Hi my question is about this orthogonal projection tensor $P^\sigma_\nu\equiv\delta^\sigma_\nu+U^\sigma U_\nu$.
It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.
$ P^\sigma_\nu V^\nu_\parallel ~~= ~~0 $
$ P^\sigma_\nu W^\nu_\perp ~~=~~ W^\sigma_\perp $
However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.
$ P^\sigma_\nu V^\nu_\parallel ~~=~~\delta^\sigma_\nu V^\nu_\parallel+U^\sigma U_\nu V^\nu_\parallel $
If $\sigma=\nu$ then $\delta^\sigma_\nu=1$.
$ P^\sigma_\nu V^\nu_\parallel~~=~~ V^\sigma_\parallel +U^\sigma U_\sigma V^\sigma_\parallel \qquad \qquad [U^\sigma U_\sigma=-1]\\ \quad\qquad=~~ V^\sigma_\parallel - V^\sigma_\parallel \\ \quad\qquad=~~0 \quad\checkmark $
That looks good, now let's check when $\sigma\neq\nu$ and $\delta^\sigma_\nu=0$.
$ P^\sigma_\nu V^\nu_\parallel~~= ~~0 +U^\sigma U_\nu V^\nu_\parallel \qquad \qquad~~~ [V\parallel U\implies U_\nu V^\nu\neq0]\\ \quad\qquad~\neq~~ 0 \qquad \qquad \qquad \qquad~~~~~~ [\mathrm{contradicts~Eq.}(2)] $
It looks like I should totally disregard the case where $\sigma\neq\nu$ but I don't see why.
Talking about $\sigma=\nu$ and $\neq \nu$ are ill defined. $\nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^\nu U_\nu$, because $\delta^\sigma_\nu V^\nu = V^\sigma$ always.