I am reading Zee's Group theory in a nutshell for physicists and came across the following theorem (Page 96):

Unitary representations

The all-important unitarity theorem states that finite groups have unitary representations, that is to say, $D^\dagger(g)D(g)=I$ for all $g$ and for all representations.

In practice, this theorem is a big help in finding representations of finite groups. As a start, we can eliminate some proposed representations by merely checking if the listed matrices are unitary or not.

According to me, the theorem states that all representations of a finite group are unitary. But I am having a hard time digesting this fact. The only take-away that I could draw from the proof is that given a finite group, we can always find a unitary representation for it. Can someone please explain me how the statement and the proof of the theorem are consistent with each other?

Also, a quick google search for "unitarity theorem" does not gives me anything relevant. Am I missing something? Or does this theorem have a more famous name?

I am just starting with group theory and have no background on abstract algebra so you may have to dumb it down a little for me.

  • 1
    $\begingroup$ As the two answer basically say: it is possible to endow the vector space carrying the representation by an inner product which makes the representation unitary. $\endgroup$ Jan 28, 2020 at 16:45

2 Answers 2

  • Every representation $(D,V)$ of a finite group $G$ is equivalent to a unitary representation.

  • It is often termed as Weyl's unitary trick. This works by simply redefining your inner product by averaging over on the space $V$. This smoothening trick works precisely because of finite number of elements and invariance of sum of finite elements (i.e. $\sum_{g \in G} D(gh) = \sum_{gh \in G} D(gh) = \sum_{g \in G} D(g)$).

$$\langle v| w \rangle = \frac{1}{|G|}\sum_{g \in G} \langle D(g)v| D(g)w \rangle$$

You can verify how $\langle v| w \rangle = \langle D(h)v| D(h)w \rangle$ for some $h \in G$. In fact, this claim is similar to that of existence of a common basis change matrix $S$ such that $D'(g) = SD(g)S^{-1}$ is a unitary representation.


They are all equivalent to unitary representations. It's not quite the same thing. Merely checking unitarity might not do the trick.

For instance \begin{align} \Gamma(e)&=\left(\begin{array}{cc} 1&0 \\ 0 & 1\end{array}\right)\, ,\qquad \Gamma(P_{12})=\left(\begin{array}{cc} 1&-1 \\ 0 & -1\end{array}\right)\, ,\qquad \Gamma(P_{13})=\left(\begin{array}{cc} 0&1 \\ 1 & 0\end{array}\right)\\ \Gamma(P_{23})&=\left(\begin{array}{cc} -1&0 \\ -1 & 1\end{array}\right)\, ,\qquad \Gamma(P_{123})=\left(\begin{array}{cc} 0&-1 \\ 1 & -1\end{array}\right)\, ,\qquad \Gamma(P_{132})=\left(\begin{array}{cc}-1&1 \\ -1 & 0\end{array}\right) \end{align} is an irreducible representation of $S_3$ but clearly not unitary. It can be conjugated to a unitary rep, i.e. there is a common $V$ so that $V\Gamma V^{-1}$ is unitary.

  • $\begingroup$ By common $V$, you mean the same $V$ for all group elements, right? $\endgroup$ Jan 27, 2020 at 16:42
  • $\begingroup$ yes exactly, the same transformation brings all $\Gamma$’s to their respective unitary forms. $\endgroup$ Jan 27, 2020 at 16:56

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