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Let $(G,K)$ be a Gelfand pair. Why, for a function $f$ $K$-binvariant with respect to a compact subgroup $K$ of a group $G$, do we have the following equality: $$ f(xy) = \int_K f(xky) \, dk$$

A function $ f : G \to \mathbb{C} $ is $ K $-binvariant if for all $ k_1, k_2 \in K $ and $ x \in G $ $$ f(k_1 x k_2) = f(x). $$

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    $\begingroup$ Replacing $x$ with $xy^{-1}$, this is equivalent to saying that $K$-biinvariance implies right invariance with respect to conjugates of $K$. There is a counterexample with $|G| = 6$. $\endgroup$
    – Sean Eberhard
    Commented Jul 14 at 10:02
  • $\begingroup$ Thank you for your help, my question is it false? $\endgroup$
    – Ryo Ken
    Commented Jul 14 at 10:07
  • $\begingroup$ MathJax note: MathJax does not obey \( \), so \(f\) comes out (f), but, perversely, does obey \\\\( \\\\) (although apparently not in comments: \\\\(f\\\\) shows up as \\\(f\\\) (not a typo!)). See Having the MO Mathjax parser recognise \( \) is a regex away. I have edited accordingly. (Also, the usual spelling is "biinvariant", three 'i's, or "bi-invariant", rather than "binvariant", two 'i's.) $\endgroup$
    – LSpice
    Commented Jul 14 at 14:52
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    $\begingroup$ @SeanEberhard, I assume that you have in mind $G = \mathrm S_3$ and $K$ an order-$2$ subgroup. Is that a Gelfand pair? $\endgroup$
    – LSpice
    Commented Jul 14 at 14:58
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    $\begingroup$ @LSpice Yes. (Also, the phrase "Gelfand pair" was added after I made my comment.) $\endgroup$
    – Sean Eberhard
    Commented Jul 14 at 22:02

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