数理解析研究所講究録
第 1754 巻 2011 年 91-96

91

Some problems about the uncountable
Specker phenomenon.
Jun Nakamura

1Introduction

We are motivated from [3]. We find a mistake in [3] and improve it
partially. The uncountable Specker phenomenon in the commutative casewas studied around 1955. J. Lo\’{s} and E.C.Zeeman [8] independently showedthat exhibits the Specker phenomenon if is less than the least measurable
cardinal. There is a similar result in the non-commutative case. S.Shelahand K.Eda [6] showed that the unrestricted free product:exhibits the Specker phenomenon if the cardinality of the index
set is less than the least measurable cardinal. Some problems occur fromthe non-commutative case and we investigate them.
$\mathbb{Z}^{\kappa}$$\kappa$$\underline{\lim}(*G,p_{XY}\backslash$$X\subseteq Y\Subset I)$$I$

2

Definitions and Basics

Definition 2.1. Let $G_{i}(i\in I)$ be groups s.t $G_{i}\cap G_{j}=\{e\}$ for any $i\neq j\in I$ .
we call elements ofletters. A word $W$ is a function$\bigcup_{i\in I}G_{i}\backslash \{e\}$$W$:$\overline{W}$$\overline{W}arrow\bigcup_{i\in I}G_{i}\backslash \{e\}$is a linearly ordered set and$\{\alpha\in\overline{W}|W(\alpha)\in$

is finite for any $i\in I$ . The class of all words is denoted by(abbreviated by ). In case the cardinality ofis countable, we say that
$W$ is a-word.
$\mathcal{W}(G_{i} :$G_{i}\}$i\in I)$$\overline{W}$$\mathcal{W}$$\sigma$

Definition 2.2.
andare isomorphic, which is denoted by $U\equiv V$ , ifthere exists an order isomorphism :s.t
. It is easily seen that becomes a set under this identification.
$U$

$V$

$\overline{U}arrow\overline{V}$$\varphi$$\forall\alpha\in\overline{U}(U(\alpha)=V(\varphi(\alpha))$$\mathcal{W}$Definition 2.3. For a subset
by the function$W_{X}:\overline{W_{X}}$$arrow$$\bigcup_{i\in I}G_{i}$$W_{X}$

where, the restricted word

$W_{X}$

of $W$ is given$\overline{W_{X}}=\{\alpha\in\overline{W}|W(\alpha)\in\bigcup_{i\in X}G_{i}\}$and. Hence $W_{X}\in$ W. If $X$ is finite, then wefor allas an element of the free product $*G$ .

$W_{X}(\alpha)=W(\alpha)$

can regard$X\subseteq I$$\alpha\in\overline{W_{X}}$92
Definition 2.4.
$U_{F}=V_{F}$for all$U$

and$F\subset\subset I$$V$are equivalent, which is denoted by $U\sim V$ , ifwhere we regard$U_{F}$

andas elements of the free

$V_{F}$

product.
“”
So, $U_{F}=V_{F}$ means that they are equal in the sense of the free product.
$*i\in pG_{i}$$*i\in FG_{i}$Let $[W]$ be the equivalent class of a word $W$ . The composition of twowords and the inverse of a word are defined naturally. Thus$\{[W]|W\in \mathcal{W}\}$ becomes a group.
$\mathcal{W}/\sim=$

Definition 2.5.
is the group
thenis isomorphic to the free product$\mathcal{W}(G_{i} :$*\in IG_{i}$$n_{\in I}G_{i}$i\in I)/\sim$$*i\in IG_{i}$. Clearly, if is finite,$I$.

