Is there a way to construct a probability measure $\mu$ on the n-torus $T^n = \mathbb R^n/\mathbb Z^n$ such that it is translation invariant, i.e. $\int_{T^n}f(x+y)d\mu(x) = \int_{T^n}f(x)d\mu(x)$ for all continuous $f$ and all $y\in T^n$ (I am viewing $T^n$ as an abelian Lie group). I have been wrapping my head around whether one could fix coordinates $x_1,\ldots,x_n$ on $\mathbb R^n$ and to somehow use the pushforward measure. But the pushforward measure should not work since the preimage (under the canonical projection $\mathbb R^n\to\mathbb R^n/\mathbb Z^n$) of any tiny square on $T^n$ are infinitely many tiny squares in $\mathbb R^n$ which has infinite measure. Is there another way, besides the pushforward measure, to construct a probability measure on the torus $T^n$? The probability measure is intuitively going to be translation invariant (i.e. left-invariant in Lie group terminology). I would also like to know why it is translation invariant in the integral-sense above.
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The solution is not to use the actual pushforward measure defined on the whole of $\mathbb R^n$.
Instead, restrict the covering map to a fundamental domain, for example $[0,1]^n$, and push the measure forward under the restricted function $[0,1]^n \to T^n$. You can also restrict further to $(0,1)^n$ and the outcome will be no different, because the boundary of $[0,1]^n$ has measure zero. One can also prove that the outcome is independent of the choice of fundamental domain, and that it is translation invariant on $T^n$.
Lee Mosher
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Hello, thank you. This helps a lot. You probably mean either $\mathbb R^n$ and $[0,1]^n$ or $T^2$, right? – Mathematics enthusiast Oct 08 '24 at 12:33
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Yes, I edited to fix those. For some reason I don't see half my errors until after I push the "save edit button", so it's a vicious cycle. – Lee Mosher Oct 08 '24 at 12:33