Weyl Tensor undefined or vanishing? So we have that the Weyl tensor in component form satisfies in dimension $n$
\begin{equation}
C_{abcd}=R_{abcd} -\frac{2}{n-2}\left(R_{a[c}g_{b]d}+R_{b[d}g_{c]a} \right)+\frac{2}{(n-1)(n-2)}Rg_{a[c}g_{b]d}
\end{equation}
The confusion is simple really. Many texts say that when $n=2$, the Weyl Tensor vanishes; but I stumbled across a text today (Gravitation: Foundations and Frontiers, By T. Padmanabhan - a great read to add) which says that it the Weyl tensor is in fact undefined in the first place for $n=2$!
After having used the fact that in two dimensions, the Riemann tensor can be expressed for $n=2$ as 
\begin{equation}
R_{abcd}=Rg_{a[c}g_{d]b}=\frac{1}{2}R(g_{ac}g_{bd}-g_{ad}g_{bc})
\end{equation}
Without substituting in the value $n=2$, The above expression for the Weyl tensor seems to reduce to (using the Riemann Tensor defined above)
\begin{equation}
C_{abcd}=\frac{1}{2}R \left( \frac{n+1}{n-1} \right)(g_{ad}g_{bc}-g_{ac} g_{bd})
\end{equation}
which does not seem to in general vanish for $n=2$. So is it just the case that, looking at the above formula, the $n=2$ case is ill defined (as mentioned in the book) for some specific reason, or that indeed the Weyl tensor does in fact always vanish?
Thanks.
 A: Your formula for the Weyl tensor is wrong. I haven't double-checked my computation, but I think the first term in parentheses should be $R_{a[c}g_{d]b}$, not $R_{a[c}g_{b]d}$, and the last term should be a multiple of $Rg_{a[c}g_{d]b}$.
In any case, there has to be something wrong with your formula, because the curvature tensor satisfies $R_{abcd}=\frac{1}{2}R(g_{ac}g_{bd}-g_{ad}g_{bc})$ for any metric with constant sectional curvature in any dimension, and for all such metrics, the Weyl tensor is identically zero.
The question of whether Weyl is zero or undefined in dimension $2$ depends on how you define it. If you define it by (the corrected version of) the formula you wrote down, then it clearly is undefined in dimension $2$ because of the $(n-2)$'s in the denominator. But there's another way to define it that does make sense in dimension $2$: Let $V$ be a finite-dimensional inner product space, and let $\mathscr R(V^*)$ denote the space of all algebraic curvature tensors on $V$ (covariant $4$-tensors that have the symmetries of the curvature tensor). Define $\operatorname{trace}\colon \mathscr R(V^*)\to \operatorname{Sym}^2(V^*)$ by
$$
(\operatorname{trace}R)_{ac} = g^{bd} R_{abcd}.
$$
Then you can define the Weyl tensor as the orthogonal projection of the Riemann tensor onto the kernel of the trace operator. This yields the usual formula in dimensions $4$ and up, but it yields zero in dimensions $2$ and $3$ because the trace operator is injective in those dimensions. So in that sense, the Weyl tensor is identically zero in dimensions $2$ and $3$.
