The Unapologetic Mathematician

Mathematics for the interested outsider

An Alternative Path

It turns out that our efforts last time were somewhat unnecessary, although they were instructive. Actually, we already had a matrix representation in our hands that would have done the trick.

The secret is to look at the block diagonal form from when we defined reducibility:

\displaystyle\left(\begin{array}{c|c}X(g)&Y(g)\\\hline{0}&Z(g)\end{array}\right)

We worked this out for the product of two group elements, finding

\displaystyle\left(\begin{array}{c|c}X(gh)&Y(gh)\\\hline{0}&Z(gh)\end{array}\right)=\left(\begin{array}{c|c}X(g)X(h)&X(g)Y(h)+Y(g)Z(h)\\\hline{0}&Z(g)Z(h)\end{array}\right)

We focused before on the upper-left corner to see that X was a subrepresentation, but we see that Z(gh)=Z(g)Z(h) as well. The thing is, if the overall representation acts on V and X acts on the submodule W, then Z acts on the quotient space V/W. That is, it’s generally not a submodule. However, it so happens that over a finite group we have V\cong W\oplus V/W as modules. That is, if we have a matrix representation that looks like the one above, then we can find a different basis that makes it look like

\displaystyle\left(\begin{array}{c|c}X(g)&0\\\hline{0}&Z(g)\end{array}\right)

This is more than Maschke’s theorem tells us — not only do we have a decomposition, but we have one that uses the exact same matrix representation in the lower right as the original one. Proving this will reprove Maschke’s theorem, and in a way that works over any field!

So, let’s look for a change-of-basis matrix that’s partitioned the same way:

\displaystyle\left(\begin{array}{c|c}I&D\\\hline{0}&I\end{array}\right)\left(\begin{array}{c|c}X(g)&Y(g)\\\hline{0}&Z(g)\end{array}\right)=\left(\begin{array}{c|c}X(g)&0\\\hline{0}&Z(g)\end{array}\right)\left(\begin{array}{c|c}I&D\\\hline{0}&I\end{array}\right)

Multiplying this out, we find

\displaystyle\left(\begin{array}{c|c}X(g)&Y(g)+DZ(g)\\\hline{0}&Z(g)\end{array}\right)=\left(\begin{array}{c|c}X(g)&X(g)D\\\hline{0}&Z(g)\end{array}\right)

which gives us the equation Y(g)+DZ(g)=X(g)D, which we rearrange to give

\displaystyle X(g)DZ\left(g^{-1}\right)=D-X(g)Y\left(g^{-1}\right)

as well as

\displaystyle X(g)DZ\left(g^{-1}\right)=D+Y(g)Z\left(g^{-1}\right)

That is, acting on the left of D by X(g) and on the right by Z\left(g^{-1}\right) doesn’t leave D unchanged, but instead adds a certain offset. We’re not looking for an invariant of these actions, but something close. Incidentally, why are these two offsets the same? Well, if we put them together we find

\displaystyle X(g)Y\left(g^{-1}\right)+Y(g)Z\left(g^{-1}\right)=Y\left(gg^{-1}\right)=Y(e)

which must clearly be zero, as desired.

Anyway, I say that things will work out if we choose

\displaystyle D=\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(g^{-1}\right)Y(g)

Indeed, we calculate

\displaystyle\begin{aligned}X(g)DZ\left(g^{-1}\right)&=X(g)\left(\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(h^{-1}\right)Y(h)\right)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(gh^{-1}\right)Y(h)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)Y(fg)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)\left(X(f)Y(g)+Y(f)Z(g)\right)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}\left(Y(g)Z\left(g^{-1}\right)+X\left(f^{-1}\right)Y(f)\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}Y(g)Z\left(g^{-1}\right)+\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)Y(f)\\&=Y(g)Z\left(g^{-1}\right)+D\end{aligned}

Just as we wanted.

Notice that just like our original proof of Maschke’s theorem, this depends on a sum that is only finite if G is a finite group.

October 28, 2010 Posted by | Algebra, Group theory, Representation Theory | 2 Comments

   

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