Harmonic exponential families on homogeneous spaces

Abstract

Exponential families play an important role in the field of information geometry. By definition, there are infinitely many exponential families. However, only a small part of them are widely used. We want to give a framework to deal with these “good” families. In the light of the observation that the sample spaces of most of them are homogeneous spaces of certain Lie groups, we propose a method to construct exponential families on homogeneous spaces G/H by taking advantage of representation theory. Families obtained by this method are G-invariant exponential families. Then the following question naturally arises: are any G-invariant exponential families on G/H obtained by this method? We give an affirmative answer to this question. More precisely, any G-invariant exponential family on G/H can be realized as a subfamily of a family obtained by our method.

Appendix

Most of the following facts are classical. However we quickly review them for the sake of readability of our paper.

1.1 Radon measure

Let X be a topological space, and ${\mathcal {B}}(X)$ the Borel algebra of X. A function $\mu :{\mathcal {B}}(X)\rightarrow {\mathbb {R}}_{\ge 0}\cup \{\infty \}$ is called a Borel measure on X if it is completely additive and $\mu (\emptyset )=0$. In this section below, we consider the Borel algebra when we regard a topological space as a measurable space.

Definition 4.1

(Image measure, support [9, Sect. 10.1, Sect. 7.1 Exercise]) Let X, Y be topological spaces, $f:X\rightarrow Y$ a measurable map and $\mu$ a Borel measure on X.

• (i) A Borel measure $f_*\mu$ on Y is defined by

  $$\begin{aligned} f_*\mu (B):=\mu (f^{-1}(B)) \quad (B\in {\mathcal {B}}(Y)). \end{aligned}$$

  This is called the image measure of the measure $\mu$ by f.

• (ii) The closed subset of X defined as follows is called the support of $\mu$.

  $$\begin{aligned} {\text {supp}}\mu :=(\cup _{U\in {\mathcal {N}}_\mu } U)^c\subset X. \end{aligned}$$

  Here, ${\mathcal {N}}_\mu :=\{U\subset X\ |\ U\text { is open, }\mu (U)=0\}$.

One of the important properties on image measure is the following:

Fact 4.2

[9, Proposition 10.1] Let $\mu$ be a Borel measure on X. For a measurable map $f:X\rightarrow Y$ and a measurable function $h:Y\rightarrow {\mathbb {R}}_{\ge 0}$, the following equality holds.

$$\begin{aligned} \int _{x\in X} h(f(x)) d\mu (x)=\int _{y\in Y} h(y) d(f_*\mu )(y). \end{aligned}$$

We define a Radon measure as a Borel measure satisfying good properties. To do this, we recall some definitions and facts we need.

Definition 4.3

([9, Sect. 7.1], [5, (3.1.5)]) We call a linear function I on $C_c(X)$ a linear functional. Moreover, a linear functional I is said to be positive if it satisfies

$$\begin{aligned} I(f)\ge 0\text { for any }f\in C_c^+(X). \end{aligned}$$

Note 4.4

Suppose a Borel measure $\mu$ on X satisfies

$$\begin{aligned} \mu (K)<\infty \text { for any compact set } K \in {\mathcal {B}}(X). \end{aligned}$$

Then the following function is well-defined, and is a positive linear functional.

$$\begin{aligned} C_c(X)\rightarrow {\mathbb {R}}, \quad f\mapsto \int _X f d\mu . \end{aligned}$$

Definition 4.5

(Locally compact space) A topological space X is said to be locally compact if for any open set $U\subset X$ and any element $x\in U$, there exists a compact subset K such that $K\subset U$ and $x\in {\text {int}}K$.

Definition 4.6

([14, Definition 2.15], [17, chapter 3, Sect. 2]) A Borel measure $\mu$ on a locally compact Hausdorff space X is called a regular Borel measure if it is regular and satisfies the following condition:

$$\begin{aligned} \mu (K) <\infty \text { for any compact set }K\text { in }X. \end{aligned}$$

Fact 4.7

(Riesz representation theorem [17, chapter 3, Sects. 1, 2, 3], [14, Theorem 2.14], [9, Theorem 7.2]) Let X be a locally compact Hausdorff space. Then the following map is bijective.

$$\begin{aligned} \{ \text {regular Borel measures on }X \}&\rightarrow \{ \text {positive linear functionals on }C_c(X) \}, \\ \mu&\mapsto \left( f\mapsto \int _X f d\mu \right) . \end{aligned}$$

Definition 4.8

(Radon measure [9, Sect. 7.1], [5, (3.1.3) Definition 2]) Let X be a locally compact Hausdorff space. We denote by ${\mathcal {R}}(X)$ the set identified by Fact 4.7, and call an element of ${\mathcal {R}}(X)$ a Radon measure on X. Here, ${\mathcal {R}}(X)$ is equipped with the weak star topology. Namely, we consider the weak star topology on the algebraic dual $C_c'(X)$ of $C_c(X)$, and introduce the relative topology on ${\mathcal {R}}(X)$ by regarding it as a subset of $C_c'(X)$. In particular, for $f\in C_c(X)$, $r\in {\mathbb {R}}$, the following subset is open in ${\mathcal {R}}(X)$:

$$\begin{aligned} \{\nu \in {\mathcal {R}}(X)\ |\ \left\langle \nu , f \right\rangle <r\}. \end{aligned}$$

1.2 Radon–Nikodým derivative

In this subsection, we introduce Radon–Nikodým derivative on a locally compact Hausdorff space X. In a general setting, Radon–Nikodým derivative is a measurable function determined only up to almost everywhere. On the other hand, we define Radon–Nikodým derivative only in the case where measures are strongly equivalent (Definition 4.10), so it is a continuous function determined uniquely on the support.

