How Is the Unit Operator Derived in Quantum Many-Body Hilbert Spaces?

[welshtill]

I am reading J.W.Negele and H.Orland's book "Quantum Many-Particle Systems". I don't know how one can derive equation (1.40) on page 6. The question is
For quantum many-body physics, suppose there are N particles. The hilbert space is

H_{N}=H\otimesH\otimes...H.

Its basis can be written as

\left|\alpha_{1}\alpha_{2}...\alpha_{N}\right)=\left|\alpha_{1}\right\rangle\otimes\left|\alpha_{2}\right\rangle\...\otimes\left|\alpha_{N}\right\rangle

with closure relation

\sum_{\alpha_{1}\alpha_{2}...\alpha_{N}}\left|\alpha_{1}\alpha_{2}...\alpha_{N}\right)\left(\alpha_{1}\alpha_{2}...\alpha_{N}\right|=1

Now introduce a symmetrization and antisymmetrization operator

P_{B,F}\psi(r_{1},r_{2}...r_{N})=\frac{1}{N!}\sum_{P}\varsigma^{P}\psi(r_{P1},r_{P2}...r_{PN})

where \varsigma=1 for bosons and -1 for fermions. \sum_{P} is sum of all permutations of coordinates.

Using this operator P_{B,F} one can obtain the sub-hilbert space for fermions H_{F} or for bosons H_{B}.

The basis in these two sub-space can be written as

P_{B,F}\left|\alpha_{1}\alpha_{2}...\alpha_{N}\right)

My question is how one can derive the following equation

\sum_{\alpha_{1}\alpha_{2}...\alpha_{N}}P_{B,F}\left|\alpha_{1}\alpha_{2}...\alpha_{N}\right)\left(\alpha_{1}\alpha_{2}...\alpha_{N}\right|P_{B,F}=1

 

[welshtill]

I now understand this relation, 1 is a unit operator in sub-hilbert space Hf or Hb.