# A proof in Howard Georgi's "Lie Algebras in Particle Physics"

I am reading the book referenced above and in the first chapter, in the proof of theorem 1.3 (fifth line of the proof), it says:

But because D2 is irreducible, P must project onto the whole space [...]

I don't see why the fact that D2 is irreducible implies that.

I know that probably this is a very stupid question, but I am new to group theory and I feel that I am missing some key point.

$P$ is defined as the projector onto a particular nonzero subspace $W$ of the vector space $V$ on which the representation acts. It is shown in the proof (see eq. 1.40) that $W$ is an invariant subspace of the representation $D_2$. Since, by definition, an irreducible representation has no invariant subspaces except for $V$ and $\{0\}$, we find immediately that $W=V$.

• @WilliamNeill Glad it helped. Commented Sep 24, 2013 at 16:29

The properties of $P$ aren't specified in your question but it's implicitly clear that $P$ is assumed or proved at that place to commute with all elements of the group (or Lie algebra), $$\forall g\in G, \quad Pg = gP$$ If the image of $P$ were not the full space $D_2$ but its proper subspace $V\subset D_2$ (different from $\{0\}$, however), then $V$ would be a representation of the group (or Lie algebra) by itself, thus proving that $D_2$ isn't irreducible. That's a proof by contradiction.

Why would $V$ be a representation of $G$? Because $$\forall v\in V,g\in G:\quad g v = gPv = Pgv \in V$$ because $Pv=v$ and $Pg=gP$. If we prove that any $gv$ for $v\in V$ is an element of $V$ – because it's of the form $P$ acting on something and $P$ on anything is in $V$ – we have proven that $V$ is closed under the action of any element $g\in G$, and is therefore a representation of $G$.

Let's assume there is a non zero vector $$|\mu\rangle \in V$$ such that $$A|\mu\rangle=0$$, meaning $$|\mu\rangle$$ belongs to the $$\ker(A)$$ i.e.

$$A|\mu\rangle=0 \implies |\mu\rangle \in \ker(A)$$

Now, let us imagine that there exists a projection operator onto this subspace i.e. $$P : V \rightarrow \ker(A)$$. By definition if $$|\mu\rangle \in \ker(A)$$, then $$P|\mu\rangle \in \ker(A)$$. $$A|\mu\rangle=0\implies AP|\mu\rangle=0$$

Now, Let's consider the case $$D_{1}(g)AP|\mu\rangle$$

\begin{align} D_{1}(g)AP|\mu\rangle &=D_{1}(g)(0) = 0\\ AD_{2}(g)P|\mu\rangle &=0 \implies D_{2}(g)P|\mu\rangle \in \ker(A) \end{align}

This means if $$P|\mu\rangle\in\ker(A)$$ then so does $$D_{2}(g)P|\mu\rangle\in\ker(A)$$ implies that $$\ker(A)$$ is this invariant subspace under the representation $$D_{2}(g)$$ since an invariant subspace is any $$W \subset V$$ such that for all $$|x\rangle \in W$$ we have $$D(g)|x\rangle\in W$$ for some representation $$D(g)$$. However, it is still not clear if the invariant subspace is a proper invariant subspace or not. In order to clarify that we need to look at the fact that $$D_{2}(g)$$ is irreducible. A representation is called irreducible if it contains no proper invariant subspace. It means $$\ker(A)$$ can't be a proper invariant subspace and the only way for that to happen is if $$\ker(A)=V$$. Thus, the projection operator is a map $$P:V\rightarrow V$$. It projects onto the whole space and the whole space is $$ker(A)$$.