Relation between Pontryagin number and Euler number for a four-dimensional closed manifold

Question

On a four-dimensional closed manifold $M^4$, we can have both a Pontryagin number and an Euler number. Are they related?

Answer 1

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 $\chi(M) = b$.