\beta=\left<\xi^{\bar{\kappa}}_\alpha\right>_{\alpha<\gamma
}$.
Let $S'\sub S$ be stationary and such that
$\beta_{\bar{\kappa}}$ is constant for $\bar{\kappa}\in S'$.

Let $d:\gamma\to\Mmm_\kappa$ be the choice function for $f$ given by
$d(\alpha)=\pi_{\emptyset\bar{\kappa}}(z^{\bar{\kappa}}_\alpha)$,
whenever $\bar{\kappa}\in S'$.
This is independent of $\bar{\kappa}\in S'$.
For if $\kappabar_0,\kappabar_1\in S'$ with $\kappabar_0<\kappabar_1$,
then for each $\alpha<\gamma$, we have
$\xi=\xi^{\bar{\kappa}_0}_\alpha=\xi^{\bar{\kappa}_1}_\alpha$,
so
\[
\pi_{\emptyset\bar{\kappa}_0}(z^{\bar{\kappa}_0}_\alpha)=
\pi_{\emptyset\bar{\kappa}_0}(g^{\bar{\kappa}_0}_\alpha(\xi))=
g_\alpha(\xi)=
\pi_{\emptyset\bar{\kappa}_1}(g^{\bar{\kappa}_1}_\alpha(\xi))=
\pi_{\emptyset\bar{\kappa}_1}(z^{\bar{\kappa}_1}_\alpha).\]

But $d\in\Mmm_\kappa$.
For given $r\in V_\kappa$, let $\bar{\kappa}\in S'$
with $r\in V_{\bar{\kappa}}$.
Then $f^{\bar{\kappa}}\in X^{\bar{\kappa}}_r$ and
$\pi^{\bar{\kappa}}_r(f^{\bar{\kappa}})=f$,
since $\pi^{\bar{\kappa}}_r\sub\pi_{\emptyset\bar{\kappa}}$.
And $X^{\bar{\kappa}}_r\in W_r$, so $f^{\bar{\kappa}}\in W_r$.
But
$<_{\bar{\kappa}}$
is in $W_r$, and so
$d^{\bar{\kappa}}=\left<z^{\bar{\kappa}}_\alpha\right>_{\alpha<\gamma}\in W_r$.
And since
$\pi^{\bar{\kappa}}_r\sub\pi_{\emptyset\bar{\kappa}}$,
$\pi^{\bar{\kappa}}_r(d^{\bar{\kappa}})=d$. Since $\pi^{\bar{\kappa}}_r\in
W_r$,
therefore $d\in W_r$.
So $d\in\Mmm_\kappa\sats$``$d$ is a choice function for $f$'', so we are done.
\end{proof}

config: arxiv


TITLE: Are all n-dimensional hypercube graphs circulant and if so what is their circulant adjacency matrix?
QUESTION [2 upvotes]: Usually the adjacency matrix representation of the n-dimensional hypercube, $Q_n$, is given as
$$Q_n=\begin{bmatrix}Q_{n-1} & I \\ I & Q_{n-1} \end{bmatrix}$$
where $Q_1$ is the adjacency matrix of $K_2$. Since the $1$-dimensional hypercube is $K_2$ and the $2$-dimensional hypercube is $C_4$ which are also circulant graphs I was wondering if all $n$-dimensional hypercubes are circulant graphs and if so what is the circulant matrix representation of their adjacency matrix.

REPLY [1 votes]: The $d$-cube is not a circulant if $d\ge3$. In fact something stronger is true.
Suppose $G$ is an abelian group acting regularly as a group of automorphisms of the $d$-cube $Q_d$. Then $Q_d$is a Cayley graph for $G$, with connection set $C$ (say).
Define two elements $g$ and $h$ of $G$ to be equivalent if they generate the same subgroup of $G$. It is known that if the eigenvalues of a Cayley graph for an abelian group are integers, $C$ must be a union of equivalence classes. (Bridges and Mena "Rational G-matrices with rational eigenvalues", JCT A, 32 (1982), 264-280.)
Assume now that the exponent of $G$ is $2^m$, where $m\ge3$. Since the $d$-cube is connected, $C$ must be a generating set for $G$ and hence it contains an element of order $2^m$, and hence $C$ contains all $2^{m-1}$
generators of some subgroup of order $2^m$. It follows that the $d$-cube must have an induced subgraph isomorphic to $K_{2^{m-1},2^{m-1}}$ and, in particular contains a copy of $K_{3,3}$.
To complete the argument we show by induction on $d$ that $Q_d$ does not contains an induced subgraph isomorphic to $K_{3,3}$. The key is that we cannot disconnect $K_{3,3}$ by deleting the edges of a perfect matching (exercise: note that all perfect matchings are equivalent under the automorhism group). But the $d$-cube consists of two copies of the
$Q_{d-1}$-cube joined by a perfect matching, and so any copy of $K_{3,3}$ must lie in one of the copies of $Q_{d-1}$, which leads to a contradiction.
So the $d$-cube cannot be a Cayley group for an abelian group of exponent
greater than four. Note that if $d=2e$, then it is a Cayley graph for
$\mathbb{Z}_4^e$.