# The Unapologetic Mathematician

## One Complete Character Table (part 1)

When we first defined the character table of a group, we closed by starting to write down the character table of $S_3$:

$\displaystyle\begin{array}{c|ccc}&e&(1\,2)&(1\,2\,3)\\\hline\chi^\mathrm{triv}&1&1&1\\\mathrm{sgn}&1&-1&1\\\vdots&\vdots&\vdots&\vdots\end{array}$

Now let’s take the consequences we just worked out of the orthonormality relations and finish the job.

We’ve already verified that the two characters we know of are orthonormal, and we know that there can be at most one more, which would make the character table look like:

$\displaystyle\begin{array}{c|ccc}&e&(1\,2)&(1\,2\,3)\\\hline\chi^\mathrm{triv}&1&1&1\\\mathrm{sgn}&1&-1&1\\\chi^{(3)}&?&?&?\end{array}$

Do we have any other representations of $S_3$ to work with? Well, there’s the defining representation. This has a character we can specify by the three values

\displaystyle\begin{aligned}\chi^\mathrm{def}(e)&=3\\\chi^\mathrm{def}\left((1\,2)\right)&=1\\\chi^\mathrm{def}\left((1\,2\,3)\right)&=0\end{aligned}

We calculate the multiplicities of the two characters we know by taking inner products:

$\displaystyle\begin{array}{cclccclcc}\langle\chi^\mathrm{triv},\chi^\mathrm{def}\rangle&=(&1\cdot1\cdot3&+&3\cdot1\cdot1&+&2\cdot1\cdot0&)/6=&1\\\langle\mathrm{sgn},\chi^\mathrm{def}\rangle&=(&1\cdot1\cdot3&+&3\cdot-1\cdot1&+&2\cdot1\cdot0&)/6=&0\end{array}$

That is, the defining representation contains one copy of the trivial representation and no copies of the signum representation. In fact, we already knew about the copy of the trivial representation, but it’s nice to see it confirmed again. Subtracting it off, we’re left with a residual character:

\displaystyle\begin{aligned}\chi^\perp(e)&=2\\\chi^\perp\left((1\,2)\right)&=0\\\chi^\perp\left((1\,2\,3)\right)&=-1\end{aligned}

Now this character might itself decompose, or it might be irreducible. We can check by calculating its inner product with itself:

$\displaystyle\begin{array}{cclccclcc}\langle\chi^\perp,\chi^\perp\rangle&=(&1\cdot2\cdot2&+&3\cdot0\cdot0&+&2\cdot-1\cdot-1&)/6=&1\end{array}$

which confirms that $\chi^\perp$ is irreducible. Thus we can write down the character table of $S_3$ as

$\displaystyle\begin{array}{c|ccc}&e&(1\,2)&(1\,2\,3)\\\hline\chi^\mathrm{triv}&1&1&1\\\mathrm{sgn}&1&-1&1\\\chi^\perp&2&0&-1\end{array}$

So, why is this just part 1? Well, we’ve calculated another character, but we still haven’t actually shown that there’s any irrep that gives rise to this character. We have a pretty good idea what it should be, but next time we’ll actually show that it exists, and it really does have the character $\chi^\perp$.

October 26, 2010 -

## 5 Comments »

1. […] Last time we wrote down the complete character table of : […]

Pingback by One Complete Character Table (part 2) « The Unapologetic Mathematician | October 27, 2010 | Reply

2. […] of two irreducible representations is not, in general, itself irreducible. Indeed, we can look at the character table for and consider the inner tensor product of two copies of […]

Pingback by Inner Tensor Products « The Unapologetic Mathematician | November 5, 2010 | Reply

3. […] we have the same three representations as in the character table of : the trivial, the signum, and the complement of the signum in the defining representation. […]

Pingback by The Character Table of S4 « The Unapologetic Mathematician | November 8, 2010 | Reply

4. […] can check this in the case of and , since we have complete character tables for both of […]

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

5. […] can use these relations to help fill out character tables. For instance, let’s consider the character table of , starting from the first two […]

Pingback by The Character Table as Change of Basis « The Unapologetic Mathematician | November 22, 2010 | Reply