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9 votes
2 answers
848 views

$\zeta$-function regularized determinants

In (mathematical) physics in order to compute path integrals one often makes an infinite dimensional change of variables and uses infinite Jacobian as a purely formal expression. This step is done in ...
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3 votes
0 answers
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Multiplicativity of $\zeta$-function regularized determinant

Let $A$ be a selfadjoint elliptic differential operator on a compact manifold. In mathematical physics and differential topology one often defines its determinant using the $\zeta$-function ...
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