# The Unapologetic Mathematician

## The Signum Representation

I’ve just noticed that there’s a very important representation I haven’t mentioned. Well, actually, I mentioned it in passing while talking about Rubik’s group, but not very explicitly. And it’s a very important one.

Way back when I defined permutation groups I talked about a permutation being even or odd. Remember that we showed that permutation can be written out as a composite of transpositions which swap two objects. In general this can be done in more than one way, but if it takes an even number of swaps to write a permutation in one way, then it will take an even number of swaps in any other way, and similarly for permutations requiring an odd number of swaps. In this way we separate out permutations into the “even” and “odd” collections.

The composite of two even permutations or two odd permutations is even, while the composite of an even and an odd permutation is odd. This is just like the multiplication table of the group $\mathbb{Z}_2$, with “even” for the group’s identity $e$ and “odd” for the other group element $o$. That is, we have a homomorphism $\mathrm{sgn}:S_n\rightarrow\mathbb{Z}_2$ for every permutation group $S_n$.

Now to make this into a representation we’re going to use a one-dimensional representation of $\mathbb{Z}_2$. We have to send the group identity $e$ to the field element $1$, but we have a choice to make for the image of $o$. We need to send it to some field element $x$, and this element must satisfy $x^2=1$ for this to be a representation. We could choose $x=1$, but this just sends everything to the identity, which is the trivial group representation. There may be other choices around, but the only one we know always exist is $x=-1$ (note we’re tacitly assuming that $1\neq-1$.

So we define the one-dimensional signum representation of $S_n$ by sending all the even permutations to the $1\times1$ matrix whose single entry is $1$, and sending all the odd permutations to the $1\times1$ matrix whose single entry is $-1$. Often we’ll just ignore the “matrix” fact in here, and just say that the signum of an even permutation is $1$ and the signum of an odd permutation is $-1$. But secretly we’re always taking this and multiplying it by something else, so we’re always using it as a linear transformation anyway.

December 18, 2008 -

## 9 Comments »

1. “Signum” representation (or the “signum” of a permutation) seems to be comparatively rare; most people just say “sign”. Is “signum” what you customarily say?

Comment by Todd Trimble | December 18, 2008 | Reply

2. Yes, mostly because when I say it out loud “sign” sounds the same as “sine”.

Comment by John Armstrong | December 18, 2008 | Reply

3. In a field $$x^2=1$$ has only two solutions, 1 and -1.

Comment by Omar | December 19, 2008 | Reply

4. Omar, I’m doing representations over fields now. I’m trying to leave my treatment open to easy modification in the future. But yes, you’re right.

Comment by John Armstrong | December 19, 2008 | Reply

5. […] on the other hand, on the other hand, will give us vectors on which the symmetric group acts by the signum representation. We use the group algebra […]

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6. […] throughout we’ve used the fact that is a representation, and so the signum of the product of two group elements is the product of their signa. We also make […]

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7. […] one we’ve seen is the signum representation of a permutation group, which sends even permutations to and odd permutations to […]

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8. […] It’s a straightforward exercise to show that is a one-dimensional submodule carrying a copy of the signum representation. […]

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9. […] know of two irreps offhand — the trivial representation and the signum representation — and so we’ll start with those and leave the table incomplete below […]

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