The Unapologetic Mathematician

Mathematics for the interested outsider

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].

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November 18, 2010 Posted by | Algebra, Group theory, Representation Theory | 1 Comment