Is this function injective?

Question

For all given ordered lists $$\mathcal A=\big\{\{a_\mu\mid\mu=1,\cdots,N\}\mid\forall\mu,\nu> \mu,\ a_\mu > a_\nu\big\},$$ the function on the quotient space $$ G_\mu(a+\mathbb R: \mathcal A / \mathbb R):= \sum_{\lambda\ne\mu}\frac1{a_\mu - a_\lambda}$$ is vector-valued. Is this function injective? If so, can we obtain the inverse function in a closed formula (with an additive constant on $a$)?

Answer 1

The domain $A:=\{x\in\mathbb R^n: x_i>x_j \text{ for } 1\le i<j\le n\}$ is an open convex subset of $\mathbb R^n$ and the map $G:A\to\mathbb R^n$ has $\partial_iG_j=\partial_jG_i=\frac1{(x_i-x_j)^2}$, therefore it is a gradient map. In fact, $G=\nabla g$ for $g(x)=\sum_{1\le i<j\le n}\log(x_i-x_j)$. The function $-g$ is convex, and strictly on every hyperplane transverse to the constants, e.g. $A_0:=A\cap \{x\in\mathbb R^n:\sum_ix_i=0\}$.

The gradient of a strictly convex function is injective, and its inverse is the gradient map of its Legendre transform (on its finiteness domain).

Answer 2

This is a result of joint efforts of myself and Fedor Petrov.

Assume, to the contrary, that the system of equations $$ G_\mu(x)=\alpha_\mu, \qquad \mu=1,2,\dots,n $$ has two different solutions $a$ and $b$ not differing by a translation along $(1,1,\dots,1)$. Notice that both $a$ and $b$ are stationary points of the function $$ f(x)=\exp\left(-\sum_\mu \alpha_\mu x_\mu\right)\prod_{\nu<\mu}(x_\mu-x_\nu). $$

Consider the line $x=a+(b-a)t$ joining $a$ and $b$. The restriction $g$ of $f$ on this line is some non-constant function which is an exponent $e^{-ct}$ multiplied by a product of linear factors, and at least one of these factors is non-constant. Notice that $f$ does not vanish on the segment $[a,b]$. Therefore, the following "dual Rolle" lemma gives a desired contradiction.

Lemma. Consider a one-variable function $g(t)=P(t)e^{-ct}$ where $P$ is a product of $k\geqslant 1$ non-constant linear factors. Then, between any two stationary points of $g$, there is a root of $g$.

Proof. Assume first that $c\neq 0$. Notice that $g'(t)=Q(t)e^{-ct}$ where $\deg Q=k$. Hence $Q$ has at most $k$ roots.

On the other hand, the $k$ roots of $P$ split $\mathbb R$ into $k-1$ intervals and two rays. On each of the intervals, $Q$ should have a root — namely, the maximum point of $g$ on that interval. Additionally, $Q$ should have a root on one of the two rays, where $e^{-ct}$ tends to $0$, by the same reason.

We have enumerated and located $k$ distinct roots of $Q$, thus all of them. This yields the desired claim.

If $c=0$, then $\deg Q=k-1$, and the same arguments work.