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I'm a physics student and I don't have a strong background of algebra, so please excuse me if I made some mistakes.

I read it here on the top of page 96 that:

Any arbitrary function $F$ can be written as a linear combination of a complete set of basis functions $f^{\Gamma_{n'}}_{j'}$
$F= \sum_{n'}\sum_{j'} f^{\Gamma_{n'}}_{j'} |\Gamma_{n'}, j'\rangle$

in which $|\Gamma_{n'}, j'\rangle$ represents the $j' -th$ basis function of the $n' -th$ irreducible representation of some arbitrary finite group.

The thing confuses me is that are there any restrictions on the function $F$?

For finite groups, both the number of invariant spaces and the dimension of each invariant space are finite, so that we have a finite set of basis functions of a finite group. (the set of functions of $|\Gamma_{n'}, j'\rangle$)

Let's say the set of basis functions is $\{1, x, x^2, x^3\}$ for some finite group. Clearly, there are a lot of functions that can not be expanded by this basis.

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    $\begingroup$ I assume you mean functions on a finite group? In that case there are no restrictions, assuming I've understood what "basis function" means here. $\endgroup$ Nov 26, 2017 at 8:41

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For any function $F:F(g) \rightarrow \mathbb{C} \mathrm{~or~} \mathbb{R} $ defined on a finite group $G=\{g_1, g_2, ... , g_N\}$, we can regard it as a column vector.

$$ \begin{align} F(G) &= \begin{bmatrix} F(g_1) \\ F(g_2) \\ \vdots \\ F(g_N) \end{bmatrix} \end{align} $$ , where $F(g)$ is a number.

Since $F(g)$ is arbitrary function,the dimension of this column vector is $|G| = N$.

Now, from the group orthogonality theorem, we have

$$ \sum_{g} D_{ij}^{\alpha}(g)^ * D_{kl}^{\beta}(g) = \frac{|G|}{d_{\alpha}} \delta_{\alpha \beta} \delta_{ik} \delta_{jl} $$

Here, you can think that each $D_{ij}^{\alpha}$ is a group function since it assigns a number for each group element.

$$ \begin{align} D_{ij}^{\alpha}(G) &= \begin{bmatrix} D_{ij}^{\alpha}(g_1) \\ D_{ij}^{\alpha}(g_2) \\ \vdots \\ D_{ij}^{\alpha}(g_N) \end{bmatrix} \end{align} $$

Now if you regard $D_{ij}^{\alpha}$ as a column vector in the $|G|$ dimensional vector space, group orthogonality says that the vector $D_{ij}^{\alpha}$ and $D_{kl}^{\beta}$ are orthoggonal.

$$ \langle D_{ij}^{\alpha} | D_{kl}^{\beta} \rangle = \sum_{g} D_{ij}^{\alpha}(g)^ * D_{kl}^{\beta}(g) = 0 $$ , if $(\alpha, i, j) \neq (\beta, k, l)$. Then, how many orthogonal vectors do we have? From the dimensionality theorem about group representation, we know

$$ \sum_\alpha d_{\alpha} ^2 = |G| $$

, thus we have |G| ortthogonal column vectors $D_{ij}^{\alpha}$ (each representation $D^{\alpha}$ is $d_{\alpha}$ dimensional matrix, so you have $d_{\alpha}^2$ group functions for each representation). Since we have |G| orthogonal vectors (so linearly independent), they can span arbitrary $F(G)$ in |G| dimension.

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