Relation between Pontryagin number and Euler number for a four-dimensional closed manifold On a four-dimensional closed manifold $M^4$, we can have both a Pontryagin number and an Euler number. Are they related?
 A: Let $M$ be a connected, closed, smooth, oriented four-dimensional manifold. 
By the Hirzebruch signature theorem, the first Pontryagin number is 
$$p_1(M) = \langle p_1(TM), [M]\rangle = 3\tau(M)$$ 
where $\tau(M) = b^+(M) - b^-(M)$ is the signature of $M$.
On the other hand, 
\begin{align*}
\chi(M) &= b_0(M) - b_1(M) + b_2(M) - b_3(M) + b_4(M)\\ 
&= 1 - b_1(M) + b^+(M) + b^-(M) - b_1(M) + 1\\ 
&= 2 - 2b_1(M) + b^+(M) + b^-(M).
\end{align*}
Now note that $p_1(M) = 3\tau(M) \equiv b^+(M) + b^-(M) \bmod 2$ and $\chi(M) \equiv b^+(M) + b^-(M) \bmod 2$, so $p_1(M)$ and $\chi(M)$ have the same parity, i.e. they are either both even, or both odd.
Aside from the fact that $p_1(M)$ and $\chi(M)$ have the same parity, and that $p_1(M)$ is divisible by three, there are no more restrictions on these two quantities - in particular, no further relations between them. To see this, it is enough to exhibit, for any integers $a$ and $b$ of the same parity, a connected, closed, smooth, oriented four-manifold $M$ with $p_1(M) = 3a$ and $\chi(M) = b$ - note, $3a$ has the same parity as $a$, so $3a$ and $b$ have the same parity. First observe that 
$$p_1(T^4\#k\mathbb{CP}^2\# l\overline{\mathbb{CP}^2}) = 3\tau(T^4\#k\mathbb{CP}^2\# l\overline{\mathbb{CP}^2}) = 3(k-l)$$
and 
$$\chi(T^4\#k\mathbb{CP}^2\# l\overline{\mathbb{CP}^2}) = k + l.$$
Taking $k = \frac{1}{2}(a + b)$ and $l = \frac{1}{2}(b - a)$, which are integers because $a$ and $b$ have the same parity, we see that $M = T^4\#k\mathbb{CP}^2\# l\overline{\mathbb{CP}^2}$ satisfies $p_1(M) = 3a$ and $\chi(M) = b$.
