# In what sense is $\sum_{r =1}^3 \epsilon^r_\mu \epsilon^r_\nu$ a projection operator?

It is mentioned in this answer that the completeness relation for the polarization vectors of a (massive) electromagnetic field $$\sum_\lambda \varepsilon^\mu(\lambda,k) \varepsilon^{\nu*}(\lambda,k) = - \eta^{\mu\nu} + \frac{k^\mu k^\nu}{M^2}.$$ can be understood since $$P_\epsilon^{\mu\nu} := \sum_\lambda \epsilon^\mu_\lambda(k)\epsilon^{*\nu}_\lambda(k)$$ is a projection operator. How can this be understood? In particular, on whom does $$P_\epsilon^{\mu\nu}$$ act in order to project out what?

Applying the operator $$P^{\mu\nu}$$ to a generic vector $$v_\nu$$,
$$-P^{\mu\nu} v_\nu = v^\mu - \frac{k^\nu v_\nu}{M^2} k_{\mu} .$$
That is, up to a sign, $$P$$ "removes" (projects out) from $$v$$ its component parallel to $$k$$ (assuming $$k_\mu k^\mu = M^2$$). Therefore, it projects any vector to the surface orthogonal to $$k$$.
• Yes, the polarisation vectors are indeed a basis of the subspace orthogonal to $k$. – fqq Nov 4 at 17:12
Hint: try to act with $$P_\epsilon^{\mu\nu}$$ on $$k_\nu$$.