= Solution
Let $\pi:T^*M\to M$ be the <cotangent bundle> projection. Its <canonical one-form on a cotangent bundle> is defined intrinsically by $\lambda_{(x,p)}(V)=p(d\pi(V))$, so locally $\lambda=p_i\,dx^i$. Choose the position-first <symplectic form>
$$
\omega=-d\lambda=dx^i\wedge dp_i.
$$
It is closed by $d^2=0$ and nondegenerate, since contraction with $a^i\partial_{x^i}+b_i\partial_{p_i}$ is $a^i dp_i-b_i dx^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this <symplectic form> independent of coordinates.
Use the convention $\iota_{X_H}\omega=dH$. Then the <Hamiltonian vector field> and the <Poisson bracket> are
$$
X_H=\frac{\partial H}{\partial p_i}\partial_{x^i}
-\frac{\partial H}{\partial x^i}\partial_{p_i},
\qquad
\{F,G\}=\frac{\partial F}{\partial x^i}\frac{\partial G}{\partial p_i}
-\frac{\partial F}{\partial p_i}\frac{\partial G}{\partial x^i},
$$
so $X_H(F)=\{F,H\}$. Choosing $\omega=d\lambda$ and $\iota_{X_H}\omega=-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.
For the <geodesic Hamiltonian>, <Hamilton's equations> give
$$
\dot x^i=g^{ij}p_j,\qquad
\dot p_i=-\frac12(\partial_i g^{jk})p_jp_k.
$$
Put $v^i=\dot x^i$, so $p_i=g_{ij}v^j$. Differentiating $g^{jk}g_{ka}=\delta^j_a$ gives
$$
(\partial_i g^{jk})p_jp_k=-(\partial_i g_{ab})v^av^b.
$$
Consequently,
$$
g_{ij}\ddot x^j+(\partial_k g_{ij})v^kv^j
=\frac12(\partial_i g_{ab})v^av^b.
$$
Symmetrizing the velocity factors and raising the first index converts this to
$$
\boxed{\ddot x^\ell+\Gamma^\ell{}_{jk}\dot x^j\dot x^k=0},
\qquad
\Gamma^\ell{}_{jk}=\frac12g^{\ell i}
(\partial_jg_{ik}+\partial_kg_{ij}-\partial_ig_{jk}).
$$
These <Christoffel symbols> are those of the <Levi-Civita connection>. Thus a Hamiltonian <integral curve of a vector field> projects to an affinely parametrized <geodesic>. Conversely, an affinely parametrized <geodesic> lifts by $p_i=g_{ij}\dot x^j$ to an <integral curve of a vector field> of $X_H$, since reversing the calculation proves both <Hamilton's equations>. This is the <geodesic flow> on the <cotangent bundle>. The conserved <Hamiltonian> is half the squared speed, and the zero-energy case gives the constant <geodesics>.
A quadratic <homogeneous polynomial> depends only on the symmetric part of its coefficient matrix. Accordingly take its unique <symmetric coefficients of a quadratic polynomial>, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with <symmetric tensors>. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean $\mathbb R^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero <Poisson bracket> with every function, whereas the lowered coefficient array is not a <Killing tensor> because it is not symmetric.
Lower the indices of the symmetric coefficient tensor using the <Riemannian metric>. Since $p_i=g_{ij}v^j$,
$$
K^{ij}p_ip_j=K_{ij}v^iv^j.
$$
Along an affinely parametrized <geodesic>, <metric compatibility> and $\nabla_vv=0$ imply
$$
\frac{d}{dt}(K_{ij}v^iv^j)
=(\nabla_aK_{bc})v^av^bv^c
=\nabla_{(a}K_{bc)}\,v^av^bv^c.
$$
Here parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the <Hamiltonian vector field> convention. Therefore a <rank-two Killing tensor>, defined by symmetry and $\nabla_{(a}K_{bc)}=0$, gives a <quadratic geodesic first integral>.
Conversely, if $\{K,H\}=0$ everywhere on $T^*M$, the last cubic expression vanishes for every $v$ at every point, because the <Riemannian metric> identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence $\nabla_{(a}K_{bc)}=0$. This proves both directions:
$$
\boxed{\{K,H\}=0\quad\Longleftrightarrow\quad
K_{ij}=K_{ji}\ \text{and}\ \nabla_{(a}K_{bc)}=0}
$$
with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require <geodesic completeness>.
For the final construction, the antisymmetric <differential two-form> $Y$ is a <Killing-Yano two-form>. Its defining equation says $\nabla_aY_{bc}=-\nabla_bY_{ac}$. Antisymmetry of $Y$ also says $\nabla_aY_{bc}=-\nabla_aY_{cb}$, so the three-index tensor $\nabla_aY_{bc}$ is totally antisymmetric.
The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot}$ and $Y_{j\cdot}$:
$$
K_{ij}=g^{k\ell}Y_{ik}Y_{j\ell}=K_{ji}.
$$
There is a useful geometric proof of its <Killing tensor> equation. Along any affinely parametrized <geodesic>, define $w_k=Y_{ik}v^i$. Then
$$
\nabla_vw_k=v^av^i\nabla_aY_{ik}+Y_{ik}\nabla_vv^i=0.
$$
The first term vanishes by antisymmetry in $a,i$, and the second by the <geodesic> equation. Thus $w$ is carried by <parallel transport>. By <metric compatibility>, its squared norm is constant, and
$$
|w|^2=g^{k\ell}Y_{ik}Y_{j\ell}v^iv^j=K_{ij}v^iv^j.
$$
Every tangent vector is the initial velocity of a local <geodesic>, so differentiation at the initial point gives $\nabla_{(a}K_{bc)}v^av^bv^c=0$ for every $v$. The same cubic-coefficient argument proves
$$
\boxed{\nabla_{(a}K_{bc)}=0}.
$$
This proves that the <square of a Killing-Yano two-form> is a <rank-two Killing tensor> and supplies a nonnegative <quadratic geodesic first integral>. The argument also explains the conserved quantity: it is the squared norm of a covector that the <Killing-Yano two-form> makes parallel along every <geodesic>.
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