Rosen's Diagonal Strict Concavity condition Consider a game with $n$ players, with strategy space $S \subset \mathbb{R}$, where $S$ is bounded set, and player's $i$ payoff function  $\pi_i:S^n \rightarrow \mathbb{R}$. 
Rosen's condition (J. B. Rosen. Existence and uniqueness of equilibrium points for concave n-person games. Econometrica, 33(3):520–534, 1965) for uniqueness of the Nash Equilibrium in n players game states that the  equlibrium will be unique when


payoff function $\pi_i(\textbf{s}) \; i \in N$  is concave in own strategy
There exists vector $\textbf{z}$ ($(\forall i \in N)(z_i \geq 0)\ \wedge (\exists i \in N) (z_i &gt;0)$ such that function $\sigma(\mathbf{s}, \mathbf{z})=\sum_{i=1}^{n}z_i\pi_i({\textbf{s}})$ is diagonally strictly concave


$N$ denotes the set of players. 

To define the concept of diagonal strict concavity, fist introduce 'pseudogradient' of function $\sigma$, defined with:
\begin{align}
g(\mathbf{s},\mathbf{z}) = \begin{pmatrix}
z_1\frac{\partial \pi_1(\mathbf{s})}{\partial s_1} \\ 
z_2\frac{\partial \pi_2(\mathbf{s})}{\partial s_2} \\ 
... \\ 
z_n\frac{\partial \pi_n(\mathbf{s})}{\partial s_n}%
\end{pmatrix}
\end{align}
Then, function $\sigma$ is said to be diagonally strictly dominant in $\mathbf{s} \in S$ for fixed $\mathbf{z} \geq 0$ if for every $\mathbf{s}^0, \mathbf{s}^1 \in S$ the following holds:
\begin{align}
(\mathbf{s}^1 - \mathbf{s}^0)'g(\mathbf{s}^{0}, \mathbf{z}) + (\mathbf{s}^0 - \mathbf{s}^1)'g(\mathbf{s}^{1}, \mathbf{z})&gt;0
\end{align}

It is shown, in the paper I cite in the beginning, that a sufficient condition for $\sigma$ to be diagonally striclty concave is that matrix $\left[G(\mathbf{x}, \mathbf{z}) +G(\mathbf{x}, \mathbf{z})' \right]$ is negative defite for $\mathbf{s} \in S$, where $G(\mathbf{x}, \mathbf{z})$ is Jacobian of pseudogradient $g$ with respect to $\mathbf{s}$. I use ' to denote transpose of a matrix. 
What is the intuition behind diagonal strict concavity condition?
So you want to find a maximum of $\sigma(s,z)$. If $\sigma$ is diagonally strictly concave you can do so by starting at any point and just following the gradient $g(s,z)$ until you find the maximum and no matter where you start, you will always end up at the same point (Start at the lower black points and follow the direction of the gradient (the direction of the steepest ascent).).


However, if $\sigma$ isn't diagonally strictly concave, you could end up at different maxima by starting at an arbitrary point and following the gradient (Follow the direction of the steepest ascent starting from the two lower black dots; you'll end up at two different points.).