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This is in reference to the answer posted to this question. The person who answered the question claims that the functional determinant of any operator $O$ is given by a product integral $$\det O = \prod_{x \in \mathcal{M}} O(x)^{\mu(x)} \label{aa}$$ where $\mu(x)$ is the measure and $x$ are the points on any $d$-dimensional manifold $\mathcal{M}$. This sort of makes sense because one can exponentiate the RHS and write $$\det O = \prod_{x \in \mathcal{M}} O(x)^{\mu(x)} = \exp(\int d^dx ~\mu(x) \ln O)$$ which is like $\det M = \exp(\operatorname{tr}\ln M)$. So, can anyone explain how to justify the first equality?

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    $\begingroup$ @Hosein you are correct, unless user44690 has abused notation. It should be something like $O(\varphi)=\lambda(\varphi)\varphi$ but the OP has denoted the eigenvalues $\lambda(\varphi)$ as $O(\varphi)$. $\endgroup$ Apr 28 at 15:00
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    $\begingroup$ Just an observation. Let's say you discretize spacetime into a set of lattice points $x_i$. Let's say we choose some reference lattice point $x_0$ to have volume $1$, and the others have volume $\mu_i$ ($\mu_i$ is sort of a degeneracy of the $i$-th lattice point). An operator is a matrix $O_{ij}$, where $i, j$ run over the lattice sites. If the operator is local, then $O_{ij} = \lambda_i \delta_{ij}$, where $\lambda_i$ is the $i$-th eigenvalue. Then, it is clear that $\det O = \prod_i \lambda_i^{\mu_i}$. For a uniform lattice, all the $\mu_i$ are the same, so $\det O = \prod_i \lambda_i$. $\endgroup$
    – Andrew
    May 2 at 1:39
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    $\begingroup$ My comment is sloppy about the precise sense in which $\mu_i$ is a degeneracy of the $i$-th lattice point, which is why I left it is as a comment instead of turning it into an answer. I wrote it in case this half-baked thought is useful. $\endgroup$
    – Andrew
    May 2 at 1:51
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    $\begingroup$ @user44690 It's more of an intuition than something I can back up rigorously. But the idea is that if you discretize space, then the volume of a region of space becomes the number of lattice points in that region after discretization. The measure factor is describing how the volume element changes in space, so intuitively it is describing how the number of lattice points changes from place to place. If you group $N$ lattice points with the same eigenvalue, you'd have an $N$-level degeneracy. $\endgroup$
    – Andrew
    May 2 at 5:07
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    $\begingroup$ To try to make this intuition slightly more concrete: First want to discretize space very finely such that the lattice spacing was uniform. Then, you would "tesselate" space such that each volume element had the same volume according to the measure. Finally, you'd group together all the lattice points from step 1 that fell in the same volume element defined in step 2. The degeneracy of volume element $i$ would be the number of lattice points in that volume element. I haven't worked through all the details, but that's a sketch of a way I think you could justify your formula. $\endgroup$
    – Andrew
    May 2 at 5:12

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