伯恩赛德定理证明求助：基里洛夫等《表示论第一卷》第9页补全

Hey there! Let's wrap up the full proof of Burnside's theorem exactly where your book left off—starting with the regular representation on $L(G) = \text{Maps}(G, K)$. I'll break this down step by step to keep it aligned with the material you're reading.

Step 1: Recap the Regular Representation & $L(G)$ #

First, let's formalize what your book introduced:

• $L(G)$ is the vector space of all functions from $G$ to $K$ (we’ll assume $K$ is an algebraically closed field of characteristic 0, like $\mathbb{C}$, which is standard for this theorem).
• The dimension of $L(G)$ is exactly $|G| = N$, since we can take the standard basis ${\delta_g}_{g \in G}$ where $\delta_g(h) = 1$ if $h = g$, and 0 otherwise.
• The regular representation $R$ of $G$ acts on $L(G)$ by left translation: for any $g \in G$ and $f \in L(G)$, $(R(g)f)(h) = f(g^{-1}h)$ for all $h \in G$.

Step 2: Decompose the Regular Representation #

By Maschke's theorem (a foundational result for finite group representations), every finite-dimensional representation of a finite group over $K$ is semisimple. That means we can decompose $L(G)$ into a direct sum of irreducible $G$-representations:
$$L(G) \cong \bigoplus_{i=1}^a m_i V_i$$
where:

• $V_1, \dots, V_a$ are all distinct irreducible representations of $G$,
• $m_i$ is the multiplicity (number of times $V_i$ appears) in the decomposition,
• $n_i = \dim V_i$.

Our goal is to show $m_i = n_i$ for each $i$, which will lead directly to Burnside's theorem.

Step 3: Compute Multiplicities Using Characters #

To find $m_i$, we’ll use character theory—this is a common tool in representation theory that links representations to functions on $G$.

First, recall the character of a representation: for a representation $W$, $\chi_W(g) = \text{tr}(W(g))$ (the trace of the linear map $W(g)$).

• For the regular representation $R$, $\chi_R(e) = \dim L(G) = N$ (since the trace of the identity map is the dimension of the space).
• For any $g \neq e$, $\chi_R(g) = 0$. Why? The trace is the sum of diagonal entries of the matrix of $R(g)$ with respect to the basis ${\delta_h}$. The diagonal entry for $\delta_h$ is $(R(g)\delta_h)(h) = \delta_h(g^{-1}h)$, which is 1 only if $g^{-1}h = h$ (i.e., $g = e$). For $g \neq e$, none of these diagonal entries are 1, so the trace is 0.

Now, the character of the decomposed regular representation is $\chi_R = \sum_{i=1}^a m_i \chi_i$, where $\chi_i$ is the character of $V_i$. To find $m_i$, we use the orthogonality relation for characters:
$$m_i = \frac{1}{N} \sum_{g \in G} \chi_R(g) \overline{\chi_i(g)}$$

Substitute the values of $\chi_R(g)$:
$$m_i = \frac{1}{N} \left( \chi_R(e)\overline{\chi_i(e)} + \sum_{g \neq e} \chi_R(g)\overline{\chi_i(g)} \right)$$
Since $\chi_R(g) = 0$ for $g \neq e$, this simplifies to:
$$m_i = \frac{1}{N} \cdot N \cdot \overline{\chi_i(e)} = \overline{\chi_i(e)}$$

But $\chi_i(e) = \dim V_i = n_i$, and over $\mathbb{C}$ (or any algebraically closed field of characteristic 0), $\chi_i(e)$ is a positive integer, so $\overline{\chi_i(e)} = n_i$. Thus, $m_i = n_i$.

Step 4: Finish the Proof #

Now, compute the dimension of $L(G)$ using the decomposition:
$$\dim L(G) = \sum_{i=1}^a m_i \cdot \dim V_i$$
We know $\dim L(G) = N$, $m_i = n_i$, and $\dim V_i = n_i$, so:
$$N = \sum_{i=1}^a n_i \cdot n_i = \sum_{i=1}^a n_i^2$$

That's the full proof of Burnside's theorem! It ties directly back to the regular representation setup your book started on page 9.

内容的提问来源于stack exchange，提问作者Apo