Let $\rho: G \rightarrow GL(V)$ be a finite dimensional complex representation of the group $G$. Show that there is an inner product on $V$ such that $G$ acts by unitary matrices.

My approach so far was to define the sought after inner product as follows:

$\sum\nolimits_{g \in G} < \rho(g) u, \rho(g) v>$ where $u,v \in V$, and $<,>$ is the standard scalar product on $V$. I need to show that this equals $\sum\nolimits_{g \in G} <u,v> = |G|<u,v>$ where $|G|$ is the cardinality of $G$. This is where I'm stuck and I don't know how to show that the left sight equals the right side. Anyone got a hint for me?


  • $\begingroup$ You cannot show $\langle u,v\rangle_G:=\sum_{g\in G}\langle\rho(g)u,\rho(g)v\rangle$ is equal to $\sum_{g\in G}\langle u,v\rangle=|G|\langle u,v\rangle$, because in general this is false (unless I misunderstand what you mean by "standard scalar product on $V$"). However, it is true that $\sum_{g\in G}\langle\rho(g)u,\rho(g)v\rangle_G=|G|\langle u,v\rangle_G$, where the $\langle\cdot,\cdot\rangle_G$'s are already averaged inner products, and this would indeed lead to $G$ acting unitarily. However it is better to just check $\langle\rho(h)u,\rho(h)v\rangle_G=\langle u,v\rangle_G$ directly. $\endgroup$ – anon Mar 20 '13 at 18:11
  • $\begingroup$ You mean $<\rho(g)u,\rho(g)v>_G = |G|<u,v>_G$ right? Or should there be an additional sum? $\endgroup$ – Howdy Ho Mar 20 '13 at 19:57
  • $\begingroup$ No, that doesn't make any sense. If you plug $g=e$ into that equation you get $\langle u,v\rangle_G=|G|\langle u,v\rangle_G$ which, unless $|G|$ is $1$ modulo the characteristic, implies $\langle u,v\rangle_G=0$ for all $u,v$. That's hardly an inner product. $\endgroup$ – anon Mar 20 '13 at 19:59
  • $\begingroup$ Oh.. yeah of course. So I calculate the following: $\sum\nolimits_{g \in G} <\rho(g)u,\rho(g)v>_G = \sum\nolimits_{g \in G}\sum\nolimits_{g \in G}<\rho(g)\rho(g)u,\rho(g)\rho(g)v> = \sum\nolimits_{g \in G}\sum\nolimits_{g \in G} <\rho(g^2)u,\rho(g^2)v>$ Does that last step lead to anything? I feel like I'm close but I can't seem to close the deal... sorry for bothering you but can you help me one last time? EDIT: OH! Since $G$ is a group $g^2$ is an element of $G$ again, so I'm still just summing over the whole group, which gives $|G|<u,v>_G$, correct ? $\endgroup$ – Howdy Ho Mar 20 '13 at 20:14
  • $\begingroup$ You can't sum over two different indices that have the same letter. It's like writing $\sum_i\sum_j ij$ as $\sum_i\sum_i ii$. More importantly: Why are you computing $\sum_{g\in G}\langle\rho(g)u,\rho(g)\rangle_G$ at all? Let's keep in our sights the goal we're after: to show that $G$'s action is unitary with respect to $\langle\cdot,\cdot\rangle_G$. This means that $\langle\rho(h)u,\rho(h)v\rangle_G=\langle u,v\rangle_G$ for all $h\in G$. This is what you're trying to show. (And BTW, no, as $g$ ranges over $G$, $g^2$ does not necessarily range over all of $G$; only the squares in $G$.) $\endgroup$ – anon Mar 20 '13 at 20:18

We need to define a $G$-invariant inner product by using an initial inner product. So pick any basis for the space and let $\langle\cdot,\cdot\rangle$ be the inner product it induces. This is pretty arbitrary. Now define

$$\langle u,v\rangle_G:=\frac{1}{|G|}\sum_{g\in G}\langle\rho(g)u,\rho(g)v\rangle.$$

It is standard to put the normalization factor $\frac{1}{|G|}$ in front, though ultimately it isn't necessary. The reason is that if $\langle\cdot,\cdot\rangle$ was $G$-invariant to begin with then $\langle\cdot,\cdot\rangle=\langle\cdot,\cdot\rangle_G$, which is aesthetically pleasing. The task is then to show that $G$ acts by unitary transformations with respect to $\langle\cdot,\cdot\rangle_G$, i.e.

$$\langle\rho(h)u,\rho(h)v\rangle_G=\langle u,v\rangle_G.$$

You need to do this by expanding the inner products (using its definition above), then employing the inherent symmetry in the situation (the summands are the exact same, only permuted).


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