Fact 4.9

[8, Proposition 2.23] For $\mu \in {\mathcal {R}}(X)$ and $f\in C^{++}(X)$, the function $\nu :{\mathcal {B}}(X)\rightarrow {\mathbb {R}}_{\ge 0}\cup \{\infty \}$ defined as follows is a Radon measure on X.

$$\begin{aligned} \nu (B)=\int _{B}f d\mu \quad (B\in {\mathcal {B}}(X)). \end{aligned}$$

(11)

Definition 4.10

[8, Sect. 2.3] For $\mu , \nu \in {\mathcal {R}}(X)$, we say $\mu$ and $\nu$ are strongly equivalent if there exists $f\in C^{++}(X)$ satisfying the equality (11). This relation is an equivalence relation on ${\mathcal {R}}(X)$.

Definition 4.11

(strictly positive [20]) A Radon measure $\mu$ on X is said to be strictly positive if $\mu (U)$ is positive for each nonempty open subset $U\subset X$.

Note 4.12

For strictly positive strongly equivalent measures $\mu$, $\nu \in {\mathcal {R}}(X)$, the function $f\in C^{++}(X)$ satisfying (11) is unique.

Definition 4.13

(Radon–Nikodým derivative [8, Sect. 2.3]) For $\mu , \nu \in {\mathcal {R}}(X)$ that are strongly equivalent, we call $f\in C^{++}(X)$ satisfying (11) the Radon–Nikodým derivative of $\nu$ by $\mu$. We denote it by $f=\frac{d\nu }{d\mu }$, $d\nu (x)=f(x)d\mu (x)$ or $\nu =f\mu$.

Fact 4.14

[9, Proposition 3.9] For strongly equivalent Radon measures $\mu ,\nu$ and $\lambda$, the following properties hold on ${\text {supp}}\mu$:

• (i) $\frac{d\nu }{d\lambda }=\frac{d\nu }{d\mu }\frac{d\mu }{d\lambda }$,

• (ii) $\frac{d\mu }{d\nu }=(\frac{d\nu }{d\mu })^{-1}$,

• (iii) for $\psi \in L^1(X,\mu )$,

  $$\begin{aligned} \int _X \psi \,d\mu =\int _X \psi \frac{d\mu }{d\nu }d\nu . \end{aligned}$$

1.3 Measure on G-space

In this subsection, we introduce the notion of G-space and consider measures on it. Then a G-action on ${\mathcal {R}}(X)$ is naturally induced and the notion of relative G-invariance is also induced as the compatibility of the G-action. The continuity of the G-action on ${\mathcal {R}}(X)$ (Proposition 4.17) is not new, but the authors could not find a reference about it, so we give a proof for the sake of completeness.

Definition 4.15

(G-space) Let X be a topological space and G a topological group. A G-action on X is a map $G\times X\rightarrow X$, $(g,x)\mapsto gx$ satisfying the following two conditions:

• (i) $ex=x$ for any $x\in X$,

• (ii) $g_1(g_2x)=(g_1g_2)x$ for any $g_1,g_2\in G, x\in X$.

A topological space equipped with a continuous G-action is called a G-space.

In this subsection below, G is a locally compact group and X is a locally compact Hausdorff G-space.

Definition 4.16

[8, Sect. 2.6] We define a G-action on ${\mathcal {R}}(X)$ by

$$\begin{aligned} G\times {\mathcal {R}}(X)&\rightarrow {\mathcal {R}}(X),\quad (g,\mu )\mapsto g\cdot \mu , \end{aligned}$$

(12)

$$\begin{aligned} where\quad (g\cdot \mu )(B)&:=\mu (g^{-1}B)\quad (B\in {\mathcal {B}}(X)). \end{aligned}$$

(13)

Although one considered the right action on ${\mathcal {R}}(X)$ in [8], we adopt the left action according to [5]. Note that for $g, g_1, g_2\in G$, $\nu , \mu \in {\mathcal {R}}(X)$,

• (i) $g_1\cdot (g_2\cdot \mu )= (g_1g_2)\cdot \mu$,

• (ii) the equality $d\nu (x)=f(x)d\mu (x)$ implies $d(g\cdot \nu )(x)=f(g^{-1}x)d(g\cdot \mu )(x)$,

• (iii) for a measurable function $f:X\rightarrow {\mathbb {R}}_{\ge 0}$, we have

  $$\begin{aligned} \int _{x\in X}f(x)d(g\cdot \mu )(x)=\int _{x\in X}f(gx)d\mu (x). \end{aligned}$$

Here, (iii) follows immediately from Fact 4.2.

Proposition 4.17

This G-action on ${\mathcal {R}}(X)$ above is continuous.

We give a proof of this proposition at the end of this subsection.

Definition 4.18

([8, Sect. 2.6], [5, (7.1.1) Definition 1]) Let $\mu \in {\mathcal {R}}(X)$.

