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By $\mathbb{T}^1$ I mean the one-dimensional Torus which is the same as the circle $S^1$.

How does one make sense of: $$\int_{\mathbb{T}^1}f dm = ?$$

(1) What is the definition of $\mathbb{T}^1$ suitable for making sense of the integral? (For example: the set $\{e^{2\pi i x}: x \in [0,1]\} \subset \mathbb{C}$ or $\mathbb{R}/\mathbb{Z}=\{x+\mathbb{Z}: x \in \mathbb{R}\}$?)

(2) What is the standard measure $m$ on the 1-Torus?

(3) What does the integral of $f:\mathbb{T}^1 \to \mathbb R$ with respect to measure $m$ mean?

Does it mean $$\int_{\mathbb{T}^1}f dm = \int_0^1f(e^{2\pi i x})dx ?$$

I'm a bit confused and would appreciate some help. A reference to a book would be nice.

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Everything suggests you are not talking about the torus but about the 1-sphere $S^1$. –  Lord_Farin Mar 28 '13 at 12:49
    
In my notation $\mathbb{T}^1=S^1$. –  Cantor Mar 28 '13 at 12:51

1 Answer 1

up vote 2 down vote accepted

For the present question, $\int_{\Bbb T^1} f \,\mathrm dm$ is probably best interpreted as the line integral in $\Bbb C$ over the unit circle (it may however depend a bit on how $f: \Bbb T^1 \to \Bbb R$ is given to you, if it is given as a periodic function from $\Bbb R$ to $\Bbb R$ you can use standard integration over any period).

The two standard measures on $\Bbb T^1$ are scalar multiples of one another; mostly I've seen the variant where $\int_{\Bbb T^1} 1\,\mathrm dm = 2\pi$ but also the scaled variant $m'$ with $\int_{\Bbb T^1} 1\,\mathrm dm' = 1$ is of use.

The above will likely satisfy your present needs. For further development of integration over manifolds I can recommend the excellent book Multidimensional Real Analysis (2 vol.) — J.J. Duistermaat, J.A.C. Kolk (Cambridge studies in advanced mathematics 86-87). The second volume addresses integration.


Addendum: In view of the request for a more thorough approach, let me make a dash for it.

First, we need to abstract away any surrounding of $\Bbb T^1$, such as its embedding into $\Bbb C$ or its construction as $\Bbb R/\Bbb Z$. It is simply a topological space. In fact, consider it as a one-dimensional real manifold. That means it has charts, homeomorphisms $\phi_\alpha: \Bbb R \supset U \to \phi_\alpha(U) \subset \Bbb T^1$ ($U \subset \Bbb R$ open, connected, i.e. open interval, $\alpha$ ranging over some index set) that we can "glue" together to "cover $\Bbb T^1$ with pieces of $\Bbb R$", as in the image (courtesy of Wikipedia):

Now we can divide $\Bbb T^1$ into small "1-rectangles" (i.e. line segments, but this way there is a clearer perspective towards higher dimensional approaches) each of which lies entirely in a domain $\phi_\alpha(U_\alpha)$ (one of the four colours in the picture) that overlap at most on their boundary (which is a null set for the measure; I'll let that rest here).

On each of these small 1-rectangles we can compute the integral as we would in $\Bbb R$ by passing through the chart $\phi_\alpha$ (and using the measure on $\Bbb R$ that we would like to use; I've only seen multiples of the standard Lebesgue measure). The complete integral is naturally the sum of the constituents so defined.

This way we can define the integral over $\Bbb T^1$ considered as a real manifold. This means in particular that the charts have been specified. Different charts (e.g. scaling all $\phi_\alpha$ by a factor) will generally yield different values for the integral.

The above will hopefully have addressed point (3) of your question in more detail and also have provided a glimpse of what happens when $\Bbb T^1$ is replaced with more general manifolds).

Now when we consider $\Bbb T^1$ as embedded in $\Bbb C$, we are de facto viewing it as a 1D submanifold of the complex plane. In this situation, we do not significantly (only at the end points) differ from the definition of a line integral in $\Bbb C$, viz

$$\int_\gamma f \,\mathrm dm = \int_0^1 f(\gamma(t)) \gamma'(t) \,\mathrm dt$$

where we have now interpreted $f$ as being defined on a subset of $\Bbb C$. Thus indeed, as long as we choose a particular embedding $\gamma: \Bbb T^1 \to \Bbb C$ where it is maybe the simplest conceptually to consider $\Bbb T^1 \cong \Bbb R / \Bbb Z$, in which case we obtain:

$$\int_{\Bbb T^1} f \,\mathrm dm = 2\pi i\int_0^1 f(e^{2\pi it})e^{2\pi it}\,\mathrm dt$$

If OTOH $f$ is given as a periodic (period $1$, say) $f: \Bbb R \to \Bbb R$, then we would have:

$$\int_{\Bbb T^1} f\,\mathrm dm = \int_0^1 f(x) \,\mathrm dx$$

and these integrals turn out to agree because the normalization is the same.


A reference of interest may be the Wikipedia article on "Manifold"; also the ref'd book is a good read when you want to continue further on this path.

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1  
Thank you for the addendum. What is the advantage of viewing the $\mathbb{T}^1$ as a manifold, while it can just be viewed as a curve in $\mathbb{C}$ and integrated according to the second part of your answer? –  Cantor Mar 28 '13 at 15:40
    
In this specific case, there isn't really (unless trying to grasp integration over manifolds by a simple example). However, I took the fact that you asked for a more thorough approach when I wrote the first paragraph as a request for a more general perspective. I tried to give that. You may now view it as a connection of your question with the Duistermaat/Kolk book I mentioned. –  Lord_Farin Mar 28 '13 at 16:03
    
Thank you. I will accept your answer, but I would've preferred something more to the point without mention of integration on manifolds. I understand that it is a generalization, but the differentiable structure is not needed to integrate on a torus (if I understood correctly) and the differential geometry point of view only makes things more complicated. I would've prefered a generalization in the direction of integral over $\mathbb{T}^2$ viewing it as a subset of $\mathbb{C^2}$ i.e. $\{(e^{2\pi i x}, e^{2\pi i y}): x,y \in [0,1]\}$. –  Cantor Mar 28 '13 at 16:21
1  
Thank you. I'm sorry that I didn't manage to correctly sense what you were looking for; if you're (for the moment) content integrating only over explicit submanifolds of some $\Bbb R^n$ or $\Bbb C^n$ then I concede that my answer is excessively general. –  Lord_Farin Mar 28 '13 at 16:28

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