Definition 2.6. $W$ is reduced if $W\equiv UXV$ implies $[X]\neq e$ for any nonempty word $X$ where is the identity, and for any contiguous elements
belong to the same .
and of , it never occurs thatand$e$$\beta$$\alpha$$\overline{W}$$W(\alpha)$$W(\beta)$$G_{i}$

Definition 2.7. $l_{i}(W)$ is the cardinality of $\{\alpha\in X|X(\alpha)\in G_{i}\}$ where $X$
is the reduced word of $W$ .
Theorem 2.1. ([1] Theoreml.4.)For any word $W$ , there exists a reducedword such that $[W]=[V]$ and is unique up to isomorphism,$V$$V$Proposition 2.1. ([1] Propositionl.9.) If
are elements ofsuch thatare finite for all $i\in I$ , then there exists a$g_{\lambda}(\lambda\in\Lambda)$$x_{\dot{n}\in I}G_{i}$$\{\lambda\in\Lambda|l_{i}(g_{\lambda})\neq 0\}$via
natural homomorphism
are copies of the integer group and is the 1 of$\delta_{\lambda}\mapsto g_{\lambda}(\lambda\in\Lambda)$$\varphi:x_{\lambda\in\Lambda}\mathbb{Z}_{\lambda}arrow x_{i\in I}G_{i}$$\mathbb{Z}_{\lambda}(\lambda\in\Lambda)$

3$\delta_{\lambda}$$\mathbb{Z}_{\lambda}$

where.

The non-commutative uncoutable Specker
phenomenon
be a n-slender group. For: $X\subseteq Y\Subset I$ ) $arrow S$ , there exist

Theorem 3.1. (S.Shelah and K. $Eda[6J)$ Let$S$

any homomorphism :-complete ultrafilterson I such thatfor any. Moreover, if the cardinality of I is less than theleast measumble cardinal, then factors through some finitely generated freegroup.
$h$$\mathcal{U}_{1},$$\omega_{1}$$u_{1}\in \mathcal{U}_{1},$$L^{m(*}i\in x^{\mathbb{Z}_{i},p_{XY}}$$\cdots,$$\cdots,\mathcal{U}_{n}$$u_{n}\in \mathcal{U}_{n}$$h$$h=\overline{h}op_{u1^{\cup\cdots\cup u_{n}}}$93$\zeta^{m(*}i\in x\mathbb{Z}_{i},p_{XY}:X\subseteq Y\Subset I)\overline{h\nearrow}S$$/$$\ovalbox{\tt\small REJECT}$$p_{u1\cup\cdots\cup u_{n}}$$/_{\exists\overline{h}}/$/$1m(*\mathbb{Z}_{i},p_{XY}:X\subseteq Y\Subset u_{1}\cup\cdots\cup u_{n})$Let. It becomes an ultra-$\mathcal{F}=\{X|\exists u_{1}\in \mathcal{U}_{1}\cdots\exists u_{n}\in \mathcal{U}_{n}(\bigcup_{i\leq n}u_{i}\subseteq X)\}$filter on . We introduce an equivalence relation
on$X\subseteq Y\Subset I)$ .
if and only if there exists$p_{u}(x)=p_{u}(y)$ . Then, we get the following diagram.
$I$$\sim \mathcal{F}$$u\in \mathcal{F}$$x\sim \mathcal{F}y$:such that$1\underline{m}(*\mathbb{Z}_{i},p_{XY}$$\underline{\lim}(*\mathbb{Z},p_{XY}:X\subseteq Y\Subset I)\backslash \overline{h\nearrow}S$$/$$\ovalbox{\tt\small REJECT}$$/_{\exists\overline{h}}/$/$\underline{\lim}(*\mathbb{Z},p_{XY}:X\subseteq Y\Subset I)/\mathcal{F}\backslash$

be the order isomorDefinition 3.1. Letwith $|F|=|G|$ andphism from to G. Then, we natumlly regard
as an isomorphism from$F,$ $G\Subset\omega$$e_{FG}$$F$$e_{FG}$. An elementneous if and only if for any $F,$$*\mathbb{Z}to*\mathbb{Z}$) is homoge.

