With this definition of a sharp threshold:
A sequence of increasing boolean functions $(\mathbf{f_n})$ undergoes a sharp threshold at $(p_n)$ if there exists a $(\delta_n)$ tending to 0 such that $f_n(p_n - \delta_n) \to 0$ and $f_n(p_n + \delta_n) \to 1$.
I have seen (see the introduction to the appendix), using their notation of transition window $\tau_n$ instead of $\delta_n$, that "having a triangle" in the random graph $G(n,p_n)$ is a coarse threshold, since $p_n \sim \tau_n = {m \choose 3}^{-1/3}$, and so $\tau_n/p_n$ does not go to zero with $n$. The window over which the transition occurs is not small compared to the scale at which the transition occurs.
But, having a triangle is a monotone graph property (adding edges cannot take away triangles). So it should have a sharp transition, via this theorem, which claims that every monotone graph property has a sharp threshold.
What is going on here? Does having a triangle undergo a sharp transition or not?
These are two different definitions of "sharp threshold".
In "Sharp thresholds of graph properties", we pick $\epsilon>0$ and let $p_0(n), p_c(n), p_1(n)$ be such that $G_{n,p_0}, G_{n,p_c}, G_{n,p_1}$ has property $P$ with probability $\epsilon, \frac12, 1-\epsilon$ respectively. The threshold is sharp if $p_1 - p_0 \ll p_c$.
In "Every monotone graph property has a sharp threshold", with the same notation, it is shown that for any monotone graph property, $p_1 - p_0 = O(\frac1{\log n})$ (with some dependency on $\epsilon$ inside the big-$O$). This would imply a sharp threshold by the previous definition if $p_c$ is, for example, constant.
However, for triangles, $p_c = \Theta(\frac1n)$ and so it is far from enough, to satisfy the previous definition, to have $p_1 - p_0 = O(\frac1{\log n})$. In fact, when $p = \frac \lambda n$, the probability of having no triangles is roughly $e^{-\lambda^3/6}$, which lets us prove that $p_1 - p_0 = \Theta(\frac1n)$. This is much better than what the Friedgut-Kalai result implies for general monotone properties, but it is still not enough.