This is a selection on from my lecture notes text, Lectures on
Forcing and Large Cardinals, which I wrote long ago and which shows
the main equivalences, including the ones you mention.
Theorem. If $\kappa^{<\kappa}=\kappa$, then the following are
equivalent.
(weak compactness property) $\kappa$ is weakly compact. That
is, $\kappa$ is uncountable and every $\kappa$-satisfiable theory
in an $L_{\kappa,\kappa}$ language of size at most $\kappa$ is
satisfiable.
(extension property) For every $A\newcommand\of{\subseteq}\of
V_\kappa$, there is a transitive structure $W$ properly extending
$V_\kappa$ and $A^*\of W$ such that $\langle
V_\kappa,{\in},A\rangle\newcommand\elesub{\prec}\elesub\langle
W,{\in},A^*\rangle$.
(tree property) $\kappa$ is inaccessible and has the tree
property.
(filter property) If $M$ is a set containing at most
$\kappa$-many subsets of $\kappa$, then there is a
$\kappa$-complete nonprincipal filter $F$ measuring every set in
$M$.
(weak embedding property) For every $A\of\kappa$ there is a
transitive set $M$ of size $\kappa$ with
$\kappa\in M$ and a transitive set $N$ with an embedding $j:M\to N$ with critical point $\kappa$.
(embedding property) For every transitive set $M$ of size
$\kappa$ with $\kappa\in M$ there is a
transitive set $N$ and an embedding $j:M\to N$ with critical point $\kappa$.
(normal embedding property) For every $\kappa$-model $M$ there
is a $\kappa$-model $N$ and an embedding
$j:M\to N$ with critical point $\kappa$, such that $N=\{j(f)(\kappa)\mid f\in M\}$.
(Hauser embedding property) For every $\kappa$-model $M$ there
is a $\kappa$-model $N$ and an embedding $j:M\to N$
with critical point $\kappa$ such that $j,M\in N$.
(partition property) $\kappa\to(\kappa)^2_2$.
Proof: (weak compactness implies extension property) Assume
that $L_{\kappa,\kappa}$ exhibits the weak compactness property
and suppose $A\of V_\kappa$. First, we argue that $\kappa$ is
inaccessible. The regularity of $\kappa$ follows from our
assumption that
$\newcommand\ltkappa{{{<}\kappa}}\kappa^\ltkappa=\kappa$ (although
even without that assumption, it follows from the weak compactness
property). Suppose now that $2^\beta\geq\kappa$ for some
$\beta<\kappa$. Let $L$ be the language having a constant symbol
$\check\alpha$ for every $\alpha<\kappa$ and a unary predicate
symbol $U$. For any $x\of\beta$, let $\sigma_x$ be the sentence
$(\bigvee_{\alpha\in x}\neg
U(\check\alpha))\vee(\bigvee_{\alpha\in\beta\setminus x}
U(\check\alpha))$. Thus, $\sigma_x$ asserts that $U$ is different
from $x$ on the set $\{\check\alpha\mid \alpha<\beta\}$. Let $S$
be the theory consisting of all $Los\sigma_x$ for $x\of\beta$. This
theory has size $2^\beta$, but the language of $S$ has size only
$\beta$. Since $2^\beta\geq\kappa$, the theory $S$ is
$\kappa$-satisfiable, since any subtheory of size less than
$\kappa$ will omit some $\sigma_x$, and then we can interpret
$\check\alpha$ as $\alpha$ and $U$ as $x$ to build a model. But
clearly $S$ is not satisfiable, since $U$ must pick out some
pattern $x=\{\alpha<\beta\mid U(\check x)\}$ under any
interpretation. So $\kappa$ is inaccessible.
Next, we show that $\kappa$ has the extension property. Let $L$ be
the language with a constant symbol $\check a$ for every $a\in
V_\kappa$, as well as a binary relation symbol $\check\in$, an
additional constant symbol $c$ and a predicate symbol $\dot A$.
