MathOverflow

The following uniformization theorems are classical.
Theorem 1 (Lusin-Souslin) If $f:X\rightarrow Y$ is a Borel map between Polish spaces, $A\subseteq X$ is Borel and $f\upharpoonright A$ is injective, then $f(A)$ is Borel.
Theorem 2 (Lusin-Novikov) Same as Theorem 1 but change "injective" to "countable-to-one"....

In a recent project we found a curious identity for simplices (Theorem 5.6).
Let $\Delta\subset\Bbb R^d$ be a $d$-simplex with facets $F_0,...,F_d$, $v_i\in\Bbb R^d$ the vertex opposite to $F_i$, $u_i\in\Bbb R^d$ the normal vector of $F_i$, and $h_i\in\Bbb R$ the height of $F_i$ over the origin.All of this can be summarized as...

Let $R$ be a Riemann curvature tensor on $\mathbb{R}^n$.By the Gauss formula, there exists an $\mathbb{R}^m$-valued symmetric bilinear form $s$ on $\mathbb{R}^n$ such that$$ \langle R(V,W)X,Y\rangle \equiv \langle s(V,Y), s(W,X) \rangle - \langle s(V...

Numerical evidence suggests that the complex zeros of:
$$f(s):=\frac{\zeta(s)}{\Gamma(s)} - \frac{\Gamma(1-s)}{\zeta(1-s)}$$
all reside on the line $\Re(s)=\frac12$, except for a finite few outside the critical strip. These zeros come in pairs that each "embrace" a non-trivial zero ($\rho$) of $\zeta(s)$ infinitely tightly. Obviously $f(\rho)$ induces a pole....