I know there exists the property:
$$Y_\ell^m(0,\phi)=\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m0}$$
but can it be applied equally as simply to
$$Y_\ell^{m-1}(0,\phi)\stackrel{?}{=}\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m1}$$
or
$$Y_\ell^{m+1}(0,\phi)\stackrel{?}{=}\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m(-1)}$$
or am I missing something?
Thanks in advance
You can definitely do so. You've only changed $m$ to $m\pm1$, shifting the value of $m$ that yields a zero upper index. The statement that
$$\begin{cases}Y_{\ell}^{0}\left(0,\varphi\right)=\sqrt{\frac{2\ell+1}{4\pi}}\\Y_{\ell}^{\rm non-zero\:number}\left(0,\varphi\right)=0\end{cases}$$
is kept unchanged.