Is this 4-vector identity $k^\mu k^\nu = \frac{1}{4}k^2\delta^{\mu\nu}$ correct?

Question

I've often seen the identity

\begin{equation} k^\mu k^\nu = \frac{1}{4}k^2\delta^{\mu\nu} \tag{1} \end{equation}

used in various derivations (for example, see Schwartz QFT pg. 625). This result doesn't make any sense to me. Let's consider the simple four-vector $$ k^\mu = (1,0,0,0) $$ Here, $k^2 = 1$ (using mostly-minus metric). Therefore, \begin{equation} k^0k^0 = 1 \neq \frac{1}{4} \quad \text{or} \quad k^1k^1 = 0 \neq \frac{1}{4} . \end{equation} What am I missing?

Answer 1

The rule (1) contains an implicitly written angular $k$-integration:

$$ \int_{\mathbb{S}^{d-1}}\! d^{d-1}\Omega ~ k^{\mu}k^{\nu} ~=~\int_{\mathbb{S}^{d-1}}\! d^{d-1}\Omega ~ k^2 \delta^{\mu\nu}/d.\tag{1'}$$ (Schwartz is working with $d=4$ dimensions and Euclidean signature.) The rule (1) is of course not true without the angular integration/averaging in $k$-space.