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I have posted a question in MSE https://math.stackexchange.com/questions/4468138/question-about-finite-dimensional-representations-of-a-semi-simple-lie-group but didn't receive any comment or answer.

My question was the following:

I've encountered the following paragraph while reading page 3 of this paper https://link.springer.com/article/10.1007/BF01232026?noAccess=true

Let $G$ be a semi-simple Lie group with a maximal compact subgroup $K$. Let $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{s}$ be the corresponding Cartan decomposition. Let $V$ be a finite dimensional representation of $G$ over $\mathbb{R}$. There exists a $K$-invariant inner product $(,)$ on $V$ such that any $X \in \mathfrak{s}$ acts as a symmetric operator.

I know that since $K$ is a compact subgroup, there exists a $K$-invariant inner product $(,)$ on $V$, but I don't know why there is such an inner product for which every $X \in \mathfrak{s}$ acts as a symmetric operator ? Any help please.

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    $\begingroup$ This is well known and is in Helgason's book. In the complexification $\mathfrak g _{\mathbb C}$ of $\mathfrak g$, you have the real subalgebra $\mathfrak k \oplus i \mathfrak p$, which can be seen to be a "compact Lie algebra"; it is the Lie algebra of a compact group $U$ in $G(\mathbb C)$. The representation $V$ extends to a complex representation of $G(\mathbb C)$; $U$ being compact, its lie algebra elements act by skew symmetric matrices with respect to a suitable $U$ invariant Hermitian metric on $V\otimes {\mathbb C}$. Hence $\mathfrak p$ is symmetric. $\endgroup$
    – Venkataramana
    Commented Jun 10, 2022 at 10:52
  • $\begingroup$ Thanks a lot for your comment @Venkataramana $\endgroup$
    – Mira
    Commented Jun 11, 2022 at 17:34

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