The trace of the regular character

Question

In this paragraph, I feel confused why $\chi_u(g)$ is equal to the number of elements $x\in G$ for which $gx = x$, where $\chi_u(g)$ is the trace of the linear transformation of U defined by g, sending $u\in U$ to $gu$, $\mathbb{C}G$ is the algebra over the f.g. group $G$

Answer 1

Regular representation is one that sends each basis vector (of a specific basis) to another vector of the same basis (instead of a linear combination). So the matrix of this representation corresponding to each element (for this basis) would be a permutation matrix.

This means we are now calculating the trace of a permutation matrix. This is the sum of diagonal entries which are either one or zero. This in turn means trace is the count of diagonal entries that are equal to 1.

The $(i,i)$th diagonal entry will be 1 iff the element $g$ (for which this is the matrix) of the group fixes the $i$ th basis vector.

So what you said is true for all permutation representation, not just regular representation.

Answer 2

The regular representation $\rho$ of a finite group $G$ has dimension $\lvert G \rvert$. The vector space $\mathbb{C}G$ has basis $\{g\in G\}$. Now if you view $\rho$ as a homomorphism from $G$ to $GL_n(\mathbb{C})$, then you will have something on the diagonal of $\rho(g)$ exactly when $\rho(g)(x) = x$.