Consider the function $(n-1)^2.$ Clearly this is $\mathcal{O}(n^2)$ since the constant for the upper bound is $1.$ However, it seems to me that it is not $\Omega (n^2)$ since this is a strictly smaller function for $n > \frac{1}{2}$ and a constant would have to be defined in terms of $n.$ However, it certainly seems close and in some textbooks discussing the cover times of paths (which is $(n-1)^2$) I have seen the function labeled as $\Theta (n^2)$ instead of $\Theta ((n-1)^2).$ Which is/are correct?
EDIT I suppose the question I am really asking at the core is, although we can drop lower order terms in a polynomial for big $\mathcal{O}$ notation, can we do it for big $\Omega$ ?
Note that $(\frac n 2 )^2\leq(n-1)^2\leq n^2$ for sufficiently large $n$, hence $(n-1)^2=\Theta(n^2)$. What is important is the dominating behaviour of the term of highest degree, therefore when $n$ is suffiently large, the terms of lower degree can be neglected by a constant portion of the term of highest degree.