To prove that there exists a unique solution to the Newton equation with the given potential, we can use the fact that the equation of motion for a particle with potential energy $U(q)$ is given by:

$m \frac{d^2 q}{d t^2}=-\nabla U(q)$

where $m$ is the mass of the particle. Substituting $U(q)=-B(q,q)$ into this equation, we get:

$m \frac{d^2 q}{d t^2}=B^{\prime}(q) \dot{q}$

where $B'(q)$ is the derivative of $B(q,q)$ with respect to $q$. Since $B$ is nonnegative definite and symmetric, we have $B'(q)$ is a symmetric matrix that is also nonnegative definite. This means that $B'(q)$ can be diagonalized by an orthogonal matrix $O$, such that $B'(q)=O^T\Lambda O$, where $\Lambda$ is a diagonal matrix with nonnegative entries. Thus, we can write the equation of motion as:

$m \frac{d^2}{d t^2}(O q)=\left(O^T \Lambda O\right)(O \dot{q})=\Lambda\left(O^T \dot{q}\right)$,

where we have used the fact that $O$ is orthogonal, so $O^TO=I$. Defining $p=O^T\dot{q}$, we obtain the system of first-order differential equations:

$\frac{d p}{d t}=\frac{1}{m} \Lambda p, \quad \frac{d q}{d t}=O p$

with initial conditions $q(t_1)=q_1$ and $q(t_2)=q_2$. Since $\Lambda$ is diagonal with nonnegative entries, the solutions to the first equation are of the form $p_i(t)=c_ie^{\omega_i t}$, where $\omega_i=\sqrt{\lambda_i/m}$ and $c_i$ are constants determined by the initial conditions. The solutions to the second equation are of the form $q_i(t)=d_ie^{\omega_i t}$, where $d_i$ are constants determined by the initial conditions. Thus, we have a unique solution for $p(t)$ and $q(t)$.

To show that this solution provides a minimum for the action, we need to compute the action:

$S=\int_{t_1}^{t_2}\left[\frac{1}{2} m \dot{q}^2+U(q)\right] d t=\int_{t_1}^{t_2}\left[\frac{1}{2} p^T \Lambda^{-1} p-B(q, q)\right] d t$

Using the solutions for $p(t)$ and $q(t)$, we can evaluate the action as:

$S=\sum_{i=1}^n\left[\frac{m \omega_i^2}{2} \int_{t_1}^{t_2} p_i^2 d t-B(q, q)\right]=-\sum_{i=1}^n \frac{m \omega_i^2}{2} c_i^2-B\left(q_1, q_2\right)$

Since $B$ is nonnegative definite, we have $B(q_1,q_2)\leq 0$. Also, since $\omega_i^2\geq 0$ and $c_i$ are constants, we have $-\sum_{i=1}^n\frac{m\omega_i^2}{2}c_i^2\geq 0$.

Let $f(z)$ be the generating function for labeled trees with 1-valent and 4-valent vertices. Then, we can decompose a 4-valent vertex into two 1-valent vertices and two 2-valent vertices, which gives us the following recursion: $$f(z) = z + z\left(f(z)\right)^2$$ The first term, $z$, accounts for the 1-valent vertex, while the second term accounts for the 4-valent vertex. The factor of $z$ in the second term accounts for the fact that the 4-valent vertex can appear anywhere in the tree.

We want to find the generating function for labeled trees with 1-valent and 4-valent vertices, but with the additional restriction that the total degree of the vertices is equal to $n$. Let $g_n$ be the number of such trees. Then, we have: $$\frac{1}{n}\sum_{k=1}^{n-1} g_k g_{n-k} = [z^n]\left(f(z)\right)^2$$ The factor of $1/n$ in the sum is to account for the fact that we’re dividing by the total number of edges in the tree. The coefficient $[z^n]\left(f(z)\right)^2$ counts the number of ways to split a tree of size $n$ into two smaller trees.

Let $g(z) = \sum_{n=0}^\infty g_n z^n$. Then, we have: $$g(z) = z + z\left(g(z)\right)^2$$ Multiplying both sides by $g(z)$ and differentiating with respect to $z$, we get: $$g(z)\left(1 + 2z g'(z)\right) = 1 + g(z)^2$$ Solving for $g(z)$, we get: $$g(z) = \frac{1 – \sqrt{1 – 4z^2}}{2z}$$ Therefore, the generating function for the labeled trees with 1-valent and 4-valent vertices, weighted by $1/n$, is given by: $$\frac{1}{z}\left(\frac{1 – \sqrt{1 – 4z^2}}{2}\right)$$

Mumford’s theorem for $g=1$ states that any algebraic curve of genus $1$ over an algebraically closed field is isomorphic to an elliptic curve in Weierstrass form.

To prove this, we start with an algebraic curve $C$ of genus $1$ defined over an algebraically closed field $K$. By definition of genus, $C$ is a smooth projective curve of degree $2$.

Since $C$ has genus $1$, it has exactly one nontrivial invertible sheaf $\mathcal{L}$ of degree $1$. By the Riemann-Roch theorem, we have $\dim_K H^0(C,\mathcal{L})=\dim_K H^0(C,K)-\deg(\mathcal{L})+1=2-1+1=2$. Thus, there exist two linearly independent global sections $f$ and $g$ of $\mathcal{L}$.

We can use these sections to define a morphism $\phi:C\to\mathbb{P}^2_K$ as follows. Let $P\in C$ be a point, and let $(f(P):g(P):1)$ be the corresponding point in $\mathbb{P}^2_K$. Then define $\phi(P)=(f(P)^2:g(P)^2:f(P)g(P))$. It can be shown that $\phi$ is a morphism of algebraic curves, and that its image is contained in the zero set of the Weierstrass equation $y^2=x^3+ax+b$, where $a$ and $b$ are constants in $K$.

To complete the proof, we need to show that $\phi$ is an isomorphism. Since $C$ and the zero set of the Weierstrass equation are both smooth and projective, it suffices to show that $\phi$ is bijective and that its differential is everywhere nonzero.

To show that $\phi$ is bijective, we note that the inverse image of a point $(x:y:1)$ in $\mathbb{P}^2_K$ under $\phi$ consists of two points if $x$ and $y$ are nonzero, and of one point if either $x$ or $y$ is zero. It can be shown that these points are distinct, so $\phi$ is injective. To show that $\phi$ is surjective, we note that any point $(x:y:z)$ in the zero set of the Weierstrass equation can be mapped to by the point $(\frac{y}{z},\frac{x}{z})$ on $C$, so $\phi$ is surjective.

Finally, we need to show that the differential $d\phi$ is everywhere nonzero. It can be shown that $d\phi$ is nonzero at the point $(0,0)$ of $C$, which corresponds to the point at infinity in $\mathbb{P}^2_K$. Since $C$ and the zero set of the Weierstrass equation are smooth, $d\phi$ is nonzero everywhere.

Therefore, $\phi$ is an isomorphism of algebraic curves, so $C$ is isomorphic to an elliptic curve in Weierstrass form.