# The Unapologetic Mathematician

## The (Left) Regular Representation

Now it comes time to introduce what’s probably the most important representation of any group, the “left regular representation”. This arises because any group $G$ acts on itself by left-multiplication. That is, we have a function $G\times G\to G$ — given by $(g,h)\mapsto gh$. Indeed, this is an action because first $(e,g_1)\mapsto g_1$; and second $(g_1,(g_2,h))\mapsto(g_1,g_2h)\mapsto g_1g_2h$, and $(g_1g_2,h)\mapsto g_1g_2h$ as well.

So, as with any group action on a finite set, we get a finite-dimensional permutation representation. The representing space $\mathbb{C}G$ has a standard basis corresponding to the elements of $G$. That is, to every element $g\in G$ we have a basis vector $\mathbf{g}\in\mathbb{C}G$. But we can recognize this as the standard basis of the group algebra $\mathbb{C}[G]$. That is, the group algebra itself carries a representation.

Of course, this shouldn’t really surprise us. After all, representations of $G$ are equivalent to modules for the group algebra; and the very fact that $\mathbb{C}[G]$ is an algebra means that it comes with a bilinear function $\mathbb{C}[G]\times\mathbb{C}[G]\to\mathbb{C}[G]$, which makes it into a module over itself.

We should note that since this is the left regular representation, there is also such a thing as the right regular representation, which arises from the action of $G$ on itself by multiplication on the right. But by itself right-multiplication doesn’t really give an action, because it reverses the order of multiplication. Indeed, for a group action as we’ve defined it first acting by $g_2$ and then acting by $g_1$ is the same as acting by the product $g_1g_2$. But if we first multiply on the right by $g_2$ and then by $g_1$ we get $hg_2h_1$, which is the same as acting by $g_2g_1$. The order has been reversed.

To compensate for this, we define the right regular representation by the function $(g,h)\mapsto hg^{-1}$. Then $(g_1,(g_2,h))\mapsto(g_1,hg_2^{-1})\mapsto hg_2^{-1}g_1^{-1}=h(g_1g_2)^{-1}$, and $(g_1g_2,h)\mapsto h(g_1g_2)^{-1}$ as well.

As an exercise, let’s work out the matrices of the left regular representation for the cyclic group $\mathbb{Z}_4$ with respect to its standard basis. We have four elements in this group: $\{g^0,g^1,g^2,g^3\}$ and $g^4=g^0$. Thus the regular representation will be four-dimensional, and we will index the rows and columns of our matrices by the exponents $0$, $1$, $2$, and $3$. Then in the matrix $\rho(g^k)$ the entry in the $i$th row and $j$th column is $\delta_{g^kg^j,g^i}$. The multiplication rule tells us that $(g^i,g^j)\mapsto g^{i+j}$, where the exponent is defined up to a multiple of four, and so the matrix entry is $1$ if $i=j+k$, and $0$ otherwise. That is:

\displaystyle\begin{aligned}\rho(g^0)&=\begin{pmatrix}1&0&0&0\\{0}&1&0&0\\{0}&0&1&0\\{0}&0&0&1\end{pmatrix}\\\rho(g^1)&=\begin{pmatrix}{0}&0&0&1\\1&0&0&0\\{0}&1&0&0\\{0}&0&1&0\end{pmatrix}\\\rho(g^2)&=\begin{pmatrix}{0}&0&1&0\\{0}&0&0&1\\1&0&0&0\\{0}&1&0&0\end{pmatrix}\\\rho(g^3)&=\begin{pmatrix}{0}&1&0&0\\{0}&0&1&0\\{0}&0&0&1\\1&0&0&0\end{pmatrix}\end{aligned}

You can check for yourself that these matrices indeed give a representation of the cyclic group.