We know that a function $f(z)$ can be decomposed into $\frac{f(z)+f(-z)}{2}$ and $\frac{f(z)-f(-z)}{2}$. These are called the even and odd components.

I have made a generalisaiton of this. Suppose $f(z)$ is our function on the complex plane, and $e^{\frac{i2\pi q}{n}}$ are the $nth$ roots of unity, $q=0,1,2,3...n-1$. We can decompose the function into $n$ functions $g_k(z)$, $k=0,1,2,3....n-1$ given by:

$g_k(z)= \frac{1}{n} \sum _{q=0} ^{n-1} e^{\frac{2\pi qk}{n}} f(e^{2\pi i \frac{q}{n}} z)$

Adding these back together:

$$\sum _{k=0}^{n-1} g_k (z)$$

$$= \sum _{k=0}^{n-1} \frac{1}{n} \sum _{q=0} ^{n-1} e^{\frac{2\pi qk}{n}} f(e^{2\pi i \frac{q}{n}} z)$$

$$= \sum _{q=0}^{n-1} \frac{1}{n}( \sum _{k=0} ^{n-1} e^{\frac{2\pi qk}{n}} f(e^{2\pi i \frac{q}{n}} z))$$

If we look at the first term of this sum corresponding to $q=0$, we get $\frac{1}{n} nf(z)=f(z)$. The terms for other values of $q$ are 0.

This means that the decomposition $g_k (z)$ adds upto $f(z)$. For $n=2$, this is the same as the decomposition of $f(z)$ into even and odd functions.

This is a generalisation of the decomposition into even and odd functions. Each of the functions $g_k(z)$ has a rotational symmetry of rotation of the argument $z$ by $e^{\frac{i2\pi k}{n}}$. Is there a name for this generalisation?

  • $\begingroup$ This is series multi-section in the context of power series. $\endgroup$
    – Somos
    Aug 21, 2023 at 12:55
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    $\begingroup$ you've got a representation of the cyclic group $C_n=\langle \zeta \rangle$ on the complex vector space of functions on the complex plane, given by $(\zeta \cdot f)(z) = f(\zeta_n z)$ where $\zeta_n$ is a complex $n$th root of 1. The rep decomposes into a direct sum of irreducibles on which $\zeta$ acts as $\zeta_n^q$ for some $q$, and the functions $g_k$ are the projections of $f$ onto the isotypic components. $\endgroup$ Aug 21, 2023 at 13:02
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    $\begingroup$ ...so a name for this generalisation would be: "representation theory" $\endgroup$ Aug 21, 2023 at 13:14
  • $\begingroup$ @MatthewTowers Thank you. If you can write an answer about the most abstract version of this idea (groups, representation and decomposition stuff), I will accept that as an answer. Please also tell how it relates to special exampels like this post and maybe Fourier transform (if that's related too) $\endgroup$
    – Ryder Rude
    Aug 21, 2023 at 14:02

1 Answer 1


Suppose you have a representation $V$ of a group $G$ (say) which you know decomposes as an internal direct sum $V_1 \oplus V_2 \oplus \cdots$. Given $v \in V$ you might want to write $v = \sum v_i$ where $v_i \in V_i$, that is, you might want to "know" in some sense the projection operations $\pi : V \to V_i$.

This problem contains your generalisation of even and odd functions. Specifically, if $V$ is the vector space of complex functions made into a rep of $G=C_n=\langle \zeta\rangle$ the cyclic group of order $n$ by $(\zeta f)(z) = f(e^{2\pi i / n}z)$, then $V$ decomposes as a direct sum of the subspaces $V_k$ on which $\zeta$ acts as the scalar $e^{2\pi ik/n}$ and you can write down formulas expressing an arbitrary $f \in V$ as a sum $\sum f_k$ with $f_k\in V_k$. The case $n=2$ is the even-odd example.

The simple version of the connection with classical Fourier analysis is that you look at the action of the circle group (think of it as the group of complex numbers with absolute value 1 under multiplication) on complex square-integrable functions on the unit circle $S^1$, observe that the functions $z\mapsto z^n$ span 1-d subreps, and prove a theorem that says the rep decomposes as a direct product of these subreps in a suitable sense. This MO question and its answers go into the technical detail and explain why Fourier analysis isn't just a special case of rep theory.

  • $\begingroup$ I took $n\rightarrow \infty$ and got this infinite expansion of complex functions: $f(z)= \sum _{k=0}^{\infty} (\int _0^1 e^{i2\pi kx}f(ze^{i2\pi x}))$. Is there a name of this specific series? $\endgroup$
    – Ryder Rude
    Aug 22, 2023 at 18:02

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