Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3\times[0, 1]}$ such that
\[
\partial A = K^{\prime}\times\{1\}\;\sqcup\; K \times\{0\}.
Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $[0]$ denote class of all knots concordant to a trivial knot.
$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $[0]$ and the inverse element of an element $[K]\in\mathscr{C}$ is $-[K]=[mK]$.
Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta)\mapsto\Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta\in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
$\alpha, \beta\in\ker(H_1(\Sigma, \mathbb{Z})\longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in\Omega$ such that $\partial A =\alpha$ and $\partial B =\beta$.
Let $B^+$ be a push off of $B$ in the positive normal direction such that
$\partial B^+=\beta^+$.
Then
$\Lk(\alpha, \beta^+)= A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+)=0$. Then the Seifert form is zero.
\Sigma\overset{\phi}\longhookrightarrow Y \overset{\psi}\longhookrightarrow\Omega.
\]
Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma\in\ker(H_1(\Sigma, \mathbb{Z})\longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma\in\ker\phi_*$ or $\gamma\in\ker\psi_*$.
Suppose $g(K)=0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z})\cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z})=2g_{\Sigma}$. Then the Seifert form $V$ on a $K$
has a subspace of dimension $g_{\Sigma}$ on which it is zero: