Define $ \mathbb{W} := (W_{i, j})_{1 \leq i, j \leq k} $ and \begin{align*}%$ \varphi_{C, k}(\mathbb{W}) := C^2 \, \boldsymbol{1}_k^T (I_k + \mathbb{W} )^{-1} \boldsymbol{1}_k + \sum_{1 \leq i < j \leq k} W_{i, j} \end{align*} or, equivalently \begin{align*}%$ \varphi_{C, k}(\mathbb{W}) := \frac{1}{2} \textrm{tr}\!\left( U_k \left[ 2 C^2 (I_k + \mathbb{W})^{-1} + \mathbb{W} \right] \right) \end{align*} where $ U_k := \boldsymbol{1}_k \boldsymbol{1}_k^T $ is the $ k \times k $ matrix with only ones.
We suppose that $ \mathbb{W} $ is symmetric with 0s on the diagonal and that $ W_{i, j} \in (0, 1) $ for all $ i < j $.
I want to estimate the integral \begin{align*}%$ I(N) := \int_{ (0, 1)^{k(k - 1)/2 } } e^{-N \varphi_{C, k}(\mathbb{W})} f(\mathbb{W}) d\mathbb{W} \end{align*} when $ N \to +\infty $, with $ f(\mathbb{W}) := \det(I_k + \mathbb{W})^{-1/2} \prod_{1 \leq i < j \leq k} (W_{i, j})^{-1/2} $.
I thought about the Laplace method that would give \begin{align*}%$ I(\color{blue}N\color{black}) \sim e^{-\color{blue}N\color{black} \varphi_{C, k}(\mathbb{W}^*)} f(\mathbb{W}^*) \sqrt{ \frac{ (2\pi)^{k(k - 1)/2} }{ \left| \det(\color{blue}N\color{black} \, D^2\varphi_{C, k}(\mathbb{W}^*)) \right| } } \end{align*} if there is a unique $ \mathbb{W}^* $ such that $ \nabla \varphi_{C, k}(\mathbb{W}^*) = 0 $ and if $ D^2\varphi_{C, k}(\mathbb{W}^*) $ (which is a $ {k \choose 2} \times {k \choose 2} $ matrix) is symmetric definite negative.
First question : Has anyone met such an integral anywhere (who knows...) ?
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Using the classical hermitian scalar product $ \left\langle A, B \right\rangle := \mathrm{tr}(A B^T) $, one has \begin{align*}%$ \varphi_{C,k}(\mathbb{W} + \mathbb{H}) = \varphi_{C,k}(\mathbb{W}) + \left\langle \nabla\varphi_{C,k}(\mathbb{W}) , \mathbb{H} \right\rangle + \frac{1}{2} \left\langle D^2\varphi_{C,k}(\mathbb{W})(\mathbb{H}) , \mathbb{H} \right\rangle + O(\mathbb{H}^3) \end{align*} and one computes (using the fact that $ \mathbb{W} $ is symmetric) \begin{align*}%$ \nabla\varphi_{C,k}(\mathbb{W}) & = \frac{1}{2} \left( -2C^2 (I_k + \mathbb{W})^{-1} U_k (I_k + \mathbb{W})^{-1} + U_k \right) \\ D^2\varphi_{C,k}(\mathbb{W})(\mathbb{H}) & = 2C^2 (I_k + \mathbb{W})^{-1} \mathbb{H}^T (I_k + \mathbb{W})^{-1} U_k (I_k + \mathbb{W})^{-1} \end{align*}
Now, setting $ X := (C \sqrt{2})^{-1} (I_k + \mathbb{W}) $, the gradient equation $ \nabla\varphi_{C, k}(\mathbb{W}) = 0 $ is equivalent to $ X U_k X = U_k $ with $ X $ symmetric with constant diagonal equal to $ \lambda := (C \sqrt{2})^{-1} $ and $ W_{i, j} \in (0, \lambda) $. Matricially speaking, these equations are equivalent to $ XU_k = U_k X = U_k $ (linear equations, hence better than the original quadratic one). Nevertheless, there are only $k$ equations for $ {k \choose 2} $ variables...
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For $ k = 3 $ (hence 3 parameters $x, y, z$, say), I can compute the extrema (I can type the reasoning if you want) and find that this is a point of the form $ (a, a, a) $ with explicit $a$. The Hessian can also be computed with Maple and gives \begin{align*}%$ & \det(D^2 \varphi_{C, 3}(x, y, z) ) \\ & \qquad = -32 \,\frac{ ( z-1 )^2 ( y-1 )^2 ( x-1 )^2 ( x-y-z+1 )^2 ( x-y+z-1 )^2 ( x+y-z-1 )^2 }{ ( -2 yxz + x^2 + y^2 + z^2-1 )^7} \end{align*}
In particular, $ \det(D^2 \varphi_{C, 3}(a, a, a) ) = \frac{32}{(2a + 1)^7 (a - 1)^2} > 0 $ and I can have the equivalent with the Laplace method.
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For $ k \geq 4 $, things start becoming extremely problematic: the gradient equation gives a full symplex of solutions, and the Hessian... is always non invertible ! Indeed, the underlying linear map is of the form $ \psi_{A, B}(H) := A H^T B $ with $ B $ of rank one and $ A $ invertible, and one has $ H \in \ker(\psi_{A, B}) $ iff $ H b = 0 $ (where $ B = b b^T $ since $B$ is symmetric of rank 1), i.e. $ b \in \ker(H) $. As a result, $ \psi_{A, B} $ is not invertible (it is iff $ A $ and $ B $ are invertible, and computing its determinant does not seem simple in this case).
I can continue computing higher order ``Taylor tensors'', but they will all be non invertible for the same reason, because the matrix $B := (I_k + \mathbb{W})^{-1} U_k (I_k + \mathbb{W})^{-1} $ is of rank one.
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Questions : What can be done in such a case where all higher order ``derivatives'' are not invertible ? What can I do if the solution of the gradient equation is a full convex set ?
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In fact, the case $ k = 3 $ should give the same as $ k = 4 $, as far as I know (the reasoning for the non invertibility of the Hessian does not depend on the dimension ; still, Maple finds the formula that I gave). So, there is maybe a mistake in one of the previous reasoning.