Second-class constraints and the Dirac bracket

If ${\sum }_a{\lambda }^{a}d{\chi }_a=0$, evaluate on each $X_{{\chi }_b}$. This gives ${\sum }_a{\lambda }^a{C}_{ab}=0$ for every $b$, so invertibility implies λ=0. The differentials are independent throughout $U$; [F3] makes each nonempty joint level embedded with tangent space $TN={\cap }_a\ker d{\chi }_a$. If $N=\mathrm{\varnothing }$, its unique empty smooth structure and restricted form satisfy the embedding and symplectic conditions vacuously.

For any $F$, set $X_F^D=X_F-X_{{\chi }_a}{C}^{ab}\{{\chi }_b,F\}$. Then $d{\chi }_c(X_F^D)=\{{\chi }_c,F\}-C_{ca}{C}^{ab}\{{\chi }_b,F\}=0$, so it is tangent to $N$. For tangent $v$, ω$(X_F^D,v)=dF(v)$, because $d{\chi }_a(v)=0$.

A vector tangent to $N$ in the radical of $\omega {\mathrm{\mid }}_N$ lies in $(TN{)}^{\omega }$, which is spanned by the independent $X_{{\chi }_a}$ because the annihilator of $TN$ is spanned by independent $d{\chi }_a$ and ${\omega }^{\mathrm{♭}}$ is an isomorphism. Tangency of a linear combination of these fields gives a coefficient vector annihilated by $C$, hence zero. Thus $\omega {\mathrm{\mid }}_N$ is nondegenerate; it is closed by pullback of $d\omega =0$.

Contracting the tangent field $X_F^D$ with (dG) gives $X_F^D(G)=\{G,F\}-\{G,{\chi }_a\}{C}^{ab}\{{\chi }_b,F\}=\{G,F{\}}_D$ with the declared convention after skewness and $C^{ab}=-C^{ba}$. Since $X_F^D$ is exactly the Hamiltonian field on $N$ proved in step 2.1, the bracket is the induced symplectic Poisson bracket and obeys Jacobi by [F2] applied to the symplectic $N$. If an extension vanishes on $N$, its tangent differential vanishes, so its restricted Hamiltonian field and bracket vanish; this proves extension independence. Substitution of $F={\chi }_c$ in the formula gives zero by $C{C}^{-1}=I$, so constraints are Casimirs.