• (i) $\mu$ is said to be strongly quasi G-invariant if $\mu$ and $g\cdot \mu$ are strongly equivalent and there exists a function $\lambda \in C^{++}(G\times X)$ satisfying

  $$\begin{aligned} \frac{d(g\cdot \mu )}{d\mu }(x)=\lambda (g^{-1},x) \quad (g\in G, x\in X). \end{aligned}$$

  Then note that $\lambda (g_1g_2,x)=\lambda (g_1,g_2x)\lambda (g_2,x)$ holds for $x\in X$, $g_1,g_2\in G$.

• (ii) $\mu$ is said to be relatively G-invariant if $\mu$ and $g\cdot \mu$ are strongly equivalent and there exists a positive function $\chi :G\rightarrow {\mathbb {R}}_{>0}$ satisfying

  $$\begin{aligned} \frac{d(g\cdot \mu )}{d\mu }(x)=\chi (g) ^{-1}\quad (g\in G, x\in X). \end{aligned}$$

Fact 4.19

([5, (7.1.1) Remark and Proposition 1]) If $\mu \in {\mathcal {R}}(X)$ is nonzero relatively G-invariant, the function $\chi$ above is unique and a continuous group homomorphism.

Definition 4.20

([5, (7.1.1) Remark]) We put

$$\begin{aligned} {\mathcal {R}}_{rel}(X):=\{ \mu \in {\mathcal {R}}(X)\ |\ \mu \text { is nonzero relatively }G\text {-invariant}\}. \end{aligned}$$

We call the function $\chi$ above the multiplier of $\mu \in {\mathcal {R}}_{rel}(X)$.

To show Proposition 4.17, we prepare the following:

Lemma 4.21

For any $f\in C_c(X)$, $\varepsilon >0$, and $\mu \in {\mathcal {R}}(X)$, there exist an open neighborhood $U\subset G$ of $e\in G$ and ${\overline{f}}\in C_c^+(X)$ satisfying

• (i) $|f(gx)-f(x)|\le {\overline{f}}(x)$ for any $g\in U$, $x\in X$,

• (ii) $\left\langle \mu , {\overline{f}} \right\rangle <\varepsilon$.

Proof

We take any $f\in C_c(X)$, $\varepsilon >0$, and $\mu \in {\mathcal {R}}(X)$. First, we put

$$\begin{aligned} f_g(x):=f(gx)-f(x)\quad (g\in G,\ x\in X). \end{aligned}$$

Since G is locally compact, there exists a compact set $S\subset G$ satisfying $e\in {\text {int}}S$. Put

$$\begin{aligned} T:=S^{-1} \cdot {\text {supp}}f\subset X. \end{aligned}$$

Here $S^{-1}:=\{s^{-1}\ |\ s\in S\}$. The set T satisfies

• T is compact,

• ${\text {supp}}f_g\subset T$ for $g\in S$.

Next, we put

$$\begin{aligned} \delta&:=\frac{\varepsilon }{1+\mu (T)}, \\ W&:=\{ g\in {\text {int}}S\ |\ |f_g(x)|<\delta \text { for any }x\in T\}. \end{aligned}$$

On the other hand, the set W satisfies

• $W\subset G$ is open (This comes from an elementary argument (Note 4.34)),

• $e\in W$ (This comes from $f_e(x)=0$ for any $x\in T$).

Since G is locally compact, there exists a compact set $K\subset W$ satisfying $e\in {\text {int}}K$.

Finally, we put

$$\begin{aligned} U&:={\text {int}}K, \\ {\overline{f}}(x)&:=\max _{g\in K}|f_g(x)|. \end{aligned}$$

We verify that U and ${\overline{f}}$ are what we desire. The continuity of ${\overline{f}}$ comes from an elementary argument (Note 4.34). From ${\text {supp}}{\overline{f}} \subset T$, we have ${\overline{f}}\in C_c^+(X)$. It is clear that the condition (i) of Lemma 4.21 holds by the definition of ${\overline{f}}$.

We show (ii).

$$\begin{aligned} \left\langle \mu , {\overline{f}} \right\rangle&=\int _X {\overline{f}} d\mu = \int _T {\overline{f}} d\mu \\&\le \delta \mu (T)<\varepsilon . \end{aligned}$$

Here, we used ${\overline{f}}(x)\le \delta$ for $x\in T$. $\square$

Hereafter, L denotes the left regular representation of G on C(X). Remark the following:

Note 4.22

For $f\in C_c(X)$, the following relations hold.

• (i) $\left\langle g\cdot \mu , f \right\rangle =\left\langle \mu , L_{g^{-1}}f \right\rangle$,

• (ii) $| \left\langle \mu , f \right\rangle |\le \left\langle \mu , |f| \right\rangle$.

Proof of Proposition 4.17

Take any $f\in C_c(X)$. By the definition of the weak star topology, it is enough to show that the following map is continuous:

$$\begin{aligned} \gamma :G\times {\mathcal {R}}(X)\rightarrow {\mathbb {R}},\quad (\mu ,g)\mapsto \left\langle g\cdot \mu , f \right\rangle . \end{aligned}$$

Take any $(g_0, \nu )\in G\times {\mathcal {R}}(X)$, $\varepsilon >0$ and put

$$\begin{aligned} r_0&:=\gamma (g_0,\nu )=\left\langle g_0\cdot \nu , f \right\rangle , \\ I_\varepsilon&:=(r_0-\varepsilon , r_0+\varepsilon ). \end{aligned}$$

It is enough to show that $(g_0, \nu )$ is an interior point of $\gamma ^{-1}(I_\varepsilon )$. From Lemma 4.21, we can take an open set $e\in U\subset G$ and ${\overline{f}}\in C_c^+(X)$ such that

• $|f(gx)-f(x)|\le {\overline{f}}(x)$ for any $g\in U$, $x\in X$;

• $\left\langle g_0\cdot \nu , {\overline{f}} \right\rangle <\frac{\varepsilon }{2}$.