Let $R$ be the first order theory $\text{Th}(\langle
V_\kappa,{\in},A,a\rangle_{a\in V_\kappa})\cup\{c\neq\check a\mid
a\in V_\kappa\}$, together with the infinitary assertion
$\sigma=\neg\exists\vec x(\wedge x_{n+1}\in x_n)$, which asserts
that $\check\in$ is well founded. This theory is
$\kappa$-satisfiable, by interpreting $c$ in the structure
$\langle V_\kappa,{\in},A\rangle$ to be one of the $\check a$
missing from the subtheory. By the weak compactness property,
there is a model $\langle W,{\in^*},A^*\rangle$ satisfying $R$.
The relation $\in^*$ on $W$ must be well founded, since the
structure satisfies $\sigma$. By taking the Mostowski collapse, we
may assume that $W$ is a transitive set and $\in^*$ is the $\in$
relation. Furthermore, since the first order theory of $\langle
V_\kappa,\in,A\rangle$ is satisfied, it follows that $V_\kappa\of
W$ and $\langle V_\kappa,{\in},A\rangle\elesub
\langle W,{\in},A^*\rangle$, as desired.
(extension implies tree property) Assume $\kappa$ has the
extension property. We show first that $\kappa$ is inaccessible.
Regularity follows from $\kappa^\ltkappa=\kappa$ (but it also
follows directly from the extension property). If
$2^\beta\geq\kappa$ for some $\beta<\kappa$, then let $\vec
a=\langle a_\alpha\mid \alpha<\kappa\rangle$ be a
$\kappa$-sequence of distinct subsets of $\beta$. By the Extension
property, there is a transitive set $W$ and $\vec a^*$ such that
$\langle V_\kappa,{\in},\vec a\rangle\elesub\langle W,{\in},\vec
a^*\rangle$. Let
$\kappa^*=W\newcommand\intersect{\cap}\intersect\newcommand\ORD{\text{Ord}}\ORD$,
and observe that $\kappa<\kappa^*$. Let $a$ be the $\kappa^{th}$
element of the sequence $\vec a^*$. Thus, $a\of\beta$ and
consequently $a\in V_\kappa$. Since $W$ satisfies that $a$ appears
on the sequence $\vec a^*$, it follows by elementarity that $a$
also appears on $\vec a$. Thus, $a=a_\alpha$ for some
$\alpha<\kappa$, and so $a$ appears twice on $\vec a^*$, at
coordinates $\alpha$ and $\kappa$, contradicting that these were
distinct subsets of $\kappa$. So we have established that $\kappa$
is inaccessible.
Now suppose that $T$ is a $\kappa$-tree. By replacing $T$ with an
isomorphic copy, if necessary, we may assume
$T\of\kappa^\ltkappa\of V_\kappa$. By the Extension property,
there is a transitive set $W$ and a subset $T^*\of W$ with
$\langle V_\kappa,{\in},T\rangle\elesub\langle
W,{\in},T^*\rangle$. It follows that $T^*$ is a tree of height
$\kappa^*=W\intersect\ORD$. Furthermore, for any $\alpha<\kappa$,
the $\alpha^{th}$ level of $T$, which is an element of $V_\kappa$,
is by elementarity also the $\alpha^{th}$ level of $T^*$. Thus,
$T^*$ is an end-extension of $T$. Let $q\in T^*$ be any node on
the $\kappa^{th}$ level of $T^*$, and consider the set of
predecessors $b=\{p\in T^*\mid p<_{T^*} q\}$. The elements of $b$
form a linearly ordered subset of $T^*$ on the levels below
$\kappa$. Thus, $b$ is a $\kappa$-branch through $T$. So $\kappa$
has the tree property.
(tree property implies filter property) Suppose that $\kappa$ is
inaccessible and has the tree property. Suppose that
$\{A_\alpha\mid \alpha<\kappa\}$ is a collection of $\kappa$ many
subsets of $\kappa$. By enlarging the collection if necessary, let
us assume that all singletons $\{\xi\}$, for $\xi<\kappa$, appear
on the list. For each $s\in 2^\beta$, where $\beta<\kappa$, let
$A_s$ be the result of intersecting all $A_\alpha$ or the
complement $\kappa\setminus A_\alpha$, chosen according to the
values of $s(\alpha)$. That is,
$A_s=(\newcommand\Intersect{\bigcap}\intersect\{A_\alpha\mid
s(\alpha)=1\})\intersect(\Intersect\{\kappa\setminus A_\alpha\mid
s(\alpha)=0\})$. Let $T=\{s\in 2^\ltkappa\mid |A_s|=\kappa\}$.
