Not sure if I can ask such fundamental problem here.


  1. $G$ be a finite group, $\sigma \in G$. Consider linear group actions fo $G$ on $\mathbb{R}^n$.
  2. $\sigma(K)=\{x\in \mathbb{R}^n: \sigma^{-1}(x) \in K\}$.
  3. $K\subseteq \mathbb{R}^n$ be $G$-invariant, i.e., $\bigcap_{\sigma\in G} \sigma(K) = K$.


  1. $K\subseteq \mathbb{R}^n$ is $G$-invariant.
  2. polynomial $h(x)\in \mathbb{R}[X]$ is $G$-invariant, i.e., $h(\sigma^{-1}(x))=h(x), \, \forall \sigma \in G$.

Note that

  1. Let $\mathcal{P}(K)$ be the set of probability measures supported on $K$.
  2. For $\mu\in \mathcal{P}(K)$, $\sigma^{-1}(K)\subseteq K, \, \forall\sigma$. $\mu$ is $G$-invariant if $\mu (B) = \mu(\sigma^{-1}(B))$ for any Borel set $B\subseteq K$.

Consider the following

\begin{equation} \begin{aligned} \int_K h(x) \mu(dx) &=_1 \int_K h(\sigma(x)) \mu(dx) \\ &=_2 \int_K h(x) \mu(\sigma^{-1}(dx)) \end{aligned} \end{equation}

  1. comes from the fact that $h(x)$ is $G$-invariant and $G$ is a group.

My questions are

  1. Is the changing variable of step 2. correct? (Note that $\sigma(K)=K,\, \forall \sigma \in G$ from assumption.)
  2. Can I say that since the $2$nd equality holds for all $dx$, so $$\mu(dx) = \mu(\sigma^{-1}(dx)),$$ i.e., $\mu$ is $G$-invariant.

I appreciate any help or comment!

  • 1
    $\begingroup$ Your assumption 3 looks different from saying $\sigma(K)=K$ for all $\sigma\in G$. $\endgroup$ – Abdelmalek Abdesselam Sep 17 '19 at 18:34
  • $\begingroup$ @AbdelmalekAbdesselam From my research, $K = \{x: g_j(x) \geq 0, j=1,\ldots, m\}$ and $g_j^{\sigma} = g_j$. I just prove that we can have $\sigma(K) = K$ for all $\sigma \in G$, if $G$ is a symmetric group. This is because its representation in $\mathbb{R}^n$ is an invertible matrix. So $\sigma(K) = K$. $\endgroup$ – sleeve chen Sep 17 '19 at 20:49
  • $\begingroup$ Use $\cup$ instead of $\cap$ in your assumption 3, otherwise what you said there is not correct. $\endgroup$ – Abdelmalek Abdesselam Sep 18 '19 at 16:17
  • $\begingroup$ @AbdelmalekAbdesselam Actually it should be $\cap$, which I check the paper I read. $\endgroup$ – sleeve chen Sep 18 '19 at 20:05
  • $\begingroup$ then the paper is wrong $\endgroup$ – Abdelmalek Abdesselam Sep 29 '19 at 20:56

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