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I am having trouble following the argument in page 21 of P. Walters, Intro. to ergodic theory, of the following statement:

Any continuous endomorphism on a compact group preserves Haar measure.

Obviously, this is not true as stated, as the trivial homomorphism does not preserve Haar measure. Even restricting ourselves to nontrivial homomorphisms, I am still unable to follow the argument, which is as follows:

Let $A : G \rightarrow G$ be the endomorphism and let $m$ denote the Haar measure. Define a probability measure on Borel sets $\mu(E) = m(A^{-1}(E))$ . Now,

$$ \mu(Ax.E) = m(A^{-1}(Ax.E)) = m(x.A^{-1}E) = \mu(E) .$$

I agree with the first equality by definition, and the last equality because of $m$ being a Haar measure. But I am unable to fathom why the middle equality is true.

Moreover, even if the said equality is true, the rest of the proof goes as follows: We see that $\mu$ is rotation invariant, and we use the uniqueness of Haar measure to prove that $\mu = m$. In this, I am unable to see how the above equality assures that $\mu$ is rotation-invariant.

Help would be appreciated.

Rudy the Reindeer
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Doubting Thomas
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    This thread is relevant: http://math.stackexchange.com/q/65565 –  Jul 21 '12 at 13:09
  • So, the moral is that the proof given by Walters is nonsense, and I should rather follow kneidell's proof at the link given? – Doubting Thomas Jul 21 '12 at 13:36
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    The version of Walter's book on Google books gives a different statement. There he writes "Any continuous endomorphism of a compact group onto itself preserves Haar measure." –  Jul 21 '12 at 13:40
  • @Byron : I didn't type the extra two words out of laziness. Isn't an endomorphism by definition a homomorphism to the same object? – Doubting Thomas Jul 21 '12 at 13:46
  • @DoubtingThomas The key word is onto. This says that the image of the map is the whole space. –  Jul 21 '12 at 13:48
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    One moral I see here is this: when quoting from a source, make sure to get the quotation entirely correct. – Pete L. Clark Jul 22 '12 at 13:08

1 Answers1

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The moral is that you have to assume surjectivity of $A$.

For the middle equality: we have $A^{-1}(Ax.E)=x.A^{-1}E$ as sets:

$y\in A^{-1}(Ax.E)$ means $Ay\in Ax.E$ means $A(x^{-1}y)\in E$, while

$y\in x.A^{-1}E$ means $x^{-1}y\in A^{-1}E$.

For the rotation-invariance: we now see $\mu(Ax.E)=\mu(E)$ for all $x\in G$. In other words, $\mu(y.E)=\mu(E)$ for all $y\in \text{im} A\subset G$. By surjectivity we have $\text{im} A=G$.

wildildildlife
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