This is clearly a tree. For any $\beta<\kappa$, we may define for
any $\gamma<\kappa$ a sequence $s_\gamma\in 2^\beta$ so that
$s_\gamma(\alpha)=1\iff\gamma\in A_\alpha$ for all $\alpha<\beta$.
In particular, $\gamma\in A_{s_\gamma}$, and so
$\kappa=\newcommand\Union{\bigcup}\Union\{A_s\mid s\in 2^\beta\}$.
Since $2^\beta<\kappa$, it must be that some $A_s$ has size
$\kappa$, and so $T$ has nodes on the $\beta^{th}$ level. Thus,
$T$ is a $\kappa$-tree. By the tree property, there is a
$\kappa$-branch $b\in[T]$. Let $F$ be the filter generated by the
$A_{b\newcommand\restrict{\upharpoonright}\restrict\beta}$ for
$\beta<\kappa$. Thus, $X\in F$ if and only if
$A_{b\restrict\beta}\of X$ for some $\beta<\kappa$. Since $b$
cannot choose to add a singleton $\{\xi\}$, since this does not
have size $\kappa$, it must choose the complement, and so the
filter $F$ is not principal. Since the sequence of
$A_{b\restrict\beta}$ is descending and $\kappa$ is regular, the
filter $F$ is $\kappa$-complete. And since either $A_\alpha\in F$
or $\kappa\setminus A_\alpha\in F$ explicitly at stage $\alpha$,
depending on whether $b(\alpha)=1$ or $0$, the filter $F$ decides
every set in our original family. So $\kappa$ has the filter
property.
(filter property implies weak embedding property) Assume the
filter property and suppose $A\of\kappa$. By the
Löwenheim-Skolem theorem and using $\kappa^\ltkappa=\kappa$,
there is a transitive set $M\elesub H_{\kappa^+}$ with $A\in M$
and $M^\ltkappa\of M$. By the filter property, there is a
$\kappa$-complete nonprincipal filter $F$ deciding every element
of $P(\kappa)^M$. Consider the ultrapower $M^\kappa/F$, where we
use only functions $f:\kappa\to M$ with $f\in M$. The relations
$f=_F g$ and $f\in_F g$ are still equivalence relations, even
though $F$ may not be an ultrafilter, because the corresponding
sets $\{\alpha\mid f(\alpha)=g(\alpha)\}$ and $\{\alpha\mid
f(\alpha)\in g(\alpha)\}$ are in $M$ if $f$ and $g$ are. Since $F$
is $\kappa$-complete and $M$ is closed under $\omega$-sequences,
it follows that $\in_F$ is well founded. Thus, the ultrapower
$N=\newcommand\Ult{\text{Ult}}\Ult(M,F)$ is a transitive set. The
proof that the canonical embedding $j:M\to N$ is elementary
proceeds as in Łos' Theorem, by induction on the complexity of the formula,
appealing to the Axiom of Choice in $M$ in the existential case.
Similarly, using $\kappa$-completeness, one shows for
$\alpha<\kappa$ that if $[f]_F\in[c_\alpha]_F$, then $f=_F
c_\beta$ for some $\beta<\alpha$, and consequently
$j(\alpha)=\alpha$; meanwhile, $\kappa\leq [id]_F<j(\kappa)$, so
the critical point of $j$ is $\kappa$. So $j:M\to N$ is as
desired.
(weak embedding implies embedding property) Assume the weak
embedding property and suppose $M$ is a transitive set of size
$\kappa$. Since $M\in H_{\kappa^+}$, we may code $M$ with a set
$A\of\kappa$, and then find a transitive set $\bar M$ with $M\in
\bar M$ and an embedding $j:\bar M\to
\bar N$ with critical point $\kappa$. The restriction
$j\restrict M:M\to j(M)$ is as desired.
(embedding implies normal embedding property) Assume that $\kappa$
has the embedding property, and suppose that $M$ is a
$\kappa$-model. By the embedding property there is an embedding
$j:M\to N$ with critical point $\kappa$. Let $X=\{j(f)(\kappa)\mid
f\in M\}$ be the seed hull of $\kappa$ via $j$. It follows that
$X\elesub N$, and so if $\pi:X\cong N_0$ is the Mostowski
collapse, we obtain the induced factor embedding, where
$k=\pi^{-1}$ and $j_0=\pi\circ j$.