$x\in k^{m(*}i\in x^{\mathbb{Z}_{i},p_{XY}:X}\subseteq Y\Subset\omega$$G\Subset\omega$with $|F|=|G|,$$e_{FG}(p_{F}(x))=p_{G}(x)$Let $H$ be the subgroup consisting of all homogeneous elements.
Theorem 3.2. Let be a measumble cardinal andultrafilter on . Then,$\mathcal{U}$$\kappa$$\kappa$

Proof. Let

be a -complete normal$\kappa$$\underline{\lim}(*\mathbb{Z},p_{XY}\backslash :X\subseteq Y\Subset\kappa)/\mathcal{U}\simeq H$.

for $n\geq 2$ and $x\in$
is a -complete
By the assumption,$1m(*\mathbb{Z}_{i},p_{XY} : X\subseteq Y\Subset\kappa)$ .
$\mathcal{U}^{n}=$$\{X \in[\kappa]^{n}|]u\in \mathcal{U}([u]^{n}\subseteq X)\}$$\mathcal{U}^{n}$$\kappa$

94
ultra filter for any . Since $[ \kappa]^{n}=\bigcup_{W\in*.Z_{i}}<n\{F|e_{Fn}(p_{F}(x))=W\}$ , thereexist
such that $\{F|e_{Fn}(p_{F}(x))=W_{n,x}\}\in \mathcal{U}$ . We define a)
momorphism:
as $h([x])(n)=W_{n,x}$ .
It is easily seen that is an isomorphism.
$n$$hoarrow$$W_{n,x}\in*\mathbb{Z}$$h:k^{m(*}i\in x\mathbb{Z}_{i},p_{XY}$$X\subseteq Y\Subset\kappa$$/\mathcal{U}arrow H$$h$$\square$Proposition 3.1. the cardinality of $H$ is$2^{\omega}$. Moreover,$H$

is not n-slender.

To prove it, we need a lemma.
Definition 3.2. Let$[x_{0}, x_{1}]=x_{0}x_{1}x_{0}^{-1}x_{1}^{-1}$$x_{n}\in\underline{]im}(*\iota\in\chi \mathbb{Z}_{i},p_{XY} :W$e^{\tau}define$$[x_{0}, \cdots , x_{n+1}]$

inductively

X\subseteq Y\Subset\omega)$$[x_{0}, \cdots,x_{n}]$for any

andas the following.
$n<\omega$$:=[x_{0}, \cdots , x_{n}]x_{n+1}[x_{0}, \cdots , x_{n}]^{-1}x_{n+1}^{-1}$Lemma 3.1. There exists $y_{n}\in H(n<\omega)$ such that $\forall i<n(y_{n}(i)=e)$ for
any $n<\omega$ .

Proof. Let$\delta_{i}$be the 1 of$\mathbb{Z}_{i}$. We define$y_{n}$

as the following.

$y_{n}(n)$$=$$[\delta_{0}, \cdots, \delta_{n-1}]$$y_{n}(n+1)$$=$$[\delta_{0}, \cdots, \delta_{n}][\delta_{0}, \cdots, \delta_{n-1}][\delta_{0}, \cdots, \delta_{n-2}, \delta_{n}]\cdots[\delta_{1}, \cdots, \delta_{n}]$

$=$

:$:$

We give a more precise definition. Let $A_{n+k,l}=$ { $f\in^{l}(n+k)|f$ is order preserving}with $n\leq l\leq n+k$ . $A_{n+k,l}=\{f_{i}|i<l\}(i<jarrow f_{i}<f_{j})$ is linear ordered by
the lexicographical order.

$[\delta_{f(0)}, \cdots\overline{\delta}_{f(l-1)}]:=[\delta_{f_{0}(0)}, \cdots\delta_{f_{0}(l-1)}]\cdots[\delta_{f_{l\cdot\cdot\cdot 1}(0)}, \cdots\delta_{f_{l1}(l-1)}]$$\prod_{f\in A_{n+k,l}}$$y_{n}(n+k)= \prod_{f\in A_{n+k,n+k}}[\delta_{f(0)}, \cdots\delta_{f(l-1)}]\cdots\prod_{f\in A_{n+k,n}}[\delta_{f(0)}, \cdots\delta_{f(l-1)}]$