We define an open subset ${\mathcal {U}}$ of ${\mathcal {R}}(X)$ as follows (see Definition 4.8).

$$\begin{aligned} {\mathcal {U}}:=\left\{ \mu \in {\mathcal {R}}(X)\ |\ \left\langle \mu , L_{{g_0}^{-1}}f \right\rangle \in I_{\frac{\varepsilon }{2}},\quad \left\langle \mu , L_{{g_0}^{-1}}{\overline{f}} \right\rangle <\tfrac{\varepsilon }{2}\right\} . \end{aligned}$$

Then our proposition follows from the following:

Claim.

• (i) $\nu \in {\mathcal {U}}$,

• (ii) $(Ug_0)\times {\mathcal {U}}\subset \gamma ^{-1}(I_\varepsilon )$.

The condition (i) comes from

• $\left\langle \nu , L_{{g_0}^{-1}}f \right\rangle =\left\langle g_0\cdot \nu , f \right\rangle =r_0\in I_{\frac{\varepsilon }{2}}$,

• $\left\langle \nu , L_{{g_0}^{-1}}{\overline{f}} \right\rangle =\left\langle g_0\cdot \nu , {\overline{f}} \right\rangle <\frac{\varepsilon }{2}$.

Next, we verify (ii). Take any $g\in U$ and $\mu \in {\mathcal {U}}$. It is enough to check $\gamma (gg_0, \mu )\in I_\varepsilon$.

$$\begin{aligned} |\left\langle (gg_0)\cdot \mu , f \right\rangle -r_0|&\le |\left\langle (gg_0)\cdot \mu , f \right\rangle -\left\langle g_0\cdot \mu , f \right\rangle |+|\left\langle g_0\cdot \mu , f \right\rangle -r_0|\\&< \left\langle g_0\cdot \mu , |L_{g^{-1}} f-f| \right\rangle +\tfrac{\varepsilon }{2}\\&\le \left\langle g_0\cdot \mu , {\overline{f}} \right\rangle +\tfrac{\varepsilon }{2} \\&<\varepsilon . \end{aligned}$$

$\square$

1.4 Measure on homogeneous space

Let G be a locally compact group and H a closed subgroup of G. Then the homogeneous space $X:=G/H$ is a locally compact Hausdorff G-space. In this subsection, we want to consider relatively G-invariant measures on X. For this purpose, we recall the definition of the modular function.

Fact 4.23

(Haar measure [5, (7.2.1) Therorem 1], [8, Theorem 2.10], [11, Theorem 3.8]) There exists a unique left Haar measure on G up to a positive scalar.

Definition 4.24

(Modular function [5, (7.1.3) Definition 3], [8, Sect. 2.4], [11, Sect. 3.2 (b)]) Let $\mu$ be a left Haar measure on G. Put $(\delta (g)\mu )(B):=\mu (Bg)$ $(B\in {\mathcal {B}}(G))$. Then $\delta (g)\mu$ is also a left Haar measure. From Fact 4.23, we obtain a function $\varDelta :G\rightarrow {\mathbb {R}}_{>0}$ by the following equation:

$$\begin{aligned} \delta (g)\mu =\varDelta (g)\mu \quad (g\in G). \end{aligned}$$

Here $\varDelta$ is independent of the choice of the Haar measure $\mu$. The function $\varDelta$ is called the modular function of G. Since the modular function is the multiplier of the right relatively G-invariant measure on G, it is automatically a continuous group homomorphism from Fact 4.19.

Fact 4.25

([13, Chapter III, Sect. 4, Theorem 1], [5, (7.2.6) Theorem 3], see also [4, Corollary 4.1]) Let $\chi :G\rightarrow {\mathbb {R}}_{>0}$ be a continuous group homomorphism.

• (i) The following conditions are equivalent:

  • (a) there exists $\mu \in {\mathcal {R}}_{rel}(X)$ with the multiplier $\chi$,

  • (b) $\varDelta _H(h)=\chi (h)\varDelta _G(h)$ for any $h\in H$.

  Here $\varDelta _G$, $\varDelta _H$ are the modular functions of G, H, respectively.

• (ii) $\mu \in {\mathcal {R}}_{rel}(X)$ with the multiplier $\chi$ is unique up to a positive scalar.

Remark that ${\mathcal {R}}_{rel}(X)$ can be empty for some homogeneous spaces X.

Example 4.26

Let $G=SL(2,{\mathbb {R}})$ and $H=\left\{ \begin{pmatrix}a&{}b\\ 0&{} a^{-1} \end{pmatrix}\ \Big |\ a\in {\mathbb {R}}^\times , b\in {\mathbb {R}}\right\}$. Then, G/H does not admit nonzero relatively G-invariant Radon measures. In fact, we have $\varDelta _G=1$ and $\varDelta _H(h)=a^{-2}$ for $h=\begin{pmatrix}a&{}b\\ 0&{} a^{-1} \end{pmatrix}$. On the other hand, any continuous group homomorphism $\chi :G\rightarrow {\mathbb {R}}_{>0}$ must be 1. Therefore, the condition Fact 4.25(i)(b) can not be satisfied.

