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arXiv 2202.12923math.SPmath.FA
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正则化Fredholm行列式的乘积公式:两种新证明

The product formula for regularized Fredholm determinants: two new proofs

Nikolaos Koutsonikos-Kouloumpis, Matthias Lesch

AI总结:

该研究给出正则化Fredholm行列式乘积公式的两种新的简单证明(解析与代数),将公式推广到多因子情形,并揭示其组合本质。

摘要:

对于可分希尔伯特空间中的m-可和算子A,高阶正则化Fredholm行列式detₘ(I+A)是经典Fredholm行列式的推广。近期,Britz等人给出了一个乘积公式的证明:detₘ((I+A)·(I+B)) = detₘ(I+A)·detₘ(I+B)·exp Tr(Xₘ(A,B)),其中Xₘ(A,B)是A、B的显式多项式,取值为迹类算子。当m=1时X₁(A,B)=0,因此该公式推广了经典的行列式乘积公式。\n本札记的目的之一是给出该公式两种极为简单的替代证明。第一种证明是先验解析的,利用了映射z↦detₘ(I+zA)是全纯函数这一性质;第二种证明则是完全代数化的。在我们看来,该代数证明在迹与交换子方面本身就有一些值得关注的特点。\n其次,我们将上述公式推广到多个因子的情形:detₘ(∏ₗ=1ʳ(I+Aₗ)) = (∏ₗ=1ʳ detₘ(I+Aₗ))·exp Tr(Xₘ,ᵣ(A₁,…,Aᵣ))。这一推广并非只是直接的拓展,因为我们将对其背后的组合结构获得更深入的理解。我们还将用形式幂级数的语言给出解析证明的代数化版本。核心结论是这两个恒等式本质上都属于组合性质。

英文摘要:

For an $m$-summable operator $A$ in a separable Hilbert space the higher regularized Fredholm determinant $\det\nolimits_m(I+A)$ generalizes the classical Fredholm determinant. Recently, Britz et al presented a proof of a product formula \[ \det\nolimits_m\bigl( (I+A)\cdot(I+B) \bigr) = \det\nolimits_m (I+A) \cdot \det\nolimits_m (I+B) \cdot \exp\operatorname{Tr}\bigl({X_m(A,B)}\bigr), \] where $X_m(A,B)$ is an explicit polynomial in $A,B$ with values in the trace class operators. If $m=1$ then $X_1(A,B)=0$, hence the formula generalizes the classical determinant product formula. One of the purposes of this note is to present two very simple alternative proofs of the formula. The first proof is a priori analytic and makes use of the fact that $z\mapsto \det\nolimits_m(I+zA)$ is holomorphic, while the second proof is completely algebraic. The algebraic proof has, in our opinion, some interesting aspects in its own about the trace and commutators. Secondly, we extend the above mentioned formula to several factors \[ \det\nolimits_m\Bigl( \prod_{l=1}^r (I+A_l) \Bigr) =\left( \prod_{l=1}^r \det\nolimits_m (I+A_l) \right) \cdot \exp\operatorname{Tr}\bigl({X_{m,r}(A_r,\ldots,A_r)}\bigr). \] The latter is more than just a straightforward generalization as we will gain more insights into the combinatorics behind it. Also we will present an algebraized version of the analytic proof in the language of formal power series. The upshot is that the two identities are just combinatorial in nature.

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