Note that $N_0=\{\pi(j(f)(\kappa))\mid f\in
M\}=\{j_0(f)(\kappa)\mid f\in M\}$. If $\vec x=\langle
x_\alpha\mid
\alpha<\beta\rangle\in N_0^\ltkappa$, then there are functions
$f_\alpha\in M$ with $x_\alpha=j_0(f_\alpha)(\kappa)$. Since $M$
is a $\kappa$-model, it follows that $\langle f_\alpha\mid
\alpha<\beta\rangle\in M$. Thus, $\langle j_0(f_\alpha)\mid
\alpha<\beta\rangle\in N_0$, and so by evaluation these functions
at $\kappa$, it follows that $\vec x=\langle j_0(f)(\kappa)\mid
\alpha<\beta\rangle\in N_0$, and so $N_0$ is a $\kappa$-model as
desired.
(normal embedding implies Hauser embedding property) Suppose that
$\kappa$ has the normal embedding property and suppose that $M$ is
a $\kappa$-model. Since $M$ has size $\kappa$, there is a relation
$E$ on $\kappa$ such that $\langle M,{\in}\rangle\cong\langle
\kappa,E\rangle$. Thus, $M$ must be the Mostowski collapse of the
relation $E$. By the Löwenheim-Skolem theorem theorem, there
is a $\kappa$-model $\bar M$ with $E\in \bar M$. By the normal
embedding property there is an embedding $j:\bar M\to \bar N$ with
critical point $\kappa$. Since
$j(E)\intersect\kappa\times\kappa=E$, it follows that $E\in\bar
N$, and since the Mostowski collapse of $E$ is unique, that $M$ is
in both $\bar M$ and $\bar N$. By elementarity, if $x\in M$ is
coded by $\alpha$ with respect to $E$, then $j(x)$ is coded by
$j(\alpha)=\alpha$ with respect to $j(E)$. The model $N$ can
therefore reconstruct $j\restrict M$ from $E$ and $j(E)$. Since
$\bar M$ can see that $M^\ltkappa\of M$, it follows that $\bar
N\models j(M)^{< j(\kappa)}\of j(M)$. In particular, $M$ and
$j\restrict M$, which have size $\kappa$ in $N$, must be in
$j(M)$. Thus, $j\restrict M:M\to j(M)$ has the desired Hauser
embedding property.
(embedding properties all equivalent) The Hauser embedding
property immediately implies the weak embedding property, since
any $A\of\kappa$ can be placed into a $\kappa$-model $M$.
(embedding implies weak compactness property) Assume that $\kappa$
has the embedding properties. We show first that $\kappa$ is
inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$
(although it also follows from the weak embedding property). If
$2^\beta\geq\kappa$ for some $\beta<\kappa$, then there is a
$\kappa$-sequence $\vec a=\langle a_\alpha\mid
\alpha<\kappa\rangle$ of distinct subsets of $\beta$. We may code
$\vec a$ with a subset of $\kappa$, and therefore find a
$\kappa$-model $M$ and an embedding $j:M\to N$ with $\vec a\in M$.
If $a=j(\vec a)(\kappa)$ be the $\kappa^{th}$ element of $j(\vec
a)$, then by $M^\ltkappa\of M$, it follows that $a\in M$. Since
$a\of\beta<\kappa$, it follows that $j(a)=a$. Since $j(a)$ appears
on $j(\vec a)$, it follows by elementarity that $a$ appears on
$\vec a$. So $a=a_\alpha$ for some $\alpha<\kappa$. But since
$j(\vec a)(\alpha)=j(a_\alpha)=a_\alpha$, this means that $a$
appears at least twice on the sequence $j(\vec a)$, at coordinates
$\alpha$ and $\kappa$, contradicting the assumption that the sets
were distinct. So $\kappa$ is inaccessible.
