The Euler characteristic does not determine the signature

The projective tangent-bundle supplier gives $H^{\ast }({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})=\mathbb{Z}[y]\mathrm{/}(y^3)$ and $⟨y^2,[{\mathbb{C}\mathbb{P}}^2]⟩=1$ (The tangent bundle of complex projective space and its Pontryagin classes). The signature supplier computes the real middle matrix $(1)$ and signature $1$ (The signature and the L-genus agree on complex projective space). Reversing orientation negates the fundamental class (Fundamental class of a compact oriented manifold), hence the middle form (The middle-dimensional intersection form of a closed oriented 4k-manifold); its matrix is $(-1)$, with signature $-1$.

By [F1], ${\mathbb{C}\mathbb{P}}^2={\mathrm{Gr}}_1({\mathbb{C}}^3)$; by [F2] its Schubert symbols are the integers $a_1\in \{1,2,3\}$, with $d(a_1)=a_1-1\in \{0,1,2\}$, so the cells have real dimensions $0,2,4$ and there are no cells in odd dimensions: $c_0=c_2=c_4=1$ and $c_1=c_3=0$.

By [F4], the signatures of the two closed oriented smooth four-manifolds are $\sigma ({\mathbb{C}\mathbb{P}}^2)=1$ and $\sigma (-{\mathbb{C}\mathbb{P}}^2)=-1$.

By [F3] applied to this finite CW structure, $\chi ({\mathbb{C}\mathbb{P}}^2)=c_0-c_1+c_2-c_3+c_4=1-0+1-0+1=3$.

Orientation reversal changes neither the underlying space nor its cells, since $-{\mathbb{C}\mathbb{P}}^2$ is ${\mathbb{C}\mathbb{P}}^2$ with the reversed orientation by [F5]; hence $c_n(-{\mathbb{C}\mathbb{P}}^2)=c_n({\mathbb{C}\mathbb{P}}^2)$ for every $n$ and $\chi (-{\mathbb{C}\mathbb{P}}^2)=3$ as well.

Thus ${\mathbb{C}\mathbb{P}}^2$ and $-{\mathbb{C}\mathbb{P}}^2$ have the same Euler characteristic $3$ but different signatures, so the Euler characteristic of a closed oriented four-manifold does not determine its signature, and the intersection form carries information beyond the alternating Betti-number count.