How to understand this formula in quantum field theory? In Condensed Matter Field Theory (page 288) by Altland and Simons, there is one mathematical formula:
\begin{equation}
\sum_\mathbf{p} \left(\mathbf{p}\boldsymbol{\cdot}\mathbf{v}\right)\left(\mathbf{p}\boldsymbol{\cdot}\mathbf{v}'\right)F\left(\mathbf{p}^2\right)\boldsymbol{=}\dfrac{\mathbf{v}\boldsymbol{\cdot}\mathbf{v}'}{d}\sum_\mathbf{p}\:\mathbf{p}^2F\left(\mathbf{p}^2\right)
\tag{01}\label{01}
\end{equation}
Here, $\mathbf{p}$ is the momentum, and $\mathbf{v}$ and $\mathbf{v}'$ are vectors. $F\left(\mathbf{p}^2\right)$ is a function of momentum which is invariant under rotational transformations; $d$ is the dimension considered (e.g., $d=3$ in $x,y,z$ coordinate system).
My question is how to arrive at this formula?
Mathematically, this formula seems not correct. For example, imagine a summation over only $\mathbf{p}\boldsymbol{=}\mathbf{p}_0$ and imagine $\mathbf{v}$ and $\mathbf{v}'$ being orthogonal: then the left hand side is $\left(\mathbf{p}_0\boldsymbol{\cdot}\mathbf{v}\right)\left(\mathbf{p}_0\boldsymbol{\cdot}\mathbf{v}'\right)F\left(\mathbf{p}_0^2\right)$ which can be non zero, while the right hand side is zero because $\left(\mathbf{v}\boldsymbol{\cdot}\mathbf{v}'\right)\boldsymbol{=}0$. I am wondering if there are some arguments in physics to "prove" this formula. For example, if the summation is replaced by an integration over the whole reciprocal space of momentum p with d dimension, will the formula stand?
 A: The formula holds, because there is a sum over all ${\bf p}$.  After that sum is performed, ${\bf v}\cdot{\bf v}'$ is essentially the only vector structure that is consistent with rotation symmetry.
To prove the formula, first rewrite it so that the ${\bf v}$ and ${\bf v}$' are outside the sum:
$$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{2})=
\sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\sum_{{\bf p}}p_{j}p_{k}F(p^{2}).$$
Now look just at the final sum, $\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$. Consider holding the index $j$ fixed (say $j=1$) and varying $k$. If $k\neq j$ (say $j=2$) then for every term
in the sum with ${\bf p}=(p_{j},p_{k},p_{3})$, there is another term in the sum with ${\bf p}'=(p_{j},-p_{k},p_{3})$. Since the sum terms $p_{j}p_{k}F(p^{2})$ and $p_{j}'p_{k}'F(p'^{2})$ are equal in magnitude but opposite in sign, they will cancel; and, if fact, every term in the sum will be canceled by a similar one with an inverted $p_{k}\rightarrow p_{k}'=-p_{k}$. Thus, the sum is $\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=0$ if $j\neq k$.
The expression $\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$ transforms like a two-index tensor.  (Any product of the components of two vectors ${\bf u}$ and ${\bf w}$ along with a scalar $\phi$—that is, $u_{j}w_{k}\phi$—transforms this way. The sum of interest is then just a sum of tensors that transform this way and so transforms this way itself.) The only two-tensor that has components that are nonzero only for $j=k$ is one that is proportional to the identity, with components $\delta_{jk}$.  [This is essentially just a long-winded way of saying that, since there is no preferred direction, $\sum_{{\bf p}}p_{1}p_{1}F(p^{2})
=\sum_{{\bf p}}p_{2}p_{2}F(p^{2})=\sum_{{\bf p}}p_{3}p_{3}F(p^{2})$.]  So we must have
$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$ be a scalar function times $\delta_{jk}$,
$$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\delta_{jk}G,$$
which leaves us to find $G$.
To get the scalar $G$, we contract the tensor with another $\delta_{jk}$, or
$$\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\delta_{jk}G,$$
which reduces to
$$\sum_{{\bf p}}p^{2}F(p^{2})=dG,$$
since $\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\delta_{jk}$ is just the dimensionality of space $d$.
Then it is just a matter of inserting this expression into the earlier ones and working backwards. Solving for $G$, we have
$$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\delta_{jk}G=\delta_{jk}\frac{1}{d}\sum_{{\bf p}}p^{2}F(p^{2}).$$  Then inserting this into the original expression with $v_{j}$ and $v_{k}'$ gives
$$\sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=
\sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\delta_{jk}\frac{1}{d}\sum_{{\bf p}}p^{2}F(p^{2}),$$
and collecting this in component-free form gives the final expression
$$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{2})=\frac{1}{d}{\bf v}\cdot{\bf v}'\sum_{{\bf p}}p^{2}F(p^{2}).$$
