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.

September 17, 2010 -

1. I know people have suggested before that you write a book, which I realize may take a while to happen, but in the meantime, how about compiling a large PDF of your blog entries? It’d be a substantially smaller responsibility (well, OK, that could depend on what your LaTeX code is like) but would be very useful to readers.

Comment by Akhil Mathew | September 18, 2010 | Reply

2. […] if , then there is one coset for each element of , and the coset representation is the same as the left regular representation. At the other extreme, if , then there is only one coset and we get the trivial […]

Pingback by Coset Representations « The Unapologetic Mathematician | September 20, 2010 | Reply

3. […] an example, let be any finite group, and let be its group algebra, which carries the left regular representation . Now, consider the subspace spanned by the […]

Pingback by Submodules « The Unapologetic Mathematician | September 22, 2010 | Reply

4. […] if it contains a nontrivial submodule. Thus our examples last time show that the left regular representation is always reducible, since it always contains a copy of the trivial representation as a nontrivial […]

Pingback by Reducibility « The Unapologetic Mathematician | September 23, 2010 | Reply

5. I think that you interchanged rows and columns: shouldn’t $\rho(g^k)$ have entry $\delta_{g^kg^j,g^i}$ in its i-th row and j-th column? I.e. entry 1 if i=j+k and 0 otherwise?

Comment by Landau | October 28, 2010 | Reply

6. You know, you’re right.. and I got this backwards when I defined permutation representations before. Fixing…

Comment by John Armstrong | October 28, 2010 | Reply

7. […] right away is the right regular representation, which (predictably enough) corresponds to the left regular representation. In fact, when I introduced that representation I even mentioned the right action in passing. At […]

Pingback by Right Representations « The Unapologetic Mathematician | November 2, 2010 | Reply

8. […] the Left Regular Representation Let’s take the left regular representation of a finite group on its group algebra and decompose it into irreducible […]

Pingback by Decomposing the Left Regular Representation « The Unapologetic Mathematician | November 17, 2010 | Reply

9. […] 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 […]

Pingback by The Endomorphism Algebra of the Left Regular Representation « The Unapologetic Mathematician | November 18, 2010 | Reply

10. […] show up in . In particular, since we know that every irreducible representation shows up in the left regular representation , the number of irreducible representations is . Thus to calculate this number , we must understand […]

Pingback by The Character Table is Square « The Unapologetic Mathematician | November 19, 2010 | Reply

11. […] The action of on these tabloids is basically the same as left-multiplication on the underlying set . And so we find the left regular representation. […]

Pingback by Permutation Representations from Partitions « The Unapologetic Mathematician | December 14, 2010 | Reply