Suppose $Q_X$ and $Q_Y$ are intersection forms of simply connected, smooth, closed 4-manifolds $X$ and $Y$. By Freedman, if $Q_X$ is equivalent to $Q_Y$, then $X$ is homeomorphic to $Y$ (though perhaps not diffeomorphic).
What is the notion of equivalence being used here? Similar matrices? That doesn't seem right. Thanks!
Let $Q_1$ and $Q_2$ be two quadratic forms on a vector space $V$. We say $Q_1$ and $Q_2$ are equivalent if there is an isomorphism $\phi : V \to V$ such that $Q_1(v, w) = Q_2(\phi(v), \phi(w))$.
If we choose a basis for $V$, and $V$ is finite-dimensional, we can represent $Q_1$ and $Q_2$ by matrices $X_1$ and $X_2$. Then $Q_1$ and $Q_2$ are equivalent if and only if there is an invertible matrix $A$ such that $X_1 = A^TX_2A$.