Proving the tensor $G_{ij}=(\hat k_i\hat k_j-\frac{1}{3}\delta_{ij})G^L$ is longitudinal

Let there be a $$3$$D matrix whose components are defined as $$G_{ij}=(\hat k_i\hat k_j-\frac{1}{3}\delta_{ij})G^L$$. Here, $$\hat k_i$$ are the eucledian unit vectors, satisfying $$\delta_{ij}\hat k^i\hat k^j=1$$. Then one has to prove $$\epsilon_{ij}^{\quad k}G_{kl}^{\quad,\text{ }jl}=0$$. I tried to prove this in the following way, $$\partial^lG_{kl}=(\hat k_k(\hat k_l\partial^l)-\frac{1}{3}\partial_k)G^L=(\hat k_k(\vec\nabla)-\frac{1}{3}\partial_k)G^L.$$ Now, the derivative along $$j$$ gives, $$\partial^j\partial^lG_{kl}=(\hat k_k\partial^j(\vec\nabla)-\frac{1}{3}\partial^j\partial_k)G^L$$. Now I contracted with the anti-symmetric three tensor, $$\epsilon_{ij}^{\quad k}\partial^j\partial^lG_{kl}=\epsilon_{ij}^{\quad k}(\hat k_k\partial^j(\vec\nabla)-\frac{1}{3}\partial^j\partial_k)G^L.$$ The second term in the parenthesis is symmetric in the exchange of $$j$$ and $$k$$ and thus when contracted with a totally anti-symmetric tensor gives exactly zero. So I am left with, $$\epsilon_{ij}^{\quad k}\partial^j\partial^lG_{kl}=\epsilon_{ij}^{\quad k}(\hat k_k\partial^j)(\vec\nabla)G^L.$$ This term should also be zero but I am not able to argue that. Please provide a valid argument why would the remaining term be zero or suggest if there are any other methods to prove this. The above problem can be found in Scott Dodelson Chapter 5, Exercise Problem 4.

• Down vote?! Is it not a legit question? – fogof mylife Sep 26 '18 at 19:47

To solve the problem one needs to consider the Fourier transformed variables. $$G^L(x,t)=\int\frac{d^3\vec{k}}{(2\pi)^3}e^{i\vec{k}\cdot\vec{x}}\tilde{G}^L(k,t).$$ Now, $$\epsilon_{ij}^{\quad k}(\hat k_k\partial^j)(\vec\nabla)G^L=\epsilon_{ij}^{\quad k}(\hat k_k\partial^j)(\vec\nabla)\int\frac{d^3\vec{k}}{(2\pi)^3}e^{i\vec{k}\cdot\vec{x}}\tilde{G}^L(k,t)=\epsilon_{ij}^{\quad k}\hat k_k\int\frac{d^3\vec{k}}{(2\pi)^3}k^j\vec{k}e^{i\vec{k}\cdot\vec{x}}\tilde{G}^L(k,t)\\=\epsilon_{ij}^{\quad k}\hat k_k\int\frac{d^3\vec{k}}{(2\pi)^3}(\vec{k}\cdot \hat{k}^j)\vec{k}e^{i\vec{k}\cdot\vec{x}}\tilde{G}^L(k,t)=\epsilon_{ij}^{\quad k}\hat k_k\hat{k}^j\int\frac{d^3\vec{k}}{(2\pi)^3}k^2e^{i\vec{k}\cdot\vec{x}}\tilde{G}^L(k,t)=0.$$ As $$\hat k_k\hat{k}^j$$ is symmetric under the exchange of $$j\leftrightarrow k$$. The whole epression turns out to be zero due to the totally antisymmetric property of Levi-Civita tensor.