The Unapologetic Mathematician

The Character Table of S4

Let’s use our inner tensor products to fill in the character table of $S_4$. We start by listing out the conjugacy classes along with their sizes:

$\displaystyle\begin{array}{cc}e&1\\(1\,2)&6\\(1\,2)(3\,4)&3\\(1\,2\,3)&8\\(1\,2\,3\,4)&6\end{array}$

Now we have the same three representations as in the character table of $S_3$: the trivial, the signum, and the complement of the signum in the defining representation. Let’s write what we have.

$\displaystyle\begin{array}{c|ccccc}&e&(1\,2)&(1\,2)(3\,4)&(1\,2\,3)&(1\,2\,3\,4)\\\hline\chi^\mathrm{triv}&1&1&1&1&1\\\mathrm{sgn}&1&-1&1&1&-1\\\chi^\perp&3&1&-1&0&-1\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\end{array}$

Just to check, we calculate

$\displaystyle\langle\chi^\perp,\chi^\perp\rangle=(1\cdot3\cdot3+6\cdot1\cdot1+3\cdot-1\cdot-1+8\cdot0\cdot0+6\cdot-1\cdot-1)/24=1$

so again, $\chi^\perp$ is irreducible.

But now we can calculate the inner tensor product of $\mathrm{sgn}$ and $\chi^\perp$. This gives us a new line in the character table:

$\displaystyle\begin{array}{c|ccccc}&e&(1\,2)&(1\,2)(3\,4)&(1\,2\,3)&(1\,2\,3\,4)\\\hline\chi^\mathrm{triv}&1&1&1&1&1\\\mathrm{sgn}&1&-1&1&1&-1\\\chi^\perp&3&1&-1&0&-1\\\mathrm{sgn}\otimes\chi^\perp&3&-1&-1&0&1\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\end{array}$

which we can easily check to be irreducible.

Next, we can form the tensor product $\chi^\perp\otimes\chi^\perp$, which has values

$\displaystyle\begin{aligned}\chi^\perp\otimes\chi^\perp\left(e\right)&=9\\\chi^\perp\otimes\chi^\perp\left((1\,2)\right)&=1\\\chi^\perp\otimes\chi^\perp\left((1\,2)(3\,4)\right)&=1\\\chi^\perp\otimes\chi^\perp\left((1\,2\,3)\right)&=0\\\chi^\perp\otimes\chi^\perp\left((1\,2\,3\,4)\right)&=1\end{aligned}$

Now, this isn’t irreducible, but we can calculate inner products with the existing irreducible characters and decompose it as

$\displaystyle\chi^\perp\otimes\chi^\perp=\chi^\mathrm{triv}+\chi^\perp+\mathrm{sgn}\otimes\chi^\perp+\chi^{(5)}$

where $\chi^{(5)}$ is what’s left after subtracting the other three characters. This gives us one more line in the character table:

$\displaystyle\begin{array}{c|ccccc}&e&(1\,2)&(1\,2)(3\,4)&(1\,2\,3)&(1\,2\,3\,4)\\\hline\chi^\mathrm{triv}&1&1&1&1&1\\\mathrm{sgn}&1&-1&1&1&-1\\\chi^\perp&3&1&-1&0&-1\\\mathrm{sgn}\otimes\chi^\perp&3&-1&-1&0&1\\\chi^{(5)}&2&0&2&-1&0\end{array}$

and we check that

$\displaystyle\langle\chi^{(5)},\chi^{(5)}\rangle=(1\cdot2\cdot2+6\cdot0\cdot0+3\cdot2\cdot2+8\cdot-1\cdot-1+6\cdot0\cdot0)/24=1$

so $\chi^{(5)}$ is irreducible as well.

Now, we haven’t actually exhibited these representations explicitly, but there is no obstacle to carrying out the usual calculations. Matrix representations for $V^\mathrm{triv}$ and $V^\mathrm{sgn}$ are obvious. A matrix representation for $V^\perp$ comes just as in the case of $S_3$ by finding a basis for the defining representation that separates out the copy of $V^\mathrm{triv}$ inside it. Finally, we can calculate the Kronecker product of these matrices with themselves to get a representation corresponding to $\chi^\perp\otimes\chi^\perp$, and then find a basis that allows us to split off copies of $V^\mathrm{triv}$, $V^\perp$, and $V^\mathrm{sgn}\otimes V^\perp$.

4 Comments	»

1. Is there a general procedure for filling out the character table, or is it a matter of trying different combinations until you find a new irreducible character orthogonal to the ones you’ve already got (and iterating until you have a full basis)?

Comment by Joe English | November 9, 2010 | Reply

2. We’ll come up with a general straightforward position for symmetric groups eventually.

Comment by | November 9, 2010 | Reply

3. [...] We can check this in the case of and , since we have complete character tables for both of them: [...]

Pingback by | November 17, 2010 | Reply

4. Small error, it’s the complement of the trivial rep, not the signum….

Comment by Charles Waldman | May 2, 2013 | Reply

« Previous | Next »

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 25, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9164782762527466, "perplexity_flag": "head"}