Under ${\mathcal {R}}_{rel}(X)\ne \emptyset$, ${\mathcal {R}}_{rel}(X)$ can be parameterized by ${\mathbb {R}}_{>0}$ and $W_0(G,H)$.

Corollary 4.27

Assume ${\mathcal {R}}_{rel}(X)\ne \emptyset$ and fix $\nu _0\in {\mathcal {R}}_{rel}(X)$. Then we have:

$$\begin{aligned} {\mathcal {R}}_{rel}(X)=\{ ce^\tau \nu _0\ |\ c\in {\mathbb {R}}_{>0}, \tau \in W_0(G,H) \}. \end{aligned}$$

(14)

Proof

Let $\chi _0$ be the multiplier of $\nu _0$. We show the inclusion “$\supset$”. For $c\in {\mathbb {R}}_{>0}$ and $\tau \in W_0(G,H)$, an easy calculation implies $ce^\tau \nu _0\in {\mathcal {R}}_{rel}(X)$ with the multiplier $e^\tau \chi _0$. Next, we show the inclusion “$\subset$”. Let $\nu \in {\mathcal {R}}_{rel}(X)$ with the multiplier $\chi$. We put $\tau :=\log \chi -\log \chi _0$, ${\tilde{\nu }}:=e^\tau \nu _0$ and only have to show the following:

Claim

• (i) $\tau \in W_0(G,H)$,

• (ii) $\nu$ is equal to ${\tilde{\nu }}$ up to scalar.

We show (i) and (ii) respectively.

(i). Since it is clear that $\tau$ is a continuous group homomorphism, it is enough to show that $\tau (h)=0$ for any $h\in H$. From Fact 4.25(i), for $h\in H$,

$$\begin{aligned} \tau (h)=\log \chi (h)-\log \chi _0(h)=\log \frac{\varDelta _H(h)}{\varDelta _G(h)}-\log \frac{\varDelta _H(h)}{\varDelta _G(h)}=0. \end{aligned}$$

(ii). From Fact 4.25(ii), it is enough to show ${\tilde{\nu }}\in {\mathcal {R}}_{rel}(X)$ with the multiplier $\chi$. For $g\in G$, $x\in X$,

$$\begin{aligned} d(g\cdot {\tilde{\nu }})(x)&=e^{\tau (g^{-1}x)}d(g\cdot \nu _0)(x)=e^{-\tau (g)+\tau (x)}\chi _0(g)^{-1}d\nu _0(x)\\&=e^{-\log \chi (g)+\log \chi _0(g)+\tau (x)}\chi _0(g)^{-1}d\nu _0(x)\\&=\chi (g)^{-1}e^{\tau (x)}d\nu _0(x)=\chi (g)^{-1}d{\tilde{\nu }}(x). \end{aligned}$$

$\square$

From Corollary 4.27, we have ${\mathcal {R}}_{rel}(X)=\emptyset$ or ${\mathbb {R}}_{>0}$ if $W_0(G,H)=\{0\}$. We would like to give a sufficient condition for $W_0(G,H)=\{0\}$ in the case where G is a Lie group.

Proposition 4.28

Let G be a Lie group with finitely many connected components and H a closed subgroup of G. We denote by $\mathfrak {g}$ and $\mathfrak {h}$ the Lie algebras of G and H, respectively. Then we have $W_0(G,H)=\{0\}$ if the pair (G, H) satisfies one of the following conditions:

• (i) $\mathfrak {g}=\mathfrak {h}+[\mathfrak {g},\mathfrak {g}]$,

• (ii) G is compact,

• (iii) G is semisimple.

Proof

• (i). Take any $\tau \in W_0(G,H)$. Since we have $d\tau |_\mathfrak {h}=0$ and $d\tau |_{[\mathfrak {g},\mathfrak {g}]}=0$, we have $d\tau =0$. Therefore, we obtain $\tau =0$ from Note 3.5 and Lemma 3.4.

• (ii). This follows from the fact that the image of a compact group by a continuous group homomorphism is also a compact group.

• (iii). This follows from $[\mathfrak {g},\mathfrak {g}]=\mathfrak {g}$ and the condition (i).

$\square$

Example 4.29

Suppose $G={\mathbb {R}}^\times < imes {\mathbb {R}}$ and $H={\mathbb {R}}^\times$. Then we have $W_0(G,H)=\{0\}$ from Proposition 4.28(i). In fact, we have

$$\begin{aligned} \mathfrak {g}={\mathbb {R}}\oplus {\mathbb {R}},\ \mathfrak {h}={\mathbb {R}}\oplus \{0\} \text { and }\ [\mathfrak {g},\mathfrak {g}]=\{0\}\oplus {\mathbb {R}}. \end{aligned}$$

1.5 Elementary notes

We review elementary notes without proofs, which are used in our proofs in this paper. In this subsection below, G is a locally compact group and V is a finite dimensional real vector space.

Throughout this paper, we consider the natural topology when we regard a finite dimensional real vector space as a topological space.

Note 4.30

Take a basis $v_1,\ldots , v_n$ of V. A topology of V is induced by the linear isomorphism $V\rightarrow {\mathbb {R}}^n$, $v_i\mapsto e_i$. This topology does not depend on the choice of the basis. Linear maps and bilinear maps between finite dimensional vector spaces are continuous with respect to this topology.

