Symmetry And Group

## Representations des groupes reductifs sur un corps local by J.-N. Bernstein, P. Deligne, D. Kazhdan, M.-F. Vigneras

Posted On March 23, 2017 at 6:32 am by / Comments Off on Representations des groupes reductifs sur un corps local by J.-N. Bernstein, P. Deligne, D. Kazhdan, M.-F. Vigneras

By J.-N. Bernstein, P. Deligne, D. Kazhdan, M.-F. Vigneras

Best symmetry and group books

Molecular Aspects of Symmetry

The contents of this e-book were used in classes given by way of the writer. the 1st used to be a one-semester direction for seniors on the college of British Columbia; it used to be transparent that strong undergraduates have been completely able to dealing with ordinary crew conception and its program to basic quantum chemical difficulties.

Additional resources for Representations des groupes reductifs sur un corps local (Travaux en cours)

Sample text

By diﬀerentiating, one ﬁnds xi1 xi2 . . xi λ ∂ ∂ ∂ ··· Z(b, λ) ∂bi1 ∂bi2 ∂bi = Z −1 (λ) . 32) b=0 We now introduce the function W(b, λ) = ln Z(b, λ). 33) In a probabilistic interpretation, W(b, λ) is the generating function of the cumulants of the distribution. 28), the perturbative expansion of cumulants is much simpler since it contains only connected contributions. 26) are contained in W(0, λ). Finally note that, in the Gaussian case, W(b) reduces to a form quadratic in b. Remark. i = , b=0 are called connected -point correlation functions.

The divergence of the series can easily be understood: if one changes the sign of λ in the integral, the maximum of the integrand becomes a minimum and the selected saddle point can no longer give the leading contribution to the integral. (ii) Often integrals have the more general form dx ρ(x) e−S(x)/λ . I(λ) = Then, provided ln ρ(x) is analytic at the saddle point, it is not necessary to take the factor ρ(x) into account in the saddle point equation. Indeed, this factor would induce a shift of order x − xc to the saddle point position, a solution of S (xc )(x − xc ) ∼ λρ (xc )/ρ(xc ), and, thus, √ of order λ while the contribution to the integral comes from a region of order λ, which is much larger than the shift.

Solution. One sets S(t) = 12 (t + 1/t) . The saddle point tc is given by S (tc ) = 12 (1 − 1/t2c ) = 0 ⇒ tc = 1 . Then, S (tc ) = 1/t3c = 1 . One concludes Kν (z) ∼ (π/2z)1/2 e−z . 7 Evaluate by the steepest descent method the integral 1 2π In (s) = +π/2 dθ (cos θ)n einθ tanh s −π/2 as a function of the real parameter s in the limit n → ∞. One will verify that the function is real. Solution. One introduces the function S(θ) = −iθ tanh s − ln cos θ . The function is analytic except at the points θ = π/2 mod (π).