On the wikipedia article for the Kervaire Invariant, it is stated that the relevant quadratic form can be defined geometrically via the self-intersections of immersions $S^{2m+1}\to M^{4m+2}$ determined by the framing.
How is this done?
References would be much appreciated!
Maybe what they refere to on wikipedia is the approach Levine takes in "lectures on groups of homotopy spheres". The construction goes as follows: Suppose $M$ is $k-1$-connected of dimension 2k, $k$ odd. (It is only in dimension $2k$, $k$ odd the Kervaire invariant is defined, and it is "easy" to show that any framed manifold is framed cobordant to a $k-1$ connected one, see Kervaire and Milnor, Groups of Homotopy Spheres). Suppose you have an immersion $f:S^k\rightarrow M$. Then consider the pullback-bundle of the stable tangent bundle of $M$,
$$
f^*(\epsilon^N\oplus TM)=\epsilon^N\oplus\nu(f)\oplus TS^k=\epsilon^{N-1}\oplus \nu(f)\oplus TD^{k+1}|_{S^k}
$$
where $\epsilon^N$ is the trivial $N$-plane bundle. The framing of $M$ gives you a framing of this bundle. The standard framing of $\epsilon^{N-1}\oplus TD^{k+1}$ gives you a framing of $TD^{k+1}|_{S^k}$. Together these two framings determine a map $S^k\rightarrow V_{N+2k,N+k}$, and the homotopy class is dependent only on the regular homtopy class of the immersion $f$. Since any two embeddings are regularly homotopic (homotopic through immersions) we get a well defined map $H_k(M)\rightarrow \pi_k(V_{N+2k,N+k})=\mathbb{Z}_2$ by defining $\phi_0(\alpha)=\phi_0(f)$ where $f$ is any embedding representing $\alpha$. This is a quadratic refinement of the mod 2 intersection pairing (it takes another page or two to show this). We define the Kervaire invariant as the Arf invariant:
$$
\Phi(M)=\sum_{i=1}^r\phi(x_i)\phi(y_i)
$$
where $\{x_i,y_i\}$ forms a symplectic basis for $H_k(M;\mathbb{Z}_2)$.