You're absolutely right about the sign on the formula. I transcribed that wrong bouncing between three different sources for the formula. Thank you. However, I'm not following your point about the signature. Isn't the signature of the metric coded in the inverse metric, $g^{\mu\nu}$? I shouldn't have to change this formula if the signature of my metric changes, should I?

@GluonSoup If you check the wiki article you linked on the perfect fluid, it mentions that if your signature is $(-+++)$ then the formula for $T^{\mu\nu}$ has a plus sign, and if the signature is $(+---)$ then the formula has a plus sign. Is that what you mean? Changing signature cannot be thought of like changing coordinates (indeed it's easy to show that coordinate changes preserve the metric signature). It is a convention that one chooses at the very beginning and then never changes.

@GluonSoup "[...] written with a space-positive signature."

Are you saying that the actual formula changes based on the kind of signature used?

@GluonSoup Yes, just as in SR.

@GluonSoup I'm afraid I don't understand the issue. $U$ is the fluid 4-velocity (field). I'm also not talking about shortcuts; If you use $x^0=ct$, then your $U^0 = c$, your $g^{00}=-1$, and so $T^{00}= \rho c^2$. If you use $x^0 = c$, then $U^0=1$, $g^{00}=-1/c^2$, and $T^{00}=\rho$, which is consistent with the way tensors transform under changes of coordinates.

@GluonSoup Firstly, that should be $d\tau^2 =- c^2 dt^2 + \ldots$, not $d\tau^2 = -c^2 t^2 dt^2$ for the FLRW metric. Secondly, for a diagonal metric like this we have $d\tau^2 = g_{\mu\nu}dx^\mu dx^\nu = g_{00} (dx^0)^2 +g_{11}(dx^1)^2 + g_{22}(dx^2)^2 + g_{33} (dx^3)^2$. If $x^0 = ct$, then $dx^0 = c dt$ and so $g_{00}=-1$. If $x^0 = t$, then $dx^0 = dt$ and $g_{00} = -c^2$. The latter is a mathematically viable choice, but it's usually not made because then $x^0$ has different units from $x^i$ and this makes dimensional analysis of tensor components trickier.

@GluonSoup $x^0=c$ doesn't make any sense - $x^0$ is the time coordinate, but $c$ is a constant. Do you mean $x^0 = t$? If so, then $U^0 = \frac{dx^0}{d\tau} = 1$ in the rest frame of the fluid. $U^0=c$ only if $x^0 = \color{red}{c}t$.

There's an additional issue - the left hand side of your line element should be $c^2 d\tau^2$ if $d\tau$ is an increment of proper time.

Here's the line element with $x^0$=ct.$$d\tau^2=-c^2dt^2+\frac{a(t)^2}{1-k\space r^2}dr^2+a(t)^2r^2d\theta^2+a(t)^2r^2sin(\theta)^2d\phi^2$$The inverse metric is: $$g_{\mu\nu}= \begin{bmatrix} -\frac{1}{c^2}&0&0&0\\ 0&\frac{1-k\space r^2}{a(t)^2}&0&0\\ 0&0&\frac{1}{a(t)^2r^2}&0\\ 0&0&0&\frac{csc(\theta)^2}{a(t)^2r^2} \end{bmatrix}$$In your earlier comment, you say that $g^{00}=−1$ when $x^0=c t$. This is the part I don't follow. What have I got wrong?

Here's a reference slideshare.net/KemalAkin/cosmology-65752111 (Page 12-13) that agrees with the metric above. I'm having trouble following your math.

@GluonSoup $g_{00}$ is whatever is multiplying $(dx^0)^2$ in the line element. If $dx^0 = d(ct) = c dt$, then $g_{00}=-1$. If $dx^0 = dt$, then $g_{00}=-c^2$. The reference you linked does the latter.

OK. We agree up to this point. Then what's the rational for making $U=\{1,0,0,0\}$? If we're working in SI units, then the point that I'm missing in a major way is, why isn't $U=\{c,0,0,0\}$? That's the velocity of something that's not moving in SI units.

@GluonSoup This isn't a matter of using SI or not. The definition of $U^0$ is $\frac{dx^0}{d\tau}$, with $\tau$ the proper time. If you choose coordinates (not units, but coordinates) in which $x^0 = ct$, then $U^0 = c \frac{dt}{d\tau} = c$ in the rest frame of the fluid. If you choose coordinates such that $x^0 = t$, then $U^0 = 1$. Note that if you do the latter, then $U^0$ will have different units from $U^i,i=1,2,3$, which is one reason why most people choose $x^0 = ct$.

"The definition of $U^0$ is $\frac{dx^0}{d\tau^0}$" That's it. That's what I was missing. Many thanks.