Clearly, these are desired elements.
$\square$Pmof of Propostion 3.1. Let$y_{n}(n<\omega)$be as Lemma 3.1. There exists anhomomorphism $h:rm(*i\in X\mathbb{Z}_{i},p_{XY}:X\subseteq Y\Subset\omega)arrow k^{m(*}i\in x\mathbb{Z}_{i},p_{XY}:X\subseteq$
tofor any $n<\omega$ . Clearly, the image of iswhich maps
contained by $H$ . Therefore, $H$ is not n-slender. By Lemma 2.6 in [4], we canconclude $|H|=2^{\omega}$ .
$Y\Subset\omega)$$\delta_{n}$$y_{n}$$h$$\square$95
Proposition 3.2. $H^{*}=$ {$W\in x_{n<\omega}\mathbb{Z}_{n}|W$is homogeneous} is n-slender.

implies $l_{i}(W)\neq 0$ for any $i<$
. Suppose the negation. Let be the least natural number such thatand take $i<\omega$ with $l_{i}(W)=0$ . If $i<n$ , then. It is a. Since $W$ is homogeneous,contradiction to the minirnality of . If $n\leq i$ , we can deduce a contaradiction
as well. Now, we show the n-slenderness of . Assume not, then there exists
for all $n<\omega$ . Bya homomorphism
such thattheorem 2.3 in [2], there exists a standard homomorphism andis finite, we can take $N$
such that. Because. On the other hand,is a non-trivial homogeneouswithword which is a contradiction.

Proof. Firstly, we claim that

$W\in H^{*}\backslash \{e\}$$n$$\omega$$W_{t^{0,\cdots,n-1\}}}=$$W_{\{0,\cdots,n-1\}}\neq e$$e_{n\backslash \{i\}n-1}(W_{n\backslash \{i\}})=W_{\{0,\cdots,n-2\}}\neq e$$W_{n\backslash \{i\}}$$n$$H^{*}$$h(\delta_{n})\neq e$$h:x_{n<\omega}\mathbb{Z}_{n}arrow H^{*}$$\overline{h}$$h=u\overline{h}u^{-1}$$u\in x_{n<\omega}\mathbb{Z}_{n}$$\{n|l_{0}(\overline{h}(\delta_{n}))\neq 0\}$$l_{0}(\overline{h}(\delta_{N}))=0$$\overline{h}(\delta_{N})$$\square$4

Problems

Question 4.1. What is the cardinality ofis an ultrafilter.
when there are only finitely such that
$1m(*\mathbb{Z},p_{XY} :X\subseteq Y\Subset\kappa)/\mathcal{U}$$\mathcal{U}^{n}$$n$

In the proof of theorem 3.2, the fact
is a -complete ultrafilter for all
is so. Therefore,
is an ultrafilter implies
is essential. It is clear thatis an ultrafilter is left. We
the case there are only finitely such thatin the case,conjecture 1$\mathcal{U}^{n}$$n$$\sigma$$\mathcal{U}^{n+1}$$\mathcal{U}^{n}$$\mathcal{U}^{n}$$n$$1\underline{m}(*\mathbb{Z}_{i},p_{XY}:X\subseteq Y\Subset\kappa)/\mathcal{U}|\geq\kappa$

Question 4.2. Is the cardinality of$H^{*}$

countable or uncountable$q$

Unfortunately, we find that the proof of [3] about this problem is wrong
and we can not improve it yet.

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and its Applications 84(1998) 283-306$\sigma$[3] K.Eda,$3F\urcorner Hffl\not\in\ovalbox{\tt\small REJECT}\dagger.\ovalbox{\tt\small REJECT}$する FSS$\overline{c}$“っ$\iota\grave$て RIMS Kokyuroku $1530(2007)$ 1-4[4] G.Conner and K.Eda, Fundamental groups having the whole infomation of spaces
Topology and its Applications $146-147(2005)317-328$[5] G.Higman, Unrestricted free pruducts and varieties of topological groups J.London
Math. Soc.27(1952) 73-81[6] S.Shelah and K.Eda, The non-commutative Specker phenomenon in the uncountablecase J.Algebra 252(2002) no.122-26

96[7] E.Specker, Additive Grupen von Folgen ganzer Zahlen Portugal.Math.9(1950) 131-140[8] E.C.Zeeman, On the direct sum of free cycles J.London Math.Soc.30(1955) 195-212