Real projective space shows coefficient-dependent perfectness

The formula $f([x])={\sum }_i(i+1)x_i^2$ is well defined on the quotient because replacing $x$ by $-x$ leaves every squared coordinate unchanged; the quotient ${\mathbb{R}\mathbb{P}}^n$ is the familiar closed smooth $n$-manifold with the standard charts $u_j=x_j\mathrm{/}{x}_i$ around the axis $[e_i]$.

Write ${\lambda }_j=j+1$. A tangent vector $v$ to the unit sphere at $x$ satisfies $x\cdot v=0$, and the differential of the lifted function is $2{\sum }_j{\lambda }_j{x}_j{v}_j$. It vanishes on all such $v$ exactly when $({\lambda }_j{x}_j{)}_j$ is a scalar multiple of $x$. Since the ${\lambda }_j$ are distinct, at most one coordinate of a critical point is nonzero. Thus the critical classes are exactly the axes $[e_i]$. In the chart $u_j=x_j\mathrm{/}{x}_i$ for $j\mathrm{\ne }i$, $f(u)=\frac{{\lambda }_i+{\sum }_{j\mathrm{\ne }i}{\lambda }_j{u}_j^2}{1+{\sum }_{j\mathrm{\ne }i}{u}_j^2}\mathrm{.}$ Expanding at zero gives $f(u)={\lambda }_i+{\sum }_{j\mathrm{\ne }i}({\lambda }_j-{\lambda }_i)u_j^2+O(\mathrm{\mid }u{\mathrm{\mid }}^4)$, so the Hessian is $\mathrm{diag}(2(j-i){)}_{j\mathrm{\ne }i}$, with exactly $i$ negative entries. Every critical point is nondegenerate of index $i$, and $M_f(t)=1+t+\cdots +t^n$.

Over ${\mathbb{F}}_2$ the cellular complex of [F3] reads $C_j={\mathbb{F}}_2$ with $d_j=2=0$ for all $j$, so $H_j({\mathbb{R}\mathbb{P}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le j\le n$ and zero otherwise, by [F4]; hence $b_k=1=m_k$ for all $k$ and $f$ is ${\mathbb{F}}_2$-perfect with correction polynomial $Q=0$.

Over a field $F$ of characteristic different from two the same complex reads $C_j=F$ with $d_j=2$ invertible for positive even $j$ and $d_j=0$ for odd $j$; taking kernels modulo images gives $H_0=F$, $H_j=0$ for $0<j<n$, and $H_n=F$ if $n$ is odd and $H_n=0$ if $n$ is even. Thus for $n\ge 2$ we have $b_1({\mathbb{R}\mathbb{P}}^n;F)=0<1=m_1(f)$, and $f$ is not $F$-perfect by [F5].

Finally, for every field the Euler identity holds: ${\sum }_{k=0}^n(-1{)}^k{m}_k(f)={\sum }_{k=0}^n(-1{)}^k$ equals $1$ for even $n$ and $0$ for odd $n$, while $\chi ({\mathbb{R}\mathbb{P}}^n)=1$ for even $n$ and $0$ for odd $n$ by [L1] applied to the cell counts of [F3] (one cell in each dimension $0,\dots ,n$). The alternating sums of the Betti numbers agree with the same value in both characteristics, consistent with the Euler identity.