# The Unapologetic Mathematician

## The Endomorphism Algebra of the Left Regular Representation

Since the left regular representation is such an interesting one — in particular since it contains all the irreducible representations — we want to understand its endomorphisms. That is, we want to understand the structure of $\mathrm{End}_G(\mathbb{C}[G])$. I say that, amazingly enough, it is anti-isomorphic to the group algebra $\mathbb{C}[G]$ itself!

So let’s try to come up with an anti-isomorphism $\mathbb{C}[G]\to\mathrm{End}_G(\mathbb{C}[G])$. Given any element $v\in\mathbb{C}[G]$, we define the map $\phi_v:\mathbb{C}[G]\to\mathbb{C}[G]$ to be right-multiplication by $v$. That is:

$\displaystyle\phi_v(w)=wv$

for every $w\in\mathbb{C}[G]$. This is a $G$-endomorphism, since $G$ acts by multiplication on the left, and left-multiplication commutes with right-multiplication.

To see that it’s an anti-homomorphism, we must check that it’s linear and that it reverses the order of multiplication. Linearity is straightforward; as for reversing multiplication, we calculate:

\displaystyle\begin{aligned}\left[\phi_u\circ\phi_v\right](w)&=\phi_u\left(\phi_v(w)\right)\\&=\phi_u\left(wv\right)\\&=wvu\\&=\phi_{vu}(w)\end{aligned}

Next we check that $v\mapsto\phi_v$ is injective by calculating its kernel. If $\phi_v=0$ then

\displaystyle\begin{aligned}v&=1v\\&=\phi_v(1)\\&=0(1)\\&=0\end{aligned}

so this is only possible if $v=0$.

Finally we must check surjectivity. Say $\theta\in\mathrm{End}_G(\mathbb{C}[G])$, and define $v=\theta(1)$. I say that $\theta=\phi_v$, since

\displaystyle\begin{aligned}\theta(g)&=\theta(g1)\\&=g\theta(1)\\&=gv\\&=\phi_v(g)\end{aligned}

Since the two $G$-endomorphisms are are equal on the standard basis of $\mathbb{C}[G]$, they are equal. Thus, every $G$-endomorphism of the left regular representation is of the form $\phi_v$ for some $v\in\mathbb{C}[G]$.