Note 4.31

Let $\pi :G\rightarrow GL(V)$ be a representation of G and $v_0\in V$. Then the continuous map $\alpha :G\rightarrow V$ defined by $\alpha (g):=\pi (g)v_0-v_0$ is a 1-cocycle of $\pi$. A 1-cocycle of this form is called a 1-coboundary.

Note 4.32

Let $W\subset V$ be a subspace. Let $w_0,\ldots , w_n\in W$, $v\in V$ and suppose that $w_0,\ldots , w_n$ affinely span W. Then ${\overline{w}}_0,\ldots , {\overline{w}}_n$ linearly span $W+{\mathbb {R}}v$. Here, ${\overline{w}}_k=w_k+v$.

In Method 1.2, some maps on G are regarded as maps on G/H. We can justify that as follows:

Note 4.33

A map f on G can be regarded as a map on G/H if it satisfies

$$\begin{aligned} f(gh)=f(g)\text { for any }g\in G, h\in H. \end{aligned}$$

For example, if a group homomorphism $\tau :G\rightarrow {\mathbb {R}}$ satisfies $\tau |_H=0$, then we have

$$\begin{aligned} \tau (gh)=\tau (g)+\tau (h)=\tau (g)\quad (g\in G, h\in H). \end{aligned}$$

Note 4.34

Let X be a topological space and K a compact space. For a continuous function $f:X\times K\rightarrow {\mathbb {R}}$, the function ${\overline{f}}(x):=\max _{k\in K}f(x,k)$ is also continuous.

1.6 Continuity of parameter map

In this subsection, we show the continuity of the parameter map (Proposition 2.8).

First, for this purpose, we prepare a definition and a lemma.

Definition 4.35

Let W be a finite dimensional normed space over ${\mathbb {R}}$ and X a topological space. Suppose that $x_0\in X$ and the map $\psi :X\rightarrow W$ satisfies $\psi (x_0)=0$. In this paper, we say $\psi$ is continuous along $w\in W$ at $x_0$ if for any $\varepsilon >0$, there exist an open cone $C\subset W$ and an open neighborhood $U\subset X$ of $x_0$ satisfying the following two conditions:

• (i) $w\in C$,

• (ii) $\psi (U)\cap C\subset B_\varepsilon$.

We are motivated to give the definition above by the following:

Lemma 4.36

In the same setting as Definition 4.35, we suppose that for any $w\in W$ with $\Vert w\Vert =1$, $\psi :X\rightarrow W$ is continuous along w at $x_0\in X$. Then $\psi$ is continuous at $x_0$.

Proof

Take any $\varepsilon >0$. It is enough to show that there exists an open neighborhood U of $x_0$ satisfying the following:

$$\begin{aligned} \psi (U)\subset B_\varepsilon . \end{aligned}$$

(15)

For each $w\in W$ with $\Vert w\Vert =1$, we take an open cone $C_w$ and an open neighborhood $U_w$ of $x_0$ satisfying conditions (i) and (ii) in Definition 4.35. Since $S:=\{w\in W\ |\ \Vert w\Vert =1\}$ is compact, we can and do take $w_1,\ldots , w_n\in S$ such that $\cup _{i=1}^n C_{w_i}\supset S$. Remark that $W=(\cup _{i=1}^n C_{w_i})\cup \{0\}$. Here $U:=\cap _{i=1}^n U_{w_i}$ is an open neighborhood of $x_0$. Then, (15) follows from the following calculation:

$$\begin{aligned} \psi (U)=\psi (U)\cap ((\cup _{i=1}^n C_{w_i})\cup \{0\}) =(\cup _{i=1}^n(\psi (U)\cap C_{w_i}))\cup \{0\}\subset B_\varepsilon . \end{aligned}$$

$\square$

Next, we prepare three lemmas to prove Proposition 2.8 by using the notion above. We state them under the following:

Setting 4.37

Let X be a locally compact Hausdorff space, ${\mathcal {P}}$ an exponential family on X, $(\mu ,V,T)$ its minimal realization and $a:{\mathcal {P}}\rightarrow V^\vee$ the parameter map (Definition 2.6). Let $\varTheta :=a({\mathcal {P}})$, $\varphi :\varTheta \rightarrow {\mathbb {R}}$ the log normalizer and for $\theta \in \varTheta$, we put

$$\begin{aligned} dp_\theta (x)=\exp (-\left\langle \theta , T(x) \right\rangle -\varphi (\theta ))d\mu (x). \end{aligned}$$

We give a norm $\Vert \cdot \Vert$ on $V^\vee$.

Lemma 4.38

We consider Setting 4.37. Then $T( {\text {supp}}\mu )$ affinely spans V.

