# Smooth structure on solution space

Broadly speaking the question in this post is :

How to check that the solution space of a set of equations is a smooth manifold? This happens in physics all the time, say solution set of Hamiltonian equations, Lagrangian equations etc.

As a prototypical example let $\{x_{i}\}$ be the solution set of the equations $f(x_{ij}) = 0$ in $\mathbb{R}^{n}$ (or more generally a manifold). Here $x_{ij} = |x_{i} - x_{j}|$ is the distance between the solution points (more generally a function of $x_{i},x_{j}$).

As a primitive case consider the solution set of the equations $\sum_{i \neq j} \frac{1}{x_{ij}} = j$ in $R^{n}$. If we've $m$ such solutions then the solution space is $(nm-m)$-dimensional. Then how do one check that the solution set is a smooth manifold and in particularly check for the singularities, if any.

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If the Jacobi matrix of your functions is of maximal rank, your solution space is a manifold. Probably an answer will come along with more details. –  shaye Oct 4 '11 at 21:17
@shaye the functions in the questions are real valued. Can you elaborate on your comment or may be an answer, explaining the simple example. –  J Verma Oct 5 '11 at 18:03

Let me see what I can do...

Let $f_1,\ldots,f_k$ be functions from $\mathbf{R}^n\longrightarrow \mathbf{R}$. (We may and do assume that $1\leq k\leq n$.) In linear algebra, i.e., if $f_i$ is a linear function for all $i=1,\ldots,k$, you have that the dimension of the solution space $Z(f_1,\ldots,f_k)$ is of dimension $n-k$. In general, this doesn't hold.

Since you're working in the realm of differential geometry what happens is the following.

The functions $f_i$ are $C^\infty$, i.e., infinitely many times differentiable.

The Jacobi matrix $J(f_1,\ldots,f_k)$ is of maximal rank $k$ (at every point in Z(f_1,\ldots,f_k)) if and only if the solution space $Z(f_1,\ldots,f_k)$ is a submanifold of $\mathbf{R}^n$ of dimension $n-k$.

For example, the solution space of the function $f_1(x,y) = y^2+x^2-1$ is a manifold because the Jacobi matrix $(2x 2y)$ is of rank 1. (The point $(0,0)$ doesn't satisfy the equation $f_1(x,y) =0$.)

Also, the solution space of the function $f_1(x,y) = y^2-x^2$ is not a manifold because the Jacobi matrix $(2y -2x)$ is singular at the point $(0,0)$. (In this case $(0,0)$ does satisfy the equation $f_1(x,y)=0$.)

I don't quite understand the functions in your example.

Let's say $n=3$ for simplicity.

Then you have three functions: $f_1(x_1,x_2,x_3) = \frac{1}{x_2}+\frac{1}{x_3}-1$, $f_2(x_1,x_2,x_3) = \frac{1}{x_1}+\frac{1}{x_3}-2$ and $f_3(x_1,x_2,x_3) = \frac{1}{x_1}+\frac{1}{x_2}-3$. The Jacobi matrix of this system is

$$\left( \begin{array}{ccc} 0 & \frac{-1}{x_2^2} & \frac{-1}{x_3^2} \\ \frac{-1}{x_1^2} & 0 & \frac{-1}{x_3^2} \\ \frac{-1}{x_1^2} & \frac{-1}{x_2^2} & 0 \end{array}\right).$$ This looks like something of rank $3$ to me...

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thanks for the answer, it clears the point. The functions I was considering were not the functions of $x_{i}$s but of the distances between them $|x_{i} - x_{j}|$, which not so easy to check. –  J Verma Oct 10 '11 at 1:39