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Let $X$ be a compact Hausdorff space and $C(X)$ be the space of all complex-valued continuous functions on $X$. Let $\tau:X\rightarrow X$ be a homeomorphism such that $\tau^2=\tau\circ\tau=identity$. I want to show that the subalgebra $$C(X,\tau)=\{f\in C(X):f(\tau(x))=\overline{f}(x)\}$$ is a real subalgebra of $C(X)$ and not a complex subalgebra. Please help.

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    $\begingroup$ What are your thoughts? Can you show it is a real subalgebra? Can you try showing it is a complex subalgebra and see what seems to go wrong? $\endgroup$ – Eric Wofsey Jan 19 '17 at 5:52
  • $\begingroup$ (Incidentally, the statement that it is not a complex subalgebra is false if $X$ is empty.) $\endgroup$ – Eric Wofsey Jan 19 '17 at 5:53
  • $\begingroup$ Oops, I found a solution. If $f\in C(X,\tau)$ then $i\ f\notin C(X,\tau)$. $\endgroup$ – Anindya Biswas Jan 19 '17 at 6:11
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Take a scalar $\alpha$ (member of the complex field) which has a non zero imaginary part and try and multiply a member $f$ of your subset. You will find that you would need to see what happens to $\alpha . f \left(\tau(x)\right)$. This unfortunately will not end up as $\overline {\left(\alpha . f \right)(x)}$ unless the imaginary part of $\alpha$ was zero.

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If $f\in C(X,\tau)$ then $i\ f\notin C(X,\tau)$ and hence $C(X,\tau)$ cannot be a complex subalgebra.

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