Lastly, we verify the weak compactness property. Suppose that $T$
is a $\kappa$-satisfiable theory in an $L_{\kappa,\kappa}$
language $L$ of size at most $\kappa$. Since $\kappa$ is
inaccessible, it follows that $T$ has size at most $\kappa$. We
may assume that the symbols of $L$ are built from ordinals below
$\kappa$, and so $T$ may be coded with a subset of $\kappa$. Thus,
by the $\kappa$-model embedding property, there is a
$\kappa$-model $M$ with $T,L\in M$ and an embedding $j:M\to N$
with critical point $\kappa$ into a transitive set $N$. Since
$V_\kappa\of M$, the model $M$ has all the subtheories of $T$ of
size less than $\kappa$, and the models that exist for them (for
this conclusion, one needs an infinitary version of the upward
Löwenheim-Skolem theorem, which can be proved by an
infinitary analogue of the classical proof). Thus, $M$ satisfies
that $T$ is $\kappa$-satisfiable. By elementarity, it follows that
$N$ satisfies that $j(T)$ is $j(\kappa)$-satisfiable. But $j$
fixes every element of $T$ individually, so $T=j(T)\intersect
V_\kappa\in N$, and so $T$ is a subtheory of $j(T)$ of size less
than $j(\kappa)$. Thus, $N$ has a model $\cal A$ of $T$.
Furthermore, the satisfaction relation is absolute to transitive
sets, so $\cal A$ really is a model of $T$, as desired.
(embedding implies partition property) Most other accounts derive
the partition property from the tree property, but the embedding
point of view, ever efficacious, seems to simplify things. Assume
that $\kappa$ has the embedding properties and suppose
$F:[\kappa]^2\to 2$. Since $F\in H_{\kappa^+}$, we may find a
$\kappa$-model $M_0$ with $F\in M_0$. By the normal embedding
property there is a $\kappa$-model $N_0$ and an embedding
$j_0:M_0\to N_0$ with critical point $\kappa$. Since the entire
embedding $j_0:M_0\to N_0$ has hereditary size $\kappa$, we may
find a $\kappa$-model $M$ with $M_0$, $N_0$ and $j_0$ all having
size $\kappa$ in $M$. By the Hauser embedding property there is an
embedding $j:M\to N$ such that $j,M\in N$. Applying $j$ to
$j_0:M_0\to N_0$ yields an embedding $h=j(j_0):j(M_0)\to j(N_0)$
and a factor diagram. Let $j^*=h\circ j\restrict M_0$ be the
composition of the diagonal. This diagram commutes because
$j_0(x)=y$ implies $j(j_0)(j(x))=j(y)$. Let
$\mu_0=\{X\of\kappa\mid X\in M_0\And\kappa\in j_0(X)\}$ be the
$M_0$-normal filter induced by $j_0$. Since $\mu_0$ has size
$\kappa$ in $M$, it follows from $P(\kappa)^M\of N$ that $\mu_0\in
N$. Using $j\in N$, it follows also that
$j\newcommand\image{''}\image\mu_0\in N$. Since $\mu_0\of M_0$, it
follows that $j\image\mu_0\of j(M_0)$. Since $M$ knows that
$M_0^\ltkappa\of M_0$, it follows that $N\models j(M_0)^\kappa\of
j(M_0)$, and consequently $j\image\mu_0\in j(M_0)$. This is a
collection of $\kappa$ many elements of $j(\mu_0)$ in $j(M_0)$,
and so by the $j(\kappa)$-completeness of $j(\mu_0)$ over
$j(M_0)$, it follows that $I=\Intersect j\image\mu_0\in j(\mu_0)$.
Fix any $\delta_0\in I$ and observe that $X\in\mu_0\iff\delta_0\in
j(X)$ for $X\in M_0$. Let $\delta_1=j(\kappa)$, and apply $j$ to
the definition of $\mu_0$ to see that $Y\in j(\mu_0)\iff
\delta_1\in h(Y)$ for all $Y\in j(M_0)$.