Proof

If we change T to the composition of T and a translation, then it is also a minimal realization of ${\mathcal {P}}$, so without loss of generality, we can assume that $0\in T({\text {supp}}\mu )$. We use proof by contradiction. Put

$$\begin{aligned} V_1:={\text {affine}}\text {-}{\text {span}}(T({\text {supp}}\mu ))={\text {span}}(T( {\text {supp}}\mu )) \end{aligned}$$

and assume $V_1\subsetneq V$. It is enough to show that $(\mu , V, T)$ is not a minimal realization of ${\mathcal {P}}$. Let $V_2$ be a complementary subspace of $V_1$ in V, and ${\text {pr}}_1:V\rightarrow V_1$ the first projection with regard to the decomposition $V=V_1\oplus V_2$. We define a continuous map $T_1:X\rightarrow V_1$ by $T_1:={\text {pr}}_1\circ T$. Then since for any $x\in {\text {supp}}\mu$, we have $T(x)\in V_1$, we have $T(x)=T_1(x)$ for any $x\in {\text {supp}}\mu$. Therefore, $(\mu , V_1, T_1)$ is also a realization of ${\mathcal {P}}$ and the inequality $\dim V_1<\dim V$ holds. Therefore $(\mu , V, T)$ is not a minimal realization of ${\mathcal {P}}$. $\square$

Lemma 4.39

We consider Setting 4.37. Then for any $\xi \in V^\vee$ with $\Vert \xi \Vert =1$, there exist a positive scalar $\tau >0$, an open cone $C\subset V^\vee$, and functions $f_k\in C_c^+(X)$ $(k=1,2)$ satisfying the following conditions:

• (i) $\xi \in C$,

• (ii) $\int _X f_k d\mu =1$    $(k=1,2)$,

• (iii) $\int _X f_1 dp_\theta \ge e^{\Vert \theta \Vert \tau }\int _X f_2dp_\theta$    for any $\theta \in \varTheta \cap C$.

Proof

Take any $\xi \in V^\vee$ with $\Vert \xi \Vert =1$. We prove this lemma through 5 steps. We take $\tau$, C and $f_k$ in Steps 2, 3 and 4, respectively.

(Step 1). We can take $t_k\in {\mathbb {R}}$ and $x_k\in X$ $(k=1,2)$ such that

$$\begin{aligned} t_1<t_2,\quad x_k\in {\text {supp}}\mu \cap T^{-1}(\xi ^{-1}(t_k)). \end{aligned}$$

We verify the statement above. From Lemma 4.38, $T({\text {supp}}\mu )$ affinely spans V. Therefore, there exist $t_1, t_2\in {\mathbb {R}}$ satisfying

$$\begin{aligned} t_1<t_2, \quad T({\text {supp}}\mu )\cap \xi ^{-1}(t_k)\ne \emptyset \quad (k=1,2). \end{aligned}$$

Thus, we can take $x_k\in {\text {supp}}\mu \cap T^{-1}(\xi ^{-1}(t_k))$, so (Step 1) was completed.

(Step 2). Put $\tau :=\frac{t_2-t_1}{5}$. Then we can take compact sets $S_k\subset X$ satisfying

• (2-1) $x_k\in {\text {int}}S_k$,

• (2-2) $\left\langle \xi , T(x) \right\rangle <t_1+\tau$ for any $x\in S_1$,

• (2-3) $\left\langle \xi , T(x) \right\rangle >t_2-\tau$ for any $x\in S_2$.

Let us verify this statement above. We put

$$\begin{aligned} U_1:=T^{-1}(\xi ^{-1}({\mathbb {R}}_{<t_1+\tau })),\ U_2:=T^{-1}(\xi ^{-1}({\mathbb {R}}_{>t_2-\tau })). \end{aligned}$$

Remark that $U_k$ is an open neighborhood of $x_k$. Since X is locally compact, we can and do take compact sets $S_k\subset X$ satisfying

• $x_k\in {\text {int}}S_k$,

• $S_k\subset U_k$.

Therefore we obtain the desired compact sets $S_1, S_2 \subset X$. We finished (Step 2).

(Step 3). Define the cone C by

$$\begin{aligned} C:=\left\{ \theta \in V^\vee \setminus \{0\}\ \Big |\ \left\langle \frac{\theta }{\Vert \theta \Vert }-\xi , T(x) \right\rangle <\tau \ \text { for any } x\in S_1\cup S_2\right\} . \end{aligned}$$

Then the cone $C\subset V^\vee$ is an open and $\xi \in C$ and any element $\theta \in \varTheta \cap C$ satisfies

• (3-1) $\left\langle \theta , T(x) \right\rangle < \Vert \theta \Vert (t_1+2\tau )$ for $x\in S_1$,

• (3-2) $\left\langle \theta , T(x) \right\rangle > \Vert \theta \Vert (t_2-2\tau )$ for $x\in S_2$.

We verify them above. It is clear that $\xi \in C$. The openness of C comes from an elementary argument on topological spaces (Note 4.34).

One can prove (3-2) in the same way as (3-1), so we prove only (3-1). Let $x\in S_1$ and $\theta \in \varTheta \cap C$. Then we have

$$\begin{aligned} \left\langle \theta , T(x) \right\rangle&=\Vert \theta \Vert \left\langle \frac{\theta }{\Vert \theta \Vert }, T(x) \right\rangle \\&=\Vert \theta \Vert \left( \left\langle \frac{\theta }{\Vert \theta \Vert }-\xi , T(x) \right\rangle +\left\langle \xi , T(x) \right\rangle \right) \\&<\Vert \theta \Vert (\tau +(t_1+\tau )). \end{aligned}$$

Here, we used (2-2) at the last inequality. (Step 3) was proved.

(Step 4). We can take functions $f_k\in C_c^+(X)$ satisfying

• (4-1) ${\text {supp}}f_k\subset S_k$,

• (4-2) $\int _X f_k d\mu =1$.

