For a closed oriented 4-manifold $X$, the bilinear intersection form $H_2(X)\times H_2(X)\to \Bbb Z$, $(a,b)\mapsto \langle PD(a)\cup PD(b), [X]\rangle$ is unimodular, which can be shown by Poincare duality. Suppose $X$ is a compact oriented 4-manifold with nonempty boundary, and with second betti number $\geq 1$. Can the intersection form of $X$ be trivial (so the intersection form is represented by the zero matrix)?
Intersection form of a 4-manifold with boundary
4-manifoldsalgebraic-topologygeometric-topologyintersection-theorylow-dimensional-topology
Related Solutions
For a $\mathbb{Z}$-module (aka abelian group) $V$ with a symmetric, unimodular, bilinear form $\mu\colon V\otimes V \to V$, a characteristic element is an element $c\in V$ such that
$$\forall v\in V\ \mu(v,v) \equiv \mu(c,v)\mod 2 $$
This means you don't want an element with $\mathbb{Z}/2$ coefficients, you want an integral class. The correct statement is that
For an oriented closed $4$-manifold $M$, an element in $H^2(M;\mathbb{Z})$ is characteristic iff it is an integral lift of $w_2(M)$.
It's not hard to see this algebraically using Wu classes (for a definition and important properties see for example Manifold Atlas or Milnor-Stasheff). In general, for a closed, oriented $4n$-manifold, a characteristic element of its intersection form is an integral lift of the $2n$-th Wu class $v_{2n}$, essentially by definition.
But Wu's formula $Sq(V) = W$ tells us that for a closed, oriented manifold we have $v_2 = w_2$, so if $M$ if $4$-dimensional then $\alpha \in H^2(M;\mathbb{Z})$ is characteristic iff it is an integral lift of $v_2$ iff it is an integral lift of $w_2$.
Letting $\gamma \in H^1(S^1) \approx \mathbb Z$ be a generator, there is a function $\phi : \langle M, S^1 \rangle \to H^1(M)$ defined by $$\phi[f] = f^*(\gamma) $$ for each continuous $f : M \to S^1$, where $[f] \in \langle M,S^1 \rangle$ denotes the homotopy class of $f$, and $f^* : H^1(S^1) \to H^1(M)$ denotes the homomorphism induced by $f$ with respect to the contravariant functor $H^1(\cdot)$.
The theorem you need is that this function $\phi$ is a bijection. As said in the comments, this theorem can be found in Hatcher's book, and also in many other algebraic topology books; I first learned it in Spanier's book.
In the special case that $M$ is an oriented manifold, you can evaluate $\phi[f]$ by first homotoping $f$ to be transverse to a point $p \in S^1$, in which case $f^{-1}(p)$ is a codimension 1 transversely oriented submanifold of $M$. By pairing that transverse orientation with the orientation of $M$ one obtains an orientation of $f^{-1}(p)$, which therefore represents an element of $H_{n-1}(M)$ where $n$ is the dimension of $M$. In this situation $\phi[f]$ is the Poincaré dual of the homology class of $f^{-1}(p)$. To put it another way, $\phi[f] \in H^1(M)$ is evaluated on each 1-dimensional homology class in $M$ by taking the intersection number of that class (i.e. the algebraic intersection number of any 1-cycle in that class) with $f^{-1}(p)$.
A similar construction of $\phi[f]$ also works for the case that $M$ is a simplicial complex: choose $f$ to be in general position with respect to $p$, in which case $f^{-1}(p)$ is a transversely oriented subcomplex with a product neighborhood. In this case $\phi[f] \in H^1(M)$ is defined by evaluating it on an arbitrary cycle in $M$ to be the algebraic intersection number of that cycle with $f^{-1}(p)$.
Best Answer
Yes, for example $S^2\times D^2$ is such a 4-manifold.