Why does the graviton polarization satisfy $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$? I am reading the paper ``Graviton Mode Function in Inflationary Cosmology'' by Ng (link here). The graviton $h_{ij}$ is here expanded (in the TT gauge) where
$$
h_{ij}(x) \sim \epsilon_{ij}(\mathbf{k},\lambda) h_{\mathbf{k}}(\lambda,x)
$$
and in equation 9 it is said that
$$
\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'} \ .
$$
Where does this come from? And how is the relation adjusted for differing momenta $\mathbf{k} \neq \mathbf{k}'$, ie. can one write down a relation for $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k}',\lambda) = \ldots$?
EDIT: Why does one need this condition? Is it so the Lagrangian is properly normalized when written in terms of $h_{\mathbf{k}}$?
 A: It is a definition, there is nothing deep going on.
We do this very often. If you have some object $a$, you can choose to expand it in some basis $u$ such that $a=\sum_\alpha a_\alpha u_\alpha$. It is often convenient to choose an orthonormal basis, such that $u_\alpha\cdot u_\beta=\delta_{\alpha\beta}$. Of course we do not have to do this, but we can, and it is useful to make such a choice when we are deciding which basis to use.
Here we are expanding $h$ in a basis of tensors $\epsilon_{ij}$, and it is useful to choose the tensors $\epsilon_{ij}$ to be orthonormal. It simplifies things. But you are not forced to do this, you could use other choices for $\epsilon_{ij}$, in which case $\epsilon_{\lambda}\cdot\epsilon_{\lambda'}$ would no longer equal $\delta_{\lambda\lambda'}$ (the factor of $2$ is a convention, you could reabsorb it away into either $\epsilon$ or $h$).
Note that this is true only at $\boldsymbol k=\boldsymbol k'$. In general there is no simple expression for $\epsilon(\boldsymbol k)\cdot\epsilon(\boldsymbol k')$. The reason is that you can choose the basis to be orthonormal for each $\boldsymbol k$, but you don't have enough degrees of freedom to ensure that this orthonormality persists at $\boldsymbol k\neq\boldsymbol k'$. In other words, you make a choice at $\boldsymbol k$, and another choice at $\boldsymbol k'$, but these choices are not in general compatible with each other -- each ensures orthonormality at their respective momenta, but not at other momenta.
There is another reason that may be slightly more convincing. It is often useful to choose the basis $\epsilon$ to be eigenvectors of the "angular momentum operator", i.e., to carry definite spin. Eigenvectors with different eigenvalues are automatically orthogonal so, with this choice, the expression $\epsilon_{\lambda}\cdot\epsilon_{\lambda'}$ has to be proportional to $\delta_{\lambda\lambda'}$. Again, this is a choice, you are not forced to use a basis of definite spin, but it is convenient.
Of course, you are free to choose whatever basis you want, but the coefficients will depend on the choice. If you use a different basis, the coefficients $h_\lambda$ will typically carry a different interpretation, so many of the conclusions in the notes you are following are in principle only true for the choice the author makes. A different choice is fine, but you will have to redo everything, as many of the properties of $h_\lambda$ really do hinge on the basis having fixed spin, etc.
A: I can't comment with certainty on $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$, but I suspect that it's because gravitons actually have only two independent polarizations, so this is a way to go from polarization labels ($\lambda$) to real space ones ($i,j$).
I can, however, say for certain that the polarizations at different $\mathbf{k}$ are unrelated, by necessity. To the extent that they are related is a question of the net polarization you'll observe over a certain bandwidth. I want to say it's related to coherence, but that's more related to the phase at different wavelengths.
A: The 2 is conventional, but it makes sense when you think of the simplest form these tensors take when $\hat{k} \propto \hat{z}$:
$$\epsilon_{ij}^+ =\begin{pmatrix}
0 & 0 & 0 & 0  \\
0 & 1 & 0 & 0  \\
0 & 0 & -1 & 0  \\
0 & 0 & 0 & 0  \\
\end{pmatrix}
$$
and
$$\epsilon_{ij}^\times =\begin{pmatrix}
0 & 0 & 0 & 0  \\
0 & 0 & 1 & 0  \\
0 & 1 & 0 & 0  \\
0 & 0 & 0 & 0  \\
\end{pmatrix}
$$
which both satisfy $\epsilon_{ij} \epsilon_{ij} = 2$; if you wanted a 1 there you'd have to rescale both of them by $ 1 / \sqrt{2}$.
As the other answers have stated, this has no physical consequence: rescaling the basis tensors by some factor will just mean the components will be rescaled by the inverse of that factor.
Still, having plain 1s in the tensor looks nice, which is why this convention is chosen.