In fact, from Urysohn’s lemma for locally compact Hausdorff spaces (see [14, 2.12 Urysohn’s Lemma] for example), we can and do take continuous functions $f_k\in C_c^+(X)$ satisfying $f_k(x_k)>0$ and (4-1). Since we have $0< \int _X f_k\, d\mu <\infty$, we can assume (4-2) by multiplying appropriate scalars to $f_k$. We finished (Step 4), so we move to the final step.

(Step 5). Check that $\tau$, C and $f_k$ are what we wanted.

We already checked the condition (i) and (ii) (Step 3, 4). Therefore, it is enough to verify the condition (iii).

Claim. For $\theta \in \varTheta \cap C$,

• (5-1) $\int _X f_1dp_\theta \ge e^{-\Vert \theta \Vert (t_1+2\tau )-\varphi (\theta )}$,

• (5-2) $\int _X f_2 dp_\theta \le e^{-\Vert \theta \Vert (t_2-2\tau )-\varphi (\theta )}$.

Since one can prove (5-2) in the same way as (5-1), we prove only (5-1).

$$\begin{aligned} \int _X f_1 dp_\theta&=\int _{x\in S_1}f_1(x)e^{-\left\langle \theta , T(x) \right\rangle -\varphi (\theta )}d\mu (x)\\&\ge e^{-\Vert \theta \Vert (t_1+2\tau )-\varphi (\theta )}\int _{S_1}f_1 d\mu =e^{-\Vert \theta \Vert (t_1+2\tau )-\varphi (\theta )}. \end{aligned}$$

In the inequality above, we used (3-1). Therefore our claim was proved. By this claim, for any $\theta \in \varTheta \cap C$, we obtain

$$\begin{aligned} \int _X f_1dp_\theta \ge e^{\Vert \theta \Vert (t_2-2\tau -t_1-2\tau )}\int _X f_2dp_\theta =e^{\Vert \theta \Vert \tau }\int _X f_2dp_\theta . \end{aligned}$$

Thus, (iii) was proved. $\square$

Lemma 4.40

Under Setting 4.37, additionally, we assume $\mu \in {\mathcal {P}}$. Then the parameter map $a:{\mathcal {P}}\rightarrow V^\vee$ is continuous at $\mu \in {\mathcal {P}}$.

Proof

Take any $\xi \in V^\vee$ with $\Vert \xi \Vert =1$. Since we have $a(\mu )=0$, from Lemma 4.36, it is enough to show that a is continuous along $\xi$ at $\mu \in {\mathcal {P}}$. We take $\tau >0$, an open cone C and $f_k\in C_c^+(X)$ $(k=1,2)$ as in Lemma 4.39. Take any $\varepsilon >0$, and put

$$\begin{aligned} r&:=e^\frac{\varepsilon \tau }{2}(>1),\\ U&:=\left\{ \nu \in {\mathcal {P}}\ \Big |\ \int _X f_1\, d\nu <r,\ \int _X f_2\, d\nu >r^{-1} \right\} . \end{aligned}$$

Then it is enough to show the following:

Claim.

• (i) U is an open neighborhood of $\mu$,

• (ii) $a(U)\cap C\subset B_\varepsilon$.

First, we verify (i). The openness of U comes from the definition of the topology of ${\mathcal {R}}(X)$. We have $\mu \in U$ from the choice of $f_k$.

Next, let us verify (ii). Take any $\theta \in a(U)\cap C$. Since we have $p_\theta \in U$ from $\theta \in \varTheta$, the following inequalities hold:

$$\begin{aligned} r>\int _X f_1\,dp_\theta \ge e^{\Vert \theta \Vert \tau }\int _X f_2\,dp_\theta > e^{\Vert \theta \Vert \tau }r^{-1}. \end{aligned}$$

Therefore, we obtain $e^{\Vert \theta \Vert \tau }<r^2$, that is, $\Vert \theta \Vert <\varepsilon$. $\square$

Finally, we end this subsection by proving the continuity of the parameter map.

Proof of Proposition 2.8

We consider under Setting 4.37. Take any $p_0\in {\mathcal {P}}$. It is enough to show that a is continuous at $p_0$. Put $\theta _0:=a(p_0)\in \varTheta$. We consider the following:

Claim. $(p_0, V,T)$ is a minimal realization of ${\mathcal {P}}$, and its parameter map $a_0:{\mathcal {P}}\rightarrow V^\vee$ is given by $a_0(p):=a(p)-\theta _0$.

Let us verify this claim. Since we have

$$\begin{aligned} dp(x)&=\exp (-\left\langle a(p), T(x) \right\rangle -\varphi (a(p)))d\mu (x)\quad (p\in {\mathcal {P}}),\\ dp_0(x)&=\exp (-\left\langle \theta _0, T(x) \right\rangle -\varphi (\theta _0))d\mu (x), \end{aligned}$$

we obtain the following equality by an elementary property of Radon–Nikodým derivative (Fact 4.14(i)).

$$\begin{aligned} dp(x)=\exp (-\left\langle a(p)-\theta _0, T(x) \right\rangle -\varphi (a(p))+\varphi (\theta _0))dp_0(x)\quad (p\in {\mathcal {P}}). \end{aligned}$$

Therefore, $(p_{0},V, T)$ is a minimal realization of ${\mathcal {P}}$ and its parameter map is given as $a_0$, so our claim was proved.

From Lemma 4.40, $a_0$ is continuous at $p_0$. Thus, a is continuous at $p_0$ from the claim above. $\square$