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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 regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants$\zeta$-function regularized determinants

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 regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants

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 regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants

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asv
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Let $A$ be a seladjointselfadjoint elliptic differential operator on a compact manifold. In mathematical physics and differential topology one often defines its determinant using the $\zeta$-function regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants

Let $A$ be a seladjoint elliptic differential operator on a compact manifold. In mathematical physics and differential topology one often defines its determinant using the $\zeta$-function regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants

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 regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants

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asv
asv
  • 21.8k
  • 6
  • 54
  • 121

Multiplicativity of $\zeta$-function regularized determinant

Let $A$ be a seladjoint elliptic differential operator on a compact manifold. In mathematical physics and differential topology one often defines its determinant using the $\zeta$-function regularization which will be denoted below by $\det_\zeta(A)$.

QUESTION. Let $A,B$ be two commuting selfadjoint elliptic differential operators. Then their product $AB$ is also selfadjoint and elliptic. Is it true that $\det_\zeta(AB)=\det_\zeta(A)\det_\zeta(B)?$

This question is closely related to this one $\zeta$-function regularized determinants