Suppose $j^*(F)(\delta_0,\delta_1)=i$. We will build an increasing
monochromatic sequence $\langle
\gamma_\alpha\mid \alpha<\kappa\rangle$ such that (i)
$\alpha<\alpha'\implies F(\gamma_\alpha,\gamma_{\alpha'})=i$; and
(ii) $\{\xi\mid F(\gamma_\alpha,\xi)=i\}\in\mu_0$ for every
$\alpha<\kappa$. Suppose that $\langle
\gamma_\alpha\mid \alpha<\beta\rangle$ is defined and we would like to
define $\gamma_\beta$. Consider the set of possible values for
$\gamma_\beta$, namely $X=\{\gamma<\kappa\mid
\forall\alpha<\beta\,\,F(\gamma_\alpha,\gamma)=i\text{ and }\{\xi\mid
F(\gamma,\xi)=i\}\in\mu_0\}$. It suffices to show that $X$ is not
empty. By (ii) for any $\alpha<\beta$, we know that
$j(F)(\gamma_\alpha,\delta_0)=i$, since $\delta_0$ is a seed for
$\mu_0$ via $j$. Similarly, $\{\xi\mid j(F)(\delta_0,\xi)=i\}\in
j(\mu_0)$, since $\delta_1$ is in $h$ of this set, in light of
$j^*(F)(\delta_0,\delta_1)=i$. Thus, $\delta_0\in j(X)$. It
follows that $X$ is unbounded in $\kappa$, and so we may choose
$\gamma_\beta\in X$ and continue the construction. So $F$ admits a
monochromatic set of size $\kappa$.
(partition implies tree property) Assume that
$\kappa\to(\kappa)^2_2$. We first show that $\kappa$ is
inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$
(but even without this hypothesis, it follows from the partition
property). If $2^\beta\geq\kappa$ for some $\beta<\kappa$, then
let $\langle a_\alpha\mid \alpha<\kappa\rangle$ be a
$\kappa$-sequence of distinct subsets of $\beta$. Let
$F(\alpha,\beta)=0$ if $a_\alpha$ precedes $a_\beta$ in the
lexical ordering of $2^\beta$, and otherwise $1$. It is not
difficult to prove that there is no monotone sequence of length
$\beta^+$ in the lexical order on $2^\beta$. Consequently, there
can be no monochromatic set for $F$ of size $\kappa$. So $\kappa$
is inaccessible.
Now suppose that $T$ is a $\kappa$-tree. We may assume that the
underlying set of $T$ is exactly $\kappa$. For any two nodes
$\alpha$ and $\beta$ in $T$, let us say that $\beta$ is to the
{\df right} of $\alpha$ in $T$, if $\alpha\perp_T\beta$ and
$\alpha'<\beta'$, where $\alpha'\leq_T\alpha$ and
$\beta'\leq_T\beta$ are least in $T$ such that
$\alpha'\perp_T\beta'$. Define $F(\alpha,\beta)=0$ if $\beta$ is
above or to the right of $\alpha$, and otherwise $1$. Suppose that
$H\of\kappa$ is a monochromatic set of size $\kappa$. Suppose
first that the monochromatic value is $0$. Thus, whenever
$\alpha<\beta$ in $H$, then $\beta$ is above or to the right of
$\alpha$ in $T$. By applying the partition property once more, we
may further assume that only one of these answers arises. If any
$\alpha<\beta$ in $H$ has $\alpha<_T\beta$, then the elements of
$H$ are linearly ordered, and $T$ has a $\kappa$-branch.
Otherwise, we assume that whenever $\alpha<\beta$ in $H$, then
$\beta$ is to the right of $\alpha$ in $T$. We may therefore
follow the ``right-most'' path through (the $T$-predecessors of
elements of) $H$. Specifically, using the fact that the levels of
$T$ have size less than $\kappa$ and $\kappa$ is regular, there is
on each level $\xi$ a right-most node $\zeta$ that occurs as a
$T$-predecessor of all sufficiently large elements of $H$. The set
of such nodes is clearly linearly ordered, and therefore forms a
$\kappa$-branch through $T$. The final case, when the
monochromatic value of $F$ on $H$ is $1$, is similar, except that
we follow the left-most branch through $H$ to provide a
$\kappa$-branch through $T$. QED
I believe that it was in these lecture notes that the terms $\kappa$-model and the "Hauser
property" were introduced, since I remember inventing that
terminology. (I knew Kai Hauser from the time I was an
undergraduate at Caltech, where he was a graduate student.)
Finally, let me mention that Sean Cox, Brent Cody, Thomas
Johnstone and myself have a paper in preparation called, "The
weakly compact embedding property," which is all about the
property when you drop the inaccessibility requirement. There are
some very interesting things to say, and I'l post a link when the
article is available. But meanwhile, I refer you to the slides for
my talk The weakly compact embedding property.
