The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith row and jth 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 Posted by | Algebra, Group theory, Representation Theory | 11 Comments

   

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