Let $x,y \in \mathbb{R}$ where $y=x-t$. Translation-invariant (or shift-invariant) kernel $\kappa(\cdot,\cdot)$ is defined as $\kappa(x,y) = \kappa(x,x-t) = \kappa(t)$.
Can I say that the function $\kappa$ is symmetric ?
I think "yes", if I can define the translation-invariance as "DIFFERENCE between $x$ and $y$" so that $y=x+t$. And $\kappa(x,y)=\kappa(x,x+t)=\kappa(-t)$.
But, wanted make sure the relation between translation-invariance and symmetry (i.e., $\kappa(x,y)=\kappa(y,x)$).
Can I simply say that translation invariant kernel is $\kappa(x,y) = \kappa(x-y) = \kappa(y-x)$ ? translation-invariant $\to$ symmetric. Is this always correct??? Or, is the symmetry required for the second equality?